Question

Given a triangle whose vertices are at $(0,0),(4,4)$ and $(10,0)$. A square is drawn in such that its base is on the x-axis and its two corners are on the $2$ sides of the triangle. The area of the square is equal to

• A. $\frac{400}{49}$
• B. $\frac{400}{25}$
• C. $\frac{625}{16}$
• D. $\frac{625}{49}$

Answer 1

The correct option is A $\frac{400}{49}$

Area of PQRS$=ar\left(OBC\right)-ar\left(BPQ\right)-ar\left(OPS\right)-ar\left(QRC\right)$

$\Rightarrow a^2=\frac{1}{2}\left(4\right)\left(10\right)-\frac{1}{2}(4-a)a-\frac{1}{2}\left(a\right)\left(os\right)-\frac{1}{2}\left(a\right)\left(RC\right)$

$\Rightarrow a^2=\frac{1}{2}(40-4a+a^2-a(OS+RC\left)\right)$

$\Rightarrow a^2=\frac{1}{2}(40-4a+a^2-a(10-a\left)\right)$

$\Rightarrow 2a^2=40-4a+a^2-10a+a^2$

$\Rightarrow 14a=40$

$\Rightarrow a=\frac{40}{14}=\frac{20}{7}$

$\Rightarrow$ Area of square$=a^2=\frac{400}{49}\Rightarrow \left(A\right)$.