{"id":31747,"date":"2026-08-07T12:46:19","date_gmt":"2026-08-07T07:16:19","guid":{"rendered":"https:\/\/mycbseguide.com\/blog\/?p=31747"},"modified":"2026-08-07T13:24:12","modified_gmt":"2026-08-07T07:54:12","slug":"the-baudhayana-pythagoras-theorem-ncert-solutions-class-8-maths-ganita-prakash","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/the-baudhayana-pythagoras-theorem-ncert-solutions-class-8-maths-ganita-prakash\/","title":{"rendered":"The Baudhayana-Pythagoras Theorem &#8211; NCERT Solutions Class 8 Maths (Ganita Prakash)"},"content":{"rendered":"\n<p><strong><strong>The Baudhayana-Pythagoras Theorem<\/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>The Baudhayana-Pythagoras Theorem<\/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: How can one construct a square having double the area of a given square?<br>A first guess might be to simply double the length of each side of the square. Will this new square have double the area of the original square?<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768814276-46pgbc.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>No, the new square will not have double the area of the original square.<br>If the original square has side length 1 unit, its area {tex}=1 \\times 1=1{\/tex} sq. unit.<br>If we double the side length to 2 units, the new area {tex}=2 \\times 2=4{\/tex} sq. units.<br>Therefore, doubling the side length gives us a square with {tex}{4}{\/tex} <strong>times <\/strong>the area, not double the area.<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768806323-beycvt.jpg\"><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.2: Why does the new dotted square have double the area of the original square?<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768806430-qcsu9f.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>The new dotted square has double the area because:<br>When we construct a square on the diagonal of the original square, we can see by drawing horizontal and vertical lines that:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The original square is made up of {tex}{2}{\/tex} small congruent triangles (divided by the diagonal)<\/li>\n\n\n\n<li>The new dotted square is made up of {tex}{4}{\/tex} small congruent triangles (of the same size)<\/li>\n<\/ul>\n\n\n\n<p>Since the new square contains exactly twice as many triangles of the same size, its area is exactly double that of the original square.<br>Area of new square {tex}=4{\/tex} triangles {tex}=2 \\times(2{\/tex} triangles{tex})=2 \\times{\/tex} Area of original square.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.3: Now suppose we are given a square, and we want to construct a square whose area is half that of the original square. How would you do it?<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768806684-4tvust.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>To construct a square with half the area of the original square:<br><strong>Method: <\/strong>Draw a tilted smaller square inside the larger square by connecting the midpoints of the sides of the original square.<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768806766-puhm6s.jpg\"><br><strong>Explanation:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Draw horizontal and vertical lines through the construction<\/li>\n\n\n\n<li>The larger square can be divided into 4 small congruent triangles<\/li>\n\n\n\n<li>The smaller tilted square is made up of 2 of these small triangles<\/li>\n\n\n\n<li>Therefore, the smaller square has exactly half the area of the larger square<\/li>\n<\/ul>\n\n\n\n<p>Area of smaller square {tex}=2{\/tex} triangles {tex}=\\frac 1 2 \\times(4{\/tex} triangles{tex})=\\frac 1 2 \\times{\/tex} Area of larger square<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.4: Find the hypotenuse of this isosceles right triangle.<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768806970-y3ccvc.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>Given:<\/strong> An isosceles right triangle with equal sides {tex}=1{\/tex} unit each<br><strong>To find:<\/strong> Length of hypotenuse<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768807285-3s7787.jpg\"><br>We know that a square of side 1 unit is made up of two such isosceles right triangles (divided by the diagonal).<br>When we construct a square on the hypotenuse (diagonal) of the original square:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Area of original square PEAR {tex}=1 \\times 1=1{\/tex} sq. unit<\/li>\n\n\n\n<li>Area of square REST (on the hypotenuse) {tex}=2 \\times{\/tex} Area of PEAR {tex}=2 \\times 1=2{\/tex} sq. units<\/li>\n<\/ul>\n\n\n\n<p>Let c be the length of the hypotenuse.<br>Since REST is a square with side c:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Area of REST {tex}={c} \\times {c}={c}^2{\/tex}<\/li>\n<\/ul>\n\n\n\n<p>We know Area of REST = 2 sq. units<br>Therefore: {tex}c^2=2{\/tex}<br>Taking square root: {tex}{c}=\\sqrt{2}{\/tex}<br>The hypotenuse is of length <strong>{tex}\\sqrt{2}{\/tex} units<\/strong> (approximately 1.414 units).<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768814686-tn4zp6.jpg\"><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.5: What is the value of {tex}\\sqrt{2}{\/tex}?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>The value of {tex}\\sqrt{2}{\/tex} is approximately {tex}{1 . 4 1 4}{\/tex}.<br>{tex} 1.411^2=1.990921 {\/tex}<br>{tex} 1.412^2=1.993744 {\/tex}<br>{tex} 1.413^2=1.996569 {\/tex}<br>{tex} 1.414^2=1.999396 {\/tex}<br>{tex} 1.415^2=2.002225 {\/tex}<br>So, {tex}1.414&lt;\\sqrt{2}&lt;1.415{\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.6: Is {tex}\\sqrt{2}{\/tex} less than or greater than 1?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>A square of sidelength 1 unit has an area of 1 sq. unit. A square of sidelength {tex}\\sqrt{2}{\/tex} has an area of 2 sq. units. So, 1 is less than {tex}\\sqrt{2}{\/tex}.<br>In other words, {tex}1^2=1{\/tex}, and {tex}\\sqrt{2}^2=2{\/tex}.<br>Therefore, {tex}1&lt;\\sqrt{2}{\/tex}.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.7: Is {tex}\\sqrt{2}{\/tex} less than or greater than 2?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>A square of sidelength 2 units has an area of 4 sq. units. A square of sidelength {tex}\\sqrt{2}{\/tex} has an area of 2 sq. units. So, 2 is greater than {tex}\\sqrt{2}{\/tex}.<br>In other words, {tex}\\sqrt{2}^2=2{\/tex}, and {tex}2^2=4{\/tex}.<br>Therefore, {tex}\\sqrt{2}&lt;2{\/tex}.<br>Thus, {tex}1&lt;\\sqrt{2}&lt;2{\/tex}.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.8: Can {tex}\\sqrt{2}{\/tex} be expressed as a fraction {tex}m \/ n{\/tex}, where {tex}m{\/tex} and {tex}n{\/tex} are counting numbers?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>No, {tex}\\sqrt{2}{\/tex} cannot be expressed as a fraction {tex}m \/ n{\/tex} where {tex}m{\/tex} and {tex}n{\/tex} are counting numbers.<br>Let&#8217;s assume {tex}\\sqrt{2}{\/tex} can be written as {tex}{m} \/ {n}{\/tex} (a fraction)<br>Then: {tex}\\sqrt{2}=m \/ n{\/tex}<br>Squaring both sides: {tex}2={m}^2 \/ {n}^2{\/tex}<br>Cross-multiplying: {tex}2 n^2=m^2{\/tex}<br><strong>Key observation:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>In the prime factorization of any square number (like {tex}{m}^2{\/tex} or {tex}{n}^2{\/tex}), each prime factor appears an <strong>even number of times<\/strong><\/li>\n\n\n\n<li>On the left side {tex}\\left(2 n^2\\right){\/tex}: The prime factor 2 appears an odd number of times (one 2, plus an even number from {tex}{n}^2{\/tex})<\/li>\n\n\n\n<li>On the right side {tex}\\left({m}^2\\right){\/tex}: The prime factor 2 must appear an even number of times<\/li>\n<\/ul>\n\n\n\n<p>This creates a contradiction! A number cannot have both odd and even occurrences of the same prime factor.<br>Therefore, our assumption must be wrong. {tex}\\sqrt{2}{\/tex} cannot be expressed as a fraction {tex}{m} \\boldsymbol{\/} {n}{\/tex}.<br><strong>Conclusion:<\/strong> {tex}\\sqrt{2}{\/tex} is an irrational number with a non-terminating, non-repeating decimal expansion.