RD Sharma Solutions For Class 7 Maths Chapter 15 Properties of Triangles Exercise 15.2

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RD Sharma Solutions for Class 7 Maths Chapter 15 – Properties of Triangles Exercise 15.2

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Access answers to Maths RD Sharma Solutions For Class 7 Chapter 15 – Properties of Triangles Exercise 15.2

Exercise 15.2 Page No: 15.12

1. Two angles of a triangle are of measures 150^o and 30^o. Find the measure of the third angle.

Solution:

Given two angles of a triangle are of measures 150^o and 30^o

Let the required third angle be x

We know that sum of all the angles of a triangle = 180^o

150^o + 30^o + x = 180^o

135^o + x = 180^o

x = 180^o – 135^o

x = 45^o

Therefore the third angle is 45^o

2. One of the angles of a triangle is 130^o, and the other two angles are equal. What is the measure of each of these equal angles?

Solution:

Given one of the angles of a triangle is 130^o

Also given that remaining two angles are equal

So let the second and third angle be x

We know that sum of all the angles of a triangle = 180^o

130^o + x + x = 180^o

130^o + 2x = 180^o

2x = 180^o – 130^o

2x = 50^o

x = 50/2

x = 25^o

Therefore the two other angles are 25^o each

3. The three angles of a triangle are equal to one another. What is the measure of each of the angles?

Solution:

Given that three angles of a triangle are equal to one another

So let the each angle be x

We know that sum of all the angles of a triangle = 180^o

x + x + x = 180^o

3x = 180^o

x = 180/3

x = 60^o

Therefore angle is 60^o each

4. If the angles of a triangle are in the ratio 1: 2: 3, determine three angles.

Solution:

Given angles of the triangle are in the ratio 1: 2: 3

So take first angle as x, second angle as 2x and third angle as 3x

We know that sum of all the angles of a triangle = 180^o

x + 2x + 3x = 180^o

6x = 180^o

x = 180/6

x = 30^o

2x = 30^o × 2 = 60^o

3x = 30^o × 3 = 90^o

Therefore the first angle is 30^o, second angle is 60^o and third angle is 90^o.

5. The angles of a triangle are (x − 40)^o, (x − 20)^o and (1/2 − 10)^o. Find the value of x.

Solution:

Given the angles of a triangle are (x − 40)^o, (x − 20)^o and (1/2 − 10)^o.

We know that sum of all the angles of a triangle = 180^o

(x − 40)^o + (x − 20)^o + (1/2 − 10)^o = 180^o

x + x + (1/2) – 40^o – 20^o – 10^o = 180^o

x + x + (1/2) – 70^o = 180^o

(5x/2) = 180^o + 70^o

(5x/2) = 250^o

x = (2/5) × 250^o

x = 100^o

Hence the value of x is 100^o

6. The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angles is 10^o. Find the three angles.

Solution:

Given that angles of a triangle are arranged in ascending order of magnitude

Also given that difference between two consecutive angles is 10^o

Let the first angle be x

Second angle be x + 10^o

Third angle be x + 10^o + 10^o

We know that sum of all the angles of a triangle = 180^o

x + x + 10^o + x + 10^o +10^o = 180^o

3x + 30 = 180

3x = 180 – 30

3x = 150

x = 150/3

x = 50^o

First angle is 50^o

Second angle x + 10^o = 50 + 10 = 60^o

Third angle x + 10^o +10^o = 50 + 10 + 10 = 70^o

7. Two angles of a triangle are equal and the third angle is greater than each of those angles by 30^o. Determine all the angles of the triangle

Solution:

Given that two angles of a triangle are equal

Let the first and second angle be x

Also given that third angle is greater than each of those angles by 30^o

Therefore the third angle is greater than the first and second by 30^o = x + 30^o

The first and the second angles are equal

We know that sum of all the angles of a triangle = 180^o

x + x + x + 30^o = 180^o

3x + 30 = 180

3x = 180 – 30

3x = 150

x = 150/3

x = 50^o

Third angle = x + 30^o = 50^o + 30^o = 80^o

The first and the second angle is 50^o and the third angle is 80^o.

