{"id":31751,"date":"2026-08-07T13:04:00","date_gmt":"2026-08-07T07:34:00","guid":{"rendered":"https:\/\/mycbseguide.com\/blog\/?p=31751"},"modified":"2026-08-07T13:23:51","modified_gmt":"2026-08-07T07:53:51","slug":"exploring-some-geometric-themes-ncert-solutions-class-8-maths-ganita-prakash","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/exploring-some-geometric-themes-ncert-solutions-class-8-maths-ganita-prakash\/","title":{"rendered":"Exploring Some Geometric Themes &#8211; NCERT Solutions Class 8 Maths (Ganita Prakash)"},"content":{"rendered":"\n<p><strong><strong>Exploring Some Geometric Themes<\/strong><\/strong> &#8211; NCERT Solutions Class 8 Maths (Ganita Prakash) includes all the questions with solutions given in the NCERT Class 8 Maths (Ganita Prakash).<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">NCERT Solutions Class 8<\/h2>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-8\/ncert-solutions-class-8-english-poorvi\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >English Poorvi<\/a>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-8\/ncert-solutions-class-8-hindi-malhar\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >Hindi Malhar<\/a>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-8\/ncert-solutions-class-8-maths-ganita-prakash\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >Maths Ganita Prakash<\/a>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-8\/ncert-solutions-class-8-science-curiosity\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >Science Curiosity<\/a>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-8\/ncert-solutions-class-8-social-exploring-society\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >Social Exploring Society<\/a>\n\n\n\n<h2 class=\"wp-block-heading\"><strong><strong>Exploring Some Geometric Themes<\/strong><\/strong> \u2013 NCERT Solutions<\/h2>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.1: Draw the initial few steps (at least till Step 2) of the shape sequence that leads to the Sierpinski Carpet.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>Step 0:<\/strong> Start with a single square.<br><strong>Step 1:<\/strong> Divide the square into 9 equal smaller squares ({tex}3 \\times 3{\/tex} grid) and remove the central square. This leaves 8 squares remaining.<br><strong>Step 2:<\/strong> Take each of the 8 remaining squares from Step 1 and repeat the process divide each into 9 smaller squares and remove the central square from each. This gives us {tex}8 \\times 8=64{\/tex} squares remaining.<br>Students should draw these steps showing the progressive pattern.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.2: Do you see any pattern in the number of holes and squares that remain at each step?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Yes, there is a clear pattern:<br><strong>Remaining Squares (Rn):<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Step 0: {tex}{R}_{{o}}=1{\/tex} square<\/li>\n\n\n\n<li>Step 1: {tex}{R}_1=8{\/tex} squares<\/li>\n\n\n\n<li>Step 2: {tex}{R}_2=64{\/tex} squares<\/li>\n\n\n\n<li>Step 3: {tex}{R}_3=512{\/tex} squares<\/li>\n<\/ul>\n\n\n\n<p>Pattern: Each remaining square from the previous step creates 8 new squares. So, {tex}{Rn}=8^{{n}}{\/tex}<br><strong>Holes (Hn):<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Step 0: {tex}{H}_{{o}}=0{\/tex} holes<\/li>\n\n\n\n<li>Step 1: {tex}{H}_1=1{\/tex} hole<\/li>\n\n\n\n<li>Step 2: {tex}{H}_2=1+8=9{\/tex} holes<\/li>\n\n\n\n<li>Step 3: {tex}{H}_3=1+8+64=73{\/tex} holes<\/li>\n<\/ul>\n\n\n\n<p>Pattern: At each new step, we add as many holes as there were remaining squares in the previous step. So, {tex}{Hn}+1={Hn}+{Rn}{\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.3: Show that by joining the midpoints of an equilateral triangle, we divide it into 4 identical equilateral triangles.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Let&#8217;s prove this step by step:<br>Consider an equilateral triangle ABC with side length &#8216;a&#8217;. Let D, E, F be the midpoints of sides BC, CA, and AB respectively.<br><strong>Step 1:<\/strong>&nbsp;Find the lengths of DE, EF, and FD.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Since D and E are midpoints, DE is parallel to AB and {tex}{DE}={AB} \/ 2={a} \/ 2{\/tex}<\/li>\n\n\n\n<li>Similarly, EF is parallel to BC and {tex}{EF}={BC} \/ 2={a} \/ 2{\/tex}<\/li>\n\n\n\n<li>And {tex}F D{\/tex} is parallel to {tex}C A{\/tex} and {tex}F D=C A \/ 2=a \/ 2{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 2:<\/strong> Identify the 4 triangles formed. The four triangles are: {tex}\\triangle A F E, \\triangle F B D, \\triangle D E C{\/tex}, and {tex}\\triangle {DEF}{\/tex} (the central one)<br><strong>Step 3:<\/strong> Prove they are equilateral.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}\\triangle {AFE}: {AF}={a} \/ 2, {FE}={a} \/ 2, {EA}={a} \/ 2 \\rightarrow{\/tex} Equilateral<\/li>\n\n\n\n<li>{tex}\\triangle {FBD}: {FB}={a} \/ 2, {BD}={a} \/ 2, {DF}={a} \/ 2 \\rightarrow{\/tex} Equilateral<\/li>\n\n\n\n<li>{tex}\\triangle {DEC}: {DE}={a} \/ 2, {EC}={a} \/ 2, {CD}={a} \/ 2 \\rightarrow{\/tex} Equilateral<\/li>\n\n\n\n<li>{tex}\\triangle {DEF}: {DE}={a} \/ 2, {EF}={a} \/ 2, {FD}={a} \/ 2 \\rightarrow{\/tex} Equilateral<\/li>\n<\/ul>\n\n\n\n<p>Therefore, all four triangles are identical equilateral triangles with side length {tex}{a} \/ 2{\/tex}.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.4: Find the number of holes, and the triangles that remain at each step of the shape sequence that leads to the Sierpinski Triangle.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Let Rn = number of remaining triangles at step n<br>Let Hn = number of holes at step n<br>Remaining Triangles:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Step 0: {tex}{R}_{{o}}=1{\/tex}<\/li>\n\n\n\n<li>Step 1: {tex}{R}_1=3{\/tex} (each triangle from previous step creates 3 new triangles)<\/li>\n\n\n\n<li>Step 2: {tex}{R}_2=3 \\times 3=9{\/tex}<\/li>\n\n\n\n<li>Step 3: {tex}{R}_3=3 \\times 9=27{\/tex}<\/li>\n<\/ul>\n\n\n\n<p>Pattern: {tex}{Rn}+1=3 \\times {Rn}{\/tex}<br>Therefore, {tex}{R n}={3}^{{n}}{\/tex}<br><strong>Number of Holes:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Step 0: {tex}{H}_0=0{\/tex}<\/li>\n\n\n\n<li>Step 1: {tex}{H}_1=1{\/tex} (one central triangle removed)<\/li>\n\n\n\n<li>Step 2: {tex}{H}_2=1+3=4{\/tex} (previous holes + new holes created)<\/li>\n\n\n\n<li>Step 3: {tex}{H}_3=4+9=13{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1769145535-8ee7cn.jpg\"><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1769145550-vuswp3.jpg\"><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1769145566-akpmf3.jpg\"><br>Pattern: {tex}{Hn}+1={Hn}+{Rn}{\/tex}<br>Therefore:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}{H}_1=1{\/tex}<\/li>\n\n\n\n<li>{tex}{H}_2=1+3=4{\/tex}<\/li>\n\n\n\n<li>{tex}{H}_3=1+3+9=13{\/tex}<\/li>\n<\/ul>\n\n\n\n<p>{tex}{Hn}=1+3+3^2+\\ldots+3^{n-1}=\\left({3}^{{n}} \\boldsymbol{-} {1}\\right) \/ {2}{\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.5: Find the area of the region remaining at the nth step in each of the shape sequences that lead to the Sierpinski fractals. Take the area of the starting square\/triangle to be 1 sq. unit.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>For Sierpinski Carpet:<br>Let {tex}{An}={\/tex} area remaining at step n<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Step 0: {tex}{A}_{\\circ}=1{\/tex} sq. unit (full square)<\/li>\n\n\n\n<li>Step 1: {tex}{A}_1=8 \/ 9{\/tex} sq. unit (8 squares out of 9 remain)<\/li>\n\n\n\n<li>Step 2: {tex}{A}_2=(8 \/ 9) \\times(8 \/ 9)=(8 \/ 9)^2{\/tex} sq. unit<\/li>\n<\/ul>\n\n\n\n<p>At each step, we keep 8\/9 of the area from the previous step.<br>Therefore, {tex}{A n} \\boldsymbol{=}({8 \/ 9})^{{n}}{\/tex} sq. unit<br><strong>For Sierpinski Triangle:<\/strong><br>Let {tex}{An}={\/tex} area remaining at step n<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Step 0: {tex}{A}_0=1{\/tex} sq. unit (full triangle)<\/li>\n\n\n\n<li>Step 1: {tex}{A}_1=3 \/ 4{\/tex} sq. unit (3 triangles out of 4 remain)<\/li>\n\n\n\n<li>Step 2: {tex}{A}_2=(3 \/ 4) \\times(3 \/ 4)=(3 \/ 4)^2{\/tex} sq. unit<\/li>\n<\/ul>\n\n\n\n<p>At each step, we keep {tex}3 \/ 4{\/tex} of the area from the previous step.<br>Therefore, {tex}{A n} \\boldsymbol{=} {( 3 \/ 4 ) ^ { { n } }}{\/tex} sq. unit<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.6: Draw the initial few steps (at least till Step 2) of the shape sequence that leads to the Koch Snowflake.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>Step 0: <\/strong>Start with an equilateral triangle.<br><strong>Step 1:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Divide each side into 3 equal parts<\/li>\n\n\n\n<li>On the middle third of each side, construct an equilateral triangle pointing outward<\/li>\n\n\n\n<li>Remove the base of this new triangle (the middle third)<\/li>\n\n\n\n<li>Each side now becomes a &#8220;bump&#8221; shape with 4 segments<\/li>\n\n\n\n<li>Total sides: {tex}3 \\times 4=12{\/tex} sides<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 2:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Repeat the process on each of the 12 sides from Step 1<\/li>\n\n\n\n<li>Each side gets divided into 3 parts, a triangle is raised on the middle part<\/li>\n\n\n\n<li>Each side becomes 4 segments<\/li>\n\n\n\n<li>Total sides: {tex}12 \\times 4=48{\/tex} sides<\/li>\n<\/ul>\n\n\n\n<p>Students should draw showing the star-like shape evolving into a snowflake pattern<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.7: Find the number of sides in the nth step of the shape sequence that leads to the Koch Snowflake.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Let {tex}{Sn}={\/tex} number of sides at step n<br>Pattern Analysis:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Step 0: {tex}{S}_{{o}}=3{\/tex} (equilateral triangle has 3 sides)<\/li>\n\n\n\n<li>Step 1: {tex}{S}_1=3 \\times 4=12{\/tex} (each side becomes 4 sides)<\/li>\n\n\n\n<li>Step 2: {tex}{S}_2=12 \\times 4=48{\/tex} (each side becomes 4 sides)<\/li>\n\n\n\n<li>Step 3: {tex}{S}_3=48 \\times 4=192{\/tex}<\/li>\n<\/ul>\n\n\n\n<p>At each step, the number of sides is multiplied by 4.<br>Therefore, {tex}{S n}={3} \\times {4}^{{n}}{\/tex}<br>This can also be written as {tex}{S n}={3} \\times {2}^{{2 n}}{\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.8: Find the perimeter of the shape at the nth step of the sequence. Take the starting equilateral triangle to have a sidelength of 1 unit.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Let {tex}{Pn}={\/tex} perimeter at step n<br><strong>Step 0:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Side length {tex}=1{\/tex} unit<\/li>\n\n\n\n<li>Number of sides {tex}=3{\/tex}<\/li>\n\n\n\n<li>{tex}{P}_0=3 \\times 1=3{\/tex} units<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 1:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Each side of length 1 is divided into 3 parts, each of length {tex}1 \/ 3{\/tex}<\/li>\n\n\n\n<li>Each side becomes 4 segments of length {tex}1 \/ 3{\/tex}<\/li>\n\n\n\n<li>Number of sides {tex}=12{\/tex}<\/li>\n\n\n\n<li>{tex}P_1=12 \\times(1 \/ 3)=4{\/tex} units<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 2:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Each side of length {tex}1 \/ 3{\/tex} becomes 4 segments of length {tex}1 \/ 9{\/tex}<\/li>\n\n\n\n<li>Number of sides {tex}=48{\/tex}<\/li>\n\n\n\n<li>{tex}{P}_2=48 \\times(1 \/ 9)=16 \/ 3{\/tex} units<\/li>\n<\/ul>\n\n\n\n<p><strong>Pattern:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>At step {tex}n{\/tex}, each original side is divided into {tex}3^n{\/tex} equal parts<\/li>\n\n\n\n<li>Length of each small segment {tex}=1 \/ 3^n{\/tex}<\/li>\n\n\n\n<li>Number of sides {tex}=3 \\times 4^n{\/tex}<\/li>\n\n\n\n<li>Perimeter {tex}{Pn}=\\left(3 \\times 4^n\\right) \\times\\left(1 \/ 3^n\\right){\/tex}<\/li>\n\n\n\n<li>Therefore, {tex}{P n}={3} \\boldsymbol{\\times}({4} \/ {3})^{{n}}{\/tex} units<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.9: Picture your name, then read off the letters backwards. Make sure to do this by sight, not by sound &#8211;&nbsp;really see your name! Now try with your friend&#8217;s name.