# NCERT Solutions For Class 10 Maths Chapter 11 Exercise 11.2

Exercise 11.2

## Question 1

Draw a circle of radius \(6 \,\rm{cm}\). From a point \(10\,\rm{cm}\) away from its centre, construct the pair of tangents to the circle and measure their lengths.

#### ### Solution

**Video Solution**

### Solution

**Video Solution**

**Steps:**

**Steps of construction:**

**(i) **Take a point \(O\) as centre and \(6 \,\rm{cm}\) radius. Draw a circle.

**(ii) **Take a point \(P\) such that \(OP =\) \(10\,\rm{cm}\).

**(iii) **With \(O\) and \(P\) as centres and radius more than half of \(OP\) draw arcs above and below \(OP\) to intersect at \(X\) and \(Y.\)

**(iv) **Join \(XY\) to intersect \(OP\) at \(M\).

**(v) **With \(M\) as centre and \(OM\) as radius draw a circle to intersect the given circle at \(Q\) and \(R\).

**(vi)** Join \(PQ\) and \(PR\).

\(PQ\) and \(PR\) are the required tangents where\(\begin{align}\rm{PQ}=\rm{PR}=8 \rm{cm.}\end{align}\)

**Proof:**

\(\angle {{PQO}} = {90^ \circ } \Rightarrow {\rm{PQ}} \bot {\rm{OQ}}\) (Angle in a semicircle)

\(OQ\) being the radius of the given circle, \(PQ\) in the tangent at \(Q.\)

In right \({{\Delta PQO,}}\)

\({{OP = 10 \,\rm{cm,} OQ = 6}}\,{\rm{cm}}\) (radius)

\[\begin{align} {{P}}{{{Q}}^{{2}}} &={{O}}{{{P}}^{{2}}}{{ - O}}{{{Q}}^{{2}}}\\ &= {{ (10}}{{{)}}^{{2}}}{{ - (6}}{{{)}}^{{2}}}\\ &= { {100 - 36}}\\ &= {{ 64}} \end{align}\]

\[\begin{align} {{PQ }}&=\sqrt {{{64}}} \\ &= {{8}}\,{\rm{cm}} \end{align}\]

Similarly, \(PR =\) \(8 \;\rm{cm}\).

## Question 2

Construct a tangent to a circle of radius \(4 \,\rm{cm}\) from a point on the concentric circle of radius \(6 \,\rm{cm}\) and measure its length. Also verify the measurement by actual calculation.

#### ### Solution

**Video Solution**

### Solution

**Video Solution**

**Steps:**

**Steps of construction:**

**(i)** Take \(‘O’\) as centre and radius \(4 \,\rm{cm}\) and \(6\,\rm{cm}\) draw two circles.

**(ii)** Take a point \(‘P’\) on the bigger circle and join \(OP\).

**(iii)** With \(‘O’\) and \(‘P’\) as centre and radius more than half of \(OP\) draw arcs above and below \(OP\) to intersect at \(X\) and \(Y\).

**(iv)** Join \(XY\) to intersect \(OP\) at \(M\).

**(v)** With \(M\) as centre and *\(OM\)* as radius draw a circle to cut the smaller circle at \(Q\) and \(R.\)

**(vi)** Join \(PQ\) and \(PR\).

\(PQ\) and \(PR\) are the required tangent where \(\begin{align}\rm{PQ}=4.5\end{align}\) (aprox)

**Proof:**

\(\angle {\rm{PQO}} = {90^ \circ }\) (Angle in a semi - circle)

\(\therefore\)\(\begin{align} P Q \perp O Q\end{align}\)

\(OQ\) being the radius of the smaller circle, \(PQ\) is the tangent at \(Q\)*.*

In right \({\rm{\Delta PQO,}}\)

\(OP = 6\,\rm{cm} \) (radius of the bigger circle)

\(OQ = 4 \,\rm{cm}\) (radius of the smaller circle)

\[\begin{align} {PQ}^{2} &=({OP})^{2}-(\mathrm{OQ})^{2} \\ &=(6)^{2}-(4)^{2} \\ &=36-16 \\ &=20 \end{align}\]

\[\begin{align} {{PQ }}&=\sqrt {{\rm{20}}} \\ &={{4.5}}\,\,\,{\rm{(approx)}} \end{align}\]

Similarly, \({\rm{PR = 4}}{\rm{.5 }}\left( {{\rm{approx}}{\rm{.}}} \right)\)

## Question 3

Draw a circle of radius \(3 \,\rm{cm}\). Take two points \(P\) and \(Q\) on one of its extended diameter each at a distance of \(7 \,\rm{cm}\) from its center. Draw tangents to the circle from these two points \(P\) and \(Q\).