8. If one angle of a triangle is equal to the sum of the other two, show that the triangle is a right triangle.

Solution:

Given that one angle of a triangle is equal to the sum of the other two

Let the measure of angles be x, y, z

Therefore we can write above statement as x = y + z

x + y + z = 180^o

Substituting the above value we get

x + x = 180^o

2x = 180^o

x = 180/2

x = 90^o

If one angle is 90^o then the given triangle is a right angled triangle

9. If each angle of a triangle is less than the sum of the other two, show that the triangle is acute angled.

Solution:

Given that each angle of a triangle is less than the sum of the other two

Let the measure of angles be x, y and z

From the above statement we can write as

x > y + z

y < x + z

z < x + y

Therefore triangle is an acute triangle

10. In each of the following, the measures of three angles are given. State in which cases the angles can possibly be those of a triangle:

(i) 63^o, 37^o, 80^o

(ii) 45^o, 61^o, 73^o

(iii) 59^o, 72^o, 61^o

(iv) 45^o, 45^o, 90^o

(v) 30^o, 20^o, 125^o

Solution:

(i) 63^o + 37^o + 80^o = 180^o

Angles form a triangle

(ii) 45^o, 61^o, 73^o is not equal to 180^o

Therefore not a triangle

(iii) 59^o, 72^o, 61^o is not equal to 180⁰

Therefore not a triangle

(iv) 45^o+ 45^o+ 90^o = 180^o

Angles form a triangle

(v) 30^o, 20^o, 125^o is not equal to 180^o

Therefore not a triangle

11. The angles of a triangle are in the ratio 3: 4: 5. Find the smallest angle

Solution:

Given that angles of a triangle are in the ratio: 3: 4: 5

Therefore let the measure of the angles be 3x, 4x, 5x

We know that sum of the angles of a triangle =180^o

3x + 4x + 5x = 180^o

12x = 180^o

x = 180/12

x = 15^o

Smallest angle = 3x

= 3 × 15^o

= 45^o

Therefore smallest angle = 45^o

12. Two acute angles of a right triangle are equal. Find the two angles.

Solution:

Given that acute angles of a right angled triangle are equal

We know that Right triangle: whose one of the angle is a right angle

Let the measure of angle be x, x, 90^o

x + x + 90^o= 180^o

2x = 90^o

x = 90/2

x = 45^o

The two angles are 45^o and 45^o

13. One angle of a triangle is greater than the sum of the other two. What can you say about the measure of this angle? What type of a triangle is this?

Solution:

Given one angle of a triangle is greater than the sum of the other two

Let the measure of the angles be x, y, z

From the question we can write as

x > y + z or

y > x + z   or

z > x + y

x or y or z > 90^o which is obtuse

Therefore triangle is an obtuse angle

14. In the six cornered figure, (fig. 20), AC, AD and AE are joined. Find ∠FAB + ∠ABC + ∠BCD + ∠CDE + ∠DEF + ∠EFA.

Solution:

We know that sum of the angles of a triangle is 180^o

Therefore in ∆ABC, we have

∠CAB + ∠ABC + ∠BCA = 180^o …….. (i)

In△ ACD, we have

∠DAC + ∠ACD + ∠CDA = 180^o …….. (ii)

In △ADE, we have

∠EAD + ∠ADE + ∠DEA =180^o ………. (iii)

In △AEF, we have

∠FAE + ∠AEF + ∠EFA = 180^o ………. (iv)

Adding (i), (ii), (iii), (iv) we get

∠CAB + ∠ABC + ∠BCA + ∠DAC + ∠ACD + ∠CDA + ∠EAD + ∠ADE + ∠DEA + ∠FAE + ∠AEF +∠EFA = 720^o

Therefore ∠FAB + ∠ABC + ∠BCD + ∠CDE + ∠DEF + ∠EFA = 720^o

15. Find x, y, z (whichever is required) from the figures (Fig. 21) given below:

Solution:

(i) In △ABC and △ADE we have,

∠ ADE = ∠ ABC [corresponding angles]

x = 40^o

∠AED = ∠ACB (corresponding angles)

y = 30^o

We know that the sum of all the three angles of a triangle is equal to 180^o

x + y + z = 180^o (Angles of △ADE)

Which means: 40^o + 30^o + z = 180^o

z = 180^o – 70^o

z = 110^o

Therefore, we can conclude that the three angles of the given triangle are 40^o, 30^o and 110^o

(ii)  We can see that in △ADC, ∠ADC is equal to 90^o.

(△ADC is a right triangle)

We also know that the sum of all the angles of a triangle is equal to 180^o.

Which means: 45^o + 90^o + y = 180^o (Sum of the angles of △ADC)

135^o + y = 180^o

y = 180^o – 135^o.

y = 45^o.

