# Exercise E10.4 Circles NCERT Solutions Class 9

## Question 1

Two circles of radii \(5 \rm{cm}\) and \(3 \rm{cm}\) intersect at two points and the distance between their centers is \(4 \rm{cm}.\) Find the length of the common chord.

### Solution

**Video Solution**

**What is known?**

Radii of two circles and distance between the centers of the circles.

**What is unknown?**

Length of common chord.

**Reasoning:**

Perpendicular bisector of the common chord passes through the centers of both the circles.

**Steps:**

Given that the circles intersect at \(2\) points, so we can draw the above figure.

Let \(AB\) be the common chord. Let \(O\) and \(O^\prime\) be the centres of the circles respectively.

\(\begin{align} &{O}^{\prime} {A}=5 {\rm{cm}}, {OA}=3 {\rm{cm}},\\ & {OO}^{\prime}=4 {\rm{cm}}\end{align}\)

Since the radius of the bigger circle is more than the distance between the \(2\) centres, we can say that the centre of smaller circle lies inside the bigger circle itself.

\({OO}’\) is the perpendicular bisector of \({AB.}\)

\(\begin{align} {\rm{So,}} \;{OA}&={OB}=3 \,\rm{cm}\\ {AB}&=3+3=6\, \rm{cm} \end{align}\)

Length of the common chord is \(6 \,\rm{cm.}\)

It is also evident that common chord is the diameter of the smaller circle.

## Question 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.

### Solution

**Video Solution**

**What is known?**

Chords are equal. Chords intersect at a point within the circle.

**What is unknown?**

Proof of corresponding segments of equal chords are equal.

**Reasoning:**

Equal chords are equidistant from the centre. Using this and Right angled-Hypotenuse-Side (**RHS**) criteria and Corresponding parts of congruent triangles (**CPCT**) we prove the statement.

**Steps:**

Let \({AB}\) and \({CD}\) be the \(2\) equal chords. \({AB = CD.}\)

Let the chords intersect at point \({E}\). Join \({OE.}\)

To prove \(\begin{align} {AE = CE \, {\rm{and}} \, BE = DE.} \end{align}\)

Draw perpendiculars from the center to the chords. Perpendicular bisects the chord \({AB}\) at \({M}\) and \({CD}\) at \({N.}\)

\(\begin{align} {AM = MB = CN = DN ……(1)} \end{align}\)

\(\begin{align} {\rm{In}} \;\Delta {OME}\; {\rm{and}} \;\Delta {ONE}\end{align}\)

\[\begin{align}\angle {M}&=\angle {N}=90^{\circ}\\ {OE}&={OE}\\{OM}&={ON} \quad\\\end{align}\]

Equal chords are equidistant from the centre.

By **RHS** criteria, \(\Delta {OME}\) and \(\Delta {ONE}\) are congruent.

So by \(\text{CPCT}, {ME}={NE} \ldots \ldots(2)\)

We know that: \({CE}={CN}+{NE}\) and

\({AE}={AM}+{ME}\)

From (**\(1\)**) and (**\(2\)**), it is evident \({CE = AE}\)

\({DE}={CD}-{CE}\) and

\({BE}={AB}-{AE}\)

\({AB}\) and \({CD}\) are equal, \({CE}\) and \({AE}\) are equal. So \({DE}\) and \({BE}\) are also equal.

It is proved corresponding segments of equal chords are equal.

## Question 4

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

### Solution

**Video Solution**

**What is known?**

Two concentric circles with centre \(\begin{align} {O.} \end{align}\)

**What is unknown?**

Proof that \({AB = CD}\)

**Reasoning:**

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

**Steps:**

Draw a perpendicular from the centre of the circle \({OM}\) to the line \({AD}\).

We can see that \({BC}\) is the chord of the smaller circle and \({AD}\) is the chord of the bigger circle.

We know that perpendicular drawn from the centre of the circle bisects the chord.

\(\begin{align}∴ {BM}={MC} \ldots(1)\end{align}\)

And, \(AM = MD ... (2)\)

Subtracting (**\(2\)**) from (**\(1\)**), we obtain

\[\begin{align}{AM - BM }= {MD} - {MC}\end{align}\]

\[\begin{align}∴ {AB = CD}\end{align}\]

## Question 4

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

### Solution

**Video Solution**

**What is known?**

Chords are equal. Chords intersect at a point within the circle.

**What is unknown?**

Line joining the point of intersection to the centre makes equal angles with the chords.

**Reasoning:**

Equal chords are equidistant from the centre. Using this and Right angled- Hypotenuse-Side (RHS) criteria and Corresponding parts of congruent triangles (CPCT) we prove the statement.

**Steps:**

Let \({AB}\) and \({CD}\) be the **\(2\)** equal chords. \({AB = CD.}\)

Let the chords intersect at point \({E}\). Join \({OE.}\)

Draw perpendiculars from the centre to the chords. Perpendicular bisects the chord \(AB\) at \(M\) and \(CD\) at \(N\).

**To prove:** \(\angle {OEN}=\angle {OEM}\)

\({\rm{In}} \;\Delta {OME}\; {\rm{and}} \;\Delta {ONE}\) \[\begin{align} \angle {M}&=\angle {N}=90^{\circ}\\ {OE}&={OE}\\ {OM}&={ON} \quad \\\end{align}\]

( Equal chords are equidistant from the centre.)