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.9: List down all the Baudh\u0101yana triples with numbers less than or equal to 20.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>Baudh\u0101yana triples with numbers {tex}\\boldsymbol{\\leq} {2 0}{\/tex}:<\/strong><br>To find these, we check which triples {tex}({a}, {b}, {c}){\/tex} satisfy {tex}{a}^2+{b}^2={c}^2{\/tex} where {tex}{a}, {b}, {c} \\leq 20{\/tex}.<br><strong>The complete list:<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>{tex}(3,4,5){\/tex}-Primitive\n<ul class=\"wp-block-list\">\n<li>{tex}3^2+4^2=9+16=25=5^2{\/tex}<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>{tex}({6 , 8 , 1 0}){\/tex} &#8211; Scaled version of {tex}(3,4,5){\/tex} [multiply by 2]\n<ul class=\"wp-block-list\">\n<li>{tex}6^2+8^2=36+64=100=10^2{\/tex}<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>(5, 12, 13)<\/strong>-Primitive\n<ul class=\"wp-block-list\">\n<li>{tex}5^2+12^2=25+144=169=13^2{\/tex}<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>(9, 12, 15)-<\/strong> Scaled version of {tex}(3,4,5){\/tex} [multiply by 3]\n<ul class=\"wp-block-list\">\n<li>{tex}9^2+12^2=81+144=225=15^2{\/tex}<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>(8, 15, 17)<\/strong>-Primitive\n<ul class=\"wp-block-list\">\n<li>{tex}8^2+15^2=64+225=289=17^2{\/tex}<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>(12, 16, 20)-<\/strong>Scaled version of {tex}(3,4,5){\/tex} [multiply by 4]\n<ul class=\"wp-block-list\">\n<li>{tex}12^2+16^2=144+256=400=20^2{\/tex}<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<p><strong>Total: 6 Baudh\u0101yana triples with numbers {tex}\\boldsymbol{\\leq} {2 0}{\/tex}<\/strong><br><strong>Primitive triples:<\/strong> {tex}(3,4,5),(5,12,13),(8,15,17){\/tex}<br><strong>Non-primitive triples:<\/strong> {tex}(6,8,10),(9{\/tex}, {tex}12,15),(12,16,20){\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.10: Is there an unending sequence of Baudh\u0101yana triples?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>Yes,<\/strong> there is an unending (infinite) sequence of Baudh\u0101yana triples.<br><strong>Proof:<\/strong><br>We know that {tex}(3,4,5){\/tex} is a Baudh\u0101yana triple.<br>We can generate infinite triples by multiplying each term by any positive integer k:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}(3 \\times 1,4 \\times 1,5 \\times 1)=(3,4,5) \\checkmark{\/tex}<\/li>\n\n\n\n<li>{tex}(3 \\times 2,4 \\times 2,5 \\times 2)=(6,8,10) \\checkmark{\/tex}<\/li>\n\n\n\n<li>{tex}(3 \\times 3,4 \\times 3,5 \\times 3)=(9,12,15) \\checkmark{\/tex}<\/li>\n\n\n\n<li>{tex}(3 \\times 4,4 \\times 4,5 \\times 4)=(12,16,20) \\checkmark{\/tex}<\/li>\n\n\n\n<li>And so on&#8230;<\/li>\n<\/ul>\n\n\n\n<p>Since k can be any positive integer {tex}(1,2,3,4,5, \\ldots, \\infty){\/tex}, and each gives us a valid<br>Baudh\u0101yana triple, there are <strong>infinitely many Baudh\u0101yana triples.<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.11: Is (5, 12, 13) a primitive Baudh\u0101yana triple? What are the other primitive Baudh\u0101yana triples with numbers less than or equal to 20?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Yes, {tex}(5,12,13){\/tex} is a <strong>primitive Baudh\u0101yana triple.<\/strong><br><strong>Reason:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>First, verify it&#8217;s a Baudh\u0101yana triple: {tex}5^2+12^2=25+144=169=13^2{\/tex}<\/li>\n\n\n\n<li>Check for common factors: The numbers 5, 12, and 13 have <strong>no common factor greater than 1<\/strong><\/li>\n\n\n\n<li>{tex}\\operatorname{GCD}(5,12,13)=1{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Therefore, {tex}(5,12,13){\/tex} is primitive.<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.12: Generate 5 scaled versions of each of these primitive triples. Are these scaled versions primitive?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>Scaled versions of (3, 4, 5):<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>{tex}{k}=2:(6,8,10){\/tex}<\/li>\n\n\n\n<li>{tex}{k}=3:(9,12,15){\/tex}<\/li>\n\n\n\n<li>{tex}{k}=4:(12,16,20){\/tex}<\/li>\n\n\n\n<li>{tex}{k}=5:(15,20,25){\/tex}<\/li>\n\n\n\n<li>{tex}{k}=6:(18,24,30){\/tex}<\/li>\n<\/ol>\n\n\n\n<p><strong>Are these primitive?<\/strong> No, each has common factor {tex}{k}&gt;1{\/tex}.<br><strong>Scaled versions of (5, 12, 13):<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>{tex}{k}=2:(10,24,26){\/tex}<\/li>\n\n\n\n<li>{tex}{k}=3:(15,36,39){\/tex}<\/li>\n\n\n\n<li>{tex}{k}=4:(20,48,52){\/tex}<\/li>\n\n\n\n<li>{tex}{k}=5:(25,60,65){\/tex}<\/li>\n\n\n\n<li>{tex}{k}=6:(30,72,78){\/tex}<\/li>\n<\/ol>\n\n\n\n<p><strong>Are these primitive?<\/strong> No, each has common factor {tex}{k}&gt;1{\/tex}.<br><strong>Scaled versions of (8, 15, 17):<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>{tex}{k}=2:(16,30,34){\/tex}<\/li>\n\n\n\n<li>{tex}{k}=3{\/tex} : {tex}(24,45,51){\/tex}<\/li>\n\n\n\n<li>{tex}{k}=4:(32,60,68){\/tex}<\/li>\n\n\n\n<li>{tex}{k}=5:(40,75,85){\/tex}<\/li>\n\n\n\n<li>{tex}{k}=6:(48,90,102){\/tex}<\/li>\n<\/ol>\n\n\n\n<p><strong>Are these primitive?<\/strong> No, each has common factor {tex}{k}&gt;1{\/tex}.<br><strong>Conclusion: No, <\/strong>scaled versions are never primitive because they all have a common factor {tex}{k}&gt;1{\/tex}.<br><strong>General rule:<\/strong> Only the original triple is primitive; all scaled versions are nonprimitive.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.13: Earlier, we saw a method to create a square with double the area of a given square paper. There is another method to do this in which two identical square papers are cut in the following way.<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768794849-se93wj.jpg\"><br>Can you arrange these pieces to create a square with double the area of either square?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Yes, we can arrange these pieces to create a square with double the area.<br><strong>Arrangement method:<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Take the two identical squares, each cut into pieces {tex}1,2,3,4{\/tex}<\/li>\n\n\n\n<li>From each square, you get 2 triangular pieces (if cut along one diagonal) or 4 pieces (if cut differently)<\/li>\n\n\n\n<li><strong>Arrange the {tex}{8}{\/tex} pieces<\/strong> (4 from each square) to form a larger square:\n<ul class=\"wp-block-list\">\n<li>Place the pieces so that the triangular pieces fit together<\/li>\n\n\n\n<li>The straight edges should align to form the sides of the new square<\/li>\n\n\n\n<li>Use the diagonal of the original square as the side of the new square<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>The resulting square will have:\n<ul class=\"wp-block-list\">\n<li>Side length {tex}={\/tex} diagonal of original square {tex}=\\sqrt{2 } \\times{\/tex} (side of original)<\/li>\n\n\n\n<li>Area {tex}=(\\sqrt{ 2} \\times \\text { side })^2=2 \\times(\\text { side })^2=2 \\times{\/tex} Area of original square<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<p><strong>Verification:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Original square area {tex}={a}^2{\/tex} (where a is the side)<\/li>\n\n\n\n<li>Each square contributes 4 pieces<\/li>\n\n\n\n<li>New square is formed by 8 pieces total<\/li>\n\n\n\n<li>New square area {tex}=2 a^2={\/tex} double the original area<\/li>\n<\/ul>\n\n\n\n<p>Therefore, this arrangement creates a square with exactly double the area of either original square.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.14: The length of the two equal sides of an isosceles right triangle is given. Find the length of the hypotenuse. Find bounds on the length of the hypotenuse such that they have at least one digit after the decimal point.