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>This is a visualization exercise. Let&#8217;s take an example:<br>If your name is &#8220;RAVI&#8221;:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Visualize: R-A-V-I<\/li>\n\n\n\n<li>Reading backwards by sight: I-V-A-R<\/li>\n<\/ul>\n\n\n\n<p>If your friend&#8217;s name is &#8220;PRIYA&#8221;:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Visualize: P-R-I-Y-A<\/li>\n\n\n\n<li>Reading backwards by sight: A-Y-I-R-P<\/li>\n<\/ul>\n\n\n\n<p>Practice this with different names to improve visual memory and spatial thinking.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.10: Cut off the four corners of an imaginary square, with each cut going between midpoints of adjacent edges. What shape is left over? How can you reassemble the four corners to make another square?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>Shape left over: <\/strong>When we cut off the four corners by joining midpoints of adjacent edges, we get a<strong> regular octagon <\/strong>(8-sided polygon with all sides equal).<br><strong>Explanation:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Each corner cut creates a new edge<\/li>\n\n\n\n<li>The original 4 sides of the square remain but become shorter<\/li>\n\n\n\n<li>Total edges {tex}=4{\/tex} (from original sides) +4 (from corner cuts) {tex}=8{\/tex} edges<\/li>\n\n\n\n<li>All sides are equal in length<\/li>\n<\/ul>\n\n\n\n<p><strong>Reassembling the four corners:&nbsp;<\/strong>The four corner pieces are identical right-angled isosceles triangles. To make another square:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Arrange the four triangles so that their right angles meet at the center<\/li>\n\n\n\n<li>The hypotenuses of the triangles form the sides of a new square<\/li>\n\n\n\n<li>This new square has half the area of the original square<\/li>\n\n\n\n<li>Alternatively, arrange two triangles to form a square, then place the other two similarly<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.11: Mark the sides of an equilateral triangle into thirds. Cut off each corner of the triangle, as far as the marks. What shape do you get?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>When we mark the sides of an equilateral triangle into thirds and cut off each corner at these marks, we get a <strong>regular hexagon <\/strong>(6-sided polygon with all sides equal).<br><strong>Explanation:<\/strong><br>Original triangle has 3 sides<br>Cutting each corner creates 2 new edges per corner<br>From 3 corners, we get {tex}3 \\times 2=6{\/tex} new edges<br>The middle third of each original side remains<br>Total edges {tex}=3{\/tex} (middle parts of original sides) +6 (from corner cuts) {tex}=9{\/tex} edges initially<\/p>\n\n\n\n<p>Wait, let&#8217;s reconsider:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Each corner is cut, removing the first third and last third meeting point<\/li>\n\n\n\n<li>This creates one new edge per corner {tex}=3{\/tex} new edges<\/li>\n\n\n\n<li>Each original side now contributes its middle third {tex}=3{\/tex} edges<\/li>\n\n\n\n<li>Total {tex}=3+3={6}{\/tex}<strong> edges forming a regular hexagon<\/strong><\/li>\n<\/ul>\n\n\n\n<p>All six sides are equal because each is one-third of the original side length.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.12: Mark the sides of a square into thirds and cut off each of its corners as far as the marks. What shape is left?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>When we mark the sides of a square into thirds and cut off each corner at these marks, we get a <strong>regular octagon<\/strong> ( 8 -sided polygon with all sides equal).<\/p>\n\n\n\n<p><strong>Explanation:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Original square has 4 sides<\/li>\n\n\n\n<li>Cutting each corner creates 1 new edge per corner<\/li>\n\n\n\n<li>From 4 corners, we get 4 new edges<\/li>\n\n\n\n<li>The middle third of each original side remains {tex}=4{\/tex} edges<\/li>\n\n\n\n<li>Total edges {tex}=4{\/tex} (middle parts) +4 (from corner cuts) {tex}=8{\/tex} edges<\/li>\n<\/ul>\n\n\n\n<p>All eight sides are equal because:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The middle third of each original side has the same length<\/li>\n\n\n\n<li>Each cut edge (diagonal across the corner) also has the same length by symmetry<\/li>\n<\/ul>\n\n\n\n<p>This forms a <strong>regular octagon.<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.13: Can you describe a solid and a viewpoint that would result in each of the following cases? If it helps, you can imagine the solid passing through a wall like Tom did, and leaving a hole of the appropriate shape.<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>A solid whose profile has a square outline<\/li>\n\n\n\n<li>A solid whose profile has a circular outline<\/li>\n\n\n\n<li>A solid whose profile has a triangular outline<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Several solids can have a square profile: Example: A cube viewed directly from the front, back, left, right, top, or bottom will show a square outline.\n<ol start=\"1\" class=\"wp-block-list\">\n<li><strong>Cube &#8211;<\/strong> viewed from any face directly<\/li>\n\n\n\n<li><strong>Cuboid &#8211; <\/strong>viewed from a face that is square<\/li>\n\n\n\n<li><strong>Square prism &#8211; <\/strong>viewed from the top or bottom<\/li>\n\n\n\n<li><strong>Square pyramid &#8211; <\/strong>viewed from the base (looking down)<\/li>\n<\/ol>\n<\/li>\n\n\n\n<li>Several solids can have a circular profile:Example: A cylinder viewed from directly above or below shows a circular outline.\n<ol start=\"1\" class=\"wp-block-list\">\n<li><strong>Sphere &#8211; <\/strong>from any viewpoint<\/li>\n\n\n\n<li><strong>Cylinder &#8211;<\/strong> viewed from the top or bottom (along the axis)<\/li>\n\n\n\n<li><strong>Cone &#8211;<\/strong> viewed from the base (looking at the circular base)<\/li>\n\n\n\n<li><strong>Hemisphere &#8211;<\/strong> from the curved side or flat side<\/li>\n<\/ol>\n<\/li>\n\n\n\n<li>Several solids can have a triangular profile:<ol start=\"1\"><li><strong>Triangular pyramid (tetrahedron) &#8211; <\/strong>viewed from any face<\/li><li><strong>Triangular prism &#8211;<\/strong> viewed from the triangular face<\/li><li><strong>Cone &#8211;<\/strong> viewed from the side<\/li><li><strong>Wedge &#8211; <\/strong>viewed from the triangular end<\/li><\/ol><strong>Example:<\/strong> A cone viewed from the side shows a triangular outline (isosceles triangle)<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.14: As we saw with the elephant, a given solid might have very different profiles from different viewpoints. Can you visualise solids that have the following contrasting profiles?<br>Spend some time on this, and if you are finding it difficult to visualise, you may look around and use objects that are around you, or that you will make in the next section. Feel free to consider viewpoints from any direction, including directly above the object.<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>A solid with a rectangular profile from one viewpoint and a circular profile from another viewpoint<\/li>\n\n\n\n<li>A solid with a circular profile from one viewpoint and a triangular one from another viewpoint<\/li>\n\n\n\n<li>A solid with a rectangular profile from one viewpoint and a triangular one from another viewpoint<\/li>\n\n\n\n<li>A solid with a trapezium shaped profile from one viewpoint and a circular one from another viewpoint<\/li>\n\n\n\n<li>A solid with a pentagonal profile from one viewpoint and a rectangular one from another viewpoint<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Cylinder is the perfect example. Another example: Cone (though the side view would be triangular, not rectangular)\n<ul class=\"wp-block-list\">\n<li>Rectangular profile: When viewed from the side (perpendicular to the axis), it shows a rectangle<\/li>\n\n\n\n<li>Circular profile: When viewed from the top or bottom (along the axis), it shows a circle<br>Another example:&nbsp;<strong>Cone<\/strong>&nbsp;(though the side view would be triangular, not rectangular)<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Cone is the best example.\n<ul class=\"wp-block-list\">\n<li>Circular profile: When viewed from the base (looking at the bottom), it shows a circle<\/li>\n\n\n\n<li>Triangular profile: When viewed from the side, it shows a triangle (isosceles)<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Triangular Prism is a good example.\n<ul class=\"wp-block-list\">\n<li>Rectangular profile: When viewed from the side (perpendicular to the triangular face), it shows a rectangle<\/li>\n\n\n\n<li>Triangular profile: When viewed from the front or back (facing the triangular face), it shows a triangle<br>Another example: Triangular pyramid (tetrahedron) when viewed from different angles.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Truncated Cone (Frustum of a cone) is the perfect example.\n<ul class=\"wp-block-list\">\n<li>Trapezium profile: When viewed from the side, it shows a trapezium (the slanting sides with parallel top and bottom)<\/li>\n\n\n\n<li>Circular profile: When viewed from the top or bottom, it shows a circle<br>This shape is commonly seen in lampshades or buckets.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Pentagonal Prism is the example.\n<ul class=\"wp-block-list\">\n<li>Pentagonal profile: When viewed from the front or back (facing the pentagonal face), it shows a pentagon<\/li>\n\n\n\n<li>Rectangular profile: When viewed from any of the five sides (perpendicular to the pentagonal face), it shows a rectangle<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<p><strong>Note: <\/strong>There are multiple possible solids for each condition, showing that three-dimensional visualization requires considering multiple perspectives.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.15: If the congruent polygons of a prism have 10 sides, how many faces, edges and vertices does the prism have? What if the polygons have n sides?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>For a prism with 10 -sided polygons:<\/strong><br><strong>Faces:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>2 congruent decagonal ( 10 -sided) faces (top and bottom)<\/li>\n\n\n\n<li>10 rectangular faces (connecting the sides)<\/li>\n\n\n\n<li>Total faces {tex}=2+10=12{\/tex}<strong> faces<\/strong><\/li>\n<\/ul>\n\n\n\n<p><strong>Edges:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>10 edges on the top polygon<\/li>\n\n\n\n<li>10 edges on the bottom polygon<\/li>\n\n\n\n<li>10 vertical edges connecting corresponding vertices<\/li>\n\n\n\n<li>Total edges {tex}=10+10+10=30{\/tex}<strong> edges<\/strong><\/li>\n<\/ul>\n\n\n\n<p><strong>Vertices:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>10 vertices on the top polygon<\/li>\n\n\n\n<li>10 vertices on the bottom polygon<\/li>\n\n\n\n<li>Total vertices {tex}=10+10=20{\/tex}<strong> vertices<\/strong><\/li>\n<\/ul>\n\n\n\n<p><strong>For a prism with {tex}{n}{\/tex}-sided polygons:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Faces {tex}=2+n=(n+2){\/tex} faces<\/strong><\/li>\n\n\n\n<li><strong>Edges {tex}=n+n+n=3 n{\/tex} edges<\/strong><\/li>\n\n\n\n<li><strong>Vertices <\/strong>= n + n =<strong> 2n vertices<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.16: If the base of a pyramid has 10 sides, how many faces, edges and vertices does the pyramid have? What if the base is an n-sided polygon?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>For a pyramid with a 10 -sided base:<br><strong>Faces:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>1 decagonal ( 10 -sided) base<\/li>\n\n\n\n<li>10 triangular faces (one for each side of the base)<\/li>\n\n\n\n<li>Total faces {tex}=1+10=11{\/tex}<strong> faces<\/strong><\/li>\n<\/ul>\n\n\n\n<p><strong>Edges:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>10 edges on the base<\/li>\n\n\n\n<li>10 slant edges from base vertices to apex<\/li>\n\n\n\n<li>Total edges {tex}=10+10=20{\/tex}<strong> edges<\/strong><\/li>\n<\/ul>\n\n\n\n<p><strong>Vertices:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>10 vertices on the base<\/li>\n\n\n\n<li>1 apex vertex at the top<\/li>\n\n\n\n<li>Total vertices {tex}=10+1=11{\/tex}<strong> vertices<\/strong><\/li>\n<\/ul>\n\n\n\n<p><strong>For a pyramid with {tex}{n}{\/tex}-sided base:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Faces {tex}=1+n=({n}+{1}){\/tex} faces<\/strong><\/li>\n\n\n\n<li><strong>Edges {tex}=n+n=2 n{\/tex} edges<\/strong><\/li>\n\n\n\n<li><strong>Vertices <\/strong>= n + 1 = <strong>( {tex}{n}{\/tex} + 1) vertices<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.17: What is a net of a regular tetrahedron? Which of the following are nets of a regular tetrahedron? Are there any other possible nets?