# Ex.11.2 Q2 Constructions Solution - NCERT Maths Class 10

## Question

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.

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