M.C.Q (1 Marks) - Maths STD 10 Questions - Vidyadip

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M.C.Q (1 Marks)

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MCQ 11 Mark

In the given figure, $RQ$ is a tangent to the circle with centre $O$. If $SQ = 6cm$ and $QR = 4cm,$ then $OR $is equal to:

A
$2.5cm$
B
$3cm$
✓
$5cm$
D
$8cm$

Answer

Correct option: C.

$5cm$

$SQ = 6cm \Rightarrow OQ = 3cm$
$QR = 4cm$
Since RQ is a tangent to the circle at Q.
$\angle\text{RQO}=90^\circ$ ....(tangent is perpendicular to the radius of a circle)
In $\triangle\text{RQO},$
By using Pythagoras theorem,
$ O R^2=R Q^2+O Q^2 $
$ =4^2+3^2 $
$ =16+9 $
$ =25 $
$ \therefore O R^2=25 $
$ \Rightarrow O R=5 {~cm}$

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MCQ 21 Mark

In the given figure, $QR$ is a common tangent to the given circle, touching externally at the point $T$. The tangent at $T$ meets $QR$ at $P$. If $PT = 3.8cm$ then the length of $QR$ is:

A
$1.9cm$
B
$3.8cm$
C
$5.7cm$
✓
$7.6cm$

Answer

Correct option: D.

$7.6cm$

We know that tangent from an external point to the circle are equal.
$PQ = PT = 3.8cm$
$PR = PT = 3.8cm$
$QR = PQ + PR$
$= 3.8 +3.8$
$=7.6cm$

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MCQ 31 Mark

In the given figure, $AB$ and $AC$ are tangent to the circle with centre $O$ such that $\angle\text{BAC}=40^\circ.$ Then, $\angle\text{BOC}$ is equal to:

A
$80^\circ $
B
$100^\circ$
C
$120^\circ$
✓
$140^\circ$

Answer

Correct option: D.

$140^\circ$

Since $AB$ and $AC$ are the tangent to the circle.
$\angle\text{OBA}=\angle\text{BAC}=90^\circ$....(tangent is perpendicular to the radius of a circle)
In $ABOC$
$\angle\text{OBA}+\angle\text{BAC}+\angle\text{OCA}+\angle\text{BOC}=360^\circ$
$\Rightarrow90^\circ+40^\circ+90^\circ+\angle\text{BOC}=360^\circ$
$\Rightarrow\angle\text{BOC}=140^\circ$

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MCQ 41 Mark

In the given figure, quadrilateral $ABCD$ is circumscribed, touching the circle at $P, Q, R $and S. If $AP = 5cm$, $BC = 7cm$ and $CS = 3cm, AB =?$

✓
$9cm$
B
$10cm$
C
$12cm$
D
$8cm$

Answer

Correct option: A.

$9cm$

We know that tangent from an external point to the circle are equal.
$AP = AQ = 5cm$
$CS = CR = 3cm$
$RB = BC - CR$
$= 7 +3$
$=4cm$
So, $BQ = RB = 4cm$
Thus,$ AB = AQ + RB = 5 + 4 = 9cm$

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MCQ 51 Mark

In the given figure, $O$ is the centre of two concentric circles of radii $5cm$ and $3cm$. From an external point $P$ tangents $PA$ and $PB$ are drawn to these circles. If PA = $12cm$ then $PB$ is equal to:

A
$5\sqrt{2}\text{cm}$
B
$3\sqrt{5}\text{cm}$
✓
$4\sqrt{10}\text{cm}$
D
$5\sqrt{10}\text{cm}$

Answer

Correct option: C.

$4\sqrt{10}\text{cm}$

Construction: Join $OB$.
We know that tangent is perpendicular to the radius of a circle.
In $\triangle\text{OPA},$
By Pythagoras theorem,
$ O P^2=O A^2+A P^2 $
$ \Rightarrow O P^2=5^2+12^2 $
$ \Rightarrow O P^2=169 $
$ \Rightarrow O P=13 \mathrm{~cm}$
In $\triangle\text{OPB},$
By Pythagoras theorem,
$ O P^2=O B^2+P B^2 $
$ \Rightarrow P B^2=O P^2-O B^2 $
$ \Rightarrow P B^2=13^2-3^2 $
$ \Rightarrow P B^2=160$
$\Rightarrow \text{PB} =4\sqrt{10}\text{cm} $

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MCQ 61 Mark

In the given figure, $O$ is the centre of the circle. $AB$ is the tangent to the circle at the point P. If $\angle\text{PAO} = 30^\circ$ then $\angle\text{CPB} + \angle\text{ACP}$ is equal to:

A
$60^\circ$
✓
$90^\circ$
C
$120^\circ$
D
$150^\circ$

Answer

Correct option: B.

$90^\circ$

Since APB is a straight line,
$\angle\text{APD}+\angle\text{DPC}+\angle\text{CPB}=180º$
We know that angles that subtend the same arc are equal.
$\Rightarrow\angle\text{APD}=\angle\text{ACP}$
$\Rightarrow\angle\text{ACP}+90^\circ+\angle\text{CPB}=180^\circ$....(Since $\angle\text{DPC}$ is inscribed in a semicircle)
$\Rightarrow\angle\text{CPB}+\angle\text{ACP}=90^\circ$

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MCQ 71 Mark

In the given figure, $O$ is the centre of the circle $AB$ is a chord and $AT$ is the tangent at $A$. If $\angle\text{AOB} = 100^\circ$ then $\angle\text{BAT} $ is equal to:

A
$40^\circ$
✓
$50^\circ$
C
$90^\circ$
D
$100^\circ$

Answer

Correct option: B.