#### ### Solution

**Video Solution**

### Solution

**Video Solution**

**Steps:**

**Steps of construction:**** **

**(i)** Draw a circle with \(O\) as centre and radius is \(3 \,\rm{cm}\).

**(ii)** Draw a diameter of it extend both the sides and take points \(P\)*,* \(Q\) on the diameter such that \(\begin{align}\rm{OP}=\rm{OQ}=7 \rm{cm}\end{align}\)

**(iii)** Draw the perpendicular bisectors of \(OP\) and \(OQ\) to intersect \(OP\) and \(OQ\) at \(M\) and \(N\) respectively.

**(iv)** With \(M\) as centre and \(OM\) as radius draw a circle to cut the given circle at \(A\) and \(C\). With \(N\) as centre and \(ON\) as radius draw a circle to cut the given circle at \(B\) and \(D\).

**(v)** Join \(PA\), \(PC\) \(QB\), \(QD\)

\(PA\),\(PC\) and \(QB\),\(QD\) are the required tangents from \(P\) and \(Q\) respectively.

**Proof:**

\(\angle {{PAO}} =\angle {{QBO}}=90^\circ \) (Angle in a semi - circle)

\(\therefore {PA} \perp {AO},\; {QB} \perp {BO}\)

Since \(OA\) and \(OB\) are the radii of the given circle, \(PA\) and QB are its tangents at \(A\) and \(B\) respectively.

In right angle triangle \(PAO\) and \(QBO\)

\(OP = OQ = 7 \,\rm{cm }\) (By construction)

\(OA = OB = 3\,\rm{cm}\) (radius of the given circle)

\[\begin{align} {{P}}{{{A}}^{{2}}}&={{ (OP}}{{{)}}^{{2}}}{{ -(OA}}{{{)}}^{{2}}}\,\,\,\,\,\,\,\\ &= {{ (7}}{{{)}}^{{2}}}{{ - (3}}{{{)}}^{{2}}}\\ &= {{ 49 - 9}}\\ &={{ 40}}\\ {{PA }}&=\sqrt {{{40}}} \\ &={{ 6}}{{.3}}\,\,{\rm{(approx)}} \end{align}\]

And

\[\begin{align} {{Q}}{{{B}}^{{2}}}&={{(OQ}}{{{)}}^{{2}}}{{ - (OB}}{{{)}}^{{2}}}\,\,\,\,\,\,\,\\ &={{(7}}{{{)}}^{{2}}}{{ - (3}}{{{)}}^{{2}}}\\ &={{ 49 - 9}}\\ &={{ 40}}\\ {{QB }}&=\sqrt {{{40}}} \\ &={{ 6}}{{.3}}\,\,{\rm{(approx)}} \end{align}\]

## Question 4

Draw a pair of tangents to a circle of radius \(5 \,\rm{cm}\) which are inclined to each other at an angle of \(60^\circ\).

#### ### Solution

**Video Solution**

### Solution

**Video Solution**

**Steps:**

**Steps of construction:**

**(i)** With \(O\) as centre and \(5 \,\rm{cm}\) as radius draw a circle.

**(ii)** Take a point \(A\) on the circumference of the circle and join \(OA\).

**(iii)** Draw \(AX\) perpendicular to \(OA\).

**(iv)** Construct \(\begin{align}\angle \rm{AOB}=120^{\circ} \end{align}\) where \(B\) lies on the circumference.

**(v)** Draw \(BY\) perpendicular to \(OB\).

**(vi)** Both \(AX\) and \(BY\) intersect at \(P\).

**(vii)** \(PA\) and \(PB\) are the required tangents inclined at \(\begin{align}60^{\circ}\end{align}\) .

**Proof:**

\(\angle {{OAP}} = \angle {{OBP}} = 90^\circ\) (By construction)

\(\angle {{AOB}} = 120^\circ\) (By construction)

In quadrilateral \(OAPB\),

\[\begin{align}\angle APB&\!=\!{{360}^{{}^\circ }}\!-\!\!\left[ \begin{array}& \angle OAP\!+\!\angle OBP+ \\ \angle AOB \\ \end{array}\!\right] \\ & \!=\!{{360}^{^\circ }}\!-\!\![{{90}^{^\circ }}+{{90}^{^\circ }}+{{120}^{{}^\circ }}] \\ &\!=\!{{360}^{^\circ }}\!-\!\!{{300}^{^\circ }} \\ & \!=\!{{60}^{^\circ }} \\ \end{align}\]

Hence \(PA\) and \(PB\) are the required tangents inclined at \(\begin{align}60^{\circ}\end{align}\).

## Question 5

Draw a line segment \(AB\) of length \(8 \,\rm{cm}\). Taking \(A\) as centre, draw a circle of radius \(4 \,\rm{cm}\) and taking \(B\) as center, draw another circle of radius \(3 \,\rm{cm}\). Construct tangent to each circle from the center of the other circle.

