Question

In the diagram below, $A$ and $B(20,0)$ lie on the $x$-axis and $C(0,30)$ lies on the $y$-axis such that $\angle ACB={90}^{\circ }.$ A rectangle $DEFG$ is inscribed in $△ABC.$ Given that the area of $△CGF$ is $351$ sq. units. Then $\frac{1}{9}\left(\text{area of rectangle}DEFG\right)$ is
(correct answer + 3, wrong answer 0)

Answer 1

Note that $AO=\frac{{30}^2}{20}=45$
Then the area of $△ABC$ is $\frac{1}{2}\times (20+45)\times 30=975$

Let the height of $△CGF$ be $h.$
Then ${\left(\frac{h}{30}\right)}^2=\frac{351}{975}={\left(\frac{3}{5}\right)}^2$
$\Rightarrow \frac{h}{30-h}=\frac{3}{2}$

Note that the rectangle $DEFG$ has the same base as $△CGF$. Then its area is $351\times \frac{2}{3}\times 2=468$

So, $\frac{1}{9}\times 468=52$