$50^\circ$

In $\triangle\text{OAB},$
$\text{OA}=\text{OB}$ ....(radii of the same circle)
$\Rightarrow\angle\text{OAB}=\angle\text{OAB}$ ....(angles opposite equal sides are equal)
$\angle\text{AOB}+\angle\text{OAB}+\angle\text{OBA}=180^\circ$ (Angle Sum Property)
$\Rightarrow\angle\text{AOB}+\angle\text{OAB}+\angle\text{OAB}=180^\circ$
$\Rightarrow100^\circ+2\angle\text{OAB}=180^\circ$
$\Rightarrow2\angle\text{OAB}=80^\circ$
$\Rightarrow\angle\text{OAB}=40^\circ$
Since AT is the tangent,
$\angle\text{OAT}=90^\circ$
$\Rightarrow\angle\text{OAB}+\angle\text{BAT}=90^\circ$
$\Rightarrow40^\circ+\angle\text{BAT}=90^\circ$
$\Rightarrow\angle\text{BAT}=50^\circ$

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MCQ 81 Mark

In the given figure, $O$ is the centre of a circle; $PQL$ and $PRM$ are the tangents at the points $Q$ and $R$ respectively, and $S$ is a point on the circle, such that $\angle\text{SQL} = 50^\circ.$ and $\angle\text{SRM} = 60^\circ.$ Find $ \angle\text{QSR}=?$

A
$40^\circ$
B
$50^\circ $
C
$60^\circ$
✓
$70^\circ$

Answer

Correct option: D.

$70^\circ$

Since $PL$ and $PM$ are the tangent to the circle.
$\angle\text{OQL}=\angle\text{ORM}=90^\circ$ (tangent is perpendicular to the radius of a circle)
So,
$\angle\text{OQL}+\angle\text{SQL}+\angle\text{OQS}$
$\Rightarrow90^\circ=50^\circ+\angle\text{OQS}$
$\Rightarrow\angle\text{OQC}=40^\circ$
Similarly, we can find $\angle\text{ORS}=30^\circ.$
In $\triangle\text{OQS},$
$\text{OQ}=\text{OS}$
$\angle\text{OQS}=\angle\text{OSQ}=40^\circ$ ...(angles opposite equal sides are equal)
In $\triangle\text{ORS},$
$\text{OR}=\text{OS}$
$\angle\text{ORS}=\angle\text{OSR}=30^\circ$
So, $\angle\text{QSR}=\angle\text{OSQ}+\angle\text{OSR}=40^\circ+30^\circ=70^\circ.$

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MCQ 91 Mark

In the given figure, $AP, AQ$ and $BC$ are tangents to the circle. If $AB = 5cm, AC = 6cm$ and $BC = 4cm$ then the length of $AP$ is:

A
$15cm$
B
$10cm$
C
$9cm$
✓
$7.5cm$

Answer

Correct option: D.

$7.5cm$

Let $BC$ intersect the circle at $D$.
We know that tangent from an external point to the circle are equal.
$BP = BD$
$CD = CQ$
$AP = AQ$
perimeter of $\triangle\text{ABC}$
$= AB + BC + AC$
$= AB + (BD + CD) + AC$
$= AB + (BP + CQ) + AC$
$= (AB + BP) + (AC + CQ)$
$= AP + AQ$
Since perimeter of $\triangle\text{ABC}$ $= AB + BC + AC = 5 + 6 + 4 = 15cm$
$\Rightarrow AP + AQ = 15$
$\Rightarrow 2AP = 15$
$\Rightarrow AP = 7.5cm$

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MCQ 101 Mark

In the given figure, PT is a tangent to a circle with centre $O$. If $OT = 6cm$ and $OP = 10cm$, then the length of tangent $PT$ is:

✓
$8cm$
B
$10cm$
C
$12cm$
D
$16cm$

Answer

Correct option: A.

$8cm$

$OT = 6cm$
$OP = 10cm$
Since $PT$ is a tangent to the circle at $T$.
$\angle\text{PTO}=90^\circ$....(tangent is perpendicular to the radius of a circle)
In $\triangle\text{PTO},$
By Pythagoras theorem,
$ \mathrm{OP} 2=\mathrm{PT}^2+\mathrm{OT}^2 $
$ \Rightarrow \mathrm{PT}^2=\mathrm{OP}^2-\mathrm{OT}^2 $
$ \Rightarrow \mathrm{PT}^2=10^2-6^2 $
$ \Rightarrow \mathrm{PT}^2=100-36 $
$ \Rightarrow \mathrm{PT}=8 \mathrm{~cm}$

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MCQ 111 Mark

In the given figure, a circle is inscribed in a quadrilateral $ABCD$ touching its sides $AB, BC, CD$ and $AD$ at $P, Q, R$ and $S$ respectively. If the radius of the circle is $10cm, BC = 38cm, PB = 27cm$ and $\text{AD} \perp \text{CD}$ then the length of $CD$ is:

A
$11cm$
B
$15cm$
C
$20cm$
✓
$21cm$

Answer

Correct option: D.

$21cm$

We know that tangles from an external point to the circle are equal.
$BQ = PB = 27cm$
So, $CQ = BC - BQ = 38 - 27 = 11cm$
$\Rightarrow CR = CQ = 11cm$
In quad. SORD,
$\angle\text{SDR}=90^\circ....(\therefore\text{AD}\perp\text{CD})$
$\Rightarrow\angle\text{OSD}=\angle\text{ORD}=90^\circ$
Also, $OS = OR$ and $SD = SR$
So, quad. $SORD$ is a square.
Thus, $DR = SO = 10cm$
Hence, $CD = DR + CR = 10 + 11 = 21cm.$

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MCQ 121 Mark

In the given figure, $O$ is the centre of a circle. $AOC$ is its diameter, such that $\angle\text{ACB} = 50^\circ.$ If AT is the tangent to the circle at the point $A$, then$\angle \text{BAT} = ?$

A
$40^\circ$
✓
$50^\circ$
C
$60^\circ$
D
$65^\circ$

Answer

Correct option: B.

$50^\circ$

Construction: Join $OC$.
Since $AC$ is a diameter of the circle.
$\angle\text{ABC}=90^\circ$ ....(angle in a semicircle is 90º)
In $\triangle\text{ABC},$
$\angle\text{ABC}+\angle\text{BCA}+\angle\text{BAC}=180^\circ$ ......(Angle Sum Property)
$\Rightarrow90^\circ+50^\circ+\angle\text{BAC}=180^\circ$
$\Rightarrow\angle\text{BAC}=40^\circ$
Since AC is a tangent to the inner circle.
$\Rightarrow\angle\text{OAT}=90^\circ$ .....(Tangent is perpendicular to the radius of a circle)
$\Rightarrow\angle\text{BAO}+\angle\text{BAT}=90^\circ$
$\Rightarrow40^\circ+\angle\text{BAT}=90^\circ$
$\Rightarrow\angle\text{BAT}=50^\circ$

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MCQ 131 Mark

$PQ$ is a tangent to a circle with centre $O$ at the point $P$. If $\triangle\text{OPQ}$ is an isosceles triangle, then $\angle\text{OQP}$ is equal to:

A
$30^\circ$
✓
$45^\circ$
C
$60^\circ$
D
$90^\circ$

Answer

Correct option: B.

