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A Tale of Three Intersecting Lines – NCERT Solutions Class 7 Maths (Ganita Prakash)

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A Tale of Three Intersecting Lines – NCERT Solutions Class 7 Maths (Ganita Prakash) includes all the questions with solutions given in the NCERT Class 7 Maths (Ganita Prakash).

NCERT Solutions Class 7

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A Tale of Three Intersecting Lines – NCERT Solutions

Q.1: What happens when the three vertices lie on a straight line?

Solution:

When the three vertices lie on a straight line, they are said to be collinear. In this case, they do not form a closed shape, and hence, a triangle cannot be formed.

Q.2: Construct a triangle with side lengths 4 cm, 4 cm, and 6 cm.

Solution:

4, 4, 6

  1. Draw the base PQ using one of the given side lengths. Let {tex}{PQ}=6 {~cm}{/tex}.
  2. From points P and Q , draw arcs of radius 4 cm each. Let the arcs intersect at point R.
  3. The point of intersection {tex}R{/tex} is the required third vertex. Join {tex}P R{/tex} and {tex}Q R{/tex} to form {tex}\triangle P Q R{/tex}.

Q.3: Construct a triangle with side lengths 3 cm, 4 cm, and 5 cm.

Solution:

3, 4, 5

  1. Draw the base PQ with one of the side lengths. Let {tex}{PQ}=5 {~cm}{/tex}.
  2. From P, draw a long arc of radius 3 cm.
  3. From Q, draw an arc of radius 4 cm intersecting the first arc at point R.
  4. The point {tex}R{/tex} is the required third vertex. Join {tex}P R{/tex} and {tex}Q R{/tex} to get {tex}\triangle P Q R{/tex}.

Q.4: Construct a triangle with side lengths 1 cm, 5 cm, and 5 cm.

Solution:

1, 5, 5

  1. Construct the base PQ with one of the side lengths. Let PQ = 5 cm.
  2. From P, draw long arc of radius 1 cm.
  3. From Q, draw arc of radius 5 cm intersecting the first arc at point R.
  4. The point {tex}R{/tex} is the required third vertex. Join {tex}P R{/tex} and {tex}Q R{/tex} to get {tex}\triangle P Q R{/tex}.

Q.5: Construct a triangle with side lengths 4 cm, 6 cm, and 8 cm.

Solution:

4, 6, 8

  1. Construct the base PQ with one of the side lengths. Let {tex}{PQ}=8 {~cm}{/tex}.
  2. From P, draw a long arc of radius 4 cm.
  3. From {tex}Q{/tex}, draw an arc of radius 6 cm intersecting the first arc at point {tex}R{/tex}.
  4. The point {tex}R{/tex} is the required third vertex. Join {tex}P R{/tex} and {tex}Q R{/tex} to get {tex}\triangle P Q R{/tex}.

Q.6: Construct an equilateral triangle with each side measuring 3.5 cm, 3.5 cm, 3.5 cm

Solution:

3.5, 3.5, 3.5

  1. Construct the base PQ of length 3.5 cm.
  2. From points P and Q , draw arcs of radius 3.5 cm each. Let the arcs intersect at point R.
  3. The point {tex}R{/tex} is the required third vertex. Join {tex}P R{/tex} and {tex}Q R{/tex} to get {tex}\triangle P Q R{/tex}.

Q.7: Construct a triangle with side lengths 3 cm, 4 cm, and 8 cm. What is happening? Are you able to construct the triangle?

Solution:


From the base PQ of length 8 cm, arcs are drawn from points P and Q with radii 3 cm and 4 cm, respectively. However, these arcs do not intersect at any point.
Therefore, it is not possible to construct a triangle with side lengths 3 cm, 4 cm, and 8 cm.

Q.8: Here is another set of lengths: 2 cm, 3 cm, and 6 cm. Check if a triangle is possible for these side lengths.

Solution:


No, it is impossible to construct a triangle with side lengths 2 cm, 3 cm and 6 cm.

Q.9: Can we say anything about the existence of a triangle having side lengths 3 cm, 3 cm and 7 cm? Verify your answer by construction.

Solution:


Consider the paths between A and B:
Direct path length {tex}=A B=3 {~cm}{/tex}
Round about path length via vertex {tex}C=A C+B C=7+3=10 {~cm}{/tex}
Consider the paths between B and C:
Direct path length {tex}={BC}=3 {~cm}{/tex}
Round about path length via vertex {tex}A=A B+A C=3+7=10 {~cm}{/tex}
Consider the paths between A and C:
Direct path length {tex}=A C=7 {~cm}{/tex}
Round about path length via vertex {tex}B=A B+C B=3+3=6 {~cm}{/tex}
Since the direct path is longer than the roundabout path here. Therefore, such a triangle doesn’t exist
Additionally, a triangle with the given side lengths is not possible as the two arcs from the end points of the base do not meet to provide a third vertex.