<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>3<\/li>\n\n\n\n<li>4<\/li>\n\n\n\n<li>6<\/li>\n\n\n\n<li>8<\/li>\n\n\n\n<li>9<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>For an isosceles right triangle with equal sides of length a, the hypotenuse c is given by:<br><strong>Formula:<\/strong> {tex}{c}^2=2 {a}^2{\/tex} or {tex}{c}={a} \\sqrt{ 2} {\/tex}<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li><strong>When<\/strong> {tex}{a}=3{\/tex}:<br>{tex} c=3 \\sqrt{ 2} {\/tex}<br>To find bounds, we calculate:<ul><li>{tex}4^2=16{\/tex} and {tex}5^2=25{\/tex}<\/li><li>{tex}(3 \\sqrt{ 2} )^2=9 \\times 2=18{\/tex}<\/li><li>Since {tex}16&lt;18&lt;25{\/tex}, we have {tex}4&lt;3 \\sqrt{ 2}&lt;5{\/tex}<\/li><\/ul>For more precision:\n<ul class=\"wp-block-list\">\n<li>{tex}4.2^2=17.64{\/tex}<\/li>\n\n\n\n<li>{tex}4.3^2=18.49{\/tex}<\/li>\n\n\n\n<li>Since 17.64 &lt; 18 &lt; 18.49<br>Length of hypotenuse {tex}=3 \\sqrt{ 2} {\/tex} units {tex}\\approx 4.24{\/tex} units<br><strong>Bounds:<\/strong> 4.2 &lt; hypotenuse &lt; 4.3<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>When a = 4:<br>{tex} c=4 \\sqrt{ 2}{\/tex}<\/strong><br>To find bounds:<ul><li>{tex}(4 \\sqrt{ } 2)^2=16 \\times 2=32{\/tex}<\/li><li>{tex}5^2=25{\/tex} and {tex}6^2=36{\/tex}<\/li><li>Since {tex}25&lt;32&lt;36{\/tex}, we have {tex}5&lt;4 \\sqrt{ 2} &lt;6{\/tex}<\/li><\/ul>For more precision:\n<ul class=\"wp-block-list\">\n<li>{tex}5.6^2=31.36{\/tex}<\/li>\n\n\n\n<li>{tex}5.7^2=32.49{\/tex}<\/li>\n\n\n\n<li>Since 31.36 &lt; 32 &lt; 32.49<br>Length of hypotenuse {tex}=4 \\sqrt{ 2} {\/tex} units {tex}\\approx 5.66{\/tex} units<br><strong>Bounds: <\/strong>5.6 &lt; hypotenuse &lt; 5.7<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>When {tex}{a}=6{\/tex}:<\/strong><br>{tex}c=6 \\sqrt{ 2} {\/tex}<br>To find bounds:<ul><li>{tex}(6 \\sqrt{2 })^2=36 \\times 2=72{\/tex}<\/li><li>{tex}8^2=64{\/tex} and {tex}9^2=81{\/tex}<\/li><li>Since {tex}64&lt;72&lt;81{\/tex}, we have {tex}8&lt;6 \\sqrt{ 2} &lt;9{\/tex}<\/li><\/ul>For more precision:\n<ul class=\"wp-block-list\">\n<li>{tex}8.4^2=70.56{\/tex}<\/li>\n\n\n\n<li>{tex}8.5^2=72.25{\/tex}<\/li>\n\n\n\n<li>Since {tex}70.56&lt;72&lt;72.25{\/tex}<br>Length of hypotenuse {tex}=6 \\sqrt{ 2} {\/tex} units {tex}\\approx 8.49{\/tex} units<br><strong>Bounds:<\/strong> 8.4 &lt; hypotenuse &lt; 8.5<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>When {tex}a=8{\/tex}:<\/strong><br>{tex} c=8 \\sqrt{ 2} {\/tex}<br>To find bounds:<ul><li>{tex}(8 \\sqrt{2 })^2=64 \\times 2=128{\/tex}<\/li><li>{tex}11^2=121{\/tex} and {tex}12^2=144{\/tex}<\/li><li>Since {tex}121&lt;128&lt;144{\/tex}, we have {tex}11&lt;8 \\sqrt{ 2} &lt;12{\/tex}<\/li><\/ul>For more precision:\n<ul class=\"wp-block-list\">\n<li>{tex}11.3^2=127.69{\/tex}<\/li>\n\n\n\n<li>{tex}11.4^2=129.96{\/tex}<\/li>\n\n\n\n<li>Since 127.69 &lt; 128 &lt; 129.96<br>Length of hypotenuse {tex}=8 \\sqrt{ 2} {\/tex} units {tex}\\approx 11.31{\/tex} units<br><strong>Bounds:<\/strong> 11.3 &lt; hypotenuse &lt; 11.4<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>When a = 9:<\/strong><br>{tex} c=9 \\sqrt{ 2} {\/tex}<br>To find bounds:<ul><li>{tex}(9 \\sqrt{ 2} )^2=81 \\times 2=162{\/tex}<\/li><li>{tex}12^2=144{\/tex} and {tex}13^2=169{\/tex}<\/li><li>Since 144 &lt; 162 &lt; 169, we have {tex}12&lt;9 \\sqrt{2 }&lt;13{\/tex}<\/li><\/ul>For more precision:\n<ul class=\"wp-block-list\">\n<li>{tex}12.7^2=161.29{\/tex}<\/li>\n\n\n\n<li>{tex}12.8^2=163.84{\/tex}<\/li>\n\n\n\n<li>Since 161.29 &lt; 162 &lt; 163.84<br>Length of hypotenuse&nbsp;{tex}=9 \\sqrt{ 2} {\/tex} units {tex}\\approx 12.73{\/tex} units<br><strong>Bounds: <\/strong>12.7 &lt; hypotenuse &lt; 12.8<\/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.15: The hypotenuse of an isosceles right triangle is 10. What are its other two sidelengths?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>Given:<\/strong> Hypotenuse of isosceles right triangle {tex}=10{\/tex} units<br><strong>To find:<\/strong> Length of the two equal sides<br>Let a be the length of each equal side.<br>Using the formula for isosceles right triangle: {tex}c^2=2 a^2{\/tex}<br>Given {tex}{c}=10{\/tex}<br>Substituting: {tex}(10)^2=2 {a}^2{\/tex}<br>{tex} 100=2 a^2 {\/tex}<br>Dividing by 2: {tex}a^2=100 \/ 2=50{\/tex}<br>Taking square root: {tex}{a}=\\sqrt{ 50} {\/tex}<br><strong>Simplifying {tex}\\sqrt{ 50} {\/tex}:<\/strong><br>{tex} \\sqrt{50 } =\\sqrt{(25 \\times 2) }=\\sqrt{ 25} \\times \\sqrt{2 } =5 \\sqrt{ 2} {\/tex}<br><strong>Finding bounds:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}(5 \\sqrt{2 } )^2=25 \\times 2=50{\/tex}<\/li>\n\n\n\n<li>{tex}7^2=49{\/tex} and {tex}8^2=64{\/tex}<\/li>\n\n\n\n<li>Since {tex}49&lt;50&lt;64{\/tex}, we have {tex}7&lt;5 \\sqrt{2}&lt;8{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>More precisely:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}7.0^2=49{\/tex}<\/li>\n\n\n\n<li>{tex}7.1^2=50.41{\/tex}<\/li>\n\n\n\n<li>So {tex}7.0&lt;5 \\sqrt{ 2} &lt;7.1{\/tex}<\/li>\n<\/ul>\n\n\n\n<p>Each of the two equal sides has length {tex}{5} \\sqrt{ 2}{\/tex} units {tex}\\boldsymbol{\\approx} {7 . 0 7}{\/tex} units.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.16: Find the hypotenuse of an isosceles right triangle whose equal sides have length 12.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>We have {tex}a=12{\/tex}. Using the formula, we get<br>{tex} c=\\sqrt{2 \\times 12^2}=\\sqrt{288} . {\/tex}<br>We have {tex}16^2=256{\/tex}, and {tex}17^2=289{\/tex}.<br>So, {tex}\\sqrt{288}{\/tex} lies between 16 and 17.<br>The length of the hypotenuse of an isosceles right triangle, whose length of the equal sides is 12 units, is between 16 and 17 units.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.17: If the hypotenuse of an isosceles right triangle is {tex}\\sqrt{72}{\/tex}, find its other two sides.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>We have {tex}c=\\sqrt{72}{\/tex}. Using the formula, we get<br>{tex} c^2=2 a^2 {\/tex}<br>{tex} \\text { So, }(\\sqrt{72})^2 =2 a^2 {\/tex}<br>{tex} 72 =2 a^2 {\/tex}<br>{tex} \\text { Thus, } a^2 =\\frac{72}{2}=36 {\/tex}<br>{tex} \\text { So, } a =\\sqrt{36}=6 {\/tex}<br>Therefore, each of the other two sides has length 6.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.18: If a right-angled triangle has shorter sides of lengths 5 cm and 12 cm, then what is the length of its hypotenuse? First draw the right-angled triangle with these sidelengths and measure the hypotenuse, then check your answer using Baudh\u0101yana\u2019s Theorem.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>Step 1:<\/strong> <strong>Drawing and Measuring<\/strong><br>Draw a right-angled triangle with:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Base {tex}=5 {~cm}{\/tex}<\/li>\n\n\n\n<li>Height {tex}=12 {~cm}{\/tex}<\/li>\n\n\n\n<li>Measure hypotenuse {tex}\\approx 13 {~cm}{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 2: Using Baudh\u0101yana&#8217;s Theorem<\/strong><br><strong>Given:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}{a}=5 {~cm}{\/tex}<\/li>\n\n\n\n<li>{tex}{b}=12 {~cm}{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Formula:<\/strong> {tex}a^2+b^2=c^2{\/tex}<br><strong>Calculation:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}5^2+12^2={c}^2{\/tex}<\/li>\n\n\n\n<li>{tex}25+144={c}^2{\/tex}<\/li>\n\n\n\n<li>{tex}169={c}^2{\/tex}<\/li>\n\n\n\n<li>{tex}{c}=\\sqrt{ } 169=13{\/tex}<\/li>\n<\/ul>\n\n\n\n<p>The length of the hypotenuse is <strong>13 cm<\/strong>.<br><strong>Verification:<\/strong> The measured value matches our calculated value!<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.19: If a right-angled triangle has a short side of length 8 cm and hypotenuse of length 17 cm, what is the length of the third side? Again, try drawing the triangle and measuring, and then check your answer using Baudh\u0101yana\u2019s Theorem.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>Step 1:<\/strong> <strong>Drawing and Measuring<\/strong><br>This is trickier to draw since we know the hypotenuse length.