NCERT Solutions for Class 9 Maths Chapter 10: Circles - Exercise 10.4

Exercise 10.4

1.Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord. 

Ans:

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Consider the radius of the circle with centre as $\text{O}$ and ${{\text{O}}^{\prime }}$ be $5~\text{cm}$ and $3~\text{cm}$ respectively.

$\text{OA}=\text{OB}=5~\text{cm}$   (Radius of same circle)

${{\text{O}}^{\prime }}\text{A}={{\text{O}}^{\prime }}\text{B}=3~\text{cm}$   (Radius of same circle)

OO' will be the perpendicular bisector of chord $\text{AB}$.

$\therefore \text{AC}=\text{CB}$

It is given that,

$\text{O}{{\text{O}}^{\prime }}=4~\text{cm}$

Let the length of OC be $\text{x}$ cm. Therefore, O'C will be ($4-\text{x}$) cm.

In $\Delta \text{OAC}$,

$\angle ACO$ is a right angle. Therefore,

Using Pythagoras theorem,

$\text{O}{{\text{A}}^{2}}=\text{A}{{\text{C}}^{2}}+\text{O}{{\text{C}}^{2}}$

$\Rightarrow {{5}^{2}}=\text{A}{{\text{C}}^{2}}+{{\text{x}}^{2}}$

$\Rightarrow 25-{{\text{x}}^{2}}=\text{A}{{\text{C}}^{2}}\quad \ldots \left( 1 \right)$

In $\Delta {{\text{O}}^{\prime }}\text{AC}$,

$\angle ACO'$ is a right angle. Therefore,

Using Pythagoras theorem,

${{\text{O}}^{\prime }}{{\text{A}}^{2}}=\text{A}{{\text{C}}^{2}}+{{\text{O}}^{\prime }}{{\text{C}}^{2}}$

$\Rightarrow {{3}^{2}}=A{{C}^{2}}+{{(4-x)}^{2}}$

$\Rightarrow 9=\text{A}{{\text{C}}^{2}}+16+{{\text{x}}^{2}}-8\text{x}$

$\Rightarrow \text{A}{{\text{C}}^{2}}=-{{\text{x}}^{2}}-7+8\text{x}\,\,\,\,\,...\left( 2 \right)$

From equations (1) and (2), we get

$25-{{x}^{2}}=-{{x}^{2}}-7+8x$

$8x=32$

$\text{x}=4$

Putting the value of $x$ in equation (1), we get

$\begin{align} & \text{A}{{\text{C}}^{2}}=25-{{4}^{2}} \\ & \text{A}{{\text{C}}^{2}}=25-16 \\ & \text{A}{{\text{C}}^{2}}=9 \\ & \text{A}{{\text{C}}^{2}}=\sqrt{9}=3\,\text{cm} \\ \end{align}$

Since,

$\begin{align} & AB=2\times AC \\ & \,\,\,\,\,\,\,\,=2\times 3=6\text{ cm} \\ \end{align}$

Therefore, the common chord of both the circle will pass through the centre of the smaller circle i.e., O' and hence, it will be the diameter of the smaller circle.

2. If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord. 

Ans: Let RS and PQ be two chords of equal lengths, of a given circle and they are intersect each other at point T.

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Construct two perpendicular lines $\text{OV}$ and $\text{OU}$ on these chords.

In $\Delta \text{OVT}$ and $\Delta \text{OUT}$,

Since equal chords of a circle are equidistant from the centre. Therefore,

$\text{OV}=\text{OU}$

$\angle OVT=\angle OUT={{90}^{{}^\circ }}$

The side OT is present in both triangles. Therefore,

$\text{OT}=\text{OT}$ (Common)

$\therefore \Delta \text{OVT}\cong \Delta $ OUT (By RHS axiom of congruency)

Hence $\text{VT}=\text{UT}$     …(1) 

As they are corresponding parts of the corresponding triangles.

It is also given that $\text{PQ}=\text{RS}$     …(2)

$\Rightarrow \frac{1}{2}\text{PQ}=\frac{1}{2}\text{RS}$

$\Rightarrow \text{PV}=\text{RU}\quad \ldots \left( 3 \right)$

On adding Equations (1) and (3), we obtain

$\text{PV}+\text{VT}=\text{RU}+\text{UT}$

$\Rightarrow \text{PT}=\text{RT}$         …(4)

On subtracting Equation (4) from Equation (2), we obtain

$\text{PQ}-\text{PT}=\text{RS}-\text{RT}$

$\Rightarrow \text{QT}=\text{ST}$        …(5)

From equation (4) and (5) we can conclude that the corresponding segments of chords PQ and RS are congruent to each other.

3.If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords. 

Ans: Let RS and PQ be two chords of equal lengths, of a given circle and they are intersect each other at point T.

(Image will be uploaded soon)

Construct two perpendicular lines $\text{OV}$ and $\text{OU}$ on these chords.