Q.10: “In the rough diagram, is it possible to assign lengths in a different order such that the direct paths are always coming out to be shorter than the roundabout paths? If this is possible, then a triangle might exist.”

Solution:

No, rearranging the sides 10 cm, 15 cm, and 30 cm won’t help to form a triangle because the largest side (30 cm) is greater than the sum of the other two sides (10 cm + 15 cm = 25 cm), which makes triangle formation impossible.

Q.11: Is such rearrangement of lengths possible in the triangle?

Solution:

No, the rearrangement of the lengths 10 cm, 15 cm, and 30 cm can’t help in the formation of a triangle because 30 cm is always going to be greater than 10 + 15 = 25 cm, regardless of the order of the sides.

Q.12: Further, for a given set of lengths, is it possible to identify which lengths will immediately be less than the sum of the other two, without calculations?

Solution:

Yes, it is possible to identify which length will be immediately less than the sum of the other two by simply arranging the direct lengths in increasing order.

Q.13: How will the two circles turn out for a set of lengths that do not satisfy the triangle inequality? Find 3 examples of sets of lengths for which the circles:

  1. touch each other at a point,
  2. do not intersect.

Solution:

For a set of lengths that do not satisfy the triangle inequality, the two circles either touch each other at a point or do not intersect internally.

  1. Circles touch each other at a point:
    sum of the two smaller lengths = longest length
    4 cm, 4 cm, 8 cm
    3 cm,3 cm, 6 cm
    5 cm, 5 cm,10 cm
  2. Circles do not intersect:
    sum of the two smaller lengths < longest length
    5 cm, 10 cm, 18 cm
    3 cm, 7 cm, 12 cm
    4 cm, 8 cm, 15 cm

Q.14: Frame a complete procedure that can be used to check the existence of a triangle.

Solution:

  1. Arrange the given lengths in increasing order
  2. Check if each length is smaller than the sum of the other two lengths.
  3. If yes, a triangle can be formed. If not, a triangle cannot be formed.

Q.15: Let us take two angles, say 60° and 70°, whose sum is less than 180°. Let the included side be 5 cm.
What could the measure of the third angle be? Does this measure change if the base length is changed to some other value, say 7 cm? Construct and find out.

Solution:

If the two angles of a triangle are {tex}60^{\circ}{/tex} and {tex}70^{\circ}{/tex} and included side is 5 cm . Then, the third angle {tex}=180^{\circ}-\left(60^{\circ}+70^{\circ}\right)=180^{\circ}-130^{\circ}=50^{\circ}{/tex}.

Also, If the two angles of a triangle are {tex}60^{\circ}{/tex} and {tex}70^{\circ}{/tex} and included side is 7 cm .
Then, the third angle {tex}=180^{\circ}-\left(60^{\circ}+70^{\circ}\right)=180^{\circ}-130^{\circ}=50^{\circ}{/tex}.
Thus, changing the base length doesn’t change the measure of the third angle of a triangle.

Q.16: Use the points on the circle and/or the centre to form isosceles triangles.

Solution:

  1. Take any two points A and B on the circle and connect them by drawing a chord.
  2. Draw lines from the center of the circle C to each of these points.
  3. The triangle formed by these two radii and the chord is an isosceles triangle with two sides equal to the radius of the circle.

Q.17: Use the points on the circles and/or their centres to form isosceles and equilateral triangles. The circles are of the same size.

Solution:

  1. Join the points of intersection and the centres of the two circles to form isosceles and equilateral triangles.
    Isosceles triangles: {tex}\triangle {ACD}{/tex} and {tex}\triangle {BCD}{/tex}. Equilateral triangles: {tex}\triangle {ACB}{/tex} and {tex}\triangle {ADB}{/tex}.
  2. Join the points of intersection and centres of the three circles to form isosceles and equilateral triangles.
    Isosceles triangles: {tex}\triangle {ADC}, \triangle {BDC}, \triangle {AEB}, \triangle {ECB}, \triangle {BAF}{/tex} and {tex}\triangle {CAF}{/tex}.
    Equilateral triangles: {tex}\triangle {DAB}, \triangle {ACB}, \triangle {AEC}{/tex} and {tex}\triangle {BCF}{/tex}.