$45^\circ$

Given that $\triangle\text{PQO}$ is an isosceles triangle.
Since $PQ$ is a tangent to the circle at $P$.
Sunce $PQ$ is a tangent to the circle at $P$.
$\angle\text{OPQ}=90^\circ$ .....(tangent is perpendicular to the radius of a circle)
In $\triangle\text{OPQ},$
$OP = OQ$
$\Rightarrow\angle\text{OQP}=\angle\text{POQ}$
Using Angle Sum Property,
$\angle\text{OQP}+\angle\text{POQ}+\angle\text{OPQ}=180^\circ$
$\Rightarrow\angle\text{OQP}+\angle\text{OQP}+90^\circ=180^\circ$
$\Rightarrow2\angle\text{OQP}=90^\circ$
$\Rightarrow\angle\text{OQP}=45^\circ$

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MCQ 141 Mark

If the angle between two radii of a circle is $130^\circ$, then the angle between the tangent at the ends of the radii is:

A
$65^\circ$
B
$40^\circ$
✓
$50^\circ$
D
$90^\circ$

Answer

Correct option: C.

$50^\circ$

In quad. $AOBP$
$\angle\text{PAO}+\angle\text{PBO}+\angle\text{AOB}+\angle\text{APB}=360^\circ$....(Angle Sum Property)
$\Rightarrow90^\circ+90^\circ+130^\circ+\angle\text{APB}=360^\circ$....(Since radius of a circle is perpendicular to the tangent)
$\Rightarrow\angle\text{APB}=50^\circ$

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MCQ 151 Mark

In the given figure, $O$ is the centre of a circle, $BOA$ is its diameter and the tangent at the point $P$ meets $BA$ extended at $T$. $\angle\text{PBO} = 30^\circ$ then $\angle\text{PTA} =?$

A
$60^\circ$
✓
$30^\circ$
C
$15^\circ$
D
$45^\circ$

Answer

Correct option: B.

$30^\circ$

In $\triangle\text{OBP},$
$\text{OB} = \text{OP} $....(radii of the circle)
$\Rightarrow\angle\text{OBP}=\angle\text{OPB}=30^\circ$ ....(angles opposite equal sides are equal)
Since $PT$ is a tangent,
$\angle\text{OPT}=90^\circ$
In $\triangle\text{BPT},$
$\angle\text{BPT}+\angle\text{PBT}+\angle\text{PTB}=180^\circ$....(Angle Sum Property)
$\Rightarrow(30^\circ+90^\circ)+30^\circ+\angle\text{PTB}=180^\circ$
$\Rightarrow\angle\text{PTB}=30^\circ$
That is, $\angle\text{PTA}=30^\circ$

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MCQ 161 Mark

$O$ is the centre of a circle of radius $5cm$. At a distance of $13cm$ from $O$, a point $P$ is taken. From this point, two tangents $PQ$ and $PR$ are drawn to the circle. Then, the area of quadrilateral $PQOR$ is:

✓
$ 60 \mathrm{~cm}^2 $
B
$ 32.5 \mathrm{~cm}^2 $
C
$ 65 \mathrm{~cm}^2 $
D
$ 30 \mathrm{~cm}^2$

Answer

Correct option: A.

$ 60 \mathrm{~cm}^2 $

In $\triangle\text{OPQ}$ and $\triangle\text{ORP},$
$\angle\text{OQP}=\angle\text{ORP}=90^\circ$ ....(Since OP and RP are tangent to the circle)
$\text{OP}=\text{OP}$ ....(common side)
$\text{OQ}=\text{OR}$ ....(radii of the same circle)
$\Rightarrow\triangle\text{OPQ}\cong\triangle\text{ORP}$ ....(RHS congruence criterion)
So, the areas of both the triangle will be the swame.
In $\triangle\text{OPQ},$
By Pythagoras theorem,
$\text{OP}^2=\text{OQ}^2+\text{PQ}^2$
$\Rightarrow\text{PQ}^2=\text{OP}^2-\text{OQ}^2$
$\Rightarrow\text{PQ}^2=\text{13}^2-\text{5}^2$
$\Rightarrow\text{PQ}^2=144$
$\Rightarrow\text{PQ}=12\text{cm}$
$\text{ar}(\triangle\text{OPQ})=\frac{1}{2}\times\text{PQ}\times\text{OQ}$
$=\frac{1}{2}\times\text{12}\times5$
$=30\text{cm}^2$
$\text{ar}(\text{quad PQOR})=\text{ar}(\triangle\text{OPQ})+\text{ar}(\triangle\text{ORP})$
$\Rightarrow\text{ar}(\text{quad PQOR})=30\text{cm}^2+30\text{cm}^2$
$\Rightarrow\text{ar}(\text{quad PQOR})=60\text{cm}^2$

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MCQ 171 Mark

In the given figure, $PA$ and $PB$ are two tangents to the circle with centre $O$. If $\angle\text{APB}=60^\circ$ then $\angle\text{OAB}$ is:

A
$15^\circ$
✓
$30^\circ $
C
$60^\circ$
D
$90^\circ$

Answer

Correct option: B.

$30^\circ $

We know that tangent from an external point to a circle are equal.
So,
$\text{PA}=\text{PB}$
$\Rightarrow\angle\text{PAB}+\angle\text{PBA}$ ....(angles opposite equal sides are equal)
Now in $\triangle\text{PAB},$
$\angle\text{PAB}+\angle\text{PBA}+\angle\text{APB}=180^\circ$ ...(angles Sum Property)
$\Rightarrow\angle\text{PAB}+\angle\text{PAB}+60^\circ=180^\circ$
$\Rightarrow2\angle\text{PAB}+60^\circ=180^\circ$
$\Rightarrow2\angle\text{PAB}=120^\circ$
$\Rightarrow\angle\text{PAB}=60^\circ$
Since AP is a tangent to the circle,
$\angle\text{OAP}=90^\circ$
$\Rightarrow\angle\text{OAB}+\angle\text{PAB}=90^\circ$
$\Rightarrow\angle\text{OAB}+60^\circ=90^\circ$
$\Rightarrow\angle\text{OAB}=30^\circ$

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MCQ 181 Mark

In the given figure, point $P$ is $26cm$ away from the centre $O$ of a circle and the length $PT$ of the tangent drawn from $P$ to the circle is $24cm$. Then the radius ofthe circle is:

✓
$10cm$
B
$12cm$
C
$13cm$
D
$15cm$

Answer

Correct option: A.

