NCERT Solutions For Class 10 Maths Chapter 11 Exercise 11.2

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Chapter 11 Ex.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

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

Chapter 11 Ex.11.2 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

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

Chapter 11 Ex.11.2 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

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

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

Chapter 11 Ex.11.2 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

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 \(\begin{align}60^{\circ}\end{align}\).

Chapter 11 Ex.11.2 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

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 \(B\).

Chapter 11 Ex.11.2 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

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 \(O\).

Chapter 11 Ex.11.2 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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