In $\Delta \text{OVT}$ and $\Delta \text{OUT}$,

Since equal chords of a circle are equidistant from the centre. Therefore,

$\text{OV}=\text{OU}$

$\angle OVT=\angle OUT={{90}^{{}^\circ }}$

The side OT is present in both triangles. Therefore,

$\text{OT}=\text{OT}$ (Common)

$\therefore \Delta \text{OVT}\cong \Delta $ OUT (By RHS axiom of congruency)

Hence $\angle \text{OTV}=\angle \text{OTU}$  

As they are corresponding parts of the corresponding triangles.

Therefore, it can be concluded from above that the line joining the point of intersection to the centre makes equal angles with the chords.

4. If a line intersects two concentric circles (circles with the same centre) with centre $\text{O}$ at $\text{A},\text{B},\text{C}$ and $D$, prove that $AB=CD$ (see figure).

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Ans: Construct a perpendicular line OM on line AD. 

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It can be observed that the chord of the smaller circle is BC and the chord of the bigger circle is AD.

Perpendicular drawn from the centre of the circle bisects the chord. 

\[\begin{align} & \therefore \,\,\,\,\,\,\,\,\,BM=MC\,\,\,\,\,\,\,\,\text{ }.\,..\text{ }\left( 1 \right) \\ & \text{and, }AM=MD\,\,\,\,\,\,\,\,\text{ }...\text{ }\left( 2 \right) \\ \end{align}\]

On subtracting Equation (2) from (1), we obtain

AM − BM = MD − MC 

∴ AB = CD

5. Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5 m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6 m each, what is the distance between Reshma and Mandip? 

Ans: Construct two perpendicular line OA and OB on line RS and SM respectively.

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$\text{AR}=\text{AS}=\frac{6}{2}=3\,\text{m}$

$\text{OR}=\text{OS}=\text{OM}=5~\text{m}$. (Radii of the same circle)

In $\Delta \text{OAR}$,

Using Pythagoras theorem,

$\text{O}{{\text{A}}^{2}}+\text{A}{{\text{R}}^{2}}=\text{O}{{\text{R}}^{2}}$

$\Rightarrow$  $\text{O}{{\text{A}}^{2}}+{{(3~\text{m})}^{2}}={{(5~\text{m})}^{2}}$

$\Rightarrow$  $\text{O}{{\text{A}}^{2}}=(25-9){{\text{m}}^{2}}=16~{{\text{m}}^{2}}$

$\Rightarrow$  $\text{OA}=4~\text{m}$

ORSM will be a kite as pair of adjacent sides are equal $(\text{OR}=\text{OM}$ and $\text{RS}=\text{SM}$ ). Since the diagonals of a kite are perpendicular and the diagonal common to both the isosceles triangles is bisected by another diagonal.

$\angle $ RCS will be of ${{90}^{{}^\circ }}$ and $\text{RC}=\text{CM}$

Area of $\Delta \text{ORS}=\frac{1}{2}\times \text{OA}\times \text{RS}$

$\Rightarrow$  $\frac{1}{2}\times \text{RC}\times \text{OS}=\frac{1}{2}\times 4\times 6$

$\Rightarrow$  $\text{RC}\times 5=24$

$\Rightarrow$  $\text{RC}=4.8$

$\text{RM}=2\text{RC}=2(4.8)=9.6$ m

Therefore, Reshma and Mandip are $9.6~\text{m}$ apart.

6. A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone. 

Ans:

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As all the sides of the triangle are equal. Therefore,

$\Delta ASD$ is an equilateral triangle.

OA (radius) $=20~\text{m}$

Since circumcentre(O) is the point of intersection of all the medians of equilateral triangle ASD. We also know that medians intersect each other in the ratio \[2:1\]. Since $\text{AB}$ is the median of equilateral triangle ASD, we can write it as,

$\Rightarrow \frac{\text{OA}}{\text{OB}}=\frac{2}{1}$

$\Rightarrow \frac{20~\text{m}}{\text{OB}}=\frac{2}{1}$

$\Rightarrow \text{OB}=\left( \frac{20}{2} \right)=10~\text{m}$

$\text{AB}=\text{OA}+\text{OB}=(20+10)\text{m}=30~\text{m}$

In $\Delta \text{ABD}$,

Using Pythagoras theorem,

$\text{A}{{\text{D}}^{2}}=\text{A}{{\text{B}}^{2}}+\text{B}{{\text{D}}^{2}}$

$\Rightarrow$ $\text{A}{{\text{D}}^{2}}={{(30)}^{2}}+{{\left( \frac{\text{AD}}{2} \right)}^{2}}$

$\Rightarrow$ $\text{A}{{\text{D}}^{2}}=900+\frac{1}{4}\text{A}{{\text{D}}^{2}}$

$\Rightarrow$  $\frac{3}{4}\text{A}{{\text{D}}^{2}}=900$

$\Rightarrow$  $\text{A}{{\text{D}}^{2}}=1200$

$\Rightarrow$  $\text{AD}=20\sqrt{3}\,\,\text{m}$

Therefore, the length of the string of each phone will be $20\sqrt{3}~\text{m}$.

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

NCERT Solutions for Class 9 Maths Chapter 10 Circles (Ex 10.4) Exercise 10.4

Opting for the NCERT solutions for Ex 10.4 Class 9 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 10.4 Class 9 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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