We can also say that in △ ABC, ∠ABC + ∠ACB + ∠BAC is equal to 180^o.

(Sum of the angles of △ABC)

40^o + y + (x + 45^o) = 180^o

40^o + 45^o + x + 45^o = 180^o (y = 45^o)

x = 180^o –130^o

x = 50^o

Therefore, we can say that the required angles are 45^o and 50^o.

(iii) We know that the sum of all the angles of a triangle is equal to 180^o.

Therefore, for △ABD:

∠ABD +∠ADB + ∠BAD = 180° (Sum of the angles of △ABD)

50^o + x + 50^o = 180^o

100^o + x = 180^o

x = 180^o – 100^o

x = 80^o

For △ ABC:

∠ABC + ∠ACB + ∠BAC = 180^o (Sum of the angles of △ABC)

50^o + z + (50^o + 30^o) = 180^o

50^o + z + 50^o + 30^o = 180^o

z = 180^o – 130^o

z = 50^o

Using the same argument for △ADC:

∠ADC + ∠ACD + ∠DAC = 180^o (Sum of the angles of △ADC)

y + z + 30^o = 180^o

y + 50^o + 30^o = 180^o (z = 50^o)

y = 180^o – 80^o

y = 100⁰

Therefore, we can conclude that the required angles are 80^o, 50^o and 100^o.

(iv) In △ABC and △ADE we have:

∠ADE = ∠ABC (Corresponding angles)

y = 50^o

Also, ∠AED = ∠ACB (Corresponding angles)

z = 40^o

We know that the sum of all the three angles of a triangle is equal to 180^o.

We can write as x + 50^o + 40^o = 180^o (Angles of △ADE)

x = 180^o – 90^o

x = 90^o

Therefore, we can conclude that the required angles are 50^o, 40^o and 90^o.

16. If one angle of a triangle is 60^o and the other two angles are in the ratio 1: 2, find the angles.

Solution:

Given that one of the angles of the given triangle is 60^o.

Also given that the other two angles of the triangle are in the ratio 1: 2.

Let one of the other two angles be x.

Therefore, the second one will be 2x.

We know that the sum of all the three angles of a triangle is equal to 180^o.

60^o + x + 2x = 180^o

3x = 180^o – 60^o

3x = 120^o

x = 120^o/3

x = 40^o

2x = 2 × 40^o

2x = 80^o

Hence, we can conclude that the required angles are 40^o and 80^o.

17. If one angle of a triangle is 100^o and the other two angles are in the ratio 2: 3. Find the angles.

Solution:

Given that one of the angles of the given triangle is 100^o.

Also given that the other two angles are in the ratio 2: 3.

Let one of the other two angle be 2x.

Therefore, the second angle will be 3x.

We know that the sum of all three angles of a triangle is 180^o.

100^o + 2x + 3x = 180^o

5x = 180^o – 100^o

5x = 80^o

x = 80/5

x = 16

2x = 2 ×16

2x = 32^o

3x = 3×16

3x = 48^o

Thus, the required angles are 32^o and 48^o.

18. In △ABC, if 3∠A = 4∠B = 6∠C, calculate the angles.

Solution:

We know that for the given triangle, 3∠A = 6∠C

∠A = 2∠C ……. (i)

We also know that for the same triangle, 4∠B = 6∠C

∠B = (6/4) ∠C …… (ii)

We know that the sum of all three angles of a triangle is 180^o.

Therefore, we can say that:

∠A + ∠B + ∠C = 180^o (Angles of △ABC)…… (iii)

On putting the values of ∠A and ∠B in equation (iii), we get:

2∠C + (6/4) ∠C + ∠C = 180^o

(18/4) ∠C = 180^o

∠C = 40^o

From equation (i), we have:

∠A = 2∠C = 2 × 40

∠A = 80^o

From equation (ii), we have:

∠B = (6/4) ∠C = (6/4) × 40^o

∠B = 60^o

∠A = 80^o, ∠B = 60^o, ∠C = 40^o

Therefore, the three angles of the given triangle are 80^o, 60^o, and 40^o.

19. Is it possible to have a triangle, in which

(i) Two of the angles are right?

(ii) Two of the angles are obtuse?

(iii) Two of the angles are acute?

(iv) Each angle is less than 60^o?

(v) Each angle is greater than 60^o?

(vi) Each angle is equal to 60^o?

