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 \[=100-4=96\,sq.cm.\]You need to login to perform this action.
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