## RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles VSAQS

Other Exercises

- RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.1
- RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.2
- RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.3
- RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.4
- RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5
- RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles VSAQS
- RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles MCQS

Question 1.

In the figure, two circles intersect at A and B. The centre of the smaller circle is O and it lies on the circumference of the larger circle. If ∠APB = 70°, find ∠ACB.

Solution:

Arc AB subtends ∠AOB at the centre and ∠APB at the remaining part of the circle

∴ ∠AOB = 2∠APB = 2 x 70° = 140°

Now in cyclic quadrilateral AOBC,

∠AOB + ∠ACB = 180° (Sum of the angles)

⇒ 140° +∠ACB = 180°

⇒ ∠ACB = 180° – 140° = 40°

∴ ∠ACB = 40°

Question 2.

In the figure, two congruent circles with centre O and O’ intersect at A and B. If ∠AO’B = 50°, then find ∠APB.

Solution:

Two congruent circles with centres O and O’ intersect at A and B

∠AO’B = 50°

∵ OA = OB = O’A = 04B (Radii of the congruent circles)

Question 3.

In the figure, ABCD is a cyclic quadrilateral in which ∠BAD = 75°, ∠ABD = 58° and ∠ADC = IT, AC and BD intersect at P. Then, find ∠DPC.

Solution:

∵ ABCD is a cyclic quadrilateral,

∴ ∠BAD + ∠BCD = 180°

⇒ 75° + ∠BCD – 180°

⇒ ∠BCD = 180°-75°= 105° and ∠ADC + ∠ABC = 180°

⇒ 77° + ∠ABC = 180°

⇒ ∠ABC = 180°-77°= 103°

∴ ∠DBC = ∠ABC – ∠ABD = 103° – 58° = 45°

∵ Arc AD subtends ∠ABD and ∠ACD in the same segment of the circle 3

∴ ∠ABD = ∠ACD = 58°

∴ ∠ACB = ∠BCD – ∠ACD = 105° – 58° = 47°

Now in ∆PBC,

Ext. ∠DPC = ∠PBC + ∠PCB

=∠DBC + ∠ACB = 45° + 47° = 92°

Hence ∠DPC = 92°

Question 4.

In the figure, if ∠AOB = 80° and ∠ABC = 30°, then find ∠CAO.

Solution:

In the figure, ∠AOB = 80°, ∠ABC = 30°

∵ Arc AB subtends ∠AOB at the centre and

∠ACB at the remaining part of the circle

∴ ∠ACB = \(\frac { 1 }{ 2 }\)∠AOB = \(\frac { 1 }{ 2 }\) x 80° = 40°

In ∆OAB, OA = OB

∴ ∠OAB = ∠OBA

But ∠OAB + ∠OBA + ∠AOB = 180°

∴ ∠OAB + ∠OBA + 80° = 180°

⇒ ∠OAB + ∠OAB = 180° – 80° = 100°

∴ 2∠OAB = 100°

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

Similarly, in ∆ABC,

∠BAC + ∠ACB + ∠ABC = 180°

∠BAC + 40° + 30° = 180°

⇒ ∠BAC = 180°-30°-40°

= 180°-70°= 110°

∴ ∠CAO = ∠BAC – ∠OAB

= 110°-50° = 60°

Question 5.

In the figure, A is the centre of the circle. ABCD is a parallelogram and CDE is a straight line. Find ∠BCD : ∠ABE.

Solution:

In the figure, ABCD is a parallelogram and

CDE is a straight line

∵ ABCD is a ||gm

∴ ∠A = ∠C

and ∠C = ∠ADE (Corresponding angles)

⇒ ∠BCD = ∠ADE

Similarly, ∠ABE = ∠BED (Alternate angles)

∵ arc BD subtends ∠BAD at the centre and

∠BED at the remaining part of the circle

Question 6.

In the figure, AB is a diameter of the circle such that ∠A = 35° and ∠Q = 25°, find ∠PBR.

Solution:

In the figure, AB is the diameter of the circle such that ∠A = 35° and ∠Q = 25°, join OP.