Solution:

(i) No, because if there are two right angles in a triangle, then the third angle of the triangle must be zero, which is not possible.

(ii) No, because as we know that the sum of all three angles of a triangle is always 180^o. If there are two obtuse angles, then their sum will be more than 180^o, which is not possible in case of a triangle.

(iii) Yes, in right triangles and acute triangles, it is possible to have two acute angles.

(iv) No, because if each angle is less than 60^o, then the sum of all three angles will be less than 180^o, which is not possible in case of a triangle.

(v) No, because if each angle is greater than 60⁰, then the sum of all three angles will be greater than 180⁰, which is not possible.

(vi) Yes, if each angle of the triangle is equal to 60⁰, then the sum of all three angles will be 180⁰, which is possible in case of a triangle.

20. In△ ABC, ∠A = 100^o, AD bisects ∠A and AD ⊥ BC. Find ∠B

Solution:

Given that in △ABC, ∠A = 100^o

Also given that AD ⊥ BC

Consider △ABD

∠BAD = 100/2   (AD bisects ∠A)

∠BAD = 50^o

∠ADB = 90^o   (AD perpendicular to BC)

We know that the sum of all three angles of a triangle is 180^o.

Thus,

∠ABD + ∠BAD + ∠ADB = 180^o (Sum of angles of △ABD)

Or,

∠ABD + 50^o + 90^o = 180^o

∠ABD =180^o – 140^o

∠ABD = 40^o

21. In △ABC, ∠A = 50^o, ∠B = 70^o and bisector of ∠C meets AB in D. Find the angles of the triangles ADC and BDC

Solution:

We know that the sum of all three angles of a triangle is equal to 180^o.

Therefore, for the given △ABC, we can say that:

∠A + ∠B + ∠C = 180^o (Sum of angles of △ABC)

50^o + 70^o + ∠C = 180^o

∠C= 180^o –120^o

∠C = 60^o

∠ACD = ∠BCD =∠C2 (CD bisects ∠C and meets AB in D.)

∠ACD = ∠BCD = 60/2= 30^o

Using the same logic for the given △ACD, we can say that:

∠DAC + ∠ACD + ∠ADC = 180^o

50^o + 30^o + ∠ADC = 180^o

∠ADC = 180^o– 80^o

∠ADC = 100^o

If we use the same logic for the given △BCD, we can say that

∠DBC + ∠BCD + ∠BDC = 180^o

70^o + 30^o + ∠BDC = 180^o

∠BDC = 180^o – 100^o

∠BDC = 80^o

Thus,

For △ADC: ∠A = 50^o, ∠D = 100^o ∠C = 30^o

△BDC: ∠B = 70^o, ∠D = 80^o ∠C = 30^o

22. In ∆ABC, ∠A = 60^o, ∠B = 80^o, and the bisectors of ∠B and ∠C, meet at O. Find

(i) ∠C

(ii) ∠BOC

Solution:

(i) We know that the sum of all three angles of a triangle is 180^o.

Hence, for △ABC, we can say that:

∠A + ∠B + ∠C = 180^o (Sum of angles of ∆ABC)

60^o + 80^o + ∠C= 180^o.

∠C = 180^o – 140^o

∠C = 40^o.

(ii)For △OBC,

∠OBC = ∠B/2 = 80/2 (OB bisects ∠B)

∠OBC = 40^o

∠OCB =∠C/2 = 40/2 (OC bisects ∠C)

∠OCB = 20^o

If we apply the above logic to this triangle, we can say that:

∠OCB + ∠OBC + ∠BOC = 180^o (Sum of angles of △OBC)

20^o + 40^o + ∠BOC = 180^o

∠BOC = 180^o – 60^o

∠BOC = 120^o

23. The bisectors of the acute angles of a right triangle meet at O. Find the angle at O between the two bisectors.

Solution:

Given bisectors of the acute angles of a right triangle meet at O

We know that the sum of all three angles of a triangle is 180^o.

