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

Exercise 10.5

1. In the given figure, $\mathbf{A}$, $\mathbf{B}$ and $\mathbf{C}$ are three points on a circle with center $\mathbf{O}$ such that $\angle \mathbf{BOC}=\mathbf{3}{{\mathbf{0}}^{\circ }}$ and \[\angle \mathbf{AOB}=\mathbf{6}{{\mathbf{0}}^{\circ }}\]. If $\mathbf{D}$ is a point on the circle other than the arc $\mathbf{ABC}$, find $\angle \mathbf{ADC}$.

Ans:

We can notice that,

$ \angle AOC=\angle AOB+\angle BOC $ 

$ ={{60}^{\circ }}+{{30}^{\circ }} $ 

$ ={{90}^{\circ }} $ 

Now, recall that a subtended angle at its centre is double the angle it has at any point on the remaining section of the circle.

Thus, 

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

$ =\dfrac{1}{2}\times {{90}^{\circ }} $ 

$ ={{45}^{\circ }}. $ 

Hence, $\angle ADC={{45}^{\circ }}$.

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.

Ans:

By the given information, in $\Delta \,OAB$,

$AB=OA=OB$, since each is a radius of the circle.

Therefore, the triangle $\Delta \,OAB$ is equilateral.

So, the value of each of the interior angles of $\Delta \,AOB$ is ${{60}^{\circ }}$.

That is, $\angle AOB={{60}^{\circ }}$.

$ \Rightarrow \angle ACB=\dfrac{1}{2}\angle AOB $ 

$ =\dfrac{1}{2}\times {{60}^{\circ }} $ 

$ ={{30}^{\circ }}. $ 

Now, in the cyclic quadrilateral $ACBD$,

$\angle ACB+\angle ADB={{180}^{\circ }}$, since the sum of the opposite angles in cyclic quadrilateral is ${{180}^{\circ }}$.

$\Rightarrow \angle ADB={{180}^{\circ }}-{{30}^{\circ }}={{150}^{\circ }}$.

Hence, the chord's angle at the points on the minor arc and on the major arc, respectively, are  ${{30}^{o}}$ and ${{150}^{o}}$.

3. In the given figure, $\angle \mathbf{PQR}=\mathbf{10}{{\mathbf{0}}^{o}}$, where $\mathbf{P}$, $\mathbf{Q}$ and $\mathbf{R}$ are points on a circle with center $\mathbf{O}$. Find $\angle \mathbf{OPR}$.

Ans:

First, take $PR$as a chord of the circle centered at $O$.

Then, consider any point $S$ on the major arc of the circle.

Therefore, we get the equilateral quadrilateral $PQRS$.

So, $\angle PQR+\angle PSR={{180}^{o}}$, since the sum of the opposite angles of a cyclic quadrilateral is ${{180}^{o}}$.

$\Rightarrow \angle PSR={{180}^{o}}-{{100}^{o}}={{80}^{o}}$.

Now, it is generally known that the angle subtended by an arc at the center is twice the angle subtended by it at any point on the remaining part of the circle.

Thus, $\angle POR=2\angle PSR=2\times {{80}^{o}}={{160}^{o}}$.

Then, in the triangle $\Delta POR$,

$OP=OR$, since each of them are radius of the circle.

$\angle OPR=\angle ORP$, angles opposite to the equal sides of the triangle.

Therefore,

$\angle OPR+\angle ORP+\angle POR={{180}^{o}}$, by the angle sum property of a triangle.

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

$\Rightarrow 2\angle OPR={{180}^{o}}-{{160}^{o}}={{20}^{o}}$.

Hence, $\angle OPR={{10}^{o}}$.

4. In fig. $\mathbf{10}.\mathbf{38}$, $\angle \mathbf{ABC}=\mathbf{6}{{\mathbf{9}}^{o}}$, $\angle \mathbf{ACB}=\mathbf{3}{{\mathbf{1}}^{o}}$, find $\angle \mathbf{BDC}$?

Ans:

In the triangle $\Delta \,ABC$,

$\angle BAC+\angle ABC+\angle ACB={{180}^{o}}$, by the angle sum property of a circle.

$\Rightarrow \angle BAC+{{69}^{o}}+{{31}^{o}}={{180}^{o}}$

$ \Rightarrow \angle BAC={{180}^{o}}-{{100}^{o}} $ 

$ \Rightarrow \angle BAC={{80}^{o}} $ 

Now, it is known that angles in the same segment of the circle are equal. So here, $\angle \text{BAC}=\angle \text{BDC}$.

Hence, $\angle BDC={{80}^{o}}$.

