Ex.11.2 Q2 Constructions Solution - NCERT Maths Class 10

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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 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)


\(\angle {\rm{PQO}} = {90^ \circ }\) (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, \({\rm{PR = 4}}{\rm{.5 }}\left( {{\rm{approx}}{\rm{.}}} \right)\)

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