RD Sharma Solutions Class 9 Mathematics Chapter 16 Exercise VSAQs for circles are provided here. In this exercise, students will revise all the concepts listed in this chapter. Thus, here we have our set of RD Sharma Solutions for Class 9 to help students in their endeavour to create a better understanding of circles.
RD Sharma Solutions for Class 9 Maths Chapter 16 Circles Exercise VSAQs
Access Answers to Maths RD Sharma Solutions for Class 9 Chapter 16 Circles Exercise VSAQs Page number 16.89
Question 1: In 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:
By degree measure theorem: ∠AOB = 2 ∠APB
so, ∠AOB = 2 × 70° = 140°
Since AOBC is a cyclic quadrilateral, we have
∠ACB + ∠AOB = 180°
∠ACB + 140° = 180°
∠ACB = 40°
Question 2: In figure, two congruent circles with centres O and O’ intersect at A and B. If ∠AO’B = 50°, then find ∠APB.
Solution:
As we are given that, both the triangle are congruent which means their corresponding angles are equal.
Therefore, ∠AOB = AO’B = 50°
Now, by degree measure theorem, we have
∠APB = ∠AOB/2 = 250
Question 3: In figure, ABCD is a cyclic quadrilateral in which ∠BAD=75°, ∠ABD=58° and ∠ADC=77°, AC and BD intersect at P. Then, find ∠DPC.
Solution:
∠DBA = ∠DCA = 580 …(1)
[Angles in same segment]ABCD is a cyclic quadrilateral :
Sum of opposite angles = 180 degrees
∠A +∠C = 1800
750 + ∠C = 1800
∠C = 1050
Again, ∠ACB + ∠ACD = 1050
∠ACB + 580 = 1050
or ∠ACB = 470 …(2)
Now, ∠ACB = ∠ADB = 470
[Angles in same segment]Also, ∠D = 770 (Given)
Again From figure, ∠BDC + ∠ADB = 770
∠BDC + 470 = 770
∠BDC = 300
In triangle DPC
∠PDC + ∠DCP + ∠DPC = 1800
300 + 580 + ∠DPC = 1800
or ∠DPC = 920
Question 4: In figure, if ∠AOB = 80° and ∠ABC=30°, then find ∠CAO.
Solution:
Given: ∠AOB = 800 and ∠ABC = 300
To find: ∠CAO
Join OC.
Central angle subtended by arc AC = ∠COA
then ∠COA = 2 x ∠ABC = 2 x 300 = 600 …(1)
In triangle OCA,
OC = OA
[same radii]∠OCA = ∠CAO …(2)
[Angle opposite to equal sides]In triangle COA,
∠OCA + ∠CAO + ∠COA = 1800
From (1) and (2), we get
2∠CAO + 600 = 1800
∠CAO = 600
RD Sharma Solutions for Class 9 Maths Chapter 16 Circles Exercise VSAQs
RD Sharma Solutions Class 9 Maths Chapter 16 Circles Exercise VSAQs is based on the following topics and subtopics:
Circular Disc
Concentric Circles
Arc of a circle
Length of an arc
Central Angle
Minor Arc
Major Arc
Degree measure of an arc
Chord and segment of a circle
Some important results on Congruent arcs and chords
Cyclic Quadrilateral and many more
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