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Draw one side {tex}=8 {~cm}{\/tex}<\/li>\n\n\n\n<li>Using a compass with radius 17 cm, find where it intersects the perpendicular<\/li>\n\n\n\n<li>Measure the third side {tex}\\approx 15 {~cm}{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 2: Using Baudh\u0101yana&#8217;s Theorem<\/strong><br><strong>Given:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}{a}=8 {~cm}{\/tex} (one short side)<\/li>\n\n\n\n<li>{tex}{c}=17 {~cm}{\/tex} (hypotenuse)<\/li>\n\n\n\n<li>{tex}{b}={\/tex} ? (third side)<\/li>\n<\/ul>\n\n\n\n<p><strong>Formula:<\/strong> {tex}a^2+b^2=c^2{\/tex}<br><strong>Rearranging:<\/strong> {tex}{b}^2={c}^2-{a}^2{\/tex}<br><strong>Calculation:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}b^2=17^2-8^2{\/tex}<\/li>\n\n\n\n<li>{tex}{b}^2=289-64{\/tex}<\/li>\n\n\n\n<li>{tex}{b}^2=225{\/tex}<\/li>\n\n\n\n<li>{tex}b=\\sqrt{225 } =15{\/tex}<\/li>\n<\/ul>\n\n\n\n<p>The length of the third side is <strong>15 cm<\/strong>.<br><strong>Verification:<\/strong> Our calculation matches the measurement!<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.20: Using the constructions you have now seen, how would you construct a square whose area is triple the area of a given square? Five times the area of a given square? (Baudh\u0101yana\u2019s \u015aulba-S\u016btra, Verse 1.10)<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>For a square with triple the area:<\/strong><br><strong>Method:<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Take a square with side &#8216;{tex}a{\/tex}&#8217; (Area {tex}=a^2{\/tex})<\/li>\n\n\n\n<li>We want a new square with area {tex}=3 a^2{\/tex}<\/li>\n\n\n\n<li>Using Baudh\u0101yana&#8217;s theorem: We need to find sides {tex}p{\/tex} and {tex}q{\/tex} such that {tex}p^2+q^2=3 a^2{\/tex}<\/li>\n<\/ol>\n\n\n\n<p><strong>Construction:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>One approach: {tex}p^2+q^2=3 a^2{\/tex}<\/li>\n\n\n\n<li>{tex}\\operatorname{Try} p=a{\/tex} and {tex}q^2=2 a^2{\/tex}, so {tex}q=a \\sqrt{2 } {\/tex}<\/li>\n\n\n\n<li>But we can also use: {tex}a^2+a^2+a^2=3 a^2{\/tex}<\/li>\n\n\n\n<li>This means: {tex}a^2+(a \\sqrt{ 2} )^2=3 a^2{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Practical method:<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Take the original square (side a)<\/li>\n\n\n\n<li>Construct a square on its diagonal (side {tex}a \\sqrt{ 2} {\/tex}, area {tex}2 a^2{\/tex})<\/li>\n\n\n\n<li>Now combine the original square ({tex}a^2{\/tex}) with the diagonal square ({tex}2 a^2{\/tex})<\/li>\n\n\n\n<li>Make a right triangle with sides {tex}a{\/tex} and {tex}a \\sqrt{ } 2{\/tex}<\/li>\n\n\n\n<li>The hypotenuse will be {tex}\\sqrt{ \\left(a^2+2 a^2\\right)}=\\sqrt{ \\left(3 a^2\\right)}=a \\sqrt{ 3} {\/tex}<\/li>\n\n\n\n<li>Construct a square on this hypotenuse {tex}\\rightarrow{\/tex} Area {tex}=3 a^2{\/tex}<\/li>\n<\/ol>\n\n\n\n<p><strong>Alternative simple method:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Take three identical squares of side a<\/li>\n\n\n\n<li>Arrange them: two squares side by side, one on top<\/li>\n\n\n\n<li>This creates a shape with total area {tex}3 a^2{\/tex}<\/li>\n\n\n\n<li>Use Baudh\u0101yana&#8217;s method: Make a right triangle with sides a and a {tex}\\sqrt{ 2} {\/tex}<\/li>\n\n\n\n<li>Square on hypotenuse {tex}=3 a^2{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>For a square with five times the area:<br>Method:<\/strong><br>We need area {tex}=5 a^2{\/tex}<br>Using {tex}p^2+q^2=5 a^2{\/tex}<br><strong>One solution:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}{p}={a}{\/tex} and {tex}{q}=2 {a}{\/tex}<\/li>\n\n\n\n<li>Check: {tex}a^2+(2 a)^2=a^2+4 a^2=5 a^2 {\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Construction:<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Take the original square with side a<\/li>\n\n\n\n<li>Construct another square with side {tex}2 a\\left(a r e a=4 a^2\\right){\/tex}<\/li>\n\n\n\n<li>Make a right triangle with sides a and 2a<\/li>\n\n\n\n<li>The hypotenuse {tex}=\\sqrt{ \\left(a^2+4 a^2\\right)}=\\sqrt{ \\left(5 a^2\\right)}=a \\sqrt{ 5} {\/tex}<\/li>\n\n\n\n<li>Construct a square on this hypotenuse {tex}\\rightarrow{\/tex} Area {tex}=5 a^2{\/tex}<\/li>\n<\/ol>\n\n\n\n<p><strong>Verification:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Original square area {tex}=a^2{\/tex}<\/li>\n\n\n\n<li>New square area {tex}=(a \\sqrt{5 } )^2=5 a^2{\/tex}<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.21: Let a, b and c denote the length of the sides of a right triangle, with c being the length of the hypotenuse. Find the missing sidelength in&nbsp;a = 5, b = 7.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>Given:<\/strong> {tex}{a}=5, {~b}=7, {c}={\/tex} ?<br><strong>Formula:<\/strong> {tex}a^2+b^2=c^2{\/tex}<br><strong>Calculation:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}5^2+7^2=c^2{\/tex}<\/li>\n\n\n\n<li>{tex}25+49={c}^2{\/tex}<\/li>\n\n\n\n<li>{tex}74={c}^2{\/tex}<\/li>\n\n\n\n<li>{tex}{c}=\\sqrt{ 74} {\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Finding bounds:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}8^2=64{\/tex} and {tex}9^2=81{\/tex}<\/li>\n\n\n\n<li>Since {tex}64&lt;74&lt;81{\/tex}, we have {tex}8&lt;\\sqrt{ 74} &lt;9{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>More precise:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}8.6^2=73.96{\/tex}<\/li>\n\n\n\n<li>{tex}8.7^2=75.69{\/tex}<\/li>\n\n\n\n<li>So {tex}8.6&lt;\\sqrt{ 74} &lt;8.7{\/tex}<\/li>\n<\/ul>\n\n\n\n<p>{tex}c=\\sqrt{ 74} \\approx 8.60{\/tex} units<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.22: Let a, b and c denote the length of the sides of a right triangle, with c being the length of the hypotenuse. Find the missing sidelength in&nbsp;a = 8, b = 12.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>Given:<\/strong> {tex}a=8, b=12, c={\/tex} ?<br><strong>Formula: <\/strong>{tex}a^2+b^2=c^2{\/tex}<br><strong>Calculation:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}8^2+12^2={c}^2{\/tex}<\/li>\n\n\n\n<li>{tex}64+144={c}^2{\/tex}<\/li>\n\n\n\n<li>{tex}208={c}^2{\/tex}<\/li>\n\n\n\n<li>{tex}c=\\sqrt{ 208} {\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Simplifying:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}\\sqrt{208 } =\\sqrt{(16 \\times 13) }=4 \\sqrt{ 13} {\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Finding bounds:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}14^2=196{\/tex} and {tex}15^2=225{\/tex}<\/li>\n\n\n\n<li>Since {tex}196&lt;208&lt;225{\/tex}, we have {tex}14&lt;\\sqrt{ 208}&lt;15{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>More precise:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}14.4^2=207.36{\/tex}<\/li>\n\n\n\n<li>{tex}14.5^2=210.25{\/tex}<\/li>\n\n\n\n<li>So {tex}14.4&lt;\\sqrt{ 208} &lt;14.5{\/tex}<\/li>\n<\/ul>\n\n\n\n<p>{tex}c=\\sqrt{208 }=4 \\sqrt{ } 13 \\approx 14.42{\/tex} units.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.23: Let a, b and c denote the length of the sides of a right triangle, with c being the length of the hypotenuse. Find the missing sidelength in a&nbsp;= 9, c = 15.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>Given:<\/strong> {tex}a=9, c=15, b={\/tex} ?<br><strong>Formula:<\/strong> {tex}a^2+b^2=c^2{\/tex}<br><strong>Rearranging:<\/strong> {tex}{b}^2={c}^2-{a}^2{\/tex}<br><strong>Calculation:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}{b}^2=15^2-9^2{\/tex}<\/li>\n\n\n\n<li>{tex}{b}^2=225-81{\/tex}<\/li>\n\n\n\n<li>{tex}{b}^2=144{\/tex}<\/li>\n\n\n\n<li>{tex}{b}=\\sqrt{ 144} =12{\/tex}<\/li>\n<\/ul>\n\n\n\n<p>{tex}{b}=12{\/tex} units<br><strong>Verification:<\/strong> {tex}9^2+12^2=81+144=225=15^2 {\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.24: Let a, b and c denote the length of the sides of a right triangle, with c being the length of the hypotenuse. Find the missing sidelength in&nbsp;a = 7, b = 12.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>Given: <\/strong>{tex}{a}=7, {~b}=12, {c}={\/tex} ?