Arc PB subtends ∠POB at the centre and

∠PAB at the remaining part of the circle

∴ ∠POB = 2∠PAB = 2 x 35° = 70°

Now in ∆OP,

OP = OB radii of the circle

∴ ∠OPB = ∠OBP = 70° (∵ ∠OPB + ∠OBP = 140°)

Now ∠APB = 90° (Angle in a semicircle)

∴ ∠BPQ = 90°

and in ∆PQB,

Ext. ∠PBR = ∠BPQ + ∠PQB

= 90° + 25°= 115°

∴ ∠PBR = 115°

Question 7.

In the figure, P and Q are centres of two circles intersecting at B and C. ACD is a straight line. Then, ∠BQD =

Solution:

In the figure, P and Q are the centres of two circles which intersect each other at C and B

ACD is a straight line ∠APB = 150°

Arc AB subtends ∠APB at the centre and

∠ACB at the remaining part of the circle

∴ ∠ACB = \(\frac { 1 }{ 2 }\) ∠APB = \(\frac { 1 }{ 2 }\) x 150° = 75°

But ∠ACB + ∠BCD = 180° (Linear pair)

⇒ 75° + ∠BCD = 180°

∠BCD = 180°-75°= 105°

Now arc BD subtends reflex ∠BQD at the centre and ∠BCD at the remaining part of the circle

Reflex ∠BQD = 2∠BCD = 2 x 105° = 210°

But ∠BQD + reflex ∠BQD = 360°

∴ ∠BQD+ 210° = 360°

∴ ∠BQD = 360° – 210° = 150°

Question 8.

In the figure, if O is circumcentre of ∆ABC then find the value of ∠OBC + ∠BAC.

Solution:

In the figure, join OC

∵ O is the circumcentre of ∆ABC

∴ OA = OB = OC

∵ ∠CAO = 60° (Proved)

∴ ∆OAC is an equilateral triangle

∴ ∠AOC = 60°

Now, ∠BOC = ∠BOA + ∠AOC

= 80° + 60° = 140°

and in ∆OBC, OB = OC

∠OCB = ∠OBC

But ∠OCB + ∠OBC = 180° – ∠BOC

= 180°- 140° = 40°

⇒ ∠OBC + ∠OBC = 40°

∴ ∠OBC = \(\frac { { 40 }^{ \circ } }{ 2 }\) = 20°

∠BAC = OAB + ∠OAC = 50° + 60° = 110°

∴ ∠OBC + ∠BAC = 20° + 110° = 130°

Question 9.

In the AOC is a diameter of the circle and arc AXB = 1/2 arc BYC. Find ∠BOC.

Solution:

In the figure, AOC is diameter arc AxB = \(\frac { 1 }{ 2 }\) arc BYC 1

∠AOB = \(\frac { 1 }{ 2 }\) ∠BOC

⇒ ∠BOC = 2∠AOB

But ∠AOB + ∠BOC = 180°

⇒ ∠AOB + 2∠AOB = 180°

⇒ 3 ∠AOB = 180°

∴ ∠AOB = \(\frac { { 180 }^{ \circ } }{ 3 }\) = 60°

∴ ∠BOC = 2 x 60° = 120°

Question 10.

In the figure, ABCD is a quadrilateral inscribed in a circle with centre O. CD produced to E such that ∠AED = 95° and ∠OBA = 30°. Find ∠OAC.

Solution:

In the figure, ABCD is a cyclic quadrilateral

CD is produced to E such that ∠ADE = 95°

O is the centre of the circle

∵ ∠ADC + ∠ADE = 180°

⇒ ∠ADC + 95° = 180°

⇒ ∠ADC = 180°-95° = 85°

Now arc ABC subtends ∠AOC at the centre and ∠ADC at the remaining part of the circle

∵ ∠AOC = 2∠ADC = 2 x 85° = 170°

Now in ∆OAC,

∠OAC + ∠OCA + ∠AOC = 180° (Sum of angles of a triangle)

⇒ ∠OAC = ∠OCA (∵ OA = OC radii of circle)

∴ ∠OAC + ∠OAC + 170° = 180°

2∠OAC = 180°- 170°= 10°

∴ ∠OAC = \(\frac { { 10 }^{ \circ } }{ 2 }\) = 5°

Hope given RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles VSAQS are helpful to complete your math homework.

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