Question : ABCD is a square. Draw an equilateral $\triangle $PBC on side BC considering BC is a base and an equilateral $\triangle $QAC on diagonal AC considering AC is a base. Find the value of $\frac{\text{area of $\triangle PBC$}}{\text{area of $\triangle QAC$}}$.
Option 1: $\frac{1}{2}$
Option 2: $1$
Option 3: $\frac{1}{3}$
Option 4: $\frac{1}{4}$
Correct Answer: $\frac{1}{2}$
Solution :
Let the side length of the square ABCD as $a$.
The area of an equilateral triangle with $s$ as side length $=\frac{\sqrt{3}}{4}s^2$
The area of $\triangle $PBC,
$=\frac{\sqrt{3}}{4}a^2$
The area of $\triangle $QAC,
$=\frac{\sqrt{3}}{4}(a\sqrt{2})^2 = \frac{\sqrt{3}}{4} \times 2a^2 = \frac{\sqrt{3}}{2}a^2$
So, $\frac{\text{Area of $\triangle$PBC}}{\text{Area of $\triangle$QAC}}=\frac{\frac{\sqrt{3}}{4}a^2}{\frac{\sqrt{3}}{2}a^2} = \frac{1}{2}$
Hence, the correct answer is $\frac{1}{2}$.
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