Q.18: We checked by construction that there are no triangles having side lengths 3 cm, 4 cm and 8 cm; and 2 cm, 3 cm and 6 cm. Check if you could have found this without trying to construct the triangle.

Solution:

  1. Let AB = 3 cm, BC = 4 cm and AC = 8 cm.
    Consider the paths between A and B:
    Direct path length {tex}=A B=3 {~cm}{/tex}
    Round about path length via vertex {tex}C=A C+B C=8+4=12 {~cm}{/tex}.
    Thus, the direct path length is shorter than the roundabout path length.
    Consider the paths between B and C:
    Direct path length {tex}=B C=4 {~cm}{/tex}
    Round about path length via vertex {tex}A=A B+A C=3+8=11 {~cm}{/tex}.
    Thus, the direct path length is shorter than the roundabout path length.
    Consider the paths between A and C:
    Direct path length {tex}={AC}=8 {~cm}{/tex}
    Round about path length via vertex {tex}B=A B+B C=3+4=7 {~cm}{/tex}.
    Here, the direct path length is longer than the roundabout path length.
    So, the triangle with the given side lengths doesn’t exist.
  2. Let AB = 2 cm, BC = 3 cm and AC = 6 cm
    Consider the paths between A and B:
    Direct path length {tex}=A B=2 {~cm}{/tex}
    Round about path length via vertex {tex}C=A C+B C=6+3=9 {~cm}{/tex}.
    Thus, the direct path length is shorter than the roundabout path length.
    Consider the paths between B and C:
    Direct path length {tex}={BC}=3 {~cm}{/tex}
    Round about path length via vertex {tex}A=A B+A C=2+6=8 {~cm}{/tex}.
    Thus, the direct path length is shorter than the roundabout path length.
    Consider the paths between A and C:
    Direct path length {tex}=A C=6 {~cm}{/tex}
    Round about path length via vertex {tex}B=A B+B C=2+3=5 {~cm}{/tex}.
    Here, the direct path length is longer than the roundabout path length.
    So, the triangle with the given side lengths doesn’t exist.

Q.19: Can we say anything about the existence of a triangle for each of the following sets of lengths?

  1. 10 km, 10 km and 25 km
  2. 5 mm, 10 mm and 20 mm
  3. 12 cm, 20 cm and 40 cm

You would have realised that using a rough figure and comparing the direct path lengths with their corresponding roundabout path lengths is the same as comparing each length with the sum of the other two lengths. There are three such comparisons to be made.

Solution:

  1. If the direct path is 25 km long, then the roundabout path is 10 km + 10 km = 20 km. Since the direct path cannot be longer than the roundabout path. Therefore, 10 km, 10 km, and 25 km can’t be the side lengths of a triangle.
  2. If the direct path is 20 mm long, then the roundabout path is 5 mm + 10 mm = 15 mm. Since the direct path cannot be longer than the roundabout path. Therefore, 5 mm, 10 mm, and 20 mm can’t be the side lengths of a triangle.
  3. If the direct path is 40 cm long, then the roundabout path is 12 cm + 20 cm = 32 cm. Since the direct path cannot be longer than the roundabout path. Therefore, 12 cm, 20 cm, and 40 cm can’t be the side lengths of a triangle.

Q.20: For each set of lengths seen so far, you might have noticed that in at least two of the comparisons, the direct length was less than the sum of the other two (if not, check again!). For example, for the set of lengths 10 cm, 15 cm and 30 cm, there are two comparisons where this happens:
10 < 15 + 30
15 < 10 + 30
But this doesn’t happen for the third length: 30 > 10 + 15. Will this always happen? That is, for any set of lengths, will there be at least two comparisons where the direct length is less than the sum of the other two? Explore for different sets of lengths.

Solution:

Yes, for any set of lengths, there will always be at least two comparisons where the direct length is less than the sum of the other two. If two out of three comparisons have a direct length smaller than the sum of the other two, then such a triangle doesn’t exist.
However, if all three comparisons have direct length smaller than the sum of the other two, then such a triangle exists.
Let us consider some examples:

  1. {tex}5 {~cm}, 7 {~cm}{/tex} and 9 cm
    {tex} 5+7>9 {/tex}
    {tex} 7+9>5 {/tex}
    {tex} 9+5>7 {/tex}
    Here, all three comparisons have a direct length smaller than the sum of the other two lengths. Hence, a triangle with the given side lengths exists.
  2. {tex}2 {~cm}, 3 {~cm}{/tex} and 6 cm
    {tex} 2+3<6 {/tex}
    {tex} 3+6>2 {/tex}
    {tex} 6+2>3 {/tex}
    Here, only two comparisons have a direct length smaller than the sum of the other two lengths. Hence, a triangle with the given side lengths doesn’t exist.
  3. {tex}7 {~cm}, 15 {~cm}{/tex} and 30 cm
    {tex} 7+5<30 {/tex}
    {tex} 15+30>7 {/tex}
    {tex} 30+7>15 {/tex}
    Here, all three comparisons have a direct length smaller than the sum of the other two lengths. Hence, a triangle with the given side lengths doesn’t exist.

