## RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.2

**Other Exercises**

- RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.1
- RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.2
- RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.3
- RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.4
- RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles VSAQS
- RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS

**Question 1.**

**In the figure, OA and OB are opposite rays:**

**(i) If x = 25°, what is the value of y?**

**(ii) If y = 35°, what is the value of x?**

**Solution:**

**(i)** If x = 25°

∴ 3x = 3 x 25° = 75°

But ∠AOC + ∠BOC = 180° (Linear pair)

⇒ ∠AOC + 75° = 180°

⇒ ∠AOC = 180° – 75°

⇒ ∠AOC = 105°

∴ 2y + 5 = 105° ⇒ 2y= 105° – 5° = 100°

⇒ y = \(\frac { { 100 }^{ \circ } }{ 2 }\) = 50°

∴ If x = 25° then y = 50°

**(ii)** If y = 35°, then ∠AOC = 2y + 5

∴ 2y + 5 = 2 x 35° + 5 = 70° + 5 = 75°

But ∠AOC + ∠BOC = 180° (Linear pair)

⇒ 75° + ∠BOC = 180°

⇒ ∠BOC = 180°-75°= 105°

∴ 3x = 105° ⇒ x = \(\frac { { 105 }^{ \circ } }{ 3 }\) = 35°

∴ x = 35°

**Question 2.**

**In the figure, write all pairs of adjacent angles and all the linear pairs.**

**Solution:**

In the given figure,

**(i)** ∠AOD, ∠COD; ∠BOC, ∠COD; ∠AOD, ∠BOD and ∠AOC, ∠BOC are the pairs of adjacent angles.

**(ii)** ∠AOD, ∠BOD and ∠AOC, ∠BOC are the pairs of linear pairs.

**Question 3.**

**In the figure, find x, further find ∠BOC, ∠COD and ∠AOD.**

**Solution:**

In the figure,

AOB is a straight line

∴ ∠AOD + ∠DOB = 180° (Linear pair)

⇒ ∠AOD + ∠DOC + ∠COB = 180°

⇒ x+ 10° + x + x + 20 = 180°

⇒ 3x + 30° = 180°

⇒ 3x= 180° -30°= 150°

⇒ x = \(\frac { { 150 }^{ \circ } }{ 3 }\) = 50°

∴ x = 50°

Now ∠BOC =x + 20° = 50° + 20° = 70°

∠COD = x = 50°

and ∠AOD = x + 10° = 50° + 10° = 60°

**Question 4.**

**In the figure, rays OA, OB, OC, OD and OE have the common end point O. Show that ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°.**

**Solution:**

Produce AO to F such that AOF is a straight line

Now ∠AOB + ∠BOF = 180° (Linear pair)

⇒ ∠AOB + ∠BOC + ∠COF = 180° …(i)

Similarly, ∠AOE + ∠EOF = 180°

⇒ ∠AOE + ∠EOD + ∠DOF = 180° …(ii)

Adding (i) and (ii)

∠AOB + ∠BOC + ∠COF + ∠DOF + ∠EOD + ∠AOE = 180° + 180°

⇒ ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°

Hence proved.

**Question 5.**

**In the figure, ∠AOC and ∠BOC form a linear pair. If a – 2b = 30°, find a and b.**

**Solution:**

∠AOC + ∠BOC = 180° (Linear pair)

⇒ a + 6 = 180° …(i)

and a – 2b = 30° …(ii)

Subtracting (ii) from (i), 3b = 150°

⇒ b = \(\frac { { 150 }^{ \circ } }{ 3 }\) = 50°

and a + 50° = 180°

⇒ a= 180°-50°= 130°

∴ a = 130°, b = 50°

**Question 6.**

**How many pairs of adjacent angles are formed when two lines intersect in a point?**

**Solution:**

If two lines AB and CD intersect each other at a point O, then four pairs of linear pairs are formed i.e.