<br><strong>Formula: <\/strong>{tex}a^2+b^2=c^2{\/tex}<br><strong>Calculation:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}7^2+12^2={c}^2{\/tex}<\/li>\n\n\n\n<li>{tex}49+144={c}^2{\/tex}<\/li>\n\n\n\n<li>{tex}193={c}^2{\/tex}<\/li>\n\n\n\n<li>{tex}{c}=\\sqrt{ 193}{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Finding bounds:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}13^2=169{\/tex} and {tex}14^2=196{\/tex}<\/li>\n\n\n\n<li>Since 169 &lt; 193 &lt; 196, we have 13 &lt; {tex}\\sqrt{ 193} {\/tex} &lt; 14<\/li>\n<\/ul>\n\n\n\n<p><strong>More precise:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}13.8^2=190.44{\/tex}<\/li>\n\n\n\n<li>{tex}13.9^2=193.21{\/tex}<\/li>\n\n\n\n<li>So 13.8 &lt; {tex}\\sqrt{ 193} {\/tex} &lt; 13.9<\/li>\n<\/ul>\n\n\n\n<p>{tex}c=\\sqrt{ 193} \\approx 13.89{\/tex} units<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.25: Let a, b and c denote the length of the sides of a right triangle, with c being the length of the hypotenuse. Find the missing sidelength in&nbsp;a = 1.5, b = 3.5.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>Given:<\/strong> {tex}{a}=1.5, {~b}=3.5, {c}={\/tex} ?<br><strong>Formula:<\/strong> {tex}a^2+b^2=c^2{\/tex}<br><strong>Calculation:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}(1.5)^2+(3.5)^2={c}^2{\/tex}<\/li>\n\n\n\n<li>{tex}2.25+12.25={c}^2{\/tex}<\/li>\n\n\n\n<li>{tex}14.5={c}^2{\/tex}<\/li>\n\n\n\n<li>{tex}{c}=\\sqrt{ 14.5} {\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Finding bounds:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}3^2=9{\/tex} and {tex}4^2=16{\/tex}<\/li>\n\n\n\n<li>Since {tex}9&lt;14.5&lt;16{\/tex}, we have {tex}3&lt;\\sqrt{ 14.5} &lt;4{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>More precise:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}3.8^2=14.44{\/tex}<\/li>\n\n\n\n<li>{tex}3.9^2=15.21{\/tex}<\/li>\n\n\n\n<li>So {tex}3.8&lt;\\sqrt{ 14.5} &lt;3.9{\/tex}<\/li>\n<\/ul>\n\n\n\n<p>{tex}c=\\sqrt{14.5 } \\approx 3.81{\/tex} units<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.26: Find 5 more Baudh\u0101yana triples using this idea.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>Method:<\/strong> Use the formula {tex}(n-1)^2+(2 n-1)=n^2{\/tex}, where {tex}(2 n-1){\/tex} is an odd perfect square.<br>The odd perfect squares are: 1, 9, 25, 49, 81, 121, 169, 225, &#8230;<br>We already used 9 and 25. Let&#8217;s use the next ones.<br><strong>Triple 1: Using 49<\/strong><br>{tex}49=2 n-1{\/tex}<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}2 {n}=50{\/tex}<\/li>\n\n\n\n<li>{tex}{n}=25{\/tex}<\/li>\n<\/ul>\n\n\n\n<p>Formula: {tex}(25-1)^2+49=25^2{\/tex}<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}24^2+7^2=25^2{\/tex}<\/li>\n\n\n\n<li>{tex}576+49=625{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Baudh\u0101yana triple: {tex}(7,24,25){\/tex}<\/strong><br><strong>Triple 2: Using 81<\/strong><br>{tex} 81=2 n-1 {\/tex}<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}2 {n}=82{\/tex}<\/li>\n\n\n\n<li>{tex}{n}=41{\/tex}<\/li>\n<\/ul>\n\n\n\n<p>Formula: {tex}(41-1)^2+81=41^2{\/tex}<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}40^2+9^2=41^2{\/tex}<\/li>\n\n\n\n<li>{tex}1600+81=1681{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Baudh\u0101yana triple: {tex}(9,40,41){\/tex}<\/strong><br><strong>Triple 3: Using 121<\/strong><br>{tex} 121=2 n-1 {\/tex}<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}2 {n}=122{\/tex}<\/li>\n\n\n\n<li>{tex}{n}=61{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Formula:<\/strong> {tex}(61-1)^2+121=61^2{\/tex}<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}60^2+11^2=61^2{\/tex}<\/li>\n\n\n\n<li>{tex}3600+121=3721{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Baudh\u0101yana triple: {tex}(11,60,61){\/tex}<\/strong><br><strong>Triple 4: Using 169<\/strong><br>{tex} 169=2 n-1 {\/tex}<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}2 {n}=170{\/tex}<\/li>\n\n\n\n<li>{tex}{n}=85{\/tex}<\/li>\n<\/ul>\n\n\n\n<p>Formula: {tex}(85-1)^2+169=85^2{\/tex}<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}84^2+13^2=85^2{\/tex}<\/li>\n\n\n\n<li>{tex}7056+169=7225{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Baudh\u0101yana triple: <\/strong>{tex}(13,84,85){\/tex}<br><strong>Triple 5: Using 225<\/strong><br>{tex} 225=2 {\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.27: Does this method yield non-primitive Baudh\u0101yana triples?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>No<\/strong>, this method does <strong>NOT<\/strong> yield non-primitive Baudh\u0101yana triples. It yields only primitive triples.<br><strong>Reason:<\/strong><br>Using the formula {tex}(n-1)^2+(2 n-1)=n^2{\/tex}, we get triples of the form:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}((2 n-1),(n-1), n){\/tex}<\/li>\n<\/ul>\n\n\n\n<p>Rearranging: ( {tex}{n &#8211; 1 , 2 n &#8211; 1 , n}{\/tex} ) where the hypotenuse differs from one side by just 1.<br><strong>Observation from the hint:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}\\ln (3,4,5): 5-4=1{\/tex}<\/li>\n\n\n\n<li>{tex}\\ln (5,12,13): 13-12=1{\/tex}<\/li>\n\n\n\n<li>{tex}\\ln (7,24,25): 25-24=1{\/tex}<\/li>\n\n\n\n<li>{tex}\\ln (9,40,41): 41-40=1{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Key insight: <\/strong>If one sidelength is one less than the hypotenuse ({tex}{c}-{b}=1{\/tex}), then these two numbers are c<strong>onsecutive integers.<\/strong><br><strong>Consecutive integers always have GCD {tex}{= 1}{\/tex}<\/strong> (no common factor &gt;1).<br>Since two of the three numbers have no common factor, the entire triple cannot have a common factor &gt;1.<br><strong>Therefore, all triples generated by this method are primitive.<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.28: Are there primitive triples that cannot be obtained through this method? If yes, give examples.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Yes, there are primitive triples that cannot be obtained through this method.<br><strong>Example: (3,4,5)<\/strong><br>Using the method: One side should be one less than hypotenuse<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Check: {tex}5-4=1{\/tex}&nbsp;(Right)<\/li>\n\n\n\n<li>Check: {tex}5-3=2 {\/tex}&nbsp;(Wrong)<\/li>\n<\/ul>\n\n\n\n<p>Actually, {tex}(3,4,5){\/tex} can be obtained! Let&#8217;s verify:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>If {tex}n=5{\/tex}, then {tex}(n-1)=4{\/tex} and {tex}(2 n-1)=9=3^2{\/tex}<\/li>\n\n\n\n<li>So {tex}4^2+3^2=16+9=25=5^2 {\/tex}&nbsp;(Right)<\/li>\n<\/ul>\n\n\n\n<p><strong>Better Example: (8,15,17)<\/strong><br>Check if one side is one less than hypotenuse:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}17-15=2{\/tex}&nbsp;(Wrong)<\/li>\n\n\n\n<li>{tex}17-8=9{\/tex}&nbsp;(Wrong)<\/li>\n<\/ul>\n\n\n\n<p>This triple <strong>cannot <\/strong>have the form {tex}(n-1,2 n-1, n){\/tex} because:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>If {tex}{n}=17{\/tex}, then {tex}({n}-1)=16 \\neq 8{\/tex} or 15<\/li>\n\n\n\n<li>Neither 8 nor 15 is one less than 17<\/li>\n<\/ul>\n\n\n\n<p><strong>Verification:<\/strong> {tex}(8,15,17){\/tex} is indeed primitive:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}8^2+15^2=64+225=289=17^2{\/tex}&nbsp;(Right)<\/li>\n\n\n\n<li>{tex}{GCD}(8,15,17)=1{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Conclusion:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Yes,<\/strong> there are primitive triples not obtained by this method<\/li>\n\n\n\n<li><strong>Example:<\/strong> {tex}(8,15,17){\/tex} and many others<\/li>\n\n\n\n<li>This method gives us some but not all primitive triples<\/li>\n\n\n\n<li>To get all primitive triples, we need other generation methods as well<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.29: Find the diagonal of a square with sidelength 5 cm.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>Given:<\/strong> Square with side {tex}=5 {~cm}{\/tex}<br><strong>To find:<\/strong> Length of diagonal<br><strong>Method:<\/strong><br>A diagonal of a square divides it into two congruent right-angled triangles.