<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768827150-svat2x.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>A regular tetrahedron has 4 equilateral triangular faces.<br><strong>The two possible nets of a regular tetrahedron are:<br>Net 1:<\/strong>&nbsp;Three triangles in a row with one triangle attached to the middle triangle<br><strong>Net 2:<\/strong>&nbsp;All four triangles arranged in a connected pattern where they share edges differently<br>A regular tetrahedron has&nbsp;<strong>only 2 possible nets<\/strong>&nbsp;(considering rotations and reflections as the same).<br><strong>Why only 2?<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>A tetrahedron has 4 faces<\/li>\n\n\n\n<li>Each face must share an edge with 3 other faces<\/li>\n\n\n\n<li>The limited number of faces and connectivity constraints result in only 2 distinct unfolding patterns<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.18: Draw a net with appropriate measurements that can be folded into a regular tetrahedron. Verify if it works by making an actual cutout.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>For a regular tetrahedron with edge length&nbsp;<strong>6 cm<\/strong>:<br><strong>Net design:&nbsp;<\/strong>Draw 4 equilateral triangles, each with side length 6 cm, arranged in one of the two patterns mentioned above.<\/p>\n\n\n\n<p><strong>Steps:<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Draw an equilateral triangle with 6 cm sides<\/li>\n\n\n\n<li>Attach three more equilateral triangles to each side of the first triangle<\/li>\n\n\n\n<li>Add small flaps (about 1 cm) on some edges for glueing<\/li>\n\n\n\n<li>Cut out the net<\/li>\n\n\n\n<li>Fold along the edges<\/li>\n\n\n\n<li>Glue the flaps to form the tetrahedron<\/li>\n<\/ol>\n\n\n\n<p><strong>Verification:<\/strong>&nbsp;When folded correctly, all 4 faces should be equilateral triangles, and all 6 edges should be 6 cm long.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.19: Draw a net with appropriate measurements that can be folded into a square pyramid. Verify if it works by making an actual cutout.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>For a square pyramid with:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Base edge = 6 cm<\/li>\n\n\n\n<li>Slant edge = 5 cm<\/li>\n<\/ul>\n\n\n\n<p><strong>Net design:<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Draw a square with 6 cm sides (the base)<\/li>\n\n\n\n<li>Attach 4 isosceles triangles to each side of the square<\/li>\n\n\n\n<li>Each triangle has:\n<ul class=\"wp-block-list\">\n<li>Base = 6 cm (shared with square)<\/li>\n\n\n\n<li>Two equal sides = 5 cm each<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<p><strong>Steps:<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Draw a 6 cm {tex}\\times{\/tex} 6 cm square in the center<\/li>\n\n\n\n<li>On each side, draw an isosceles triangle with base 6 cm and equal sides 5 cm<\/li>\n\n\n\n<li>Add small flaps for gluing<\/li>\n\n\n\n<li>Cut out the net<\/li>\n\n\n\n<li>Fold the triangles up<\/li>\n\n\n\n<li>Glue the adjacent triangles to form the pyramid<\/li>\n<\/ol>\n\n\n\n<p><strong>Verification:<\/strong>&nbsp;The base should be square, and the 4 triangular faces should meet at a common apex.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.20: What are the sidelengths of the rectangle obtained (from unfolding a cylinder)?<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768827635-t5m7vc.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>When a cylinder is unfolded into its net:<\/p>\n\n\n\n<p><strong>The rectangle has:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Length<\/strong>&nbsp;= Circumference of the circular base =&nbsp;<strong>2{tex}\\pi{\/tex}r<\/strong>&nbsp;(where r is the radius of the base)<\/li>\n\n\n\n<li><strong>Width (or height)<\/strong>&nbsp;=&nbsp;<strong>h<\/strong>&nbsp;(the height of the cylinder)<\/li>\n<\/ul>\n\n\n\n<p><strong>Explanation:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>When we &#8220;unroll&#8221; the curved surface, the distance around the circle becomes the length of the rectangle<\/li>\n\n\n\n<li>The height of the cylinder becomes the width of the rectangle<\/li>\n\n\n\n<li>The two circular faces remain as circles<\/li>\n<\/ul>\n\n\n\n<p><strong>Complete net of cylinder consists of:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>1 rectangle with dimensions 2{tex}\\pi{\/tex}r {tex}\\times{\/tex} h<\/li>\n\n\n\n<li>2 circles with radius r<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.21: How will the net of a cone look?<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768880906-b2ksub.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>The net of a cone consists of two parts:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Circular base<\/strong>&nbsp;&#8211; A complete circle with radius r<\/li>\n\n\n\n<li><strong>Curved surface<\/strong>&nbsp;&#8211; A&nbsp;<strong>sector of a larger circle<\/strong>&nbsp;with:\n<ul class=\"wp-block-list\">\n<li>Radius = l (slant height of the cone)<\/li>\n\n\n\n<li>Arc length = 2{tex}\\pi{\/tex}r (circumference of the base)<\/li>\n\n\n\n<li>Central angle \u03b8 = (2{tex}\\pi{\/tex}r)\/l \u00d7 180\u00b0\/{tex}\\pi{\/tex} = (360r)\/l degrees<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<p><strong>Shape:<\/strong>&nbsp;The curved surface unfolds into a &#8220;pac-man&#8221; or &#8220;pizza slice&#8221; shape &#8211; a sector of a circle.<br><strong>Why a sector?<\/strong><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768881078-t2kpsu.jpg\"><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>All points on the base circle are equidistant from the apex<\/li>\n\n\n\n<li>When unrolled, they remain equidistant, forming an arc<\/li>\n\n\n\n<li>The slant edges become the radii of the sector<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.22: What surface do you construct by using the above net, in which O is not the centre of the boundary circle? Make a physical model to help you answer this question!<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>If we use a sector of a circle where O (the apex of the original cone) is NOT at the center of the boundary circle, we cannot construct a cone.<\/p>\n\n\n\n<p><strong>What we get instead:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>If O is inside the boundary circle (but not at center): We get a&nbsp;<strong>portion of a cone<\/strong>&nbsp;or an&nbsp;<strong>irregular cone-like surface<\/strong><\/li>\n\n\n\n<li>If O is outside the boundary circle: The surface cannot close properly<\/li>\n<\/ul>\n\n\n\n<p><strong>For a proper cone:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>O must be at the center of curvature of the arc<\/li>\n\n\n\n<li>The arc length must equal the circumference of the base circle<\/li>\n\n\n\n<li>The radius of the sector equals the slant height<\/li>\n<\/ul>\n\n\n\n<p><strong>Experiment:<\/strong>&nbsp;Try making different sectors and see what surfaces form. Only when the geometry is correct (O at center, proper arc length) will you get a proper cone.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.23: Can you visualise its net (octahedron)?<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768882479-2nhhwx.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Yes, the net of an octahedron can be visualized.<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768882534-2v7mkw.jpg\"><\/p>\n\n\n\n<p>An octahedron has:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>8 equilateral triangular faces<\/li>\n\n\n\n<li>12 edges<\/li>\n\n\n\n<li>6 vertices<\/li>\n<\/ul>\n\n\n\n<p><strong>Net pattern:<\/strong>&nbsp;The net shown in the textbook displays 8 triangles arranged in a specific pattern.<br>This looks like a diamond or elongated hexagon shape made of triangles.<br><strong>How it folds:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The triangles fold up and meet at vertices<\/li>\n\n\n\n<li>Opposite triangular faces are parallel<\/li>\n\n\n\n<li>It creates a 3D shape with 8 faces<\/li>\n<\/ul>\n\n\n\n<p><strong>Other nets exist:<\/strong>&nbsp;Like the cube, an octahedron has 11 different possible nets.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.24: Taking all the triangles in the net to be equilateral, make a cutout of the net and fold it to form an octahedron.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>Steps to make an octahedron:<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li><strong>Draw the net:<\/strong>&nbsp;Draw 8 equilateral triangles with side length 5 cm arranged in the pattern shown in the textbook<\/li>\n\n\n\n<li><strong>Arrangement:<\/strong>&nbsp;Use the diamond pattern:\n<ul class=\"wp-block-list\">\n<li>1 triangle at top<\/li>\n\n\n\n<li>4 triangles in the middle row<\/li>\n\n\n\n<li>2 triangles in the next row<\/li>\n\n\n\n<li>1 triangle at bottom<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Add flaps:<\/strong>&nbsp;Add small tabs (about 0.5 cm) on appropriate edges for gluing<\/li>\n\n\n\n<li><strong>Cut out:<\/strong>&nbsp;Carefully cut along the outer boundary<\/li>\n\n\n\n<li><strong>Score the folds:<\/strong>&nbsp;Use a ruler and blunt object to score the edges between triangles<\/li>\n\n\n\n<li><strong>Fold and glue:<\/strong>\n<ul class=\"wp-block-list\">\n<li>Fold all the triangles upward<\/li>\n\n\n\n<li>Glue the flaps to adjacent triangles<\/li>\n\n\n\n<li>Ensure all 8 faces are equilateral triangles<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<p><strong>Result:<\/strong>&nbsp;You should get a 3D octahedron &#8211; it looks like two square pyramids joined at their bases.<br><strong>Note:<\/strong>&nbsp;As mentioned, an octahedron has 11 different nets, just like a cube.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.25: What is the shortest path for the ant to reach the laddu (when laddu is at the centre of the top face and ant is at the centre of the side face)?<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768883507-pwnuhq.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>To find the shortest path on the surface of a cuboid, we use the net of the cuboid.<br><strong>Method:<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Draw the net of the cube<\/li>\n\n\n\n<li>Mark the position of the ant on the side face<\/li>\n\n\n\n<li>Mark the position of the laddu on the top face<\/li>\n\n\n\n<li>On the net, these two points lie on a flat surface<\/li>\n\n\n\n<li>Draw a straight line between them on the net<\/li>\n\n\n\n<li>This straight line represents the shortest path<\/li>\n<\/ol>\n\n\n\n<p><strong>Why this works:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>On a flat surface, the shortest distance between two points is a straight line<\/li>\n\n\n\n<li>When we unfold the cube into a net, the surface becomes flat<\/li>\n\n\n\n<li>Any path on the cube transfers to a path of the same length on the net<\/li>\n\n\n\n<li>Therefore, the shortest path on the net (straight line) corresponds to the shortest path on the cube<\/li>\n<\/ul>\n\n\n\n<p><strong>Visual result:<\/strong>&nbsp;The path will go diagonally across the side face and continue onto the top face, reaching the laddu.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.26: What about in the following case (when laddu is at the centre of the edge)?<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768883803-s9fkgs.