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

Exercise 9.3

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 $\angle \mathbf{ACP}=\angle \mathbf{QCD}$.

Ans:

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.

Conclusion

NCERT Class 9 Maths Chapter 9 Exercise 9.3 Solutions focuses on the concepts of angles subtended by an arc and properties of cyclic quadrilaterals. It is important to understand how angles are formed and related within circles and cyclic quadrilaterals. Pay attention to the key formulas and theorems used to solve these problems. Practicing these exercises will help strengthen your understanding of these geometric properties and improve your problem-solving skills. Vedantu provides detailed solutions to help you grasp these concepts effectively.

Class 9 Maths Chapter 9: Exercises Breakdown

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Chapter-Specific NCERT Solutions for Class 9 Maths

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