Ex.11.2 Q1 Constructions Solution - NCERT Maths Class 10

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Question

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.

 

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

  
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