$10cm$

Construction: Join $OT$.
$PT = 24cm$
$OP = 26cm$
Sunce $PT$ is a tangent to the circle at $T$.
$\angle\text{PTO}=90^\circ$ .....(tangent is perpendicular to the radius of a circle)
In $\triangle\text{PTO},$
By Pythagoras theorem,
$ \mathrm{OP}^2=\mathrm{PT}^2+\mathrm{OT}^2 $
$ \Rightarrow \mathrm{OT}^2=\mathrm{OP}^2-\mathrm{PT}^2 $
$ \Rightarrow \mathrm{OT}^2=26^2-24^2 $
$ \Rightarrow \mathrm{OT}^2=676-576 $
$ \Rightarrow \mathrm{OT}^2=100 $
$ \Rightarrow \mathrm{OT}^2=10 \mathrm{~cm}$

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MCQ 191 Mark

In the given figure, $AD$ and $AE$ are the tangents to a circle with centre $O$ and $BC$ touches the circle at F. If $AE = 5cm$, then perimeter of $\triangle\text{ABC}$ is:

A
$15cm$
✓
$10cm$
C
$22.5cm$
D
$20cm$

Answer

Correct option: B.

$10cm$

We know that tangent from an external point to the circle equal.
So,
$AE = AD = 5cm$
$BF = BE$
$CF = CD$
Perimeter of $\triangle\text{ABC}$
$= AB + BC + AC$
$= AB + (BE + DC) + AC$
$= AB + (BE + DC) + AC$
$= (AB + BE) + (AC + DC)$
$= AE + AD$
$= 5 + 5$
$= 10cm$

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MCQ 201 Mark

Which of the following statements is not true?

A
A line which intersects a circle at two points, is called a secant of the circle.
B
A line intersecting a circle at one point only, is called a tangent to the circle.
C
The point at which a line touches the circle, is called the point of contact.
✓
A tangent to the circle can be drawn from a point inside the circle.

Answer

Correct option: D.

A tangent to the circle can be drawn from a point inside the circle.

Options (a), (b) and (c) are all true.
However, Option (d) is false since it is not possible to draw a tangent from a point inside a circle.

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MCQ 211 Mark

In the given figure, $AB$ and $AC$ are tangents to a circle with centre $O$ and radius $8cm$. If $OA = 17cm$, then the length of $AC$ (in cm) is:

A
$9$
✓
$15$
C
$\sqrt{353}$
D
$25$

Answer

Correct option: B.

$15$

Construction: Join $OC$.
Since AC is a tangent to the circle.
$\angle\text{OCA}=90^\circ$ ....(tangent is perpendicular to the radius of a circle)
In $\triangle\text{OCA},$
By Pythagoras theorem,
$ {OA}^2=O C^2+\mathrm{AC}^2 $
$ \Rightarrow 17^2=8^2+A C^2 $
$ \Rightarrow A C^2=289-64 $
$ \Rightarrow A C^2=225 $
$ \Rightarrow A C=15 \mathrm{~cm}$

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MCQ 221 Mark

In a circle of radius $7cm$, tangent $PT$ is drawn from a point $P$, such that $PT = 24cm$. If $O$ is the centre of the circle, then $OP =?$

A
$30cm$
B
$28cm$
✓
$25cm$
D
$18cm$

Answer

Correct option: C.

$25cm$

$PT = 24cm, OT = 7cm.$
Since $PT$ is a tangent to the circle at $T$.
$\angle\text{PTO}=90^\circ$ ....(tangent is perpendicular to the radius of a circle)
In $\triangle\text{PTO},$
By using Pythagoras theorem,
$ \mathrm{OP}^2= {PT}^2+ {OT}^2 $
$ \Rightarrow {OP}^2=24^2+7^2 $
$ \Rightarrow O P^2=576+49 $
$ \Rightarrow O P^2=625 $
$ \Rightarrow {OP}=25 {~cm}$

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MCQ 231 Mark

In the given figure, two circles touch each other at $C$ and $AB$ is a tangent to both the circles. The measure of $\angle\text{ACB}$ is:

A
$45^\circ$
B
$60^\circ$
✓
$90^\circ$
D
$120^\circ$

Answer

Correct option: C.

$90^\circ$

Draw a tangent to the circle at point $C$ meet $AB$ at $P$.
Then,
$PA = PC$
$\Rightarrow $ $\angle\text{PAC}=\angle\text{PCA}$
And $PB = PC$
$\Rightarrow\angle\text{PBC}=\angle\text{PCB}$
$\therefore\angle\text{PAC}+\angle\text{PBC}=\angle\text{PCA}+\angle\text{PCB}=\angle\text{ACB}$
$\Rightarrow\angle\text{PAC}+\angle\text{PBC}+\angle\text{ACB}=2\angle\text{ACB}$
$\Rightarrow180^\circ=2\angle\text{ACB}$
$\Rightarrow\angle\text{ACB}=90^\circ$

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MCQ 241 Mark

In the given figure, $O$ is the centre of the circle, $PQ$ is a chord and $PT$ is the tangent at $P$. If $ \angle\text{POQ} = 70^\circ$then $\angle\text{TPQ} $ is equal to:

✓
$35^\circ$
B
$45^\circ$
C
$55^\circ$
D
$70^\circ$

Answer

Correct option: A.

$35^\circ$

In $\triangle\text{OPQ},$
$\text{OP}=\text{OQ}$ ....(radii of the same circle)
$\Rightarrow\angle\text{OQP}=\angle\text{OPQ}$ ....(angles opposite equal sides are equal)
$\triangle\text{OPQ},$
$\angle\text{OQP}+\angle\text{OPQ}+\angle\text{POQ}=180^\circ$ ......(Angle Sum Property)
$\Rightarrow\angle\text{OPQ}+\angle\text{OPQ}+70^\circ=180^\circ$
$\Rightarrow2\angle\text{OPQ}=110^\circ$
$\Rightarrow\angle\text{OPQ}=55^\circ$
Since PT is a tangent to the inner circle.
$\Rightarrow\angle\text{OPT}=90^\circ$ .....(Tangent is perpendicular to the radius of a circle)
$\Rightarrow\angle\text{OPQ}+\angle\text{TPQ}=90^\circ$
$\Rightarrow55^\circ+\angle\text{TPQ}=90^\circ$
$\Rightarrow\angle\text{TPQ}=35^\circ$

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MCQ 251 Mark

The number of tangents that can be drawn from an external point to a circle is:

A
$1$
✓
$2$
C
$3$
D
$4$

Answer

Correct option: B.