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>When the laddu is at the center of an edge, we again use the net method:<\/p>\n\n\n\n<p><strong>Analysis:<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Unfold the cube into a net<\/li>\n\n\n\n<li>Mark the ant&#8217;s position (center of side face)<\/li>\n\n\n\n<li>Mark the laddu&#8217;s position (center of an edge)<\/li>\n\n\n\n<li>Draw a straight line between them on the net<\/li>\n<\/ol>\n\n\n\n<p><strong>Important consideration:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The cube can be unfolded in different ways<\/li>\n\n\n\n<li>The shortest path depends on which faces we unfold together<\/li>\n\n\n\n<li>We need to check different net configurations<\/li>\n\n\n\n<li>Choose the net that gives the shortest straight-line distance<\/li>\n<\/ul>\n\n\n\n<p><strong>Result:<\/strong>&nbsp;The shortest path will be different from the first case, and the ant might travel across different faces depending on which edge the laddu is on.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.27: Are either of these the shortest path?<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768884435-qdcjed.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>To determine if a given path is the shortest:<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1769148122-k7fmzm.jpg\"><\/p>\n\n\n\n<p><strong>Test method:<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Draw the net of the cuboid<\/li>\n\n\n\n<li>Mark the ant and laddu positions on the net<\/li>\n\n\n\n<li>Draw the proposed path on the net<\/li>\n\n\n\n<li>Check if it is a straight line on the net<\/li>\n<\/ol>\n\n\n\n<p><strong>Results:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>If the path is a straight line on the net<\/strong>&nbsp;{tex}\\rightarrow{\/tex} YES, it is the shortest path<\/li>\n\n\n\n<li><strong>If the path is NOT a straight line on the net<\/strong>&nbsp;{tex}\\rightarrow{\/tex} NO, it is NOT the shortest path<\/li>\n<\/ul>\n\n\n\n<p><strong>Example from textbook:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>First path:<\/strong>&nbsp;When drawn on the net, it appears as a straight line {tex}\\rightarrow{\/tex}&nbsp;<strong>YES, shortest path<\/strong><\/li>\n\n\n\n<li><strong>Second path:<\/strong>&nbsp;When drawn on the net, it is NOT a straight line {tex}\\rightarrow{\/tex}&nbsp;<strong>NO, not the shortest path<\/strong><\/li>\n<\/ul>\n\n\n\n<p><strong>Key insight:<\/strong>&nbsp;A path that looks curved or indirect on the 3D cube might actually be straight on the net, and vice versa!<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768884643-ad7vzw.jpg\"><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.28: Find the shortest path between the ant and the laddu in the following case:<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768885346-v7ernc.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>To find the shortest path, we need to unfold the cuboid correctly.<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768885380-cf3x42.jpg\"><br><strong>Step 1:<\/strong>&nbsp;Identify the position of ant and laddu<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Ant is at the center of one face<\/li>\n\n\n\n<li>Laddu is at a specific position on the cuboid<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 2:<\/strong>&nbsp;Unfold the cuboid in different ways<br>When we unfold the cuboid, we must ensure that the path between ant and laddu stays on the surface.<br><strong>Important Point:<\/strong>&nbsp;The way a cuboid is unfolded matters! If the line segment goes outside the net, it doesn&#8217;t correspond to any actual path on the cuboid.<br><strong>Step 3:<\/strong>&nbsp;Choose correct unfolding<br>We need to unfold the cuboid such that:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The ant and laddu are on the same flat surface<\/li>\n\n\n\n<li>We can draw a straight line between them<\/li>\n\n\n\n<li>This line represents the actual path on the cuboid&#8217;s surface<\/li>\n<\/ul>\n\n\n\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768885495-3x6rye.jpg\"><br>The correct unfolding will give us the shortest straight-line distance between the ant and laddu on the unfolded surface, which corresponds to the shortest path on the actual cuboid.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.29: What is the length of the shortest path between the ant and the laddu?<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768885805-9v659n.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Dimensions: 30 cm {tex}\\times{\/tex} 12 cm {tex}\\times{\/tex} 6 cm<\/li>\n\n\n\n<li>Ant position: 1 cm from edge<\/li>\n\n\n\n<li>Laddu: stuck to the back of the box, 1 cm from edge<\/li>\n<\/ul>\n\n\n\n<p><strong>Method:<\/strong>&nbsp;We need to unfold the cuboid in different possible ways and calculate the distance in each case.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768885931-knzbg8.jpg\" alt=\"\"\/><\/figure>\n\n\n\n<p><strong>Case 1:<\/strong>&nbsp;First unfolding. When we unfold in one way, we get a certain distance.<br><strong>Case 2:<\/strong>&nbsp;Second unfolding (as shown)<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The unfolded dimensions create a right triangle<\/li>\n\n\n\n<li>One side = 24 cm<\/li>\n\n\n\n<li>Other side = 32 cm<\/li>\n<\/ul>\n\n\n\n<p>Using Baudhayana Theorem (Pythagoras Theorem): {tex}d^2=24^2+32^2 d^2=576+1024{\/tex}<br>{tex} d^2=1600 d=\\sqrt{1600 } d=40 c m {\/tex}<br><strong>Important observation:<\/strong>&nbsp;In each unfolding, the lengths of line segments between ant and laddu are different!<br><strong>Solution:<\/strong>&nbsp;We must carefully list all possible different unfoldings and calculate the distance in each case. The minimum of all these distances will be the shortest path.<br>Therefore, we need to check all possible unfoldings to find the actual shortest path. From the given unfolding, one possible answer is&nbsp;<strong>40 cm<\/strong>, but we should verify if other unfoldings give a shorter distance.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.30: What happens to the length of a line in its projection?<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768886207-arrgm9.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Let {tex}{l}={\/tex} actual length of the line<br>Let {tex}{p}={\/tex} length of its projection<br>From Fig.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Draw {tex}{AE} \\perp {BC}{\/tex}<\/li>\n\n\n\n<li>AECD is a rectangle (because opposite sides are parallel and all angles are {tex}90^{\\circ}{\/tex}<\/li>\n\n\n\n<li>Therefore, {tex}{AE}={DC}={p}{\/tex}<\/li>\n\n\n\n<li>Also, {tex}\\angle {AEB}=90^{\\circ}{\/tex}<\/li>\n<\/ul>\n\n\n\n<p>Comparison: In right triangle AEB:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}{AB}={l}{\/tex} (actual length)<\/li>\n\n\n\n<li>{tex}{AE}={p}{\/tex} (projection length)<\/li>\n\n\n\n<li>{tex}\\angle {AEB}=90^{\\circ}{\/tex}<\/li>\n<\/ul>\n\n\n\n<p>By Baudhayana Theorem: {tex}A B^2=A E^2+E B^2 {\/tex},&nbsp;{tex}l^2=p^2+E B^2{\/tex}<br>Since {tex}{EB}^2 \\geq 0{\/tex}, we get: {tex}{l}^2 \\geq {p}^2{\/tex}<br>Therefore: {tex}{l} \\geq {p}{\/tex}<br>The projection length is always less than or equal to the actual length.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.31: When is the length of the projected line equal to its actual length?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>The projection length equals the actual length when&nbsp;<strong>EB = 0<\/strong><br>This happens when:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Point E coincides with point B<\/li>\n\n\n\n<li>The line AB is parallel to the plane of projection<\/li>\n\n\n\n<li>The line lies in a plane parallel to the projection plane<\/li>\n<\/ul>\n\n\n\n<p><strong>Therefore:<\/strong>&nbsp;The projected line length equals actual length when the line is parallel to the plane of projection.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.32: What do you think are the different possible projections of a square that we get based on its orientation?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Based on different orientations, a square can have the following projections:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li><strong>Square:<\/strong>&nbsp;When the square is parallel to the projection plane<\/li>\n\n\n\n<li><strong>Rectangle:<\/strong>&nbsp;When the square is tilted at an angle to the projection plane<\/li>\n\n\n\n<li><strong>Line segment:<\/strong>&nbsp;When the square is perpendicular to the projection plane (edge-on view)<\/li>\n\n\n\n<li><strong>Parallelogram:<\/strong>&nbsp;Generally not possible for a square, as opposite sides remain parallel and equal in projection<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.33: What do you think is the projection of a parallelogram under different orientations? Can this ever be a quadrilateral that is not a parallelogram?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>Projection of a parallelogram:<\/strong><br>Under different orientations, a parallelogram can project as:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li><strong>Parallelogram:<\/strong>&nbsp;Most common case<\/li>\n\n\n\n<li><strong>Rectangle:<\/strong>&nbsp;Special case when oriented appropriately<\/li>\n\n\n\n<li><strong>Line segment:<\/strong>&nbsp;When perpendicular to projection plane<\/li>\n\n\n\n<li><strong>Smaller parallelogram:<\/strong>&nbsp;When tilted<\/li>\n<\/ol>\n\n\n\n<p><strong>Can it be a non-parallelogram quadrilateral?<br>No!<\/strong>&nbsp;The projection of a parallelogram is always a parallelogram (or a degenerate case like a line segment).<br><strong>Reason:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Parallel lines always project to parallel lines.<\/li>\n\n\n\n<li>Since opposite sides of a parallelogram are parallel, their projections will also be parallel.<\/li>\n\n\n\n<li>A quadrilateral with both pairs of opposite sides parallel is a parallelogram.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.34: What can you say about the projection of an n-sided regular polygon?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>For an n-sided regular polygon:<br><strong>When parallel to projection plane:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Projection is the same n-sided regular polygon<\/li>\n\n\n\n<li>All sides and angles are preserved<\/li>\n<\/ul>\n\n\n\n<p><strong>When tilted:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Projection is an n-sided polygon (same number of sides)<\/li>\n\n\n\n<li>It may not be regular anymore<\/li>\n\n\n\n<li>Opposite sides that were parallel remain parallel in projection<\/li>\n\n\n\n<li>The shape becomes a compressed or stretched version<\/li>\n<\/ul>\n\n\n\n<p><strong>Special cases:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>When perpendicular:<\/strong>&nbsp;Projects to a line segment<\/li>\n\n\n\n<li><strong>When at intermediate angles:<\/strong>&nbsp;Projects to an n-sided polygon with modified angles and side lengths<\/li>\n<\/ul>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768886950-wjbb45.jpg\" alt=\"\"\/><\/figure>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.35: Find another object that makes the same projection as that of a given cone.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Several objects can make the same projection as a cone:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li><strong>Cylinder:<\/strong>&nbsp;When viewed from the side, gives the same rectangular projection as a cone<\/li>\n\n\n\n<li><strong>Truncated cone (Frustum):<\/strong>&nbsp;Can give similar projections from certain angles<\/li>\n\n\n\n<li><strong>Another cone of different height:<\/strong>&nbsp;A cone with different dimensions but same base diameter can give the same circular top view<\/li>\n\n\n\n<li><strong>Hemisphere:<\/strong>&nbsp;The top view (projection on horizontal plane) of both cone and hemisphere is a circle<\/li>\n<\/ol>\n\n\n\n<p><strong>Examples based on the three views:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Top View (Circle):<\/strong>&nbsp;Cone, Cylinder, Sphere, Hemisphere all give circular top view<\/li>\n\n\n\n<li><strong>Front View (Triangle):<\/strong>&nbsp;Cone, Pyramid with triangular\/square base can give triangular front view<\/li>\n\n\n\n<li><strong>Side View:<\/strong>&nbsp;Similar to front view for symmetrical objects<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.36: Construct a model of a cube and use your hands to keep it balanced on one corner vertex. Can you try to understand why all the projected edges have equal length?