# Ex.11.2 Q1 Constructions Solution - NCERT Maths Class 10

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