$2$

We can draw only two tangents from an external point to a circle.

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MCQ 261 Mark

In the given figure, if $\angle\text{AOD}=135^\circ$ then $\angle\text{BOC}$ is equal to:

A
25º
✓
45º
C
52.5º
D
62.5º

Answer

Correct option: B.

45º

We know that sum of the angles subtended by opposite sides of a quadrilateral having a circumscribed circle is 180º.
$\Rightarrow\angle\text{AOD}+\angle\text{BOC}=180^\circ$
$\Rightarrow135^\circ+\angle\text{BOC}=180^\circ$
$\Rightarrow\angle\text{BOC}=45^\circ$

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MCQ 271 Mark

In the given figure, the length of $BC$ is:

A
$7cm$
✓
$10cm$
C
$14cm$
D
$15cm$

Answer

Correct option: B.

$10cm$

We know that tangent from an external point to a circle are equal.
So,
$AF = AE = 4cm$
$\Rightarrow EC = AC - AE = 11 - 4 = 7cm$
Now,
$CD = CE = 7cm$
and $BF = BD = 3cm$
$BD = BD + CD$
$\Rightarrow BD = 3 + 7$
$\Rightarrow BD = 10cm$

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MCQ 281 Mark

In the given figure, $PQ$ is a tangent to a circle with centre $O$. $A$ is the point of contact. If $\angle\text{PAB} = 67^\circ$then the measure of $\angle\text{AQB} $ is:

A
$73^\circ$
B
$64^\circ$
C
$53^\circ$
✓
$44^\circ$

Answer

Correct option: D.

$44^\circ$

Since $\angle\text{BAC}$ is inscribed in a semicircle, $\angle\text{BAC}=90^\circ.$
Since PAQ is a straight line,
$\angle\text{PAB}+\angle\text{BAC}+\angle\text{CAQ}=180º$
$\Rightarrow67^\circ+90^\circ+\angle\text{CAQ}=180^\circ$
$\Rightarrow\angle\text{CAQ}=23^\circ$
We know that angles that subtend the same arc are equal.
$\Rightarrow\angle\text{CBA}=\angle\text{CAQ}=23^\circ$
In $\triangle\text{BAQ},$
$\angle\text{BAQ}+\angle\text{QBA}+\angle\text{AQB}=180º$....(Angle Sum Property)
$\Rightarrow(90^\circ+23^\circ)+23^\circ+\angle\text{AQB}=180^\circ$
$\Rightarrow\angle\text{AQB}=44^\circ$

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MCQ 291 Mark

Which of the following statment is not true?

A
If a point P lies inside a circle, no tangent can be drawn to the circle passing through P.
B
If a point P lies on a circle, then one and only one tangent can be drawn to the circle at P.
C
If a point P lies outside a circle, then only two tangents can be drawn to the circle from P.
✓
A circle can have more than two parallel tangents parallel to a given line.

Answer

Correct option: D.

A circle can have more than two parallel tangents parallel to a given line.

Options (a), (b) and (c) are all true.
However, Option (d) is false since we can draw only parallel tamngent on either side of the diameter, which would be parallel to a given line.

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MCQ 301 Mark

To draw a pair of tangents to a circle, which are inclined to each other at angle of $45^\circ $, we have to draw the tangents at the end points of those two radii, the angle between which is:

A
$105^\circ$
✓
$135^\circ$
C
$140^\circ$
D
$145^\circ$

Answer

Correct option: B.

$135^\circ$

Since $AB$ and $AC$ are the tangent to the circle.
$\angle\text{OBA}=\angle\text{OCA}=90^\circ$ (tangent is perpendicular to the radius of a circle)
In ACOB,
$\angle\text{OBA}+\angle\text{BAC}+\angle\text{OCA}+\angle\text{BOC}=360^\circ$
$\Rightarrow90^\circ+45^\circ+45^\circ+\angle\text{BOC}=360^\circ$
$\Rightarrow\angle\text{BOC}=135^\circ$

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MCQ 311 Mark

If two tangents inclined at an angle of $60^\circ $ are drawn to a circle of radius 3cm, then the length of each tangent is:

A
$3\text{cm}$
B
$\frac{3\sqrt{3}}{2}\text{cm}$
✓
$3\sqrt{3}\text{cm}$
D
$6\text{cm}$

Answer

Correct option: C.

$3\sqrt{3}\text{cm}$

In $\triangle\text{BAO}$ and $\triangle\text{CAO},$
$\angle\text{OBA}=\angle\text{OCA}=90^\circ$ ....(Since $AB$ and $AC$ are tangent to the circle)
$\text{OA}=\text{OA}$ ....(common side)
$\text{OB}=\text{OC}$ ....(radii of the same circle)
$\Rightarrow\triangle\text{BAO}\cong\triangle\text{CAO}$ ....($RHS$ congruence criterion)
$\angle\text{OAB}=\angle\text{OAC}$ ....(cpct)
$\Rightarrow\angle\text{OAB}=\frac{1}{2}\angle\text{BAC}=30^\circ$
So, $\triangle\text{BAO}$ is a $30-60-90$ triangle.
side opposite $30^\circ=\frac{1}{2}$ hypotenuse
$\Rightarrow\text{OB}=\frac{1}{2}$ hypotenuse
$\Rightarrow hypotene = 2OB = 2(3) = 6cm$
side opposite $60^\circ=\frac{\sqrt{3}}{2}$ hypotenuse
$\Rightarrow\text{AB}=\frac{\sqrt{3}}{2}(6)=3\sqrt{3}\text{cm}$
$\text{AB}=\text{AC}=3\sqrt{3}\text{cm}$ ....(Since tangents from an external point to the circle are equal)
Hence, the length of each tangent is $3\sqrt{3}\text{cm}.$

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MCQ 321 Mark

In the given figure, $PQ$ and $PR$ are tangents to a circle with centre $A$. If $\angle\text{QPA}=27^\circ$ then $\angle\text{QAR}$ equals:

A
$63^\circ$
B
$117^\circ$
✓
$126^\circ$
D
$153^\circ$

Answer

Correct option: C.