Q.21: Which of the following lengths can be the sidelengths of a triangle? Explain your answers. Note that for each set, the three lengths have the same unit of measure.

  1. 2, 2, 5
  2. 3, 4, 6
  3. 2, 4, 8
  4. 5, 5, 8
  5. 10, 20, 25
  6. 10, 20, 35
  7. 24, 26, 28

We observe from the previous problems that whenever there is a set of lengths satisfying the triangle inequality (each length < sum of the other two lengths), there is a triangle with those three lengths as side lengths.

Solution:

  1. {tex}2,2,5{/tex}
    {tex} 2+2<5 {/tex}
    {tex} 2+5>2 {/tex}
    {tex} 5+2>2 {/tex}
    Since, the lengths don’t follow the triangle inequality. Therefore, they can’t be the side lengths of a triangle.
  2. 3, 4, 6
    {tex} 3+4>6 {/tex}
    {tex} 4+6>3 {/tex}
    {tex} 6+3>4 {/tex}
    Since, the lengths follow the triangle inequality. Therefore, they can be the side lengths of a triangle.
  3. {tex}2,4,8{/tex}
    {tex} 2+4<8 {/tex}
    {tex} 4+8>2 {/tex}
    {tex} 8+2>4 {/tex}
    Since, the lengths don’t follow the triangle inequality. Therefore, they can’t be the side lengths of a triangle.
  4. {tex}5,5,8{/tex}
    {tex} 5+5>8 {/tex}
    {tex} 5+8>5 {/tex}
    {tex} 8+5>5 {/tex}
    Since, the lengths follow the triangle inequality. Therefore, they can be the side lengths of a triangle.
  5. {tex}10,20,25{/tex}
    {tex} 10+20>25 {/tex}
    {tex} 20+25>10 {/tex}
    {tex} 25+10>20 {/tex}
    Since, the lengths follow the triangle inequality. Therefore, they can be the side lengths of a triangle.
  6. 10, 20, 35
    {tex} 10+20<35 {/tex}
    {tex} 20+35>10 {/tex}
    {tex} 35+10>20 {/tex}
    Since, the lengths don’t follow the triangle inequality. Therefore, they can’t be the side lengths of a triangle.
  7. 24, 26, 28
    {tex} 24+26>28 {/tex}
    {tex} 26+28>24 {/tex}
    {tex} 28+24>26 {/tex}
    Since, the lengths follow the triangle inequality. Therefore, they can be the side lengths of a triangle.

Q.22: Check whether a triangle can be formed with the side lengths 1, 100, and 100.

Solution:

1, 100, 100
{tex} 1+100>100 {/tex}
{tex} 100+100>1 {/tex}
{tex} 100+1>100 {/tex}
Since, the lengths follow the triangle inequality. Therefore, they can be the side lengths of a triangle.

Q.23: Check whether a triangle can be formed with the side lengths 3, 6, 9

Solution:

3, 6, 9
{tex} 3+6=9 {/tex}
{tex} 6+9>3 {/tex}
{tex} 9+3>6 {/tex}
Since, the lengths don’t follow the triangle inequality. Therefore, they can’t be the side lengths of a triangle.

Q.24: Check whether a triangle can be formed with the side lengths 1, 1, 5

Solution:

1, 1, 5
{tex} 1+1<5 {/tex}
{tex} 1+5>1 {/tex}
{tex} 5+1>1 {/tex}
Since, the lengths don’t follow the triangle inequality. Therefore, they can’t be the side lengths of a triangle.

Q.25: Check whether a triangle can be formed with the side lengths 5, 10, 12

Solution:

{tex} 5+10>12 {/tex}
{tex} 10+12>5 {/tex}
{tex} 12+5>10 {/tex}
Since, the lengths follow the triangle inequality. Therefore, they can be the side lengths of a triangle.

Q.26: Does there exist an equilateral triangle with sides 50, 50, 50? In general, does there exist an equilateral triangle of any sidelength? Justify your answer.