Hence, for △ABC, we can say that:

∠A + ∠B + ∠C = 180^o

∠A + 90^o + ∠C = 180^o

∠A + ∠C = 180^o – 90^o

∠A + ∠C = 90^o

For △OAC:

∠OAC = ∠A/2 (OA bisects ∠A)

∠OCA   = ∠C/2 (OC bisects ∠C)

On applying the above logic to △OAC, we get

∠AOC + ∠OAC + ∠OCA = 180^o (Sum of angles of △AOC)

∠AOC + ∠A/2 + ∠C/2 = 180^o

∠AOC + (∠A + ∠C)/2 = 180^o

∠AOC + 90/2 = 180^o

∠AOC = 180^o – 45^o

∠AOC = 135^o

24.  In △ABC, ∠A = 50^o and BC is produced to a point D. The bisectors of ∠ABC and ∠ACD meet at E. Find ∠E.

Solution:

In the given triangle,

∠ACD = ∠A + ∠B. (Exterior angle is equal to the sum of two opposite interior angles.)

We know that the sum of all three angles of a triangle is 180^o.

Therefore, for the given triangle, we know that the sum of the angles = 180^o

∠ABC + ∠BCA + ∠CAB = 180^o

∠A + ∠B + ∠BCA = 180^o

∠BCA = 180^o – (∠A + ∠B)

But we know that EC bisects ∠ACD

Therefore ∠ECA = ∠ACD/2

∠ECA = (∠A + ∠B)/2 [∠ACD = (∠A + ∠B)]

But EB bisects ∠ABC

∠EBC = ∠ABC/2 = ∠B/2

∠EBC = ∠ECA + ∠BCA

∠EBC = (∠A + ∠B)/2 + 180^o – (∠A + ∠B)

If we use same steps for △EBC, then we get,

∠B/2 + (∠A + ∠B)/2 + 180^o – (∠A + ∠B) + ∠BEC = 180^o

∠BEC = ∠A + ∠B – (∠A + ∠B)/2 – ∠B/2

∠BEC = ∠A/2

∠BEC = 50^o/2

= 25^o

25. In ∆ABC, ∠B = 60°, ∠C = 40°, AL ⊥ BC and AD bisects ∠A such that L and D lie on side BC. Find ∠LAD

Solution:

We know that the sum of all angles of a triangle is 180°

Consider △ABC, we can write as

∠A + ∠B + ∠C = 180^o

∠A + 60^o + 40^o = 180^o

∠A = 80⁰

But we know that ∠DAC bisects ∠A

∠DAC = ∠A/2

∠DAC = 80^o/2

If we apply same steps for the △ADC, we get

We know that the sum of all angles of a triangle is 180°

∠ADC + ∠DCA + ∠DAC = 180^o

∠ADC + 40^o + 40^o = 180^o

∠ADC = 180^o – 80^o = 100⁰

We know that exterior angle is equal to the sum of two interior opposite angles

Therefore we have

∠ADC = ∠ALD+ ∠LAD

But here AL perpendicular to BC

100^o = 90^o + ∠LAD

∠LAD = 10^o

26. Line segments AB and CD intersect at O such that AC ∥ DB. It ∠CAB = 35^o and ∠CDB = 55^o. Find ∠BOD.

Solution:

We know that AC parallel to BD and AB cuts AC and BD at A and B, respectively.

∠CAB = ∠DBA (Alternate interior angles)

∠DBA = 35^o

We also know that the sum of all three angles of a triangle is 180^o.

Hence, for △OBD, we can say that:

∠DBO + ∠ODB + ∠BOD = 180°

35^o + 55^o + ∠BOD = 180^o (∠DBO = ∠DBA and ∠ODB = ∠CDB)

∠BOD = 180^o – 90^o

∠BOD = 90^o

27. In Fig. 22, ∆ABC is right angled at A, Q and R are points on line BC and P is a point such that QP ∥ AC and RP ∥ AB. Find ∠P

Solution:

In the given triangle, AC parallel to QP and BR cuts AC and QP at C and Q, respectively.

∠QCA = ∠CQP (Alternate interior angles)

Because RP parallel to AB and BR cuts AB and RP at B and R, respectively,

∠ABC = ∠PRQ (alternate interior angles).

We know that the sum of all three angles of a triangle is 180^o.

Hence, for ∆ABC, we can say that:

∠ABC + ∠ACB + ∠BAC = 180^o

∠ABC + ∠ACB + 90^o = 180^o (Right angled at A)

∠ABC + ∠ACB = 90^o

Using the same logic for △PQR, we can say that:

∠PQR + ∠PRQ + ∠QPR = 180^o

∠ABC + ∠ACB + ∠QPR = 180^o (∠ACB = ∠PQR and ∠ABC = ∠PRQ)

Or,

90^o+ ∠QPR =180^o (∠ABC+ ∠ACB = 90^o)

∠QPR = 90^o