$126^\circ$

In $\triangle\text{PAQ}$ and $\triangle\text{PAR},$
$\angle\text{PQA}=\angle\text{PRA}$ ....(Since $PQ$ and $PR$ are tangent to the circle)
$\text{AP}=\text{AP}$ ....(common side)
$\text{AQ}=\text{AR}$ ....(radii of the same circle)
$\Rightarrow\triangle\text{PAQ}\cong\triangle\text{PAR}$ ....($RHS$ congruence criterion)
$\angle\text{OAP}=\angle\text{RAP}$ ....(cpct)
In $\triangle\text{PAQ},$
$\angle\text{QAP}+\angle\text{PQA}+\angle\text{APQ}=180^\circ$ ....(Angle Sum property)
$\Rightarrow\angle\text{QAP}+90^\circ+27^\circ=180^\circ$.....$(\angle\text{PQA}=90^\circ,$ since radius is perpendicular to the tangent $)$
$\Rightarrow\angle\text{QAP}=63^\circ$
So, $\angle\text{QAR}=\angle\text{QAP}+\angle\text{RAP}$
$\Rightarrow\angle\text{QAR}=63^\circ+63^\circ$
$\Rightarrow\angle\text{QAR}=126^\circ$

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MCQ 331 Mark

In the given figure,$ \triangle\text{ABC}$ is right-angled at $B$, such that $BC = 6cm$ and $AB = 8cm$. $A$ circle with centre $O$ has been inscribed in the triangle. $\text{OP} \perp \text{AB}, \text{OQ} \perp \text{BC} $ and $\text{OR} \perp \text{AC}.$ If $OP = OQ = OR = x cm$, then $x =?$

✓
$2cm$
B
$2.5cm$
C
$3cm$
D
$3.5cm$

Answer

Correct option: A.

$2cm$

In right $\triangle\text{ABC},$
$\text{AC}^2=\text{AB}^2+\text{BC}^2$ ....(By Pythagoras theorem)
$\Rightarrow\text{AC}^2=8^2+6^2$
$\Rightarrow\text{AC}^2=64+36$
$\Rightarrow\text{AC}=100$
$\Rightarrow\text{AC}=10\text{cm}$
We know that tangent from an external poit to the circle are equal.
$\Rightarrow\text{CR}=\text{CQ}=\text{BC}-\text{BQ}=(6-\text{x})\text{cm}$
$\Rightarrow\text{AR}=\text{AP}=\text{AB}-\text{BP}=(8-\text{x})\text{cm}$
$\text{AC}=(\text{AR}+\text{CR})=(8-\text{x})+(6-\text{x})=(14-2\text{x})\text{cm}$
$\Rightarrow14-2\text{x}=10$
$\Rightarrow2\text{x}=4$
$\Rightarrow\text{x}=2$

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MCQ 341 Mark

In the given figure, $DE$ and $DF$ are tangents from an external point $D$ to a circle with centre $A$. If $DE = 5cm$. and $\text{DE}\perp\text{DF}$ then the radius of the circle is:

A
$3cm$
B
$4cm$
✓
$5cm$
D
$6cm$

Answer

Correct option: C.

$5cm$

Construction: Join $AE$ and $AF$.
Since DE and DF are tangent to the circle,
$\angle\text{AED}=\angle\text{AFD}=\angle\text{EDF}=90^\circ$
Also, $AE = AF .....$(radii of the same circle)
and $ED = EF ....$(Since tangent drawn from an external point to the circle are equal)
So, quadrilateral $AEDF$ is a square.
Thus, $AE = DF = 5cm$
Hence, the length of the radius is $5cm$.

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MCQ 351 Mark

In the given figure, $O$ is the centre of a circle, $PQ$ is a chord and the tangent $PT$ at $P$ makes an angle of $50^\circ $ with $PQ$. Then,$ \angle\text{POQ} = ?$

A
$130^\circ$
✓
$100^\circ $
C
$90^\circ$
D
$75^\circ$

Answer

Correct option: B.

$100^\circ $

Since $PT$ is the tangent to the circle.
$\angle\text{OPT}=90^\circ$
$\Rightarrow\angle\text{TPQ}+\angle\text{OPQ}=90^\circ$
$\Rightarrow50^\circ+\angle\text{OPQ}=90^\circ$
$\Rightarrow\angle\text{OPQ}=40^\circ$
In $\triangle\text{OPQ},$
$\text{OP}=\text{OQ}$ ....(radii of the same circle)
$\Rightarrow\angle\text{OPQ}=\angle\text{OQP}=40^\circ$....(angles opposite equal sides are equal)
Now,
$\angle\text{OPQ}+\angle\text{OQP}+\angle\text{POQ}=180^\circ$....(Angle Sum Property)
$\Rightarrow40^\circ+40^\circ+\angle\text{POQ}=180^\circ$
$\Rightarrow\angle\text{POQ}=100^\circ$

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MCQ 361 Mark

Which of the following pairs of lines in a circle cannot be parallel?

A
Two chords
B
A chord and a tangent
C
Two tangents
✓
Two diameters

Answer

Correct option: D.

Two diameters

Two diameter of the circle always passes through the centre. This means all the diameters of a given circle will intersect at the centre, and hence they cannot be parallel.

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MCQ 371 Mark

In the given figure, a triangle $PQR$ is drawn to circumscribe a circle of radius 6cm such that the segments $QT$ and $TR$ into which $QR$ is divided by the point of contact $T$ are of lengths $12cm$ and $9cm$ respectively. If the area of $\triangle\text{PQR} = 189 \text{cm}^2 $ then the length of side $PQ$ is:

A
$17.5cm$
B
$20cm$
✓
$22.5cm$
D
$25cm$

Answer

Correct option: C.