Solution:

Yes, an equilateral triangle with sides 50, 50, and 50 can exist because each side (50) is less than the sum of the other two sides (50 + 50 = 100), which satisfies the triangle inequality. Yes, an equilateral triangle can be constructed of any side length, satisfying the triangle inequality.

Q.27: For each of the following, give at least 5 possible values for the third length so there exists a triangle having these as side lengths (decimal values could also be chosen):

  1. 1, 100
  2. 5, 5
  3. 3, 7

Solution:

  1. For a triangle to exist, the sum of the two smaller lengths > longest length.
    For a triangle with sides 1 and 100, five valid possible values for the third side are: 99.1, {tex}99.7,100.3,100.6,100.8{/tex}.
    Because:
    {tex} 1+99.1>100 ; 1+99.7>100 ; 1+100>100.3 ;{/tex}{tex} 1+100>100.6 \text { and } 1+100>100.8 {/tex}
  2. For a triangle to exist, the sum of the two smaller lengths > longest length.
    For a triangle with sides 5 and 5, five valid possible values for the third side are: 1, 3, 4.5, 7, 9.9.
    Because:
    1 + 5 > 5; 3 + 5 > 5; 5 + 4.5 > 5; 5 + 5 > 7 and 5 + 5 > 9.9.
  3. For a triangle to exist, the sum of the two smaller lengths > longest length.
    For a triangle with sides 3 and 7, five valid possible values for the third side are: 4.1, 5, 6.5, 8, 9.9.
    Because:
    3 + 4.1 > 7; 3 + 5 > 7; 3 + 6.5 > 7; 3 + 7 > 8 and 3 + 7 > 9.9.

Q.28: Construct a triangle in which two sides are 3 cm and 7 cm, and the included angle between them is 75°.

Solution:

3 cm, 75o, 7 cm

  1. Construct side {tex}A B{/tex} of length 7 cm .
  2. At point {tex}A{/tex}, draw a ray {tex}A X{/tex} making an angle of {tex}75^{\circ}{/tex} with side {tex}A B{/tex}.
  3. With {tex}A{/tex} as the centre and radius 3 cm , draw an arc intersecting ray {tex}A X{/tex} at point {tex}C{/tex}.
  4. Join points {tex}B{/tex} and {tex}C{/tex} to form triangle {tex}\triangle A B C{/tex}.

Q.29: Construct a triangle in which two sides are 6 cm and 3 cm, and the included angle between them is 25o.

Solution:

6 cm, 25o, 3 cm

  1. Construct side AB of length 6 cm .
  2. At point {tex}A{/tex}, draw a ray {tex}A X{/tex} making an angle of {tex}25^{\circ}{/tex} with side {tex}A B{/tex}.
  3. With {tex}A{/tex} as the centre and radius 3 cm, draw an arc intersecting ray {tex}A X{/tex} at point {tex}C{/tex}.
  4. Join points B and C to form triangle {tex}\triangle {ABC}{/tex}.

Q.30: Construct a triangle in which two sides are 3 cm and 8 cm, and the included angle between them is 120o.

Solution:

3 cm, 120o, 8 cm

  1. Construct side AB of length 8 cm.
  2. At point {tex}A{/tex}, draw a ray {tex}A X{/tex} making an angle of {tex}120^{\circ}{/tex} with side {tex}A B{/tex}.
  3. With A as the centre and radius 3 cm, draw an arc intersecting ray AX at point C.
  4. Join points {tex}B{/tex} and {tex}C{/tex} to form triangle {tex}\triangle A B C{/tex}.

Q.31: Construct a triangle for the following measurements:
75°, 5 cm, 75°

Solution:

  1. Draw the base {tex}A B{/tex} of length 5 cm.
  2. Draw {tex}\angle A{/tex} and {tex}\angle B{/tex} both of measures {tex}45^{\circ}{/tex} each.
  3. The point of intersection of the two new line segments is the third vertex C.

Q.32: Construct triangle for the following measurements:
25°, 3 cm, 60°

Solution:

  1. Draw the base AB of length 3 cm .
  2. Draw {tex}\angle A{/tex} and {tex}\angle B{/tex} of measures {tex}25^{\circ}{/tex} and {tex}60^{\circ}{/tex} respectively.
  3. The point of intersection of the two new line segments is the third vertex C.

Q.33: Construct triangle for the following measurements:
120°, 6 cm, 30°

Solution:

  1. Draw the base {tex}A B{/tex} of length 6 cm.
  2. Draw {tex}\angle A{/tex} and {tex}\angle B{/tex} of measures {tex}120^{\circ}{/tex} and {tex}30^{\circ}{/tex} respectively.
  3. The point of intersection of the two new line segments is the third vertex C.

