Question

$ABCD$ is a rectangle of dimensions $8$ units and $6$ units. $AEFC$ is a rectangle drawn in such a way that diagonal $AC$ of the first rectangle is one side and side opposite to it is touching the first rectangle at D as shown in the figure. What is the ratio of the area of rectangle ABCD to that of AEFC?

• A. $2:1$
• B. $1:1$
• C. $25:24$
• D. $8:9$

Answer 1

The correct option is C $25:24$

From the above figure, we can say that $AC=10cm$ by Pythagoras theorem.

So, length of rectangle $EACF=10cm$.

Let $DF=x$ cm so, then $DE=(10-x)$ cm.

In $△CDF$,

${CD}^2={CF}^2+{FD}^2$

$⟹64={CF}^2+x^2$

$⟹{CF}^2=64-x^2$

Similarly, in $△ADE$,

${AD}^2={AE}^2+{ED}^2$

$⟹36={AE}^2+(10-x{)}^2$

$⟹{AE}^2=36-(10-x{)}^2$

$⟹{AE}^2=36-(100+x^2-20x)$

$⟹{AE}^2=-64-x^2+20x$

Equating, squares of $AE$ and $CF$,

$-64-x^2+20x=64-x^2$

$⟹20x=128orx=6.4$

$DF=6.4$ cm so, $DE=3.6$ cm.

That gives us $CF=4.8cm=AE$.

The area of ABCD $=8\times 6=48$ sq cm.

The area of $EACF=4.8\times 10=48$ sq cm.

The ratio of the areas $=1$.