NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.5

NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.5 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.5.

• Circles Class 9 Ex 10.1
• Circles Class 9 Ex 10.2
• Circles Class 9 Ex 10.3
• Circles Class 9 Ex 10.4
• Circles Class 9 Ex 10.5
• Circles Class 9 Ex 10.6

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 10
Chapter Name	Quadrilaterals
Exercise	Ex 10.5
Number of Questions Solved	11
Category	NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.5

Question 1.
In the figure, A, B, and C are three points on a circle with center O such that ∠BOC = 30° and ∠AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC.
Solution.

Here, ∠AOC = ∠AOB + ∠BOC = 60° + 30° = 90°
Since, arc ABC makes an angle of 90° at the center  the circle.
∴ ∠ADC =   ∠AOC
[since, the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle]
= x 90° = 45°

Question 2.
A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
Solution.

Let PQ be chord. Join OP and OQ.
It is given that PQ = OP = OQ (∵  Chord = radius
Δ OPQ is equilateral.
⇒ ∠POQ = 60°
Since arc PBQ makes reflex ∠POQ = 360 °-60 ° = 300 ° at the center of the circle and ∠PBQ at a point in the minor arc of the circle.
∠PBC = (reflex ∠POQ)

Question 3.
In the figure, ∠PQR = 100°, where P, Q, and R are points on a circle with center O. Find ∠OPR.
Solution.

Question 4.
In figure, ∠ABC = 69° and ∠ACB = 31°. Find ∠BDC.
Solution.

Question 5.
In the figure, A, B, C, and D are four points on a circle. AC and BD intersect at a point E such that
∠BEC = 130° and ∠ECD = 20°. Find ∠BAC.

Solution.


Question 6.
ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find q ∠ECD.

Solution.

Question 7.
If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, then prove that it is a rectangle.
Solution.

Given: Diagonals NP and QM of a cyclic
quadrilateral NQPM are diameters of the circle passing through the vertices M, P, Q, and N.
To prove : Quadrilateral NQPM is a rectangle.
Here, ON = OP = OQ = OM [radii of same circle]
Then,  ON = OP = NP
and  OM = OQ = MQ
∴  NP = MQ
Hence, the diagonals of the quadrilateral NQPM are equal and bisect each other. So, quadrilateral NQPM is a rectangle.
Hence proved

Question 8.
If the non-parallel sides of a trapezium are equal, then prove that it is cyclic.
Solution.
Given: Non-parallel sides PS and QR of a trapezium PQRS are equal. To prove: PQRS is a cyclic trapezium.
Construction: Draw SM ⊥  PQ and RN ⊥  PQ.

Question 9.
Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D, P and Q respectively (see figure). Prove that ∠ACP = ∠QCD.

Solution.
Given: Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn which intersect the circles at A, D, P and Q, respectively

Question 10.
If circles are drawn taking two sides of a triangle as diameters, then prove that the point of intersection of these circles lies on the third side.
Solution.

Given: Two circles are drawn with sides AC and
AB of Δ ABC as diameters. Both circles intersect each other at D.
To prove : D lies on BC.
Construction: Join AD.
Proof : Since, AC and AB are the diameters of the two circles.
∴  ∠ADB = 90°     …(i) [angle in semi-circle]
and  ∠ADC = 90°  …(ii) [angle in semi-circle]
On adding Eqs. (i) and (ii), we get
∠ADB + ∠ADC = 90°+ 90° = 180°
∠BDC = 180°
Hence, BDC is a straight line.
So, point of intersection D lies on the third side.
Hence proved.

Question 11.
ABC and ADC are two right angled triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.
Solution.
Since Δ ADC and Δ ABC are right angled triangles with common hypotenuse.

Draw a circle with AC as diameter passing through B and D. Join BD.
∵ Angles in the same segment are equal.
∴ ∠CBD = ∠CA

Question 12.
Prove that a cyclic parallelogram is a rectangle.
Solution.
Given : PQRS is a parallelogram inscribed in a circle.
To prove: PQRS is a rectangle.
Proof : Since, PQRS is a cyclic quadrilateral.
∠P + ∠R = 180°   …(i)
[since, sum of pair of opposite angles in a cyclic quadrilateral is 180°]
∠P=∠R  …(ii) [since, in a parallelogram, opposite angles are equal]

 

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