5. In the given figure, $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$ and $\mathbf{D}$ are four points on a circle. $\mathbf{AC}$ and $\mathbf{BD}$ intersect at a point $\mathbf{E}$ such that $\angle \mathbf{BEC}=\mathbf{13}{{\mathbf{0}}^{o}}$ and $\angle \mathbf{ECD}=\mathbf{2}{{\mathbf{0}}^{o}}$. Find$\angle \mathbf{BAC}$.

Ans:

It is known that, the exterior angle equals the sum of the opposite interior angles.

So, in the triangle $\Delta \,CDE$,

$\angle CDE+\angle DCE=\angle CEB$

Therefore,

$ \angle CDE+{{20}^{o}}={{130}^{o}} $ 

$ \Rightarrow \angle CDE={{110}^{o}}. $ 

Again, since the angles $\angle BAC$ and $\angle CDE$ are in the same segment of the circle, so $\angle BAC=\angle CDE$.

Hence, $\angle BAC={{110}^{o}}$.

6. $\mathbf{ABCD}$ is a cyclic quadrilateral whose diagonals intersect at a point $\mathbf{E}$. If $\angle \mathbf{DBC}=\mathbf{7}{{\mathbf{0}}^{o}}$, $\angle \mathbf{BAC}=\mathbf{3}{{\mathbf{0}}^{o}}$, find $\angle \mathbf{BCD}$. Further, if $\mathbf{AB}=\mathbf{BC}$, find $\angle \mathbf{ECD}$.

Ans:

Note that, here the angles $\angle CBD$ and $\angle CAD$ are in the same segment of the circle.

So, $\angle CBD=\angle CAD={{70}^{o}}$.

Now, $\angle BAD=\angle BAC+\angle CAD={{30}^{o}}+{{70}^{o}}={{100}^{o}}$.

Also, since sum of the opposite angles in a cyclic quadrilateral is ${{180}^{o}}$,so

$\angle BCD+\angle BAD={{180}^{o}}$

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

$\Rightarrow \angle BCD={{80}^{o}}$.

Again, in the tringle $\Delta \,ABC$,

$AB=BC$,

$\angle BCA=\angle CAB$, since angles opposite to equal sides of a triangle are equal.

Therefore, $\angle BCA={{30}^{o}}$.

Now, $\angle BCD={{80}^{o}}$

$\Rightarrow \angle BCA+\angle ACD={{80}^{o}}$

$\Rightarrow {{30}^{o}}+\angle ACD={{80}^{o}}$

$\Rightarrow \angle ACD={{50}^{o}}$

Hence, $\angle ECD={{50}^{o}}$.

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

Ans:

Suppose that $ABCD$ is a cyclic quadrilateral whose diagonals $BD$ and $AC$ intersecting each other at point the point $O$.

Then, $\angle BAD=\dfrac{1}{2}\angle BOD=\dfrac{{{180}^{o}}}{2}={{90}^{o}}$ (The angle at the circumference in a semicircle).

Also, $\angle BCD$ and $\angle BAD$ are opposite angles in the cyclic quadrilateral.

So, $\angle BCD+\angle BAD={{180}^{o}}$

$\Rightarrow \angle BCD={{180}^{o}}-{{90}^{o}}={{90}^{o}}$.

Therefore,

$\angle ADC=\dfrac{1}{2}\angle AOC=\dfrac{1}{2}\times {{180}^{o}}={{90}^{o}}$. (The angle at the circumference in a semicircle).

Also, $\angle ADC+\angle ABC={{180}^{o}}$, since $\angle ADC$, $\angle ABC$ are the opposite angles in the cyclic quadrilateral.

$\Rightarrow {{90}^{o}}+\angle ABC={{180}^{o}}$

$\Rightarrow \angle ABC={{90}^{o}}$.

Hence, all the interior angles of a cyclic quadrilateral is of ${{90}^{o}}$, and so it is a rectangle.

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

Ans:

Let $ABCD$ is a trapezium such that $AB\parallel CD$ and $BC=AD$.

Now, draw perpendicular lines $AM$ and $BN$ on the line $CD$.

Then, in the triangles $\Delta \,AMD$ and $\Delta \,BNC$,

$AD=BC$ (provided)

$\angle AMD=BNC$, by the formation, each angle is ${{90}^{o}}$.

$AM=BN$, since $AB\parallel CD$.

Thus, $\Delta \,AMD\cong \Delta \,BNC$ (By S-A-S congruence rule)

Therefore,

$\angle ADC=\angle BCD$ (Corresponding angle of congruence triangle) ……(i)

Also, $\angle BAD+\angle ADC={{180}^{o}}$ (The angles are on the same side)

$\Rightarrow \angle BAD+\angle BCD={{180}^{o}}$, by the equation (i).