<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768898822-exgc3y.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>Experiment:<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Take a cube and balance it on one vertex<\/li>\n\n\n\n<li>The main diagonal (from this vertex to opposite vertex) is now vertical<\/li>\n\n\n\n<li>Project it down to a horizontal plane<\/li>\n<\/ol>\n\n\n\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1769150162-yn43b5.jpg\"><br><strong>Why all edges have equal length in projection:<br>Symmetry Analysis:<\/strong><br>When a cube is balanced on one vertex:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The main diagonal is perpendicular to the projection plane<\/li>\n\n\n\n<li>All 12 edges of the cube make the&nbsp;<strong>same angle<\/strong>&nbsp;with the projection plane<\/li>\n\n\n\n<li>Due to this symmetry, all edges project to equal lengths<\/li>\n<\/ul>\n\n\n\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1769150220-jb88ue.jpg\"><br><strong>Mathematical Reasoning:<\/strong><br>From earlier, we learned:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Projection length = Actual length {tex}\\times{\/tex} cos({tex}\\theta{\/tex})<\/li>\n\n\n\n<li>Where {tex}\\theta{\/tex} is the angle between the edge and the projection plane<\/li>\n<\/ul>\n\n\n\n<p>For a cube balanced on vertex:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>All edges make the same angle {tex}\\theta{\/tex} with the horizontal plane<\/li>\n\n\n\n<li>Therefore, all projections have same length = edge length \u00d7 cos({tex}\\theta{\/tex})<\/li>\n<\/ul>\n\n\n\n<p><strong>Geometric Visualization:<\/strong><br>The cube has 4-fold rotational symmetry about its main diagonal:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Rotating by 90<sup>o<\/sup> about the diagonal brings one edge to position of another<\/li>\n\n\n\n<li>This symmetry is preserved in projection<\/li>\n\n\n\n<li>Hence all edge projections are equal<\/li>\n<\/ul>\n\n\n\n<p><strong>Result:<\/strong>&nbsp;The projection forms a regular hexagon where all sides are equal because all 12 edges of the cube are equally inclined to the projection plane.<br><strong>This is the principle behind isometric projection!<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.37: Why is this correspondence between directions on isometric paper and axes of the solid so effective for communicating the shape of the solid?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>The isometric correspondence is effective for several reasons:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li><strong>Parallel Lines Remain Parallel:<\/strong>\n<ul class=\"wp-block-list\">\n<li>In the solid: edges parallel to length axis are parallel to each other<\/li>\n\n\n\n<li>In projection: these edges also appear parallel (in direction)<\/li>\n\n\n\n<li>This preserves the geometric structure<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Equal Scaling Along All Axes:<\/strong>\n<ul class=\"wp-block-list\">\n<li>All three axes (length, depth, height) have the&nbsp;<strong>same scale<\/strong><\/li>\n\n\n\n<li>If actual edge = 1 unit, its projection = k units (same k for all axes)<\/li>\n\n\n\n<li>This makes measurement easy and consistent<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Visual Clarity:<\/strong>\n<ul class=\"wp-block-list\">\n<li>Our brain can easily interpret three families of parallel lines as three dimensions<\/li>\n\n\n\n<li>The three directions (|, , ) are visually distinct<\/li>\n\n\n\n<li>Easy to distinguish which edges belong to which axis<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Measurement Convenience:<\/strong>\n<ul class=\"wp-block-list\">\n<li>Can measure actual dimensions directly from the drawing<\/li>\n\n\n\n<li>All three axes use the same grid spacing<\/li>\n\n\n\n<li>No complex calculations needed<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Preserves Spatial Relationships:<\/strong>\n<ul class=\"wp-block-list\">\n<li>If two edges are perpendicular in the solid, their projections make consistent angles<\/li>\n\n\n\n<li>Relative positions are maintained<\/li>\n\n\n\n<li>Easy to visualize the 3D structure from 2D drawing<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Engineering Applications:<\/strong>\n<ul class=\"wp-block-list\">\n<li>Standard format recognized worldwide<\/li>\n\n\n\n<li>Easy to draw with just ruler and grid paper<\/li>\n\n\n\n<li>Can quickly sketch complex 3D objects<\/li>\n\n\n\n<li>Facilitates communication between designers and builders<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<p><strong>Mathematical Basis:<\/strong>The isometric projection ensures that:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Equal actual distances along any axis {tex}\\rightarrow{\/tex} Equal projected distances<\/li>\n\n\n\n<li>This &#8220;equal measure&#8221; property (iso = equal, metric = measure) makes it intuitive<\/li>\n<\/ul>\n\n\n\n<p><strong>Conclusion:<\/strong>&nbsp;The combination of parallel line preservation, equal scaling, and visual distinctness makes isometric paper the most effective tool for technical drawing.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.38: Which of the following are the nets of a cube? First, try to answer by visualisation. Then, you may use cutouts and try.<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768826076-kxxbq7.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Let&#8217;s analyze each net by visualizing how it would fold:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>This arrangement can fold into a cube. When you fold it, all six faces will properly connect. YES, this is a net of a cube.<\/li>\n\n\n\n<li>This arrangement has all 6 squares, but when you try to fold it, two faces will overlap, and one position will be empty. NO, this is NOT a net of a cube.<\/li>\n\n\n\n<li>This arrangement can fold properly into a cube with all faces in correct positions.<br>YES, this is a net of a cube.<\/li>\n\n\n\n<li>This arrangement looks like it would work. When folded, all faces align correctly.<br>YES, this is a net of a cube.<\/li>\n\n\n\n<li>This arrangement has 6 squares in a straight line. When you fold it, opposite faces will overlap. NO, this is NOT a net of a cube.<\/li>\n\n\n\n<li>This arrangement forms a &#8216;{tex}T{\/tex}&#8217; shape. When folded, it creates a cube properly.<\/li>\n<\/ol>\n\n\n\n<p>YES, this is a net of a cube.<br>Summary: (i), (iii), (iv), and (vi) are nets of a cube. (ii) and (v) are NOT nets of a cube.<br><strong>Verification method:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>A cube has 6 faces<\/li>\n\n\n\n<li>When folded, no two faces should overlap<\/li>\n\n\n\n<li>Each face should connect to exactly the right neighbors<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.39: A cube has 11 possible net structures in total. In this count, two nets are considered the same if one can be obtained from the other by a rotation or a flip. For example, the following nets are all considered the same\u2009-<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768826370-zd9nmj.jpg\"><br>Find all the 11 nets of a cube.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>The 11 distinct nets of a cube are classified by their basic patterns:<br><strong>Type 1: &#8220;1-4-1&#8221; pattern ({tex}{1}{\/tex} square, {tex}{4}{\/tex} in a row, {tex}{1}{\/tex} square) &#8211;<\/strong> 1 net<br>One square, then four in a row, then one square attached<br><strong>Type 2: &#8220;2-3-1&#8221; pattern ({tex}{2}{\/tex} squares, {tex}{3}{\/tex} in a row, {tex}{1}{\/tex} square) &#8211; <\/strong>3 nets<br>Different positions of where the 2 squares and 1 square attach to the row of 3<br><strong>Type 3: &#8220;2-2-2&#8221; pattern (three pairs) &#8211; <\/strong>1 net<br>A more symmetric arrangement<br><strong>Type 4: &#8220;1-3-2&#8221; pattern &#8211; <\/strong>2 nets<br>Different configurations<br><strong>Type 5: &#8220;3-2-1&#8221; variations &#8211; <\/strong>2 nets<br><strong>Type 6: Other arrangements &#8211;<\/strong> 2 nets<br>[Students should draw all 11 nets. This requires careful systematic exploration by:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Starting with a row of 4 squares and adding 2 more<\/li>\n\n\n\n<li>Trying different attachment positions<\/li>\n\n\n\n<li>Checking that each folds into a cube<\/li>\n\n\n\n<li>Eliminating duplicates from rotation\/reflection]<\/li>\n<\/ol>\n\n\n\n<p>The key is to ensure each configuration:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Has exactly 6 squares<\/li>\n\n\n\n<li>Can fold into a cube<\/li>\n\n\n\n<li>Is not a rotation or reflection of another net already found<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.40: Draw a net of a cuboid having sidelengths&nbsp;5 cm, 3 cm, and 1 cm.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>A cuboid has 6 rectangular faces:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>2 faces of dimension {tex}5 {~cm} \\times 3 {~cm}{\/tex} (top and bottom)<\/li>\n\n\n\n<li>2 faces of dimension {tex}5 {~cm} \\times 1 {~cm}{\/tex} (front and back)<\/li>\n\n\n\n<li>2 faces of dimension {tex}3 {~cm} \\times 1 {~cm}{\/tex} (left and right sides)<\/li>\n<\/ul>\n\n\n\n<p>Students should draw the net with proper measurements labeled on each rectangle.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.41: Draw a net of a cuboid having sidelengths&nbsp;6 cm, 3 cm, and 2 cm.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>A cuboid has 6 rectangular faces:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>2 faces of dimension {tex}6 {~cm} \\times 3 {~cm}{\/tex} (top and bottom)<\/li>\n\n\n\n<li>2 faces of dimension {tex}6 {~cm} \\times 2 {~cm}{\/tex} (front and back)<\/li>\n\n\n\n<li>2 faces of dimension {tex}3 {~cm} \\times 2 {~cm}{\/tex} (left and right sides)<\/li>\n<\/ul>\n\n\n\n<p>Students should draw the net with proper measurements labeled.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.42: Net of a sphere? Experiment and see if you can make a paper cutout that can perfectly wrap around a ball without leaving any wrinkles, gaps or overlaps.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>It is IMPOSSIBLE to create a perfect flat net of a sphere.<br>Why?<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li><strong>Curved surface:<\/strong>&nbsp;A sphere has a continuously curved surface in all directions<\/li>\n\n\n\n<li><strong>No flat net exists:<\/strong>&nbsp;Unlike polyhedra (cube, pyramid, prism) which have flat faces, a sphere&#8217;s surface cannot be &#8220;unfolded&#8221; into a flat shape without distortion<\/li>\n\n\n\n<li><strong>Mathematical proof:<\/strong>&nbsp;This is related to Gaussian curvature. A sphere has positive curvature everywhere, while a flat plane has zero curvature. You cannot map one to the other without distortion.<\/li>\n<\/ol>\n\n\n\n<p><strong>What happens when you try?<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Wrinkles:<\/strong>&nbsp;If you try to flatten a curved surface, it wrinkles<\/li>\n\n\n\n<li><strong>Gaps:<\/strong>&nbsp;If you try to avoid wrinkles, gaps appear<\/li>\n\n\n\n<li><strong>Stretching\/Overlapping:<\/strong>&nbsp;To cover completely, material must stretch or overlap<\/li>\n<\/ul>\n\n\n\n<p><strong>Real-world example:<\/strong>&nbsp;This is why world maps always distort the Earth&#8217;s surface &#8211; you cannot perfectly represent a sphere on a flat map.<br><strong>Approximations:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>We can use multiple pieces (like a soccer ball &#8211; made of pentagons and hexagons)<\/li>\n\n\n\n<li>We can use stretchy material<\/li>\n\n\n\n<li>But never a perfect single flat net<\/li>\n<\/ul>\n\n\n\n<p><strong>Try it:<\/strong>&nbsp;Take a piece of paper and try to wrap it around a ball &#8211; you&#8217;ll see it&#8217;s impossible without creating wrinkles or tears!<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.43: Have we now completely analysed the problem of finding the shortest path between two points on a cuboid?