NCERT Solutions for Class 9 Maths Chapter 9 Exercise 9.3 - Circles

Q1 In Fig. 10.36 , A,B and C are three points on a circle with centre O such that ∠BOC=30∘ and ∠AOB=60∘ . If D is a point on the circle other than the arc ABC, find ∠ADC .

Solution:
∠ AOC = ∠ AOB + ∠ BOC = 60∘+30∘=90∘

∠ AOC = 2 ∠ ADC (angle subtended by an arc at the centre is double the angle subtended by it at any)

∠ADC=12∠AOC

⇒∠ADC=1290∘=45∘

Q2 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:
Given: A chord of a circle is equal to the radius of the circle i.e. OA = AB.
To find: ∠ADB and ∠ ACB.

In △ OAB,OA = AB (Given)

OA = OB (Radii of circle)

So, OA = OB = AB

⇒ AOB is an equilateral triangle.

So, ∠ AOB = 60∘

∠ AOB = 2 ∠ ADB

⇒∠ADB=12∠AOB

⇒∠ADB=1260∘=30

ACBD is a cyclic quadrilateral.

So, ∠ ACB+ ∠ ADB = 180∘

⇒∠ACB+30∘=180∘

⇒∠ACB=180∘−30∘=150∘

Q3 In Fig. 10.37 , ∠PQR=100∘ , where P, Q and R are points on a circle with centre O. Find ∠OPR .

Solution:

Construction: Join PO and OR.
PQSR is a cyclic quadrilateral.

So, ∠ PSR + ∠ PQR = 180∘

⇒∠PSR+100∘=180∘

⇒∠PSR=180∘−100∘=80∘

Here, ∠ POR = 2 ∠ PSR

⇒∠POR=2×80∘=160∘

In △ OPR,

OP = OR (Radii)

∠ ORP = ∠ OPR (the angles opposite to equal sides)

In △ OPR,

∠ OPR+ ∠ ORP + ∠ POR = 180∘

⇒2∠OPR+160∘=180∘

⇒2∠OPR=180∘−160∘

⇒2∠OPR=20∘

⇒∠OPR=10∘

Q4 In Fig. 10.38 , ∠ABC=69∘,∠ACB=31∘, find ∠BDC

Solution:

In △ ABC,

∠ A+ ∠ ABC+ ∠ ACB= 180∘

⇒∠A+69∘+31∘=180∘

⇒∠A+100∘=180∘

⇒∠A=180∘−100∘

⇒∠A=80∘

∠ A = ∠ BDC = 80∘ (Angles in same segment)

Q5 In Fig. 10.39 , 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:

∠ DEC+ ∠ BEC = 180∘ (linear pairs)

⇒ ∠ DEC+ 130∘ = 180∘ ( ∠ BEC = 130∘ )

⇒ ∠ DEC = 180∘ - 130∘

⇒ ∠ DEC = 50∘

In △ DEC,

∠ D+ ∠ DEC+ ∠ DCE = 180∘

⇒∠D+50∘+20∘=180∘

⇒∠D+70∘=180∘

⇒∠D=180∘−70∘=110∘

∠ D = ∠ BAC (angles in same segment are equal )

∠ BAC = 110∘

Q6 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 ∠ECD .

Solution:

∠BDC=∠BAC (angles in the same segment are equal )

∠BDC=30∘

In △BDC,

∠BCD+∠BDC+∠DBC=180∘

⇒∠BCD+30∘+70∘=180∘

⇒∠BCD+100∘=180∘

⇒∠BCD=180∘−100∘=80∘

If AB = BC ,then

∠BCA=∠BAC

⇒∠BCA=30∘

Here, ∠ECD+∠BCE=∠BCD

⇒∠ECD+30∘=80∘

⇒∠ECD=80∘−30∘=50∘

Q7 If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Solution:

AC is the diameter of the circle.

Thus, ∠ADC=90∘ and ∠ABC=90∘ ............................1(Angle in a semi-circle is a right angle)

Similarly, BD is the diameter of the circle.

Thus, ∠BAD=90∘ and ∠BCD=90∘ ............................2(Angle in a semi-circle is a right angle)

From 1 and 2, we get

∠BCD=∠ADC=∠ABC=∠BAD=90∘

Hence, ABCD is a rectangle.

Q8 If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Solution:

Given: ABCD is a trapezium.

Construction: Draw AD || BE.

Proof: In quadrilateral ABED,

AB || DE (Given )

AD || BE ( By construction )

Thus, ABED is a parallelogram.

AD = BE (Opposite sides of parallelogram )

AD = BC (Given )

so, BE = BC

In △ EBC,

BE = BC (Proved above )

Thus, ∠C=∠2 ...........1(angles opposite to equal sides )

∠A=∠1 ...............2(Opposite angles of the parallelogram )

From 1 and 2, we get

∠1+∠2=180∘ (linear pair)

⇒∠A+∠C=180∘

Thus, ABED is a cyclic quadrilateral.

Q10 If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

Answer:

Given: circles are drawn taking two sides of a triangle as diameters.

Construction: Join AD.

Proof: AB is the diameter of the circle and ∠ ADB is formed in a semi-circle.

∠ ADB = 90∘ ........................1(angle in a semi-circle)

Similarly,

AC is the diameter of the circle and ∠ ADC is formed in a semi-circle.

∠ ADC = 90∘ ........................2(angle in a semi-circle)

From 1 and 2, we have

∠ ADB+ ∠ ADC= 90∘ + 90∘ = 180∘

∠ ADB and ∠ ADC are forming a linear pair. So, BDC is a straight line.

Hence, point D lies on this side.

Q12 Prove that a cyclic parallelogram is a rectangle.
Solution:

Given: ABCD is a cyclic quadrilateral.

To prove: ABCD is a rectangle.

Proof :

In cyclic quadrilateral ABCD.

∠A+∠C=180∘ .......................1(sum of either pair of opposite angles of a cyclic quadrilateral)

∠A=∠C ........................................2(opposite angles of a parallelogram are equal )

From 1 and 2,

∠A+∠A=180∘

⇒2∠A=180∘

⇒∠A=90∘

We know that a parallelogram with one angle a right angle is a rectangle.

Hence, ABCD is a rectangle.