∠AOC, ∠BOC; ∠BOC, ∠BOD; ∠BOD, ∠AOD and ∠AOD, ∠AOC

**Question 7.**

**How many pairs of adjacent angles, in all, can you name in the figure.**

**Solution:**

In the given figure 10 pairs of adjacent angles are formed as given below:

∠AOB, ∠BOC; ∠AOB, ∠BOD; ∠AOC, ∠COD; ∠AOD, ∠BOE; ∠AOB, ∠BOE; ∠AOC, ∠COF; ∠BOC, ∠COD; ∠BOC, ∠COE; ∠COD, ∠DOE and ∠BOD, ∠DOE

**Question 8.**

**In the figure, determine the value of x.**

**Solution:**

In the figure

∠AOC + ∠COB + ∠BOD +∠AOD = 360 (angles at a point )

⇒ 3x + 3x + x + 150° = 360°

⇒ 7x – 360° – 150° = 210°

⇒ x = \(\frac { { 210 }^{ \circ } }{ 7 }\) = 30°

∴ x = 30°

**Question 9.**

**In the figure, AOC is a line, find x.**

**Solution:**

In the figure,

∠AOB + ∠BOC = 180° (Linear pair)

⇒ 70° + 2x = 180°

⇒ 2x = 180° – 70°

⇒ 2x = 110°⇒x = \(\frac { { 110 }^{ \circ } }{ 2 }\) = 55°

∴ x = 55°

**Question 10.**

**In the figure, POS is a line, find x.**

**Solution:**

In the figure, POS is a line

∴ ∠POQ + ∠QOS = 180° (Linear pair)

⇒ ∠POQ + ∠QOR + ∠ROS = 180°

⇒ 60° + 4x + 40° = 180°

⇒ 4x + 100° – 180°

⇒ 4x = 180° – 100° = 80°

⇒ x = \(\frac { { 80 }^{ \circ } }{ 4 }\) =20°

∴ x = 20°

**Question 11.**

**In the figure, ACB is a line such that ∠DCA = 5x and ∠DCB = 4x. Find the value of x.**

**Solution:**

ACB is a line, ∠DCA = 5x and ∠DCB = 4x

∠ACD + ∠DCB = 180° (Linear pair)

⇒ 5x + 4x = 180° ⇒ 9x = 180°

⇒ x = \(\frac { { 180 }^{ \circ } }{ 9 }\) = 20°

∴ x = 20°

∴ ∠ACD = 5x = 5 x 20° = 100° and ∠DCB = 4x = 4 x 20° = 80°

**Question 12.**

**Given ∠POR = 3x and ∠QOR = 2x + 10°, find the value ofx for which POQ will be a line.**

**Solution:**

∠POR = 3x and ∠QOR = 2x + 10°

If POQ is a line, then

∠POR + ∠QOR = 180° (Linear pair)

⇒ 3x + 2x + 10° = 180°

⇒ 5x = 180° – 10° = 170°

⇒ x = \(\frac { { 170 }^{ \circ } }{ 5 }\) = 34°

∴ x = 34°

**Question 13.**

**What value ofy would make AOB, a line in the figure, if ∠AOC = 4y and ∠BOC = (6y + 30).**

**Solution:**

In the figure,

AOB is a line if

∠AOC + ∠BOC = 180°

⇒ 6y + 30° + 4y= 180°

⇒ 10y= 180°-30°= 150°

150°

⇒ y = \(\frac { { 150 }^{ \circ } }{ 10 }\) = 15°

∴ y = 15°

**Question 14.**

**In the figure, OP, OQ, OR and OS are four rays. Prove that: ∠POQ + ∠QOR + ∠SOR + ∠POS = 360° [NCERT]**

**Solution:**

In the figure, OP, OQ, OR and OS are the rays from O

To prove : ∠POQ + ∠QOR + ∠SOR + ∠POS = 360°

Construction : Produce PO to E

**Proof:** ∠POQ + ∠QOE = 180° (Linear pair)

Similarly, ∠EOS + ∠POS = 180°

Adding we get,

∠POQ + ∠QOR + ∠ROE + ∠EOS + ∠POS = 180° + 180° ,

⇒ ∠POQ + ∠QOR + ∠ROS + ∠POS = 360° Hence ∠POQ + ∠QOR + ∠SOR + ∠POS = 360°

**Question 15.**

**In the figure, ray OS stand on a line POQ. Ray OR and ray OT are angle bisectors of ∠POS and ∠SOQ respectively. If ∠POS = x, find ∠ROT. [NCERT]**

**Solution:**

Ray OR stands on a line POQ forming ∠POS and ∠QOS

OR and OT the angle bisects of ∠POS and ∠QOS respectively. ∠POS = x

But ∠POS + ∠QOS = 180° (Linear pair)