Q.34: For each of the following angles, find another angle for which a triangle is (a) possible, (b) not possible. Find at least two different angles for each category:

  1. {tex} 30^{\circ} {/tex}
  2. {tex} 70^{\circ} {/tex}
  3. {tex} 54^{\circ} {/tex}
  4. {tex} 144^{\circ} {/tex}

Solution:

  1. {tex}30^{\circ}{/tex}
    A triangle is possible when another angle is less than {tex}150^{\circ}{/tex}. Examples of angles are {tex}120^{\circ}{/tex} and {tex}85^{\circ}{/tex}.
    A triangle is not possible when another angle is greater than or equal to {tex}150^{\circ}{/tex}. Examples of angles are {tex}165^{\circ}{/tex} and {tex}170^{\circ}{/tex}.
  2. {tex}70^{\circ}{/tex}
    A triangle is possible when another angle is less than {tex}110^{\circ}{/tex}. Examples of angles are {tex}100^{\circ}{/tex} and {tex}67^{\circ}{/tex}.
    A triangle is not possible when another angle is greater than or equal to {tex}110^{\circ}{/tex}. Examples of angles are {tex}135^{\circ}{/tex} and {tex}150^{\circ}{/tex}.
  3. {tex}54^{\circ}{/tex}
    A triangle is possible when another angle is less than {tex}126^{\circ}{/tex}. Examples of angles are {tex}105^{\circ}{/tex} and {tex}95^{\circ}{/tex}.
    A triangle is not possible when another angle is greater than or equal to 126°. Examples of angles are 139° and 145°.
  4. {tex}144^{\circ}{/tex}
    A triangle is possible when another angle is less than {tex}36^{\circ}{/tex}. Examples of angles are {tex}20^{\circ}{/tex} and {tex}35^{\circ}{/tex}.
    A triangle is not possible when another angle is greater than or equal to {tex}36^{\circ}{/tex}. Examples of angles are {tex}65^{\circ}{/tex} and {tex}45^{\circ}{/tex}.

Q.35: Determine which of the following pairs can be the angles of a triangle and which cannot:

  1. {tex}35^{\circ}, 150^{\circ}{/tex}
  2. {tex}70^{\circ}, 30^{\circ}{/tex}
  3. {tex}90^{\circ}, 85^{\circ}{/tex}
  4. {tex}50^{\circ} .150^{\circ}{/tex}

Solution:

  1. {tex}35^{\circ}+150^{\circ}=185^{\circ}{/tex}
    {tex}185^{\circ}>180^{\circ}{/tex}.
    Since the sum of these two angles is greater than {tex}180^{\circ}{/tex}, they can’t be the angles of a triangle.
  2. {tex}70^{\circ}+30^{\circ}=100^{\circ}{/tex}
    {tex}100^{\circ}<180^{\circ}{/tex}.
    Since the sum of these two angles is less than {tex}180^{\circ}{/tex}, they can be the angles of a triangle.
  3. {tex}90^{\circ}+85^{\circ}=175^{\circ}{/tex}
    {tex}175^{\circ}<180^{\circ}{/tex}.
    Since the sum of these two angles is less than {tex}180^{\circ}{/tex}, they can be the angles of a triangle.
  4. {tex}50^{\circ}+150^{\circ}=200^{\circ}{/tex}
    {tex}200^{\circ}>180^{\circ}{/tex}.
    Since the sum of these two angles is greater than {tex}180^{\circ}{/tex}, they can’t be the angles of a triangle.

Q.36: Find the third angle of a triangle (using a parallel line) when two of the angles are 36°, 72°

Solution:


Since, {tex}X Y \| B C{/tex}, then:
{tex} \angle A B C=\angle X A B=36^{\circ} {/tex} {tex}\ldots \text { (Alternate interior angles) } {/tex}
{tex} \angle B C A=\angle Y A C=72^{\circ}{/tex} {tex}\ldots \text { (Alternate interior angles) } {/tex}
{tex} \angle X A B+\angle B A C+\angle Y A C=180^{\circ}{/tex} {tex}\ldots \text { (Sum of angles on a straight line) } {/tex}
{tex} 36^{\circ}+\angle B A C+72^{\circ}=180^{\circ} {/tex}
{tex} 108^{\circ}+\angle B A C=180^{\circ} {/tex}
{tex} \angle B A C=180^{\circ}-108^{\circ} {/tex}
{tex} \angle B A C=72^{\circ} . {/tex}

