NCERT Solutions for Exercise 10.5 Class 9 Maths Chapter 10 - Circles

Circles Class 9 Chapter 10 Exercise: 10.5

Q1 In Fig. $\small 10.36$ , A,B and C are three points on a circle with centre O such that $\small \angle BOC=30^{\circ}$ and $\small \angle AOB=60^{\circ}$ . If D is a point on the circle other than the arc ABC, find $\small \angle ADC$ .

Answer:

$\angle$ AOC = $\angle$ AOB + $\angle$ BOC= $60 ^\circ+30 ^\circ=90 ^\circ$

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

$\angle ADC=\frac{1}{2}\angle AOC$

$\Rightarrow \angle ADC=\frac{1}{2}90 ^\circ= 45 ^\circ$

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.

Answer:

Given: A chord of a circle is equal to the radius of the circle i.e. OA=OB.

To find: ADB and $\angle$ ACB.

Solution :

In $\triangle$ OAB,

OA = AB (Given )

OA = OB (Radii of circle)

So, OA=OB=AB

$\Rightarrow$ ABC is a equilateral triangle.

So, $\angle$ AOB = $60 ^\circ$

$\angle$ AOB = 2 $\angle$ ADB

$\Rightarrow \angle ADB=\frac{1}{2}\angle AOB$

$\Rightarrow \angle ADB=\frac{1}{2}60 ^\circ=30$

ACBD is a cyclic quadrilateral .

So, $\angle$ ACB+ $\angle$ ADB = $180 ^\circ$

$\Rightarrow \angle ACB+30 ^\circ= 180 ^\circ$

$\Rightarrow \angle ACB= 180 ^\circ-30 ^\circ=150 ^\circ$

Q3 In Fig. $\small 10.37$ , $\small \angle PQR=100^{\circ}$ , where P, Q and R are points on a circle with centre O. Find $\small \angle OPR$ .

Answer:

Construction: Join PS and RS.

PQRS is a cyclic quadrilateral.

So, $\angle$ PSR + $\angle$ PQR = $180 ^\circ$

$\Rightarrow \angle PSR+100 ^\circ=180 ^\circ$

$\Rightarrow \angle PSR=180 ^\circ-100 ^\circ=80 ^\circ$

Here, $\angle$ POR = 2 $\angle$ PSR

$\Rightarrow \angle POR=2\times 80 ^\circ=160 ^\circ$

In $\triangle$ OPR ,

OP=OR (Radii )

$\angle$ ORP = $\angle$ OPR (the angles opposite to equal sides)

In $\triangle$ OPR ,

$\angle$ OPR+ $\angle$ ORP+ $\angle$ POR= $180 ^\circ$

$\Rightarrow 2\angle OPR+160 ^\circ= 180 ^\circ$

$\Rightarrow 2\angle OPR=180 ^\circ- 160 ^\circ$

$\Rightarrow 2\angle OPR=20 ^\circ$

$\Rightarrow \angle OPR=10 ^\circ$

Q4 In Fig. $\small 10.38$ , $\small \angle ABC=69^{\circ}, \angle ACB=31^{\circ},$ find $\small \angle BDC$

Answer:

In $\triangle$ ABC,

$\angle$ A+ $\angle$ ABC+ $\angle$ ACB= $180^\circ$

$\Rightarrow \angle A+69 ^\circ+31 ^\circ=180^\circ$

$\Rightarrow \angle A+100 ^\circ=180^\circ$

$\Rightarrow \angle A=180 ^\circ-100^\circ$

$\Rightarrow \angle A=80 ^\circ$

$\angle$ A = $\angle$ BDC = $80 ^\circ$ (Angles in same segment)

Q5 In Fig. $\small 10.39$ , A, B, C and D are four points on a circle. AC and BD intersect at a point E such that $\small \angle BEC=130^{\circ}$ and $\small \angle ECD=20^{\circ}$ . Find $\small \angle BAC$

.