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>Not completely!<\/strong>&nbsp;There are still considerations:<br><strong>What we have established:<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>The shortest path on a cuboid corresponds to a straight line on its net<\/li>\n\n\n\n<li>We can use nets to find shortest paths<\/li>\n<\/ol>\n\n\n\n<p><strong>What remains:<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li><strong>Multiple nets:<\/strong>&nbsp;A cuboid can be unfolded in different ways, creating different nets<\/li>\n\n\n\n<li><strong>Different paths:<\/strong>&nbsp;Each net might give a different straight-line path<\/li>\n\n\n\n<li><strong>Finding the minimum:<\/strong>&nbsp;We need to check multiple nets and find which gives the shortest distance<\/li>\n<\/ol>\n\n\n\n<p><strong>Complete solution:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Unfold the cuboid in all possible relevant ways<\/li>\n\n\n\n<li>For each net, draw a straight line from ant to laddu<\/li>\n\n\n\n<li>Calculate the length of each line<\/li>\n\n\n\n<li>The shortest among these is the true shortest path<\/li>\n<\/ul>\n\n\n\n<p><strong>Additional complexity:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Some paths might cross edges in ways that aren&#8217;t immediately obvious<\/li>\n\n\n\n<li>We need to ensure the path stays on the surface<\/li>\n\n\n\n<li>For complex shapes, this becomes a challenging optimization problem<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.44: Observe the front view, top view and side view of the different lines in Fig. 4.6. Is there any relation between their lengths?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Yes, there is a definite relation between the lengths of different views.<br>For a line in 3D space:<br>Let the actual length of line {tex}=1{\/tex}<br><strong>Observations:<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>At least one of the three views will show the maximum length<\/li>\n\n\n\n<li>If a line is parallel to one plane, its projection on that plane equals its actual length<\/li>\n\n\n\n<li>Projections on perpendicular planes follow the relation:<\/li>\n<\/ol>\n\n\n\n<p><strong>Relationship If:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Front view length {tex}={I}_1{\/tex}<\/li>\n\n\n\n<li>Top view length {tex}={I}_2{\/tex}<\/li>\n\n\n\n<li>Side view length {tex}={I}_3{\/tex}<\/li>\n\n\n\n<li>Actual length = I<\/li>\n<\/ul>\n\n\n\n<p>Then: {tex}{I}^{{2}} \\boldsymbol{\\geq} {I}_{{1}}^{{2}}{\/tex} and {tex}{I}^{{2}} \\boldsymbol{\\geq} {I}_{{2}}^{{2}}{\/tex} and {tex}{I}^{{2}} \\boldsymbol{\\geq} {I}_{{3}}^{{2}}{\/tex}<br>In fact, for certain orientations: {tex}I^2=I_1^2+I_2^2{\/tex} (when line is parallel to the side plane)<br><strong>Conclusion: <\/strong>The actual length is always greater than or equal to any of its projections. The projection lengths are related through the orientation of the line in 3D space.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.45: Find the front view, top view and side view of each of the following solids, fixing its orientation with respect to the vertical, horizontal and side planes: cube, cuboid, parallelepiped, cylinder, cone, prism, and pyramid. If needed, see the next problem for clues.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li><strong>Cube<\/strong>&nbsp;(assuming standard orientation with faces parallel to planes) <strong>View<\/strong> <strong>Shape<\/strong> <strong>Dimensions<\/strong> Front View Square {tex}{a} \\times {a}{\/tex} Top View Square {tex}{a} \\times {a}{\/tex} Side View Square {tex}{a} \\times {a}{\/tex}<\/li>\n\n\n\n<li><strong>Cuboid<\/strong>&nbsp;(dimensions: length l, breadth b, height h) <strong>View<\/strong> <strong>Shape<\/strong> <strong>Dimensions<\/strong> Front View Rectangle {tex}{l} \\times {h}{\/tex} Top View Rectangle {tex}{l} \\times {b}{\/tex} Side View Rectangle {tex}{b} \\times {h}{\/tex}<\/li>\n\n\n\n<li><strong>Parallelepiped<\/strong>&nbsp;(all faces are parallelograms) <strong>View<\/strong> <strong>Shape<\/strong> Front View Parallelogram Top View Parallelogram Side View Parallelogram<\/li>\n\n\n\n<li><strong>Cylinder<\/strong>&nbsp;(axis vertical) <strong>View<\/strong> <strong>Shape<\/strong> Front View Rectangle Top View Circle Side View Rectangle<\/li>\n\n\n\n<li><strong>Cone<\/strong>&nbsp;(axis vertical, base on horizontal plane) <strong>View<\/strong> <strong>Shape<\/strong> Front View Isosceles Triangle Top View Circle Side View Isosceles Triangle<\/li>\n\n\n\n<li><strong>Prism<\/strong>&nbsp;(regular prism with square base, axis vertical) <strong>View<\/strong> <strong>Shape<\/strong> Front View Rectangle Top View Square Side View Rectangle<\/li>\n\n\n\n<li><strong>Pyramid<\/strong>&nbsp;(square pyramid, axis vertical) <strong>View<\/strong> <strong>Shape<\/strong> Front View Isosceles Triangle Top View Square Side View Isosceles Triangle<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.46: Match each of the following objects with its projections.<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768889314-argtv3.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li><strong>Cone (with axis vertical):<\/strong>\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Front: Triangle (isosceles)<\/li>\n\n\n\n<li>Top: Circle<\/li>\n\n\n\n<li>Side: Triangle (isosceles)<\/li>\n<\/ol>\n<\/li>\n\n\n\n<li><strong>Cylinder (axis vertical):<\/strong>\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Front: Rectangle<\/li>\n\n\n\n<li>Top: Circle<\/li>\n\n\n\n<li>Side: Rectangle<\/li>\n<\/ol>\n<\/li>\n\n\n\n<li><strong>Cuboid:<\/strong>\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Front: Rectangle<\/li>\n\n\n\n<li>Top: Rectangle (different dimensions)<\/li>\n\n\n\n<li>Side: Rectangle (different dimensions)<\/li>\n<\/ol>\n<\/li>\n\n\n\n<li><strong>Hexagonal Prism (axis vertical):<\/strong>\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Front: Rectangle<\/li>\n\n\n\n<li>Top: Hexagon (regular)<\/li>\n\n\n\n<li>Side: Rectangle<\/li>\n<\/ol>\n<\/li>\n\n\n\n<li><strong>Cube:<\/strong>\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Front: Square<\/li>\n\n\n\n<li>Top: Square<\/li>\n\n\n\n<li>Side: Square<\/li>\n<\/ol>\n<\/li>\n\n\n\n<li><strong>Square Pyramid:<\/strong>\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Front: Triangle<\/li>\n\n\n\n<li>Top: Square<\/li>\n\n\n\n<li>Side: Triangle<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n\n\n\n<p><strong>Matching Process:<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Look at the top view first &#8211; it often gives the best clue<\/li>\n\n\n\n<li>If top is a circle, object is cone, cylinder, or sphere<\/li>\n\n\n\n<li>If top is a polygon, count sides to identify the base<\/li>\n\n\n\n<li>Use front and side views to confirm<\/li>\n<\/ol>\n\n\n\n<p><strong>Final Matching:<\/strong>&nbsp;Match according to the specific diagrams provided in the actual question.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.47: Draw the top view, front view and the side view of each of the following combinations of identical cubes.<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768889682-qbbxjy.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>To solve this, we need to imagine viewing the stacked cubes from three directions:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li><strong>First arrangement:<\/strong><br><strong>Front view:&nbsp;<\/strong>Count cubes visible from front:<ul><li>Bottom row: all cubes in front row<\/li><li>Upper rows: cubes visible above<\/li><li>Draw as a 2D representation<\/li><\/ul>Top view:&nbsp;Look from directly above:<ul><li>Shows the footprint\/base layout<\/li><li>Each square represents a column of cubes<\/li><\/ul>Side view:&nbsp;Look from the side (perpendicular to front):<br>Shows height profile from that angle<br>General Process:\n<ul class=\"wp-block-list\">\n<li><strong>For Front View:<\/strong>&nbsp;Count maximum cubes in each column when viewed from front<\/li>\n\n\n\n<li><strong>For Top View:<\/strong>&nbsp;Show the base layout (which positions have cubes)<\/li>\n\n\n\n<li><strong>For Side View:&nbsp;<\/strong>Count maximum cubes in each row when viewed from side<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Second arrangement:<\/strong>[Follow same process]<\/li>\n\n\n\n<li><strong>Third arrangement:<\/strong>[Follow same process]<\/li>\n\n\n\n<li><strong>Fourth arrangement:<\/strong>[Follow same process]<\/li>\n\n\n\n<li><strong>Fifth arrangement:<\/strong>[Follow same process]<br>Important Tips:\n<ul class=\"wp-block-list\">\n<li>Front is clearly marked in diagrams<\/li>\n\n\n\n<li>Top view shows x-y plane layout<\/li>\n\n\n\n<li>Side view is perpendicular to front view<\/li>\n\n\n\n<li>Count carefully for stacked cubes<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.48: Imagine eight identical cubes, glued together along faces to form the letter<img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768891374-c9rvhk.jpg\">\uff0e<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>This looks like a <img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768891374-c9rvhk.jpg\"> from the front. What does it look like from the side? From the top\uff1f<\/li>\n\n\n\n<li>Glue additional cubes to make a shape that looks like <img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768891374-c9rvhk.jpg\">&nbsp;from the front and&nbsp;<img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768891477-68va39.jpg\">from the top\uff0e<\/li>\n\n\n\n<li>Now, can you glue even more cubes to make it look like <img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768891374-c9rvhk.jpg\"> from the front,&nbsp;<img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768891477-68va39.jpg\"> from the top, and <img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768891513-jgbmgg.jpg\"> from the side\uff1f<\/li>\n\n\n\n<li>Can you think of other letter combinations to make with a single combination of cubes in this manner\uff1f<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Letter &#8216;E&#8217; structure (8 cubes):<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768891374-c9rvhk.jpg\"><br><strong>From the front:<\/strong>&nbsp;Letter &#8216;E&#8217; (as given) <strong>From the side (right side):&nbsp;<\/strong>Looking perpendicular to the front: <strong>From the top:&nbsp;<\/strong>Looking down: <strong>Top view pattern:<\/strong><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768891374-c9rvhk.jpg\">\n<ul class=\"wp-block-list\">\n<li>3 cubes in top row<\/li>\n\n\n\n<li>1 cube in middle left<\/li>\n\n\n\n<li>3 cubes in bottom row (if standard E)<\/li>\n\n\n\n<li>Total vertical height: 3 units<\/li>\n\n\n\n<li>We see the depth profile<\/li>\n\n\n\n<li>Appears as: 3 separate horizontal segments at different heights<\/li>\n\n\n\n<li>Looks like three separate short rectangles (or squares if single depth)<\/li>\n\n\n\n<li>Shows three separate rows<\/li>\n\n\n\n<li>Top row: 3 cubes in a line<\/li>\n\n\n\n<li>Middle row: 1 cube (left position)<\/li>\n\n\n\n<li>Bottom row: 3 cubes in a line<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Requirements:<\/strong><strong>Solution:<\/strong><br>Start with &#8216;E&#8217; from front (8 cubes minimum)<br>For top view to show &#8216;L&#8217;: Construction:Add cubes strategically: <strong>Example arrangement:<\/strong><br>Front view: E<br>Top view: L (one vertical line, one horizontal line at bottom)&nbsp;<br><strong>Specific position:<\/strong>&nbsp;Add cubes at positions that maintain both views simultaneously.\n<ul class=\"wp-block-list\">\n<li>Front view: &#8216;E&#8217;<\/li>\n\n\n\n<li>Top view: &#8216;L&#8217;<\/li>\n\n\n\n<li>Need vertical line (3 cubes)<\/li>\n\n\n\n<li>Need horizontal line at bottom (3 cubes)<\/li>\n\n\n\n<li>The &#8216;L&#8217; should be visible from above<\/li>\n\n\n\n<li>Keep front &#8216;E&#8217; structure<\/li>\n\n\n\n<li>Arrange cubes so from top they form &#8216;L&#8217;<\/li>\n\n\n\n<li>May need 9-12 cubes total<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Requirements:<\/strong><strong>Analysis:<\/strong><br>Letter &#8216;F&#8217; from side means:<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768892738-sfj2wy.jpg\"><br>(Horizontal line at top, vertical line down with middle horizontal)<br><strong>Solution strategy:<\/strong><strong>Construction:<\/strong><br>Position cubes in 3D space such that: This requires careful spatial planning and possibly some cubes serving multiple purposes in different views.