By **RHS** criteria, \(\Delta {OME}\) and \(\Delta {ONE}\) are congruent.

\(\text{So by CPCT, }\angle {OEN}=\angle {OEM}\)

Hence proved that line joining the point of intersection of **\(2\)** equal chords to the centre makes equal angles with the chords.

## Question 5

Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius \(5\rm\,m\) 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 \(6\,\rm m\) each, what is the distance between Reshma and Mandip?

### Solution

**Video Solution**

**What is known?**

Three players are standing in a circle. Distance between two pairs is given. Radius of circle is given.

**What is unknown?**

Distance between Reshma and Mandip

**Reasoning:**

Perpendicular from center to either of the chord bisects the chord.Using this fact and by pythagoras theorem we can find value of \(OA\). After that we can find area of \(∆ORS\) by two ways like \(RS\) as base and \(OA\) as height or \(OS\) as base and \(RN\) as height. From this we will get the value of \(RN\) and double of value \(RN\) will will give the distance between Reshma and Mandip.

Steps:

Let \(O\) be the centre of the circle and \(R, M\) and \(S\) denote Reshma, Mandip and Salma respectively.

Draw a perpendicular \({OA}\) to \({RS}\) from \({O.}\) Then \(\begin{align}{RA = AS = 3}\,\rm {m.} \end{align}\)

Using Pythagoras theorem we get, \(\begin{align}OA = 4 \text{m.} \end{align}\)

We can see that quadrilateral \(ORSM\) takes that shape of a kite.

(Because \(OR = OM\) and \(RS = SM\)).

We know that the diagonals of a kite are perpendicular and the main diagonal bisects the other diagonal.

\(\angle {RNS}\) will be \(90^\circ\) and \(\begin{align}{RN = NM} \end{align}\)

\(\text{Area of } ∆ORS\) \[\begin{align} &= \frac{1}{2} \times RS \times OA\\ &=\frac{1}{2} \times 6 \times4\\&= 12 \dots(1)\end{align}\]

Also

\(\text{Area of} ∆ORS\) \[\begin{align}& =\frac{1}{2} \times OS \times RN\\&= \frac{1}{2} \times 5 \times RN\dots(2)\end{align}\]

From equation (\(1\)) and (\(2\))

\[\begin{align}\frac{1}{2} \times 5 \times RN\,= 12\end{align}\]

\[\begin{align}RN = \frac{{24}}{5} = 4.8\, \rm m\end{align}\]

\[\begin{align}RM = 2 × RN = 2 × 4.8 = 9.6 \, \rm m\end{align}\]

Distance between Reshma and Salma is \(9.6\, \rm{ m.}\)

## Question 6

A circular park of radius \(20\,\rm m\) 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 to each other. Find the length of string of each phone.

### Solution

**Video Solution**

**What is known?**

Three boys are on boundary of a circular park. Distance between them is equal. Radius of circular park is given.

**What is unknown?**

Length of string of each phone.

**Reasoning:**

Centre and Centroid are the same for an equilateral triangle. And it divides the median in the ratio \(\begin{align}2:1.\end{align}\) The median is also the perpendicular bisector for the opposite side.

**Steps:**

Let \(\begin{align}{A, D, S}\end{align}\) denote the positions of Ankur, David and Syed respectively.

\(\begin{align}\Delta{ADS}\end{align}\) is an equilateral triangle since all the **\(3\)** boys are on equidistant from one another.

Let \({B}\) denote the mid-point of \({DS}\) and hence \({AB}\) is the median and perpendicular bisector of \({DS.}\) Hence \( \Delta {.ABS}\) is a right angled triangle with \(\begin {align} \angle {ABS}=90^{\circ} \end {align}\)

**\(O\)** (centroid) divides the line \({AB}\) in the ratio \(\text{2:1.}\) So \({OA : OB = 2 : 1.}\)

\(\begin{align}\frac{{OA}}{{OB}} &= \frac{2}{1}\,\end{align}\)

\(\begin{align} \text{Since } OA &= 20 \; \rm {then}\\OB &=10\rm{m}\end{align}\)

\(\begin {align} {AB}&={OA}+{OB}\\ &=20+10 \\&=30 \, \rm{m}\qquad \ldots .(1) \end {align}\)

Let the side of equilateral triangle \(\Delta {ADS}\) be \(2x.\)

\(\begin{align}{AD}&={DS} \\ &={SA}\\ &=2 x \ldots .(2) \end {align}\)

Since \({B}\) is the mid-point of \({DS,}\) we get

\(\begin {align}{BS}={BD}=x \dots(3)\end {align}\)

Applying Pythagoras theorem to \(\Delta {ABS,}\) we get:

\[\begin {align}{AD}^{2}&={AB}^{2}+{BD}^{2} \\ {(2 x)^{2}}&={30^{2}+x^{2}} \\ 4 x^{2}&=900+x^{2}\\ 3 x^{2}&=900 \\ x^{2}&=\frac{900}{3}=300 \\ x&=\sqrt{300} \\ &=17.32 \, \rm {m}\end{align}\]

\[\begin {align}{AD}={DS}={SA}=2 x=34.64 \, \mathrm{m} \end {align}\]

Length of the string \(=\) Distance between them

\(= AD\) or \(DS\) or \(SA = 34.64\,\rm m.\)

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