$22.5cm$

We know that tangent from an external point to the circle are equal.
So,
$QT = QN = 12cm$
$TR = RM = 9cm$
Now,
$\text{ar}(\triangle\text{PQR})=\frac{1}{2}(\text{Perimeter of }\triangle\text{PQR})\times\text{r}$
$\Rightarrow\text{ar}(\triangle\text{PQR})=\frac{1}{2}(12+12+9+9+\text{x}+\text{x})\times\text{r}$
$\Rightarrow\text{ar}(\triangle\text{PQR})=\frac{1}{2}(42+2\text{x})\times6$
$\Rightarrow189=3(42+2\text{x})$
$\Rightarrow63=42+2\text{x}$
$\Rightarrow2\text{x}=21$
$\Rightarrow\text{x}=10.5$
So, $PQ = 12 + 10.5 = 22.5cm$

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MCQ 381 Mark

If tangents $PA$ and $PB$ from a point $P$ to a circle with centre $O$ are drawn, so that $\angle\text{APB} = 80^\circ, $ then $\angle\text{POA} = ?$

A
$40^\circ$
✓
$50^\circ$
C
$80^\circ$
D
$60^\circ$

Answer

Correct option: B.

$50^\circ$

In $\triangle\text{PAO}$ and $\triangle\text{PBO},$
$\angle\text{PAO}+\angle\text{PBO}=90^\circ$....(Since PQ and PR are tangent to the circle)
$\text{OP}=\text{OP}$ ....(Common side)
$\text{AO}=\text{OB}$ ....(radii of the same circle)
$\Rightarrow\triangle\text{PAO}\cong\triangle\text{PBO}$ ....(RHS congruence criterion)
$\angle\text{POA}=\angle\text{BOP}$ ....(cpct)
In quad. AOBP
$\angle\text{PAO}+\angle\text{PBO}+\angle\text{AOB}+\angle\text{APB}=360^\circ$
$\Rightarrow90^\circ+90^\circ+\angle\text{AOB}+80^\circ=360^\circ$
$\Rightarrow\angle\text{AOB}=100^\circ$
So, $\angle\text{POA}=50^\circ$

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MCQ 391 Mark

In the given figure, $O$ is the centre of a circle and $PT$ is the tangent to the circle. If $PQ$ is a chord such that $\angle\text{QPT}=50^\circ$ then $\angle\text{POQ}=?$

✓
$100^\circ$
B
$90^\circ$
C
$80^\circ$
D
$75^\circ$

Answer

Correct option: A.

$100^\circ$

Since $PT$ is the tangent to the circle,
$\angle\text{TPQ}=90^\circ$
$\Rightarrow\angle\text{TPQ}+\angle\text{OPQ}=90^\circ$
$\Rightarrow50^\circ+\angle\text{OPQ}=90^\circ$
$\Rightarrow\angle\text{OPQ}=40^\circ$
In $\triangle\text{OPQ},$
$\text{OP}=\text{OQ}$ ....(radii of the same circle)
$\Rightarrow\angle\text{OPQ}=\angle\text{OQP}=40^\circ$ ....(angles opposite equal sides are equal)
Now,
$\angle\text{OPQ}+\angle\text{OQP}+\angle\text{POQ}=180^\circ$ ....(Angle Sum Property)
$\Rightarrow40^\circ+40^\circ+\angle\text{POQ}=180^\circ$
$\Rightarrow\angle\text{POQ}=100^\circ$

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MCQ 401 Mark

In the given figure, a circle touches the side $DF$ of $\triangle\text{EDF}$ at H and touches $ED$ and $EF$ produced at $K$ and $M$ respectively. If $EK = 9cm$ then the perimeter of $\triangle\text{EDF}$ is:

A
$9cm$
B
$12cm$
C
$13.5cm$
✓
$18cm$

Answer

Correct option: D.

$18cm$

We know that tangents from an external point to the circle are equal.
So,
$EK = EM = 9cm$
$DK = DH$
$FH = FM$
perimeter of $\triangle\text{EDF}$
$= ED + EF + DF$
$= ED + EF + DH + HF$
$= (ED + DH) + (EF + HF)$
$= (ED + DK) + (EF + FM)$
$= EK + EM$
$= 9 + 9$
$= 18cm.$

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MCQ 411 Mark

In the given figure, $O$ is the centre of two concentric circles of radii $6cm$ and $10cm$. AB is a chord of outer circle which touches the inner circle. The length of $AB$ is:

A
$8cm$
B
$14cm$
✓
$16cm$
D
$\sqrt{136}\text{cm}$

Answer

Correct option: C.

$16cm$

Since $AB$ is a tangent to the circle.
$\angle\text{OPA}=90^\circ$ ....(tangent is perpendicular to the radius of a circle)
AB is a chord of the outer circle
We know that of the perpendicular drawn from the centre to a chord of a circle of a circle, bisects the chord.
In $\triangle\text{OPA},$
$ A O^2=O P^2+A P^2 $
$ \Rightarrow 10^2=6^2+A P^2 $
$ \Rightarrow A P^2=10^2-6^2 $
$ \Rightarrow P A^2=64 $
$ \Rightarrow P A=8 \mathrm{~cm} $
$ A B=2 A P=2(8)=16 \mathrm{~cm}$

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MCQ 421 Mark

In the given figure, $AT$ is a tangent to the circle with centre $O$, such that $OT = 4cm$ and $\angle\text{OTA} = 30^\circ. $Then, $AT =?$

A
$4\text{cm}$
B
$2\text{cm}$
✓
$2\sqrt{3}\text{cm}$
D
$4\sqrt{3}\text{cm}$

Answer

Correct option: C.

$2\sqrt{3}\text{cm}$

Since $\angle\text{OAT}=90^\circ$ and $\angle\text{OTQ}=30^\circ$
Clearly, $\angle\text{AOT}=60^\circ$
So, $\triangle\text{AOT}$ is a $30^\circ - 60^\circ - 90^\circ $ triangle.
Side opposite $60^\circ=\frac{\sqrt{3}}{2}$ hypotenuse
$\Rightarrow\text{AT}=\frac{\sqrt{3}}{2}(\text{OT})$
$\Rightarrow\text{AT}=\frac{\sqrt{3}}{2}(4)$
$\Rightarrow\text{AT}=2\sqrt{3}\text{cm}$

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MCQ 431 Mark

Which of the following statements is not true?

A
A tangent to a circle intersects the circle exactly at one point.
B
The point common to a circle and its tangent is called the point of contact.
C
The tangent at any point of a circle is perpendicular to the radius of the circle through the point of contact.
✓
A straight line can meet a circle at one point only.

Answer

Correct option: D.

A straight line can meet a circle at one point only.