⇒ x + ∠QOS = 180°

⇒ ∠QOS = 180° – x

∵ OR and OT are the bisectors of angle

**Question 16.**

**In the figure, lines PQ and RS intersect each other at point O. If ∠POR : ∠ROQ = 5 : 7. Find all the angles. [NCERT]**

**Solution:**

Lines PQ and PR, intersect each other at O

∵ Vertically opposite angles are equal

∴ ∠POR = ∠QOS and ∠ROQ = ∠POS

∠POR : ∠ROQ = 5:7

Let ∠POR = 5x and ∠ROQ = 7x

But ∠POR + ∠ROQ = 180° (Linear pair)

∴ 5x + 7x = 180° ⇒ 12x = 180°

⇒ x = \(\frac { { 180 }^{ \circ } }{ 12 }\) = 15°

∴ ∠POR = 5x = 5 x 15° = 75°

and ∠ROQ = 7x = 7 x 15° = 105°

But ∠QOS = POR = 75° (Vertically opposite angles)

and ∠POS = ∠ROQ = 105°

**Question 17.**

**In the figure, a is greater than b by one third of a right-angle. Find the values of a and b.**

**Solution:**

In the figure,

∠AOC + ∠BOC = 180° (Linear pair)

⇒ a + b =180° …(i)

But a = b + \(\frac { 1 }{ 3 }\) x 90° = b + 30°

⇒ a – b = 30° …(ii)

Adding (i) and (ii)

210°

2a = 210° ⇒ a = \(\frac { { 210 }^{ \circ } }{ 2 }\) = 105°

and 105° + b = 180°

⇒ b = 180° – 105°

∴ b = 75°

Hence a = 105°, b = 75°

**Question 18.**

**In the figure, ∠AOF and ∠FOG form a linear pair.**

**∠EOB = ∠FOC = 90° and ∠DOC = ∠FOG = ∠AOB = 30°**

**(i) Find the measures of ∠FOE, ∠COB and ∠DOE.**

**(ii) Name all the right angles.**

**(iii) Name three pairs of adjacent complementary angles.**

**(iv) Name three pairs of adjacent supplementary angles.**

**(v) Name three pairs of adjacent angles.**

**Solution:**

In the figure,

∠AOF and ∠FOG form a linear pair

∠EOB = ∠FOC = 90°

∠DOC = ∠FOG = ∠AOB = 30°

**(i)** ∠BOE = 90°, ∠AOB = 30°

But ∠BOE + ∠AOB + ∠EOG = 180°

⇒ 30° + 90° + ∠EOG = 180°

∴∠EOG = 180° – 30° – 90° = 60°

But ∠FOG = 30°

∴ ∠FOE = 60° – 30° = 30°

∠COD = 30°, ∠COF = 90°

∴ ∠DOF = 90° – 30° = 60°

∠DOE = ∠DOF – ∠EOF

= 60° – 30° = 30°

∠BOC = BOE – ∠COE

= 90° – 30° – 30° = 90° – 60° = 30°

**(ii)** Right angles are,

∠AOD = 30° + 30° + 30° = 90°

∠BOE = 30° + 30° + 30° = 90°

∠COF = 30° + 30° + 30° = 90°

and ∠DOG = 30° + 30° + 30° = 90°

**(iii)** Pairs of adjacent complementaiy angles are ∠AOB, ∠BOD; ∠AOC, ∠COD; ∠BOC, ∠COE

**(iv)** Pairs of adjacent supplementary angles are ∠AOB, ∠BOG; ∠AOC, ∠COG and ∠AOD, ∠DOG

**(v)** Pairs of adjacent angles are ∠BOC, ∠COD; ∠COD, ∠DOE and ∠DOE, ∠EOF.

**Question 19.**

**In the figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ROS = \(\frac { 1 }{ 2 }\)(∠QOS – ∠POS). [NCERT]**

**Solution:**

Ray OR ⊥ ROQ. OS is another ray lying between OP and OR.

To prove : ∠ROS = \(\frac { 1 }{ 2 }\)(∠QOS – ∠POS)

Proof : ∵ RO ⊥ POQ

∴ ∠POR = 90°

⇒ ∠POS + ∠ROS = 90° …(i)

⇒ ∠ROS = 90° – ∠POS

But ∠POS + ∠QOS = 180° (Linear pair)

= 2(∠POS + ∠ROS) [From (i)]

∠POS + ∠QOS = 2∠ROS + 2∠POS

⇒ 2∠ROS = ∠POS + ∠QOS – 2∠POS

= ∠QOS – ∠POS

∴ ROS = \(\frac { 1 }{ 2 }\) (∠QOS – ∠POS)

Hope given RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.2 are helpful to complete your math homework.

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