Question : ABCD is a square. Draw a triangle QBC on side BC considering BC as a base and draw a triangle PAC on AC as its base such that $\Delta$QBC$\sim\Delta$PAC. Then, $\frac{\text{Area of $\Delta$QBC}}{\text{Area of $\Delta$PAC}}$ is equal to:
Option 1: $\frac{1}{2}$
Option 2: $\frac{2}{1}$
Option 3: $\frac{1}{3}$
Option 4: $\frac{2}{3}$
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Correct Answer: $\frac{1}{2}$
Solution : We have, $\Delta$QBC$\sim\Delta$PAC Since ABCD is a square, AB = BC = CD = DA In $\Delta$ABC, $ ⇒AC=\sqrt{(AB)^2+(BC)^2}$ $⇒ AC=\sqrt{(2BC)^2}$ $⇒AC=\sqrt{2}BC$ In similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides. $⇒\frac{\text{Area of $\Delta$QBC}}{\text{Area of $\Delta$PAC}}=(\frac{BC}{AC})^2$ $⇒\frac{\text{Area of $\Delta$QBC}}{\text{Area of $\Delta$PAC}}=(\frac{BC}{\sqrt{2}BC})^2$ $⇒\frac{\text{Area of $\Delta$QBC}}{\text{Area of $\Delta$PAC}}=\frac{1}{2}$ Hence, the correct answer is $\frac{1}{2}$.
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