Thus, the sum of the opposite angles is ${{180}^{o}}$.

Hence, $ABCD$ is a cyclic quadrilateral.

9. Two circles intersect at two points $\mathbf{B}$ and $\mathbf{C}$. Through $\mathbf{B}$, two-line segments $\mathbf{ABD}$ and $\mathbf{PBQ}$ are drawn to intersect the circles at $\mathbf{A}$, $\mathbf{D}$ and $\mathbf{P}$, $\mathbf{Q}$ respectively (see the given figure). Prove that $\angle \mathbf{ACP}=\angle \mathbf{QCD}$.

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Ans:

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First, join the chords $AP$, $DQ$.

Then, $\angle PBA=\angle ACP$, (The angles are on the chord $AP$) …… (i)

and $\angle DBQ=\angle QCD$ (The angles are on the chord $DQ$)   …… (ii)

Now, the line segments $ABD$ and $PBQ$ intersects at the point $B$.

So, $\angle PBA=\angle DBQ$                                                            …… (iii)

(Since, $\angle PBA$, $\angle DBQ$ are vertically opposite angles)

Then the equations (i), (ii), and (iii) yields

$\angle ACP=\angle QCD$.

10. 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.

Ans:

Let $\Delta \,ABC$ be a triangle.

By the given information, two circles intersect each other in such a way that $AB$ and $AC$ are diameters. 

If possible, let the circles intersect at the point $D$, that does not lie on the line $BC$.

Now, join the line $AD$.

Then, $\angle ADB={{90}^{o}}$, since it is an angle subtended by semicircle.

Similarly, $\angle ADC={{90}^{o}}$.

Therefore, $\angle BDC=\angle ADB+\angle ADC={{90}^{o}}+{{90}^{o}}={{180}^{o}}$.

Thus, $BDC$ is a straight line and so, it is a contradiction.

Hence, the point of intersection of the circles lies on the third side $BC$ of the triangle $\Delta \,ABC$.

11. $\mathbf{ABC}$ and $\mathbf{ADC}$ are two right triangles with common hypotenuse $\mathbf{AC}$. Prove that $\angle \mathbf{CAD=}\angle \mathbf{CBD}$.

Ans:

In triangle $\Delta \,ABC$, we have

$\angle ABC+\angle BCA+\angle CAB={{180}^{o}}$, by the angle sum property.

$\Rightarrow {{90}^{o}}+\angle BCA+\angle CAB={{180}^{o}}$

$\Rightarrow \angle BCA+\angle CAB={{90}^{o}}$                                                     …… (i)

Again, in the triangle $\Delta \,ADC$, we obtain

$\angle CDA+\angle ACD+\angle DAC={{180}^{o}}$, by the angle sum property.

$\Rightarrow {{90}^{o}}+\angle ACD+\angle DAC={{180}^{o}}$

$\Rightarrow \angle ACD+\angle DAC={{90}^{o}}$                                                …… (ii)

Now, adding (i) and (ii), yields, 

$\angle BCA+\angle CAB+\angle ACD+\angle DAC={{180}^{o}}$

$\Rightarrow \left( \angle BCA+\angle ACD \right)+\left( \angle CAB+\angle DAC \right)={{180}^{o}}$

$\Rightarrow \angle BCD+\angle DAB={{180}^{o}}$                                               …… (iii)

Also, we are provided that,

$\angle B+\angle D={{90}^{o}}+{{90}^{o}}={{180}^{o}}$                                            …… (iv)

By observing the equations (iii) and (iv), it can be concluded that since the sum of the measures of opposite angles of quadrilateral $ABCD$ is ${{180}^{o}}$, so it must be a cyclic quadrilateral.

Thus, $\angle CAD=\angle CBD$, since both the angles are on the chord $CD$.

Hence, the required result is proved.

12. Prove that a cyclic parallelogram is a rectangle.

Ans:

Consider that cyclic parallelogram $ABCD$.

Then, $\angle A+\angle C={{180}^{o}}$, (Opposite angles of the cyclic quadrilateral) …… (i)

Again, in a parallelogram, opposite angles are the same.

Therefore, $\angle A=\angle C$ and $\angle B=\angle D$.

So, the equation (i) can be written as

$\angle A+\angle C={{180}^{o}}$

$\Rightarrow \angle A+\angle A={{180}^{o}}$

$\Rightarrow 2\angle A={{180}^{o}}$

$\Rightarrow \angle A={{90}^{o}}$.

