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

Circles Class 9 Chapter 10 Exercise: 10.5

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

Answer:

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

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

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

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

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 $\mathrm{\angle }$ ACB.

Solution :

In $\mathrm{△}$ OAB,

OA = AB (Given )

OA = OB (Radii of circle)

So, OA=OB=AB

$\Rightarrow$ ABC is a equilateral triangle.

So, $\mathrm{\angle }$ AOB = ${60}^{\circ }$

$\mathrm{\angle }$ AOB = 2 $\mathrm{\angle }$ ADB

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

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

ACBD is a cyclic quadrilateral .

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

$\Rightarrow \mathrm{\angle }ACB+{30}^{\circ }={180}^{\circ }$

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

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

Answer:

Construction: Join PS and RS.

PQRS is a cyclic quadrilateral.

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

$\Rightarrow \mathrm{\angle }PSR+{100}^{\circ }={180}^{\circ }$

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

Here, $\mathrm{\angle }$ POR = 2 $\mathrm{\angle }$ PSR

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

In $\mathrm{△}$ OPR ,

OP=OR (Radii )

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

In $\mathrm{△}$ OPR ,

$\mathrm{\angle }$ OPR+ $\mathrm{\angle }$ ORP+ $\mathrm{\angle }$ POR= ${180}^{\circ }$

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

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

$\Rightarrow 2\mathrm{\angle }OPR={20}^{\circ }$

$\Rightarrow \mathrm{\angle }OPR={10}^{\circ }$

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

Answer:

In $\mathrm{△}$ ABC,

$\mathrm{\angle }$ A+ $\mathrm{\angle }$ ABC+ $\mathrm{\angle }$ ACB= ${180}^{\circ }$

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

$\Rightarrow \mathrm{\angle }A+{100}^{\circ }={180}^{\circ }$

$\Rightarrow \mathrm{\angle }A={180}^{\circ }-{100}^{\circ }$

$\Rightarrow \mathrm{\angle }A={80}^{\circ }$

$\mathrm{\angle }$ A = $\mathrm{\angle }$ BDC = ${80}^{\circ }$ (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 $\mathrm{\angle }BEC={130}^{\circ }$ and $\mathrm{\angle }ECD={20}^{\circ }$ . Find $\mathrm{\angle }BAC$

.

Answer:

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

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

$\Rightarrow$ $\mathrm{\angle }$ DEC = ${180}^{\circ }$ - ${130}^{\circ }$

$\Rightarrow$ $\mathrm{\angle }$ DEC = ${50}^{\circ }$

In $\mathrm{△}$ DEC,

$\mathrm{\angle }$ D+ $\mathrm{\angle }$ DEC+ $\mathrm{\angle }$ DCE = ${180}^{\circ }$

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

$\Rightarrow \mathrm{\angle }D+{70}^{\circ }={180}^{\circ }$

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

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

$\mathrm{\angle }$ BAC = ${110}^{\circ }$

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

Answer:

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

$\mathrm{\angle }BDC={30}^{\circ }$

In $\mathrm{△}BDC,$

$\mathrm{\angle }BCD+\mathrm{\angle }BDC+\mathrm{\angle }DBC={180}^{\circ }$

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

$\Rightarrow \mathrm{\angle }BCD+{100}^{\circ }={180}^{\circ }$

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

If AB = BC ,then

$\mathrm{\angle }BCA=\mathrm{\angle }BAC$

$\Rightarrow \mathrm{\angle }BCA={30}^{\circ }$

Here, $\mathrm{\angle }ECD+\mathrm{\angle }BCE=\mathrm{\angle }BCD$

$\Rightarrow \mathrm{\angle }ECD+{30}^{\circ }={80}^{\circ }$

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

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

Similarly, BD is the diameter of the circle.

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

From 1 and 2, we get

$\mathrm{\angle }BCD=\mathrm{\angle }ADC=\mathrm{\angle }ABC=\mathrm{\angle }BAD={90}^{\circ }$

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 $\mathrm{△}$ EBC,

BE = BC (Proved above )

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

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

From 1 and 2, we get

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

$\Rightarrow \mathrm{\angle }A+\mathrm{\angle }C={180}^{\circ }$

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

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

Similarly,

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

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

From 1 and 2, we have

$\mathrm{\angle }$ ADB+ $\mathrm{\angle }$ ADC= ${90}^{\circ }$ + ${90}^{\circ }$ = ${180}^{\circ }$

$\mathrm{\angle }$ ADB and $\mathrm{\angle }$ 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.

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

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

From 1 and 2,

$\mathrm{\angle }A+\mathrm{\angle }A={180}^{\circ }$

$\Rightarrow 2\mathrm{\angle }A={180}^{\circ }$

$\Rightarrow \mathrm{\angle }A={90}^{\circ }$

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

Hence, ABCD is a rectangle.