Options (a), (b) and (c) are all true.
However, Option (d) is false since a straight line can meet a circle at two points even as shown below.

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MCQ 441 Mark

Quadrilateral $ABCD$ is circumscribed to a circle. If $AB = 6cm, BC = 7cm$ and $CD = 4cm$, then the length of $AD$ is:

✓
$3cm$
B
$4cm$
C
$6cm$
D
$7cm$

Answer

Correct option: A.

$3cm$

Using the property, tangent from an external point to the circle are equal.
We can say, $AB + CD = AD + BC$
$\Rightarrow AD = AB + CD - BC$
$\Rightarrow AD = 6 + 4 - 7$
$\Rightarrow AD = 3cm$

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MCQ 451 Mark

The chord of a circle of radius $10cm$ subtends a right angle at its centre. The length of the chord (in cm) is:

A
$\frac{5}{\sqrt{2}}\text{cm}$
B
$5\sqrt{2}\text{cm}$
✓
$10\sqrt{2}\text{cm}$
D
$10\sqrt{3}\text{cm}$

Answer

Correct option: C.

$10\sqrt{2}\text{cm}$

In $\triangle\text{POQ},$
By using Pythagoras theorem,
$ P Q^2=P O^2+O Q^2 $
$ \Rightarrow P Q^2=10^2+10^2 $
$ \Rightarrow P Q^2=100+100 $
$ \Rightarrow P Q^2=200 $
$ \Rightarrow P Q^2=200$
$\Rightarrow\text{PQ}=10\sqrt{2}\text{cm}$
So, the length of the chord is $10\sqrt{2}\text{cm}$

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MCQ 461 Mark

In the given figure, $PQR$ is a tangent to the circle at $Q$, whose centre is $O$ and $AB$ is a chord parallel to $PR$, such that$ \angle\text{BQR} = 70^\circ.$ Then, $\angle\text{AQB}=?$

A
$20^\circ$
B
$35^\circ$
✓
$40^\circ$
D
$45^\circ$

Answer

Correct option: C.

$40^\circ$

Since $AB || PQ$
$\angle\text{BQR}=\angle\text{ABQ}=70^\circ$ ....(alternate angles)
and $\angle\text{PQA}=\angle\text{BAQ}=70^\circ$....(alternate angles)
In $\triangle\text{ABQ},$
$\angle\text{ABQ}+\angle\text{BAQ}+\angle\text{AQB}=180^\circ$....(angle Sum Property)
$\Rightarrow70^\circ+70^\circ+\angle\text{AQB}=180^\circ$
$\Rightarrow\angle\text{AQB}=40^\circ$

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MCQ 471 Mark

In the given figure, three circles with centres $A, B, C,$ respectively, touch each other externally. If $AB = 5cm, BC = 7cm$ and $CA= 6cm,$ the radius of the circle with centre $A$ is:

A
$1.5cm$
✓
$2cm$
C
$2.5cm$
D
$3cm$

Answer

Correct option: B.

$2cm$

Let the radii of the circle with centres $A, B$ and $C$ be $x, y,$ and $z$ respectively.
We know that radii of the same circle are equal.
$x + y = 5$
$y + z = 7$
$z + x = 6$
Adding the three equation, we get
$2(x + y + z) = 18$
$\Rightarrow x + y + z = 9$
$\Rightarrow x = 2$
So, the radius of the circle with centre $A$ is $2cm$.

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MCQ 481 Mark

In the given figure, $PA$ and $PB$ are two tangents drawn from an external point $P$ to a circle with centre $C$ and radius $4cm$. If $\text{PA}\perp \text{PB}$ then the length of each tangent.

A
$3cm$
✓
$4cm$
C
$5cm$
D
$6cm$

Answer

Correct option: B.

$4cm$

Construction: Join $CA$ and $CB$.
Since $AP$ and $PB$ are tangent to the circle,
$\angle\text{CAP}=\angle\text{CBA}=90^\circ$
Given that $\angle\text{APB}=90^\circ$
We know that tangents drawn from an external point to the circle are equal.
$\Rightarrow\text{AP}=\text{PB}$
Also, $\text{CA}=\text{CB}$ ....(radii of the same circle)
So, quadrilateral APBC is a square.
Thus,$AP = PB = CA = CB = 4cm$.

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MCQ 491 Mark

If a chord $AB$ subtends an angle of $60^\circ$ at the centre of a circle, then the angle between the tangents to the circle drawn from $A$ and $B$ is:

A
$30^\circ$
B
$60^\circ$
C
$90^\circ$
✓
$120^\circ$

Answer

Correct option: D.

$120^\circ$

Since $AD$ and $DB$ are the tangent to the circle.
$\angle\text{OAD}=\angle\text{OBD}=90^\circ$....(tangent is perpendicular to the radius of a circle)
In $ABOD$
$\angle\text{OAD}+\angle\text{ADB}+\angle\text{OBD}+\angle\text{AOB}=360^\circ$
$\Rightarrow90^\circ+\angle\text{ADB}+90^\circ+60^\circ=360^\circ$
$\Rightarrow\angle\text{ADB}=120^\circ$

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MCQ 501 Mark

In a right triangle $ABC,$ right angled at $B, BC = 12cm$ and $AB = 5cm$. The radius of the circle inscribed in the triangle is:

A
$1cm$
✓
$2cm$
C
$3cm$
D
$4cm$

Answer

Correct option: B.

$2cm$

In right $\triangle\text{ABC},$
$\text{AC}^2=\text{AB}^2+\text{BC}^2$ ....(By Pythagoras theorem)
$\Rightarrow\text{AC}^2=5^2+12^2$
$\Rightarrow\text{AC}^2=169$
$\Rightarrow\text{AC}=13\text{cm}$
We know that,
$\text{ar}(\triangle\text{ABC})=\frac{1}{2}\times(\text{perimeter of }\triangle\text{ABC})\times\text{r}$
$\Rightarrow\frac{1}{2}\times\text{base}\times\text{height}=\frac{1}{2}\times(\text{perimeter of }\triangle\text{ABC})\times\text{r}$
$\Rightarrow\frac{1}{2}\times12\times5=\frac{1}{2}\times(5+12)\times\text{r}$
$\Rightarrow\frac12\times5=30\times\text{r}$
$\Rightarrow\text{r}=2\text{cm}$

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M.C.Q (1 Marks) - Maths STD 10 Questions - Vidyadip