Thus, one interior angle of the parallelogram $ABCD$ is of ${{90}^{o}}$. That means, it is a rectangle.

Hence, it is proved that a cyclic parallelogram is a rectangle.

NCERT Solutions for Class 9 Maths

• Chapter 1 - Number System

• Chapter 2 - Polynomials 

• Chapter 3 - Coordinate Geometry

• Chapter 4 - Linear Equations in Two Variables

• Chapter 5 - Introductions to Euclid's Geometry

• Chapter 6 - Lines and Angles

• Chapter 7 - Triangles

• Chapter 8 - Quadrilaterals

• Chapter 9 - Areas of Parallelogram and Triangles

• Chapter 10 - Circles

• Chapter 11 - Constructions

• Chapter 12 - Heron's formula

• Chapter 13 - Surface area and Volumes

• Chapter 14 - Statistics

• Chapter 15 - Probability

NCERT Solution Class 9 Maths of Chapter 10 All Exercises

Exercise 10.5 (Ex 10.5) Class 9 Maths Chapter 10 NCERT Solutions - Free PDF Download

Vedantu provides you with the option to download the PDF of Class 9 Chapter Circles Exercise 10.5 Solutions. The PDF makes it easier to study the concepts and revise them quickly. The notes if downloaded come in handy, allowing the students to study anytime they want, and even makes the process of group studies simpler. 

Exercise 10.5 Class 9 Maths Solutions

The questions in Exercise 10.5 are based on all the concepts included in the chapter. It mainly covers the concept of the angle subtended by an arc of a circle and the cyclic quadrilaterals. The exercise consists of 12 questions, including a variety of short answers as well as comprehensive ones. Let us give you an overview of the exercise. 

Question 1:

The first question is based on the theorem, ‘The angle subtended by an arc at the centre of the circle is twice the angle subtended by the arc at any other remaining point in the circle.’ The students can solve this easily by applying this concept. 

Question 2:

In the question, the given condition is that the length of a chord of the circle is equal to its radius. This means that AB = OA = OB, forming an equilateral triangle AOC. This will help you find the angles in the circle easily. Moreover, ACBD is a cyclic quadrilateral. The angles can be found out by applying properties of opposite angles of a quadrilateral. 

Question 3:

In this question, the students need to find out the angle OPR in the circle. This can be done implementing the theorem - the angle subtended by an arc at the centre is twice the angle on remaining parts of the circle, followed by the angle sum property of a triangle.

Question 4:

In this question, two angles, ABC and ACB are given, respectively. Ths students need to follow the property that the angles in the same segment of a circle are equal. Following this, they can use the angle sum property of a triangle to find the angles.

Question 5:

This question follows the same approach as the previous one - that the angles in the same segment are equal. Following this, the students can solve it by applying the exterior angle property of a triangle.

Question 6:

In this question, a cyclic quadrilateral is given. So, the students can solve it using the opposite angle property of a cyclic quadrilateral. This can further be solved by the hint that AB=BC. So, the angles opposite these sides will also be equal.

Question 7:

In this question, a cyclic quadrilateral is given in which the diagonals are equal to the diameter of the circle. Also, according to the property of semi-circle, angles in a semi-circle are equal. Therefore, all the internal angles will be right angles.

Question 8:

In the given question, the non-parallel sides of the trapezium are equal. The students have to prove that it is cyclic. They can do so by drawing two perpendiculars to the base. Following this, the RHS congruence rule can be implied. An equation is formed proving that opposite angles are supplementary. This will help in proving that it is a cyclic quadrilateral.

Question 9:

In this question, the students need to prove that the angles ACP and QCD are equal. They can do so by joining the two chords AP and DQ. It can further be solved by the property of angles in the same segment. Again, the vertically opposite angles are also equal. These will help in proving the given question.

Question 10:

In this question, two sides of a triangle are considered as the diameter of the circles. The students need to prove that the point of intersection lies on the third side. This can be done using the property that angles in a semi-circle are equal. These angles, when added form a straight line, proving the question.

Question 11:

In the given question, two right triangles have a common hypotenuse. Students are required to prove that angle CAD and CBD are equal. The points ABCD are concyclic as they lie in the same semi-circle. Also, this can be proved using the statement that ‘angles in the same segment are equal’.

Question 12:

In this question, a cyclic parallelogram is given and asks for it to be proven to be a rectangle. It can be done by considering that the opposite angles of a cyclic quadrilateral are equal to 180° and the opposite angles in a parallelogram are equal. Therefore, it proves the given statement.

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