Answer:

$\angle$ DEC+ $\angle$ BEC = $180 ^\circ$ (linear pairs)

$\Rightarrow$ $\angle$ DEC+ $130 ^\circ$ = $180 ^\circ$ ( $\angle$ BEC = $130 ^\circ$ )

$\Rightarrow$ $\angle$ DEC = $180 ^\circ$ - $130 ^\circ$

$\Rightarrow$ $\angle$ DEC = $50 ^\circ$

In $\triangle$ DEC,

$\angle$ D+ $\angle$ DEC+ $\angle$ DCE = $180 ^\circ$

$\Rightarrow \angle D+50 ^\circ+20 ^\circ= 180 ^\circ$

$\Rightarrow \angle D+70 ^\circ= 180 ^\circ$

$\Rightarrow \angle D= 180 ^\circ-70 ^\circ=110 ^\circ$

$\angle$ D = $\angle$ BAC (angles in same segment are equal )

$\angle$ BAC = $110 ^\circ$

Q6 ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If $\small \angle DBC=70^{\circ}$ , $\small \angle BAC$ is $\small 30^{\circ}$ , find $\small \angle BCD$ . Further, if $\small AB=BC$ , find $\small \angle ECD$ .

Answer:

$\angle BDC=\angle BAC$ (angles in the same segment are equal )

$\angle BDC= 30 ^\circ$

In $\triangle BDC,$

$\angle BCD+\angle BDC+\angle DBC= 180 ^\circ$

$\Rightarrow \angle BCD+30 ^\circ+70 ^\circ= 180 ^\circ$

$\Rightarrow \angle BCD+100 ^\circ= 180 ^\circ$

$\Rightarrow \angle BCD=180 ^\circ- 100 ^\circ=80 ^\circ$

If AB = BC ,then

$\angle BCA=\angle BAC$

$\Rightarrow \angle BCA=30 ^\circ$

Here, $\angle ECD+\angle BCE=\angle BCD$

$\Rightarrow \angle ECD+30 ^\circ=80 ^\circ$

$\Rightarrow \angle ECD=80 ^\circ-30 ^\circ=50 ^\circ$

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

Answer:

AC is the diameter of the circle.

Thus, $\angle ADC=90 ^\circ$ and $\angle ABC=90 ^\circ$ ............................1(Angle in a semi-circle is a right angle)

Similarly, BD is the diameter of the circle.

Thus, $\angle BAD=90 ^\circ$ and $\angle BCD=90 ^\circ$ ............................2(Angle in a semi-circle is a right angle)

From 1 and 2, we get

$\angle BCD=\angle ADC=\angle ABC=\angle BAD =90 ^\circ$

Hence, ABCD is a rectangle.

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

Answer:

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 $\triangle$ EBC,

BE = BC (Proved above )

Thus, $\angle C = \angle 2$ ...........1(angles opposite to equal sides )

$\angle A= \angle 1$ ...............2(Opposite angles of the parallelogram )

From 1 and 2, we get

$\angle 1+\angle 2=180 ^\circ$ (linear pair)

$\Rightarrow \angle A+\angle C=180 ^\circ$

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 $\angle$ ADB is formed in a semi-circle.

$\angle$ ADB = $90 ^\circ$ ........................1(angle in a semi-circle)

Similarly,

AC is the diameter of the circle and $\angle$ ADC is formed in a semi-circle.

$\angle$ ADC = $90 ^\circ$ ........................2(angle in a semi-circle)

From 1 and 2, we have

$\angle$ ADB+ $\angle$ ADC= $90 ^\circ$ + $90 ^\circ$ = $180 ^\circ$

$\angle$ ADB and $\angle$ 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.
Answer:

Given: ABCD is a cyclic quadrilateral.

To prove: ABCD is a rectangle.

Proof :

In cyclic quadrilateral ABCD.

$\angle A + \angle C = 180 ^\circ$ .......................1(sum of either pair of opposite angles of a cyclic quadrilateral)

$\angle A = \angle C$ ........................................2(opposite angles of a parallelogram are equal )

From 1 and 2,

$\angle A + \angle A = 180 ^\circ$

$\Rightarrow 2\angle A = 180 ^\circ$

$\Rightarrow \angle A = 90 ^\circ$

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

Hence, ABCD is a rectangle.