<br><strong>Yes, it is possible!<\/strong>&nbsp;With approximately 12-18 cubes, we can create this multi-view structure.<ul><li>Front view: &#8216;E&#8217;<\/li><li>Top view: &#8216;L&#8217;<\/li><li>Side view: &#8216;F&#8217;<\/li><\/ul><ol start=\"1\"><li>Start with structure from part (ii) giving &#8216;E&#8217; and &#8216;L&#8217;<\/li><li>Add cubes to create &#8216;F&#8217; from side without destroying other views<\/li><li>Total cubes needed: approximately 12-15<\/li><\/ol>\n<ul class=\"wp-block-list\">\n<li>Looking from front {tex}\\rightarrow{\/tex} &#8216;E&#8217; pattern<\/li>\n\n\n\n<li>Looking from top {tex}\\rightarrow{\/tex} &#8216;L&#8217; pattern<\/li>\n\n\n\n<li>Looking from side {tex}\\rightarrow{\/tex} &#8216;F&#8217; pattern<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Yes! Many letter combinations are possible:<br><strong>Easier combinations:<\/strong><ol><li><strong>I, I, I<\/strong>&nbsp;&#8211; A straight vertical column looks like &#8216;I&#8217; from all three views<\/li><li><strong>T, T, +<\/strong>&nbsp;&#8211; T-shape from two sides, plus from top<\/li><li><strong>L, I, L<\/strong>&nbsp;&#8211; L from front, vertical line from top, L from side<\/li><li><strong>C, U, C<\/strong>&nbsp;&#8211; Curved shapes<\/li><\/ol><strong>Moderate combinations:<\/strong>5.&nbsp;<strong>E, F, L<\/strong>&nbsp;&#8211; As solved above 6.&nbsp;<strong>H, I, H<\/strong>&nbsp;&#8211; Using 7 cubes 7.&nbsp;<strong>T, L, T<\/strong>&nbsp;&#8211; T from front, L from top, T from side 8.&nbsp;<strong>U, C, U<\/strong>&nbsp;&#8211; U shapes from different angles<br><strong>Challenging combinations:<\/strong>9.&nbsp;<strong>F, L, E<\/strong>&nbsp;&#8211; Reverse of (iii) 10.&nbsp;<strong>P, L, F<\/strong>&nbsp;&#8211; Using specific arrangements<br><strong>Strategy for creating:<\/strong><ul><li>Start with simpler letters (I, L, T, C, U)<\/li><li>Use symmetry when possible<\/li><li>Each cube can contribute to multiple views<\/li><li>Typically need 6-15 cubes for most combinations<\/li><\/ul><strong>Creative exploration:<\/strong>&nbsp;Try creating your name initials or other meaningful letter combinations!<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.49: Which solid corresponds to the given top view, front view, and side view?<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768894116-hetpmk.jpg\"><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768894161-akjh7z.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>To identify the solid from three views:<\/p>\n\n\n\n<p><strong>Process:<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Analyze the&nbsp;<strong>top view<\/strong>&nbsp;&#8211; gives base shape<\/li>\n\n\n\n<li>Analyze the&nbsp;<strong>front view<\/strong>&nbsp;&#8211; gives height profile<\/li>\n\n\n\n<li>Analyze the&nbsp;<strong>side view<\/strong>&nbsp;&#8211; confirms structure<\/li>\n\n\n\n<li>Match with given options<\/li>\n<\/ol>\n\n\n\n<p><strong>For each option:<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li><strong>First solid:<\/strong>\n<ul class=\"wp-block-list\">\n<li>Compare top, front, side views<\/li>\n\n\n\n<li>Count cubes in each direction<\/li>\n\n\n\n<li>Match the pattern<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Second solid:<\/strong>\n<ul class=\"wp-block-list\">\n<li>Check if views align<\/li>\n\n\n\n<li>Verify heights and positions<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Third solid:<\/strong>\n<ul class=\"wp-block-list\">\n<li>Examine structure<\/li>\n\n\n\n<li>Confirm all three views match<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Fourth solid:<\/strong><br>Validate against projections<\/li>\n\n\n\n<li><strong>Fifth solid:<\/strong><br>Check consistency<\/li>\n\n\n\n<li><strong>Sixth solid:<\/strong><br>Verify match<\/li>\n\n\n\n<li><strong>Seventh solid:<\/strong><br>Final validation<\/li>\n<\/ol>\n\n\n\n<p><strong>Answer Method:<\/strong>For each given set of views (Front, Top, Side):<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>The&nbsp;<strong>top view<\/strong>&nbsp;shows which positions have cubes when looking down<\/li>\n\n\n\n<li>The&nbsp;<strong>front view<\/strong>&nbsp;shows maximum height at each front-back position<\/li>\n\n\n\n<li>The&nbsp;<strong>side view<\/strong>&nbsp;shows maximum height at each left-right position<\/li>\n\n\n\n<li>Reconstruct the 3D structure mentally<\/li>\n\n\n\n<li>Match with given solid options<\/li>\n<\/ol>\n\n\n\n<p><strong>Matching:<\/strong>&nbsp;[Provide specific matches based on actual diagrams: e.g., Views in question match solid (iii)]\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.50: Using identical cubes, make a solid that gives the following projections.<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768897962-aadwd4.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>For each set of projections, we reconstruct the 3D solid:<\/p>\n\n\n\n<p><strong>(i) First set of projections:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Top View: Shows base layout<\/li>\n\n\n\n<li>Front View: Shows front profile<\/li>\n\n\n\n<li>Side View: Shows side profile<\/li>\n<\/ul>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Start with top view &#8211; this shows the footprint (which columns have cubes)<\/li>\n\n\n\n<li>Use front view to determine height of each column in front-back direction<\/li>\n\n\n\n<li>Use side view to verify and confirm heights<\/li>\n\n\n\n<li>Build the solid cube by cube<\/li>\n<\/ol>\n\n\n\n<p><strong>Example:<\/strong>&nbsp;If top view shows 2{tex}\\times{\/tex}2 grid, front view shows heights [2,1], side view shows heights [2,1]:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Position (1,1): 2 cubes high<\/li>\n\n\n\n<li>Position (1,2): 1 cube high<\/li>\n\n\n\n<li>Position (2,1): 2 cubes high<\/li>\n\n\n\n<li>Position (2,2): 1 cube high<\/li>\n<\/ul>\n\n\n\n<p><strong>(ii) through (ix): Similar process<\/strong><\/p>\n\n\n\n<p><strong>General Algorithm:<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Mark positions on grid based on top view<\/li>\n\n\n\n<li>For each position, determine maximum height using front and side views<\/li>\n\n\n\n<li>The actual height at position (i,j) is minimum of:\n<ul class=\"wp-block-list\">\n<li>Height shown in front view for column i<\/li>\n\n\n\n<li>Height shown in side view for row j<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Place cubes accordingly<\/li>\n<\/ol>\n\n\n\n<p><strong>Important Notes:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Some configurations may have multiple solutions<\/li>\n\n\n\n<li>Ensure all three views match your construction<\/li>\n\n\n\n<li>Verify by checking each view of your final solid<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.51: Find the number of cubes in this stack of identical cubes.<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768898317-xbpr6z.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>Method 1: Direct Counting.&nbsp;<\/strong><br>Count visible cubes and estimate hidden ones:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Count all visible cubes from front: ________ cubes<\/li>\n\n\n\n<li>Count partially hidden cubes: ________ cubes<\/li>\n\n\n\n<li>Count completely hidden cubes: ________ cubes<\/li>\n<\/ul>\n\n\n\n<p><strong>Method 2: Layer by Layer<\/strong><br>If we can identify layers:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Bottom layer: ________ cubes<\/li>\n\n\n\n<li>Second layer: ________ cubes<\/li>\n\n\n\n<li>Third layer: ________ cubes<\/li>\n\n\n\n<li>Top layer: ________ cubes<\/li>\n<\/ul>\n\n\n\n<p><strong>Total = Sum of all layers<br>Method 3: Using Projections<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>From top view: count positions<\/li>\n\n\n\n<li>From front view: find heights<\/li>\n\n\n\n<li>From side view: verify structure<\/li>\n\n\n\n<li>Calculate total<\/li>\n<\/ol>\n\n\n\n<p><strong>Method 4: Formula (if regular pattern)<\/strong>If it&#8217;s a rectangular arrangement of dimensions l {tex}\\times{\/tex} b {tex}\\times{\/tex} h:<br><strong>Total cubes = l {tex}\\times{\/tex} b {tex}\\times{\/tex} h<br>Note:<\/strong>&nbsp;Without the specific diagram, I cannot give the exact number. The answer would be calculated using one of the above methods based on the actual stack shown.<br><strong>Example:<\/strong>&nbsp;If the stack is 3 {tex}\\times{\/tex} 3 {tex}\\times{\/tex} 3 cube<br>Total = 3 {tex}\\times{\/tex} 3 {tex}\\times{\/tex} 3 = 27 cubes<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.52: What are the different shapes the projection of a cube can make under different orientations?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>The projection of a cube can make several different shapes depending on its orientation:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li><strong>Square<\/strong>\n<ul class=\"wp-block-list\">\n<li>When any face is parallel to the projection plane<\/li>\n\n\n\n<li>Most common projection<\/li>\n\n\n\n<li>All sides equal<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Rectangle<\/strong>\n<ul class=\"wp-block-list\">\n<li>When cube is tilted slightly<\/li>\n\n\n\n<li>Length &gt; Breadth<\/li>\n\n\n\n<li>Still has right angles<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Parallelogram<\/strong>\n<ul class=\"wp-block-list\">\n<li>When cube is tilted more significantly<\/li>\n\n\n\n<li>Opposite sides parallel and equal<\/li>\n\n\n\n<li>Angles not 90\u00b0<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Regular Hexagon<\/strong>\n<ul class=\"wp-block-list\">\n<li>When cube is balanced on one vertex<\/li>\n\n\n\n<li>This is the&nbsp;<strong>isometric projection<\/strong><\/li>\n\n\n\n<li>All six sides equal<\/li>\n\n\n\n<li>Occurs when cube is oriented such that main diagonal is perpendicular to plane<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Non-regular Hexagon<\/strong>\n<ul class=\"wp-block-list\">\n<li>At certain intermediate orientations<\/li>\n\n\n\n<li>Six sides but not all equal<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Line segment<\/strong>\n<ul class=\"wp-block-list\">\n<li>Degenerate case<\/li>\n\n\n\n<li>When viewing exactly along one edge<\/li>\n\n\n\n<li>Very unlikely in practice<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<p>Most Important Projections:<br><strong>Square:<\/strong>&nbsp;Standard view (face parallel to plane)<br><strong>Rectangle:<\/strong>&nbsp;Slightly tilted<br><strong>Regular Hexagon:<\/strong>&nbsp;Isometric view (balanced on vertex)<br><strong>Parallelogram:<\/strong>&nbsp;General tilted view<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.53: In addition to the 5 ways shown in Fig. 4.8, are there any additional ways of gluing four cubes together along faces? Can you visualise and draw these as well?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>The 5 Tetris shapes shown in Fig. 4.8 are:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Straight line (I-shape): 4 cubes in a row<\/li>\n\n\n\n<li>L-shape: 3 cubes in a row + 1 perpendicular<\/li>\n\n\n\n<li>T-shape: 3 cubes in a row + 1 on top of middle<\/li>\n\n\n\n<li>Z-shape (S-shape): Two pairs offset<\/li>\n\n\n\n<li>Square (O-shape): 2 {tex}\\times{\/tex} 2 arrangement<\/li>\n<\/ol>\n\n\n\n<p><strong>Additional ways to glue 4 cubes:<\/strong><br>Actually, these are the&nbsp;<strong>only 5 ways<\/strong>&nbsp;to arrange 4 cubes face-to-face in 2D (called Tetrominos).<br>However, in&nbsp;<strong>3D<\/strong>, we can have additional arrangements:<\/p>\n\n\n\n<p><strong>6. 3D L-shape:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>3 cubes in one plane (L-shape)<\/li>\n\n\n\n<li>1 cube attached perpendicular to the plane<\/li>\n\n\n\n<li>Different from 2D L-shape<\/li>\n<\/ul>\n\n\n\n<p><strong>7. 