#### ### Solution

**Video Solution**

### Solution

**Video Solution**

**Steps:**

**Steps of construction:**

**(i)** Draw \(\begin{align}\rm{AB}=8\, \rm{cm}\end{align}\).With \(A\) and \(B\) as centers \(4 \,\rm{cm}\) and \(3 \,\rm{cm}\) as radius respectively draw two circles.

**(ii)** Draw the perpendicular bisector of \(AB\), intersecting \(AB\) at \(O\).

**(iii)** With \(O\) as center and *\(OA\)* as radius draw a circle which intersects the two circles at \(P\), \(Q\), \(R\) and \(S\)*.*

**(iv)** Join \(BP\), \(BQ\), \(AR\) and \(AS\).

**(v)** \(BP\) and \(BQ\) are the tangents from \(B\) to the circle with center \(A\). \(AR\) and \(AS\) are the tangents from \(A\) to the circle with center \(B\).

** ****Proof:**

**\(\angle {APB}=\angle {AQB}=90^{\circ}\)** (Angle in a semi-circle)

\(\therefore \rm{AP} \perp \rm{PB}\) and \({AQ} \perp {QB}\)

Therefore, \(BP\) and \(BQ\) are the tangents to the circle with center \(A\)*.*

Similarly, \(AR\) and \(AS\) are the tangents to the circle with center \(B\).

## Question 6

Let *\(ABC\)* be a right triangle in which \(AB = 6 \,\rm{cm},\) \(BC = 8 \,\rm{cm}\) and \(\begin{align}\angle \rm{B}=90^{\circ}.\end{align}\) \(BD\) is the perpendicular to \(AC\). The circle through \(B\), \(C\) and \(D\) is drawn. Construct the tangents from \(A\) to this circle.

#### ### Solution

**Video Solution**

### Solution

**Video Solution**

**Steps:**

**Steps of construction:**

**(i)** Draw \(\begin{align}\rm{BC}=8 \rm{cm}\end{align}\). Draw the perpendicular at \(B\) and cut \(\begin{align}\rm{BA}=6 \rm{cm}\end{align}\)on it. Join \(AC\) right \(\begin{align}\triangle {ABC}\end{align}\) is obtained.

**(ii)** Draw \(BD\) perpendicular to \(AC\).

**(iii)** Since \(\begin{align}\angle \rm{BDC}=90^{\circ}\end{align}\) and the circle has to pass through \(B, C\) and \(D.\) \(BC\) must be a diameter of this circle. So, take \(O\) as the midpoint of \(BC\) and with \(O\) as centre and \(OB\) as radius draw a circle which will pass through \(B\), \(C\) and \(D\).

**(iv)** To draw tangents from \(A\) to the circle with center \(O\).

**(a) **Join \(OA\), and draw its perpendicular bisectors to intersect \(OA\) at \(E\).

**(b) **With \(E\) as center and \(EA\) as radius draw a circle which intersects the previous circle at \(B\) and \(F\)*.*

**(c)** Join \(AF\).

\(AB\) and \(AF\) are the required tangents.

**Proof:**

\(\angle {{ABO}} = \angle {{AFO}} = 90^\circ \) ( Angle in a semi \(-\) circle)

\(\therefore\) \({AB} \bot {OB}\) and \({AF} \bot {OF}\)

Hence \(AB\) and \(AF\) are the tangents from \(A\) to the circle with centre \(O\)*.*

## Question 7

Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.

### Solution

**Video Solution**

**Steps:**

**Steps of construction:**

**(i)** Draw any circle using a bangle.

To find its centre

**(a)** Draw any two chords of the circle say \(AB\) and \(CD\)*.*

**(b)** Draw the perpendicular bisectors of \(AB\) and \(CD\) to intersect at \(O\).

Now, \(‘O’\) is the centre of this circle (since the perpendiculars drawn from the centre of a circle to any chord bisect the chord and vice versa).

To draw the tangents from a point \(‘P’\) outside the circle.

**(ii)** Take any point \(P\) outside the circle and draw the perpendicular bisector of \(OP\) which meets at \(OP\) at \(O’\).

**(iii)** With \(O’\) as center and \(OO’\) as radius draw a circle which cuts the given circle at \(Q\) and \(R\).

**(iv)** Join \(PQ\) and \(PR\).

\(PQ\) and \(PR\) are the required tangents.

**Proof:**

**\(\angle {\rm{OQP}} = \angle {\rm{ORP}} = 90^\circ \)** (Angle in a semi \(-\) circle)

\(\therefore \,{OQ} \,\bot\, {QP}\) and \({OR}\, \bot\, {RP}\)

Hence, \(PQ\) and \(PR\) are the tangents to the given circle.

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