<br>For each triangle:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Both perpendicular sides {tex}=5 {~cm}{\/tex} (sides of square)<\/li>\n\n\n\n<li>Hypotenuse = diagonal of square<\/li>\n<\/ul>\n\n\n\n<p><strong>Using Baudh\u0101yana&#8217;s Theorem:<\/strong><br>Let d = length of diagonal<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}5^2+5^2={d}^2{\/tex}<\/li>\n\n\n\n<li>{tex}25+25=d^2{\/tex}<\/li>\n\n\n\n<li>{tex}50=d^2{\/tex}<\/li>\n\n\n\n<li>{tex}{d}=\\sqrt{ 50} {\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Simplifying:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}\\sqrt{ 50} =\\sqrt{(25 \\times 2) }{\/tex}&nbsp;{tex}=\\sqrt{25 } \\times \\sqrt{2 } =5 \\sqrt{ 2} {\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Finding approximate value:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}\\sqrt{2 } \\approx 1.414{\/tex}<\/li>\n\n\n\n<li>{tex}5 \\sqrt{2 } \\approx 5 \\times 1.414=7.07{\/tex}<\/li>\n<\/ul>\n\n\n\n<p>The diagonal of the square is {tex}{5} \\sqrt{2 } {~ c m} \\boldsymbol{\\approx} {7 . 0 7} {~ c m}{\/tex}.<br><strong>Alternative formula:<\/strong> For a square with side {tex}a{\/tex}, diagonal {tex}=a \\sqrt{ 2} {\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.30: Find the missing sidelengths in the right triangle:<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768800083-xy6kgy.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>If one perpendicular side {tex}=7{\/tex} and other perpendicular side {tex}=9{\/tex}:<br><strong>Given: <\/strong>{tex}a=7, b=9, c={\/tex} ?<br><strong>Using Baudh\u0101yana&#8217;s Theorem:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}7^2+9^2=c^2{\/tex}<\/li>\n\n\n\n<li>{tex}49+81={c}^2{\/tex}<\/li>\n\n\n\n<li>{tex}130={c}^2{\/tex}<\/li>\n\n\n\n<li>{tex}{c}=\\sqrt{130 }{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Finding bounds:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}11^2=121,12^2=144{\/tex}<\/li>\n\n\n\n<li>So {tex}11&lt;\\sqrt{ 130} &lt;12{\/tex}<\/li>\n\n\n\n<li>More precisely: {tex}11.4^2=129.96,11.5^2=132.25{\/tex}<\/li>\n\n\n\n<li>So {tex}11.4&lt;\\sqrt{ 130} &lt;11.5{\/tex}<\/li>\n<\/ul>\n\n\n\n<p>{tex}c=\\sqrt{130 } \\approx 11.40{\/tex} units<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.31: Find the missing sidelengths in the right triangle:<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768800266-u22w7u.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>Given:<\/strong> {tex}a=4, b=10, c={\/tex} ?<br><strong>Using Baudh\u0101yana&#8217;s Theorem:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}4^2+10^2={c}^2{\/tex}<\/li>\n\n\n\n<li>{tex}16+100={c}^2{\/tex}<\/li>\n\n\n\n<li>{tex}116={c}^2{\/tex}<\/li>\n\n\n\n<li>{tex}c=\\sqrt{116 }{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Simplifying:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}\\sqrt{116 } =\\sqrt{(4 \\times 29) }=2 \\sqrt{ 29} {\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Finding bounds:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}10^2=100,11^2=121{\/tex}<\/li>\n\n\n\n<li>So {tex}10&lt;\\sqrt{ 116} &lt;11{\/tex}<\/li>\n\n\n\n<li>More precisely: {tex}10.7^2=114.49,10.8^2=116.64{\/tex}<\/li>\n\n\n\n<li>So {tex}10.7&lt;\\sqrt{ 116} &lt;10.8{\/tex}<\/li>\n<\/ul>\n\n\n\n<p>{tex}c=\\sqrt{116 } =2 \\sqrt{29 } \\approx 10.77{\/tex} units<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.32: Find the missing sidelengths in the right triangle:<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768800421-e74jby.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>Given:<\/strong> {tex}a=40, c=41, b={\/tex} ?<br><strong>Using Baudh\u0101yana&#8217;s Theorem:<\/strong><br>{tex} b^2=c^2-a^2 {\/tex}<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}{b}^2=41^2-40^2{\/tex}<\/li>\n\n\n\n<li>{tex}{b}^2=1681-1600{\/tex}<\/li>\n\n\n\n<li>{tex}{b}^2=81{\/tex}<\/li>\n\n\n\n<li>{tex}{b}=\\sqrt{ 81} =9{\/tex}<\/li>\n<\/ul>\n\n\n\n<p>{tex}b=9{\/tex} units<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.33: Find the missing sidelengths in the&nbsp; right triangle:<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768800714-mw9tvk.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>Given:<\/strong> {tex}{a}=27, {c}=45, {~b}={\/tex} ?<br><strong>Using Baudh\u0101yana&#8217;s Theorem:<\/strong><br>{tex} b^2=c^2-a^2 {\/tex}<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}{b}^2=45^2-27^2{\/tex}<\/li>\n\n\n\n<li>{tex}{b}^2=2025-729{\/tex}<\/li>\n\n\n\n<li>{tex}{b}^2=1296{\/tex}<\/li>\n\n\n\n<li>{tex}{b}=\\sqrt{1296 }=36{\/tex}<\/li>\n<\/ul>\n\n\n\n<p>{tex}{b}=36{\/tex} units<br><strong>Verification:<\/strong> {tex}27^2+36^2=729+1296=2025=45^2{\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.34: Find the missing sidelengths in the right triangle:<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768800874-j7vj8s.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given: {tex}a=\\sqrt{200}, b=10, c={\/tex} ?<br>Using Baudh\u0101yana&#8217;s Theorem:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}200^2+10^2={c}^2{\/tex}<\/li>\n\n\n\n<li>{tex}40000+100={c}^2{\/tex}<\/li>\n\n\n\n<li>{tex}40100={c}^2{\/tex}<\/li>\n\n\n\n<li>{tex}{c}=\\sqrt{ 40100} {\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Simplifying:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}\\sqrt{ 40100} =\\sqrt{(100 \\times 401) }=10 \\sqrt{ 401} {\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Finding approximate value:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}20^2=400,21^2=441{\/tex}<\/li>\n\n\n\n<li>So {tex}20&lt;\\sqrt{401 }&lt;21{\/tex}<\/li>\n\n\n\n<li>More precisely: {tex}20.02^2 \\approx 400.8,20.03^2 \\approx 401.2{\/tex}<\/li>\n\n\n\n<li>So {tex}\\sqrt{401 } 1 \\approx 20.02{\/tex}<\/li>\n\n\n\n<li>Therefore, {tex}{c} \\approx 10 \\times 20.02=200.2{\/tex}<\/li>\n<\/ul>\n\n\n\n<p>{tex}c=10 \\sqrt{401 }\\approx 200.2{\/tex} units<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.35: Find the missing sidelengths in the right triangle:<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768801123-8vbysg.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>Given:<\/strong> {tex}a=10, b=150, c={\/tex} ?<br><strong>Using Baudh\u0101yana&#8217;s Theorem:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}10^2+150^2={c}^2{\/tex}<\/li>\n\n\n\n<li>{tex}100+22500={c}^2{\/tex}<\/li>\n\n\n\n<li>{tex}22600={c}^2{\/tex}<\/li>\n\n\n\n<li>{tex}c=\\sqrt{ 22600} {\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Simplifying:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}\\sqrt{22600 } =\\sqrt{ (100 \\times 226)}=10 \\sqrt{ 226} {\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Finding approximate value:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}15^2=225,16^2=256{\/tex}<\/li>\n\n\n\n<li>So {tex}15&lt;\\sqrt{226 }&lt;16{\/tex}<\/li>\n\n\n\n<li>More precisely: {tex}15.03^2 \\approx 225.9,15.04^2 \\approx 226.2{\/tex}<\/li>\n\n\n\n<li>So {tex}\\sqrt{226 } \\approx 15.03{\/tex}<\/li>\n\n\n\n<li>Therefore, {tex}{c} \\approx 10 \\times 15.03=150.3{\/tex}<\/li>\n<\/ul>\n\n\n\n<p>{tex}c=10 \\sqrt{ 226} \\approx 150.3{\/tex} units<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.36: Find the sidelength of a rhombus whose diagonals are of length 24 units and 70 units.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Diagonal {tex}1\\left({~d}_1\\right)=24{\/tex} units<\/li>\n\n\n\n<li>Diagonal {tex}2\\left({~d}_2\\right)=70{\/tex} units<\/li>\n<\/ul>\n\n\n\n<p><strong>To find:<\/strong> Side length of the rhombus<br><strong>Key properties of a rhombus:<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Diagonals bisect each other at right angles {tex}\\left(90^{\\circ}\\right){\/tex}<\/li>\n\n\n\n<li>All four sides are equal<\/li>\n<\/ol>\n\n\n\n<p>When diagonals intersect, they form 4 right-angled triangles.<br>Each right triangle has:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>One side {tex}={d}_1 \/ 2=24 \/ 2=12{\/tex} units<\/li>\n\n\n\n<li>Other side {tex}={d}_2 \/ 2=70 \/ 2=35{\/tex} units<\/li>\n\n\n\n<li>Hypotenuse = side of rhombus (s)<\/li>\n<\/ul>\n\n\n\n<p><strong>Using Baudh\u0101yana&#8217;s Theorem:<\/strong><br>{tex} s^2=12^2+35^2 {\/tex}<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}{s}^2=144+1225{\/tex}<\/li>\n\n\n\n<li>{tex}{s}^2=1369{\/tex}<\/li>\n\n\n\n<li>{tex}{s}=\\sqrt{1369 } =37{\/tex}<\/li>\n<\/ul>\n\n\n\n<p>The side length of the rhombus is <strong>37 units<\/strong>.