question_answer

Points P and Q lie on the sides AB and AC of triangle ABC respectively such that line segment PQ is parallel to side BC. If the ratio of AP : PB is 1 : 4 and area of A APQ is 4 square cm, what is the area of trapezium PQCB?

A)  60 sq. cm.

B)  16 sq. cm.

C)  96 sq. cm.

D)  21 sq.cm.

Correct Answer: C

Solution :

\[PQ\,||\,BC\] \[\therefore \]\[\angle APQ=\angle ABC\] \[\angle AQP=\angle ACB\] By AA-similarity,             \[\Delta ABC\tilde{\ }\Delta APQ\] \[\frac{AP}{PB}=\frac{1}{4}\Rightarrow \frac{PB}{AP}=\frac{4}{1}\] \[\Rightarrow \]\[\frac{AP+PB}{AP}=\frac{4+1}{1}\] \[\Rightarrow \]\[\frac{AB}{AP}=\frac{5}{1}\] \[\therefore \]\[\frac{\text{Area}\,\text{of}\,\Delta APQ}{\text{Area}\,\text{of}\Delta \Alpha \Beta C}\] \[=\frac{A{{P}^{2}}}{A{{B}^{2}}}=\frac{1}{25}\] \[\Rightarrow \]Area of \[\Delta \Alpha \Beta C=4\times 25\] \[=100\,sq.\,cm.\] \[\therefore \]Area of trapezium PQCB \[=100-4=96\,sq.cm.\]