Q.37: Find the third angle of a triangle (using a parallel line) when two of the angles are 150°, 15°

Solution:


90o, 30o
Since, {tex}X Y \| B C{/tex}, then:
{tex}\angle {ABC}=\angle {XAB}=150^{\circ}{/tex} …(Alternate interior angles)
{tex}\angle {BCA}=\angle {YAC}=15^{\circ}{/tex} …(Alternate interior angles)
{tex}\angle {XAB}+\angle {BAC}+\angle {YAC}=180^{\circ}{/tex}. …Sum of angles on a straight line)
{tex}150^{\circ}+\angle {BAC}+15^{\circ}=180^{\circ}{/tex}
{tex} 165^{\circ}+\angle B A C=180^{\circ} {/tex}
{tex} \angle B A C=180^{\circ}-165^{\circ} {/tex}
{tex} \angle B A C=15^{\circ} . {/tex}

Q.38: Find the third angle of a triangle (using a parallel line) when two of the angles are 90°, 30°

Solution:


75°, 45°
Since, {tex}X Y \| B C{/tex}, then:
{tex}\angle {ABC}=\angle {XAB}=90^{\circ}{/tex} …(Alternate interior angles)
{tex}\angle {BCA}=\angle {YAC}=30^{\circ}{/tex} …(Alternate interior angles)
{tex}\angle {XAB}+\angle {BAC}+\angle {YAC}=180^{\circ}{/tex} …(Sum of angles on a straight line)
{tex}90^{\circ}+\angle {BAC}+30^{\circ}=180^{\circ}{/tex}
{tex} 120^{\circ}+\angle B A C=180^{\circ} {/tex}
{tex} \angle B A C=180^{\circ}-120^{\circ} {/tex}
{tex} \angle B A C=60^{\circ} . {/tex}

Q.39: Find the third angle of a triangle (using a parallel line) when two of the angles are 75°, 45°

Solution:


Since, {tex}X Y \| B C{/tex}, then:
{tex}\angle {ABC}=\angle {XAB}=75^{\circ}{/tex} …(Alternate interior angles)
{tex}\angle {BCA}=\angle {YAC}=45^{\circ}{/tex} …(Alternate interior angles)
{tex}\angle {XAB}+\angle {BAC}+\angle {YAC}=180^{\circ}{/tex} …(Sum of angles on a straight line)
{tex} 75^{\circ}+\angle {BAC}+45^{\circ}=180^{\circ} {/tex}
{tex} 120^{\circ}+\angle B A C=180^{\circ} {/tex}
{tex} \angle B A C=180^{\circ}-120^{\circ} {/tex}
{tex} \angle B A C=60^{\circ} . {/tex}

Q.40: Can you construct a triangle all of whose angles are equal to 70°? If two of the angles are 70° what would the third angle be? If all the angles in a triangle have to be equal, then what must its measure be? Explore and find out.

Solution:

No, it is not possible to construct a triangle with all angles equal to 70°.

Let base angles, {tex}\angle {B}{/tex} and {tex}\angle {C}=70^{\circ}{/tex}.
Since, {tex}X Y \| B C{/tex}, then:
{tex} \angle A B C=\angle X A B=70^{\circ} {/tex}{tex}\ldots\text { (Alternate interior angles) } {/tex}
{tex} \angle B C A=\angle Y A C=70^{\circ} {/tex}{tex}\ldots\text { (Alternate interior angles) } {/tex}
{tex} \angle X A B+\angle B A C+\angle Y A C=180^{\circ} {/tex}{tex}\ldots \text { (Sum of angles on a straight line) } {/tex}
{tex} 70^{\circ}+\angle B A C+70^{\circ}=180^{\circ} {/tex}
{tex} 140^{\circ}+\angle B A C=180^{\circ} {/tex}
{tex} \angle B A C=180^{\circ}-140^{\circ} {/tex}
{tex} \angle B A C=40^{\circ} . {/tex}
Therefore, the third angle would be {tex}40^{\circ}{/tex}.
If all the angles in a triangle have to be equal, then each angle must be {tex}60^{\circ}{/tex}. Such triangle with all angles equal to {tex}60^{\circ}{/tex} is known as an equilateral triangle.

Q.41: Here is a triangle in which we know {tex}\angle B=\angle C{/tex} and {tex}\angle A=50^{\circ}{/tex}. Can you find {tex}\angle B{/tex} and {tex}\angle C{/tex} ?