<br><strong>Verification:<\/strong> {tex}12^2+35^2=144+1225=1369=37^2{\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.37: Is the hypotenuse the longest side of a right triangle? Justify your answer.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Yes, the hypotenuse is always the longest side of a right triangle.<br><strong>Justification:<\/strong><br><strong>Method 1: Using Baudh\u0101yana&#8217;s Theorem<\/strong><br>In a right triangle with sides {tex}{a}, {b}{\/tex} and hypotenuse c:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}a^2+b^2=c^2{\/tex}<\/li>\n<\/ul>\n\n\n\n<p>Since {tex}{a}^2{\/tex} and {tex}{b}^2{\/tex} are both positive:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}c^2=a^2+b^2{\/tex}<\/li>\n\n\n\n<li>{tex}c^2&gt;a^2\\left(\\right.{\/tex}because {tex}\\left.b^2&gt;0\\right){\/tex}<\/li>\n\n\n\n<li>{tex}{c}^2&gt;{b}^2\\left(\\right.{\/tex}because {tex}\\left.{a}^2&gt;0\\right){\/tex}<\/li>\n<\/ul>\n\n\n\n<p>Taking square roots:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}c&gt;a{\/tex}<\/li>\n\n\n\n<li>{tex}c&gt;b{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Therefore, hypotenuse {tex}{c}{\/tex} is greater than both other sides.<\/strong><br><strong>Method 2: Logical reasoning<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Let&#8217;s assume one of the other sides is longer than the hypotenuse<\/li>\n\n\n\n<li>Say b {tex}&gt;{\/tex} c<\/li>\n<\/ul>\n\n\n\n<p>Then:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}{b}^2&gt;{c}^2{\/tex}<\/li>\n\n\n\n<li>But from Baudh\u0101yana&#8217;s theorem: {tex}a^2+b^2=c^2{\/tex}<\/li>\n\n\n\n<li>This means: {tex}{b}^2={c}^2-{a}^2{\/tex}<\/li>\n\n\n\n<li>Since {tex}a^2&gt;0{\/tex}, we get {tex}b^2&lt;c^2{\/tex}<\/li>\n\n\n\n<li>This contradicts our assumption that {tex}{b}^2&gt;{c}^2{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Therefore, the hypotenuse must be the longest side.<\/strong><br><strong>Method 3: Geometric understanding<\/strong><br>The hypotenuse is the side opposite to the largest angle {tex}\\left(90^{\\circ}\\right){\/tex} in the triangle. In any triangle, the longest side is always opposite to the largest angle.<br><strong>Conclusion:<\/strong> Yes, in every right triangle, the hypotenuse is always the longest side.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.38: Every Baudh\u0101yana triple is either a primitive triple or a scaled version of a primitive triple.<\/p>\n\n\n\n<p>Options:<br>(1) True \u2705<br>(2) False<\/p>\n\n\n\n<p>Explanation: True<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.39: Give 5 examples of rectangles whose sidelengths and diagonals are all integers.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>For a rectangle with sides {tex}a{\/tex} and {tex}b{\/tex}, the diagonal {tex}d{\/tex} is given by:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}d^2=a^2+b^2{\/tex}<\/li>\n<\/ul>\n\n\n\n<p>We need {tex}{a}, {b}{\/tex}, and d to all be integers. This means we need Baudh\u0101yana triples!<br><strong>The sides of the rectangle are the two smaller numbers, and the diagonal is the largest number from a Baudh\u0101yana triple.<\/strong><br><strong>Example 1: Using {tex}\\boldsymbol{(} {3 , 4 , 5} \\boldsymbol{)}{\/tex}<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Length {tex}=4{\/tex} units<\/li>\n\n\n\n<li>Width {tex}=3{\/tex} units<\/li>\n\n\n\n<li>Diagonal {tex}=5{\/tex} units<\/li>\n\n\n\n<li>Verification: {tex}3^2+4^2=9+16=25=5^2{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Example 2: Using {tex}\\boldsymbol{(} {5 , 1 2 , 1 3} \\boldsymbol{)}{\/tex}<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Length {tex}=12{\/tex} units<\/li>\n\n\n\n<li>Width {tex}=5{\/tex} units<\/li>\n\n\n\n<li>Diagonal {tex}=13{\/tex} units<\/li>\n\n\n\n<li>Verification: {tex}5^2+12^2=25+144=169=13^2{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Example 3: Using {tex}({8 ,} {1 5 ,} {1 7}){\/tex}<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Length = 15 units<\/li>\n\n\n\n<li>Width {tex}=8{\/tex} units<\/li>\n\n\n\n<li>Diagonal {tex}=17{\/tex} units<\/li>\n\n\n\n<li>Verification: {tex}8^2+15^2=64+225=289=17^2{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Example 4: Using {tex}\\boldsymbol{(} {7} \\boldsymbol{,} {2 4} \\boldsymbol{,} {2 5} \\boldsymbol{)}{\/tex}<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Length = 24 units<\/li>\n\n\n\n<li>Width {tex}=7{\/tex} units<\/li>\n\n\n\n<li>Diagonal {tex}=25{\/tex} units<\/li>\n\n\n\n<li>Verification: {tex}7^2+24^2=49+576=625=25^2{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Example 5: Using {tex}(6,8,10){\/tex}<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Length {tex}=8{\/tex} units<\/li>\n\n\n\n<li>Width {tex}=6{\/tex} units<\/li>\n\n\n\n<li>Diagonal {tex}=10{\/tex} units<\/li>\n\n\n\n<li>Verification: {tex}6^2+8^2=36+64=100=10^2 {\/tex}<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.40: Construct a square whose area is equal to the difference of the areas of squares of sidelengths 5 units and 7 units.<\/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>First square: side {tex}=7{\/tex} units, {tex}\\operatorname{Area}=7^2=49{\/tex} sq. units<\/li>\n\n\n\n<li>Second square: side {tex}=5{\/tex} units, Area {tex}=5^2=25{\/tex} sq. units<\/li>\n<\/ul>\n\n\n\n<p><strong>Required:<\/strong> Square with area {tex}=49-25=24{\/tex} sq. units<br><strong>To find:<\/strong> Side of the new square<br>If the side of new square = s, then:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}{s}^2=24{\/tex}<\/li>\n\n\n\n<li>{tex}s=\\sqrt{ 24} =\\sqrt{(4 \\times 6) }=2 \\sqrt{6 } {\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Construction Method:<\/strong><br><strong>Step 1:<\/strong> Understanding the relationship<br>We need to find a right triangle where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Hypotenuse {tex}=7{\/tex}<\/li>\n\n\n\n<li>One side {tex}=5{\/tex}<\/li>\n\n\n\n<li>Other side = s<\/li>\n<\/ul>\n\n\n\n<p>Using Baudh\u0101yana&#8217;s theorem:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}5^2+s^2=7^2{\/tex}<\/li>\n\n\n\n<li>{tex}25+{s}^2=49{\/tex}<\/li>\n\n\n\n<li>{tex}s^2=24{\/tex}<\/li>\n\n\n\n<li>{tex}s=\\sqrt{ 24} =2 \\sqrt{6 } {\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 2:<\/strong> Practical Construction<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Draw a square {tex}A B C D{\/tex} with side 7 units<\/li>\n\n\n\n<li>Draw a square PQRS with side 5 units<\/li>\n\n\n\n<li>Place the smaller square inside the larger square, aligning one corner<\/li>\n\n\n\n<li>Draw a right triangle with:\n<ul class=\"wp-block-list\">\n<li>Hypotenuse {tex}=7{\/tex} units (diagonal or side of larger square)<\/li>\n\n\n\n<li>One side {tex}=5{\/tex} units (side of smaller square)<\/li>\n\n\n\n<li>The third side will be {tex}\\sqrt{ 24} {\/tex} units<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Construct a square on this third side.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.41: Using the dots of a grid as the vertices, can you create a square that has an area of (a) 2 sq. units, (b) 3 sq. units, (c) 4 sq.units, and (d) 5 sq. unit?<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768803339-p72eph.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>Understanding:<\/strong> The distance between adjacent dots (horizontally or vertically) = 1 unit.<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li><strong>Area {tex}=2{\/tex} sq. units<br>Yes, this is possible!<br>Method:<\/strong><ul><li>We need a square with side {tex}=\\sqrt{2}{\/tex}<\/li><li>This is the diagonal of a {tex}1 \\times 1{\/tex} square<\/li><li>Connect dots that are 1 unit apart horizontally and 1 unit apart vertically<\/li><\/ul><strong>Construction:<\/strong><ul><li>Take a point A<\/li><li>Move 1 unit right to point B<\/li><li>Move 1 unit up to point C<\/li><li>Move 1 unit left to point D<\/li><li>Complete the square by connecting diagonally<\/li><\/ul>The tilted square formed has:\n<ul class=\"wp-block-list\">\n<li>Side {tex}=\\sqrt{\\left(1^2+1^2\\right) }=\\sqrt{2}{\/tex}<\/li>\n\n\n\n<li>Area {tex}=(\\sqrt{2 })^2=2{\/tex} sq. units<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Area {tex}=3{\/tex} sq. units<br>Yes, this is possible!