3D T-shape:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>3 cubes in a row<\/li>\n\n\n\n<li>1 cube attached perpendicular (not in same plane as T)<\/li>\n<\/ul>\n\n\n\n<p><strong>Total possibilities:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>In 2D (planar): 5 tetrominos (as shown)<\/li>\n\n\n\n<li>In 3D (allowing depth): 7-8 different arrangements<\/li>\n<\/ul>\n\n\n\n<p><strong>Drawing on Isometric Grid:<\/strong><br>For each arrangement:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Choose orientation (along which axis)<\/li>\n\n\n\n<li>Draw systematically, cube by cube<\/li>\n\n\n\n<li>Add shading for clarity<\/li>\n<\/ol>\n\n\n\n<p><strong>Examples:<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li><strong>Straight line along depth:&nbsp;<\/strong>[Draw 4 cubes in a row along direction]<\/li>\n\n\n\n<li><strong>L-shape along length-height:&nbsp;<\/strong>[Draw L in the length-height plane]<\/li>\n\n\n\n<li><strong>T-shape along depth-length with height:&nbsp;<\/strong>[Draw T configuration]<\/li>\n\n\n\n<li><strong>Z-shape:&nbsp;<\/strong>[Draw offset configuration]<\/li>\n\n\n\n<li><strong>Square 2<\/strong>{tex}\\times{\/tex}<strong>2:&nbsp;<\/strong>[Draw 2{tex}\\times{\/tex}2 arrangement]<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.54: Draw the following figures on the isometric grid.<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1769150494-skcrem.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>General Process for Drawing on Isometric Grid:<br>Step 1: Identify the 3 primary axes<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Height: | (vertical)<\/li>\n\n\n\n<li>Depth: (upper left diagonal)<\/li>\n\n\n\n<li>Length: (upper right diagonal)<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 2: Choose a starting point<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Usually start from the frontmost, lowest, leftmost corner<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 3: Draw edge by&nbsp;<\/strong><strong>edge.&nbsp;<\/strong>For&nbsp;each edge:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Determine its direction (height\/depth\/length)<\/li>\n\n\n\n<li>Determine if it goes in positive or negative direction<\/li>\n\n\n\n<li>Count units along that axis<\/li>\n\n\n\n<li>Draw accordingly<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 4: Add shading (optional)<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Shade different faces with different patterns<\/li>\n\n\n\n<li>Helps visualize the 3D structure<\/li>\n<\/ul>\n\n\n\n<p><strong>Specific Drawings:<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li><strong>First figure:<\/strong>\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Analyze the structure<\/li>\n\n\n\n<li>Identify base shape and height<\/li>\n\n\n\n<li>Draw base on isometric grid using and directions<\/li>\n\n\n\n<li>Add vertical edges using | direction<\/li>\n\n\n\n<li>Complete top face<\/li>\n\n\n\n<li>Add shading<\/li>\n<\/ol>\n<\/li>\n\n\n\n<li><strong>Second figure:<\/strong>[Follow same process]<\/li>\n\n\n\n<li><strong>Third figure:<\/strong>[Follow same process]<\/li>\n<\/ol>\n\n\n\n<p><strong>Tips:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Draw lightly first, then darken visible edges<\/li>\n\n\n\n<li>Hidden edges can be drawn with dotted lines<\/li>\n\n\n\n<li>Use the hint: track whether going up (|) or down (opposite to |)<\/li>\n\n\n\n<li>Count grid units carefully<\/li>\n<\/ul>\n\n\n\n<p><strong>Example Process for an L-shaped figure:<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Draw base: 3 units , 1 unit<\/li>\n\n\n\n<li>Draw vertical edges: 2 units |<\/li>\n\n\n\n<li>Draw top horizontal: 3 units &nbsp;(but at height)<\/li>\n\n\n\n<li>Complete remaining edges<\/li>\n\n\n\n<li>Add shading to distinguish faces<\/li>\n<\/ol>\n\n\n\n<p>Actual drawings would be provided on isometric grid paper based on specific figures in the textbook.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.55: Is there anything strange about the path of this ball? Recreate it on the isometric grid.<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768899974-r84k5f.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>Observation:<\/strong><br>Yes, there is something strange! The ball appears to be going in a path that seems to&nbsp;<strong>defy gravity<\/strong>&nbsp;or creates an&nbsp;<strong>optical illusion<\/strong>.<\/p>\n\n\n\n<p><strong>The strangeness:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The ball appears to be rolling &#8220;uphill&#8221; in a continuous loop, OR<\/li>\n\n\n\n<li>The path seems impossible in real 3D space, OR<\/li>\n\n\n\n<li>The starting and ending points don&#8217;t match physically<\/li>\n<\/ul>\n\n\n\n<p><strong>Why this happens:<\/strong><br>This is an&nbsp;<strong>impossible figure<\/strong>&nbsp;or&nbsp;<strong>optical illusion<\/strong>:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>It looks correct in 2D (isometric projection)<\/li>\n\n\n\n<li>But cannot exist in real 3D space<\/li>\n\n\n\n<li>Similar to Escher&#8217;s famous impossible staircases<\/li>\n<\/ul>\n\n\n\n<p><strong>Analysis using the hint:<br>Step 1: Identify physically realizable portion<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Some parts of the path are normal<\/li>\n\n\n\n<li>At some point, there&#8217;s an impossible connection<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 2: Identify 3 primary directions<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Height: |<\/li>\n\n\n\n<li>Depth:<\/li>\n\n\n\n<li>Length:<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 3: Recreate on isometric grid<\/strong><br>Start from a physical portion:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Draw the track segment that goes in direction<\/li>\n\n\n\n<li>Add the segment going in | direction<\/li>\n\n\n\n<li>Add the segment going in direction<\/li>\n\n\n\n<li>Continue&#8230;<\/li>\n\n\n\n<li>At some point, try to connect back &#8211; this reveals the impossibility!<\/li>\n<\/ol>\n\n\n\n<p><strong>The Trick:<\/strong><br>The isometric drawing cleverly makes edges that should be at different heights&nbsp;<strong>appear<\/strong>&nbsp;to connect. Our brain interprets it as a continuous path, but in reality:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Edge A is at height h<sub>1<\/sub><\/li>\n\n\n\n<li>Edge B is at height h<sub>2<\/sub>&nbsp;{tex}\\neq{\/tex} h<sub>1<\/sub><\/li>\n\n\n\n<li>But the drawing makes them look connected!<\/li>\n<\/ul>\n\n\n\n<p><strong>Conclusion:<\/strong>&nbsp;The path is an&nbsp;<strong>optical illusion<\/strong>. It looks continuous in the isometric projection but would be discontinuous in real 3D space. This demonstrates that isometric projections can create impossible figures!<br><strong>Recreation:<\/strong>&nbsp;[Draw the ball path on isometric grid, showing where the impossible connection occurs]\n\n\n\n<h2 class=\"wp-block-heading\">Class 8 Maths Ganita Prakash Solutions<\/h2>\n\n\n\n<ol class=\"wp-block-list\">\n<li><a href=\"https:\/\/mycbseguide.com\/blog\/a-square-and-a-cube-ncert-solutions-class-8-maths-ganita-prakash\/\">A Square and A Cube<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/mycbseguide.com\/blog\/power-play-ncert-solutions-class-8-maths-ganita-prakash\/\">Power Play<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/mycbseguide.com\/blog\/a-story-of-numbers-ncert-solutions-class-8-maths-ganita-prakash\/\">A Story of Numbers<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/mycbseguide.com\/blog\/quadrilaterals-ncert-solutions-class-8-maths-ganita-prakash\/\">Quadrilaterals<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/mycbseguide.com\/blog\/number-play-ncert-solutions-class-8-maths-ganita-prakash\/\">Number Play<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/mycbseguide.com\/blog\/we-distribute-yet-things-multiply-ncert-solutions-class-8-maths-ganita-prakash\/\">We Distribute Yet Things Multiply<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/mycbseguide.com\/blog\/proportional-reasoning-1-ncert-solutions-class-8-maths-ganita-prakash\/\">Proportional Reasoning-1<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/mycbseguide.com\/blog\/fractions-in-disguise-ncert-solutions-class-8-maths-ganita-prakash\/\">Fractions In Disguise<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/mycbseguide.com\/blog\/the-baudhayana-pythagoras-theorem-ncert-solutions-class-8-maths-ganita-prakash\/\">The Baudhayana-Pythagoras Theorem<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/mycbseguide.com\/blog\/proportional-reasoning-2-ncert-solutions-class-8-maths-ganita-prakash\/\">Proportional Reasoning-2<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/mycbseguide.com\/blog\/exploring-some-geometric-themes-ncert-solutions-class-8-maths-ganita-prakash\/\">Exploring Some Geometric Themes<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/mycbseguide.com\/blog\/tales-by-dots-and-lines-ncert-solutions-class-8-maths-ganita-prakash\/\">Tales by dots and lines<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/mycbseguide.com\/blog\/algebra-play-ncert-solutions-class-8-maths-ganita-prakash\/\">Algebra Play<\/a><\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"<p>Exploring Some Geometric Themes &#8211; NCERT Solutions Class 8 Maths (Ganita Prakash) includes all the questions with solutions given in the NCERT Class 8 Maths (Ganita Prakash). NCERT Solutions Class 8 Exploring Some Geometric Themes \u2013 NCERT Solutions Q.1: Draw the initial few steps (at least till Step 2) of the shape sequence that leads &#8230; <a title=\"Exploring Some Geometric Themes &#8211; NCERT Solutions Class 8 Maths (Ganita Prakash)\" class=\"read-more\" href=\"https:\/\/mycbseguide.com\/blog\/exploring-some-geometric-themes-ncert-solutions-class-8-maths-ganita-prakash\/\" aria-label=\"More on Exploring Some Geometric Themes &#8211; NCERT Solutions Class 8 Maths (Ganita Prakash)\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[281,2083,2111],"tags":[216],"class_list":["post-31751","post","type-post","status-publish","format-standard","hentry","category-ncert-solutions","category-ncert-solutions-class-8","category-ncert-solutions-class-8-maths-ganita-prakash","tag-ncert-solutions"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.0 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Exploring Some Geometric Themes - NCERT Solutions Class 8 Maths (Ganita Prakash) | myCBSEguide<\/title>\n<meta name=\"description\" content=\"Exploring Some Geometric Themes - NCERT Solutions Class 8 Maths (Ganita Prakash) includes all the questions with solutions\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mycbseguide.com\/blog\/exploring-some-geometric-themes-ncert-solutions-class-8-maths-ganita-prakash\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Exploring Some Geometric Themes - NCERT Solutions Class 8 Maths (Ganita Prakash) | myCBSEguide\" \/>\n<meta property=\"og:description\" content=\"Exploring Some Geometric Themes - NCERT Solutions Class 8 Maths (Ganita Prakash) includes all the questions with solutions\" \/>\n<meta property=\"og:url\" content=\"https:\/\/mycbseguide.com\/blog\/exploring-some-geometric-themes-ncert-solutions-class-8-maths-ganita-prakash\/\" \/>\n<meta property=\"og:site_name\" content=\"myCBSEguide\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/mycbseguide\/\" \/>\n<meta property=\"article:published_time\" content=\"2026-08-07T07:34:00+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2026-08-07T07:53:51+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1769145535-8ee7cn.jpg\" \/>\n<meta name=\"author\" content=\"myCBSEguide\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@mycbseguide\" \/>\n<meta name=\"twitter:site\" content=\"@mycbseguide\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"myCBSEguide\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"49 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/exploring-some-geometric-themes-ncert-solutions-class-8-maths-ganita-prakash\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/exploring-some-geometric-themes-ncert-solutions-class-8-maths-ganita-prakash\/\"},\"author\":{\"name\":\"myCBSEguide\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/#\/schema\/person\/10b8c7820ff29025ab8524da7c025f65\"},\"headline\":\"Exploring Some Geometric Themes &#8211; 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