question_answer

PQR is a right angles triangle Q being the right angle. Mid-points of QR and PR are respectively Q' and P? Area of \[\Delta \mathbf{P}'\mathbf{Q}'\mathbf{R}'\] is

A)  \[\frac{1}{2}\times area\text{ }of\text{ }\Delta PQR\] 

B)  \[\frac{2}{3}\times area\text{ }of\text{ }\Delta PQR\]

C)  \[\frac{1}{4}\times area\text{ }of\text{ }\Delta PQR\] 

D)  \[\frac{1}{8}\times area\text{ }of\text{ }\Delta PQR\]

Correct Answer: C

Solution :

(c): In \[\Delta PQR\] and \[\Delta {{P}_{1}}{{Q}_{1}}R\] \[{{P}_{1}}{{Q}_{1}}\parallel PQ\] \[\angle {{Q}_{1}}=\angle Q,\angle {{P}_{1}}=\angle P\] \[\therefore \Delta PQR\parallel \Delta {{P}_{1}}{{Q}_{1}}R\] \[\Delta {{P}_{1}}{{Q}_{1}}=\frac{1}{2}PQ.\] Area of \[{{P}_{1}}{{Q}_{1}}R=\frac{1}{2}\times {{Q}_{1}}R\times {{P}_{1}}{{Q}_{1}}\] \[=\frac{1}{2}\times \frac{1}{2}QR\times \frac{1}{2}pQ\] \[=\frac{1}{4}\left( \frac{1}{2}\times QR\times PR \right)\] \[=\frac{1}{4}\times (Area\,of\Delta PQR)\]