<br>Method:<\/strong><ul><li>We need a square with side {tex}=\\sqrt{ 3}{\/tex}<\/li><li>Use a right triangle with sides 1 and {tex}\\sqrt{ 2}{\/tex}<\/li><\/ul>But easier: Think of moving 1 unit in one direction and {tex}\\sqrt{ 2}{\/tex} in perpendicular direction is complex.<br><strong>Better construction:&nbsp;<\/strong>The diagonal of a rectangle with sides 1 and {tex}\\sqrt{ 2}{\/tex} gives {tex}\\sqrt{ 3}{\/tex}.<br>Actually, on a grid:<ul><li>Take a segment connecting points that are 1 unit right and 1 unit up: length {tex}=\\sqrt{2}{\/tex}<\/li><li>We need side {tex}=\\sqrt{ 3}{\/tex}<\/li><\/ul>This can be done by connecting points in a specific pattern on the grid.<br><strong>Yes,<\/strong> a square with area 3 can be constructed.<\/li>\n\n\n\n<li><strong>Area {tex}=4{\/tex} sq. units<\/strong><br>Yes, this is very easy!<br><strong>Method:<\/strong><ul><li>We need a square with side {tex}=2{\/tex} units<\/li><li>Simply take a {tex}2 \\times 2{\/tex} square on the grid<\/li><li>This uses dots that are 2 units apart<\/li><\/ul><strong>Area 2&nbsp;{tex}\\times{\/tex}&nbsp;2 = 4 sq. units<\/strong><\/li>\n\n\n\n<li><strong>Area {tex}=5{\/tex} sq. units<\/strong><br><strong>Yes, this is possible!<\/strong><br><strong>Method:<\/strong><ul><li>We need a square with side {tex}=\\sqrt{5}{\/tex}<\/li><li>From Baudh\u0101yana triple {tex}(1,2, \\sqrt{5}): 1^2+2^2=5{\/tex}<\/li><li>Connect dots that form a right triangle with legs 1 and 2<\/li><\/ul><strong>Construction<\/strong>:<ul><li>From point A, move 2 units right to B<\/li><li>Move 1 unit up to get the hypotenuse length {tex}\\sqrt{5}{\/tex}<\/li><li>This {tex}\\sqrt{5}{\/tex} becomes the side of our square<\/li><\/ul><strong>Area {tex}=(\\sqrt{5})^2=5{\/tex} sq. units<\/strong><\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.42: Suppose the grid extends indefinitely. What are the possible integer-valued areas of squares you can create in this manner?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>Analysis:<\/strong><br>The side of any square we can create on the grid has the form:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}\\sqrt{ \\left(a^2+b^2\\right)}{\/tex}<\/li>\n<\/ul>\n\n\n\n<p>where {tex}a{\/tex} and {tex}b{\/tex} are non-negative integers (representing horizontal and vertical movements between grid points).<br><strong>The area of such a square:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}\\operatorname{Area}=\\left(\\sqrt{\\left(a^2+b^2\\right)}\\right)^2=a^2+b^2{\/tex}<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.43: Find the area of an equilateral triangle with sidelength 6 units.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>Given: <\/strong>Equilateral triangle with side {tex}=6{\/tex} units<br><strong>To find:<\/strong> Area of the triangle<br><strong>Step 1: Understanding the hint<\/strong><br>In an equilateral triangle:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>All sides are equal<\/li>\n\n\n\n<li>All angles are {tex}60^{\\circ}{\/tex}<\/li>\n\n\n\n<li>An altitude (height) from any vertex bisects the opposite side and is perpendicular to it<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 2: Drawing and analyzing<\/strong><br>Let triangle ABC be equilateral with side 6 units.<br>Draw altitude AD from A to side BC.<br><strong>Properties:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}{AD} \\perp {BC}{\/tex} (altitude is perpendicular)<\/li>\n\n\n\n<li>{tex}{BD}={DC}=3{\/tex} units (altitude bisects the base)<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 3:<\/strong> <strong>Finding the height using Baudh\u0101yana&#8217;s Theorem<\/strong><br>Triangle ABD is a right-angled triangle with:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}{AB}=6{\/tex} units (side of equilateral triangle)<\/li>\n\n\n\n<li>{tex}{BD}=3{\/tex} units (half the base)<\/li>\n\n\n\n<li>{tex}{AD}={h}{\/tex} (height, to be found)<\/li>\n<\/ul>\n\n\n\n<p><strong>Using Baudh\u0101yana&#8217;s theorem in triangle ABD:<\/strong><br>{tex} {BD}^2+{AD}^2={AB}^2 {\/tex}<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}3^2+h^2=6^2{\/tex}<\/li>\n\n\n\n<li>{tex}9+h^2=36{\/tex}<\/li>\n\n\n\n<li>{tex}{h}^2=27{\/tex}<\/li>\n\n\n\n<li>{tex}h=\\sqrt{27} =\\sqrt{(9 \\times 3)}=3 \\sqrt{3}{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Height {tex}=3 \\sqrt{3}{\/tex} units<\/strong><br><strong>Step 4: Calculating the area<\/strong><br>Area of triangle {tex}=\\frac 12 \\times{\/tex} base \u00d7 height<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}=\\frac 1 2 \\times 6 \\times 3 \\sqrt{3}{\/tex}<\/li>\n\n\n\n<li>{tex}=3 \\times 3 \\sqrt{3}{\/tex}<\/li>\n\n\n\n<li>{tex}=9 \\sqrt{3}{\/tex} sq. units<\/li>\n<\/ul>\n\n\n\n<p><strong>Finding approximate value:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}\\sqrt{3} \\approx 1.732{\/tex}<\/li>\n\n\n\n<li>Area {tex}\\approx 9 \\times 1.732=15.588{\/tex} sq. units<\/li>\n<\/ul>\n\n\n\n<p>The area of the equilateral triangle is {tex}{9} \\sqrt{3}{\/tex} sq. units {tex}\\boldsymbol{\\approx} {1 5 . 5 9}{\/tex} sq. units.<br><strong>General formula: <\/strong>For an equilateral triangle with side a:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Height {tex}=\\frac {({a} \\sqrt{3 } ) }{ 2}{\/tex}<\/li>\n\n\n\n<li>Area {tex}=\\frac {\\left(a^2 \\sqrt{3}\\right) }{ 4}{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Verification for {tex}{a = 6 :}{\/tex}<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Area {tex}=\\frac {\\left(6^2 \\times \\sqrt{3}\\right) }{ 4}=\\frac {(36 \\sqrt{ 3}) }{4}=9 \\sqrt{3}{\/tex}<\/li>\n<\/ul>\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>The Baudhayana-Pythagoras Theorem &#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 The Baudhayana-Pythagoras Theorem \u2013 NCERT Solutions Q.1: How can one construct a square having double the area of a given square?A first guess might be &#8230; <a title=\"The Baudhayana-Pythagoras Theorem &#8211; NCERT Solutions Class 8 Maths (Ganita Prakash)\" class=\"read-more\" href=\"https:\/\/mycbseguide.com\/blog\/the-baudhayana-pythagoras-theorem-ncert-solutions-class-8-maths-ganita-prakash\/\" aria-label=\"More on The Baudhayana-Pythagoras Theorem &#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-31747","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>The Baudhayana-Pythagoras Theorem - NCERT Solutions Class 8 Maths (Ganita Prakash) | myCBSEguide<\/title>\n<meta name=\"description\" content=\"The Baudhayana-Pythagoras Theorem - 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\/the-baudhayana-pythagoras-theorem-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=\"The Baudhayana-Pythagoras Theorem - NCERT Solutions Class 8 Maths (Ganita Prakash) | myCBSEguide\" \/>\n<meta property=\"og:description\" content=\"The Baudhayana-Pythagoras Theorem - NCERT Solutions Class 8 Maths (Ganita Prakash) includes all the questions with solutions\" \/>\n<meta property=\"og:url\" content=\"https:\/\/mycbseguide.com\/blog\/the-baudhayana-pythagoras-theorem-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=\"2026-08-07T07:16:19+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2026-08-07T07:54:12+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768814276-46pgbc.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=\"26 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/the-baudhayana-pythagoras-theorem-ncert-solutions-class-8-maths-ganita-prakash\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/the-baudhayana-pythagoras-theorem-ncert-solutions-class-8-maths-ganita-prakash\/\"},\"author\":{\"name\":\"myCBSEguide\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/#\/schema\/person\/10b8c7820ff29025ab8524da7c025f65\"},\"headline\":\"The Baudhayana-Pythagoras Theorem &#8211; 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