question_answer

ABCD is a square. Draw a triangle QBC on side BC considering BC as base and draw a triangle PAC on AC as its base such that \[\Delta QBC\sim PAC\]. Then,\[\frac{Area\,of\,\Delta QBC}{Area\,of\ \Delta PAC}\]is equal to

A)  \[\frac{1}{2}\] 

B)  \[\frac{2}{1}\]

C)  \[\frac{1}{3}\]                         

D)  \[\frac{2}{3}\]

Correct Answer: A

Solution :

(a): From \[\Delta ABC\] \[AC=\sqrt{A{{B}^{2}}+B{{C}^{2}}}\] \[=\sqrt{A{{B}^{2}}+B{{C}^{2}}}\] \[~=\sqrt{2}BC\] \[\Delta QBC\parallel \Delta PAC\] \[\therefore \]\[\frac{Area\text{ }of\text{ }\Delta QBC}{Area\text{ }of\text{ }\Delta PAC\text{ }}\text{=}\frac{B{{C}^{2}}}{A{{C}^{2}}}\] \[=\frac{B{{C}^{2}}}{{{\left( \sqrt{2}BC \right)}^{2}}}\] \[=\frac{B{{C}^{2}}}{2B{{C}^{2}}}=\frac{1}{2}\]