Solution:


Given, {tex}\angle B=\angle C{/tex} and {tex}\angle A=50^{\circ}{/tex}
Draw a line {tex}X Y{/tex} parallel to {tex}B C{/tex}, then:
{tex}\angle B=\angle X A B{/tex} …(Alternate interior angles)
{tex}\angle {C}=\angle {YAC}{/tex} …(Alternate interior angles)
{tex}\angle {XAB}+\angle {A}+\angle {YAC}=180^{\circ}{/tex} …(Sum of angles on a straight line)
{tex}\angle B+50^{\circ}+\angle C=180^{\circ}{/tex}
{tex} \angle C+\angle C=180^{\circ}-50^{\circ} {/tex}
{tex} 2 \angle C=130^{\circ} {/tex}
{tex} \angle C=130^{\circ} / 2=65^{\circ} . {/tex}
Therefore, {tex}\angle B=\angle C=65^{\circ}{/tex}.

Q.42: Construct a triangle ABC with BC = 5cm, AB = 6cm, CA = 5cm. Construct an altitude from A to BC.

Solution:

  1. Draw the base {tex}{BC}=5 {~cm}{/tex}.
  2. From B, draw long arc of radius 6 cm .
  3. From C, draw arc of radius 5 cm intersecting the first arc at point A .
  4. The point A is the required third vertex. Join AB and AC to get {tex}\triangle {ABC}{/tex}.
  5. Keep the ruler aligned to the base BC. Place the set square on the ruler such that one of the edges of the right angle touches the ruler.
  6. Slide the set square along the ruler till the vertical edge of the set square touches the vertex A.
  7. Draw the altitude to BC through A using the vertical edge of the set square.

    Therefore, in {tex}\triangle {ABCAC}{/tex} is the required altitude from point A to BC.

Q.43: Construct a triangle TRY with RY = {tex}4 {~cm}, T R=7 {~cm}, \angle R=140^{\circ}{/tex}. Construct an altitude from T to RY.

Solution:

  1. Construct side {tex}{TR}=7 {~cm}{/tex}.
  2. At point {tex}R{/tex}, draw a ray {tex}R A{/tex} making an angle of {tex}140^{\circ}{/tex} with side {tex}T R{/tex}.
  3. With {tex}R{/tex} as the centre and radius 4 cm , draw an arc intersecting ray {tex}R A{/tex} at point {tex}Y{/tex}.
  4. Join points {tex}T{/tex} and {tex}Y{/tex} to form triangle {tex}\Delta T R Y{/tex}.
  5. Keep the ruler aligned to the side RY. Place the set square on the ruler such that one of the edges of the right angle touches the ruler.
  6. Slide the set square along the ruler till the vertical edge of the set square touches the vertex T.
  7. Draw the altitude to BC through A using the vertical edge of the set square.

    Therefore, in {tex}\Delta T R Y T B{/tex} is the required altitude from point {tex}T{/tex} to {tex}R Y{/tex}.

Q.44: Construct a right-angled triangle {tex}\triangle {ABC}{/tex} with {tex}\angle {B}=90^{\circ}, {AC}=5 {~cm}{/tex}. How many different triangles exist with these measurements?

Solution:

In a right-angled triangle, if {tex}\angle {B}=90^{\circ}{/tex} and {tex}{AC}=5 {~cm}{/tex}, then by the angle sum property of triangles, {tex}\angle A+\angle C=90^{\circ}{/tex}. Since {tex}\angle A{/tex} and {tex}\angle C{/tex} can take various pairs of values that sum to {tex}90^{\circ}{/tex}, this results in infinitely many right-angled triangles of different shapes.
For example:
{tex}\triangle A B C{/tex} is a right-angled triangle with {tex}\angle B=90^{\circ}, A C=5 {~cm}, \angle A=50^{\circ}{/tex}, and {tex}\angle C=40^{\circ}{/tex}, such that {tex}\angle A+\angle C=50^{\circ}+40^{\circ}=90^{\circ}{/tex}.

Q.45: Through construction, explore if it is possible to construct an equilateral triangle that is (i) right-angled (ii) obtuse-angled. Also construct an isosceles triangle that is (i) right-angled (ii) obtuse-angled.

Solution:

An equilateral triangle has each of its angles equal to {tex}60^{\circ}{/tex}, so it is impossible to construct a right-angled or obtuse-angled equilateral triangle.
An isosceles triangle can be right-angled, with one angle of {tex}90^{\circ}{/tex} and the other two angles of {tex}45^{\circ}{/tex} each.
An isosceles triangle can also be obtuse-angled, with one angle of {tex}100^{\circ}{/tex} and the other two angles of {tex}40^{\circ}{/tex} each.

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