Question

Four congruent rectangles are placed as shown in the figure. Area of the outer square is $4$ times that of the inner square. What is the ratio of length to breadth of the congruent rectangles?

Answer 1

Let the length of rectangles be ${}^{\prime }{a}^{\prime }$ and breadth be ${}^{\prime }{b}^{\prime }$
Side of outer square $(a+b)$ units
Side of inner square $(a-b)$ units

Now,
Area of outer square = $4$ $\times$ Area of inner square
Area of $ABCD$ $=$ $4\times$Area of $PQRS$
$\Rightarrow (a+b{)}^2$ = $4(a-b{)}^2$
$\Rightarrow (a+b)=2(a-b)$
$\Rightarrow 2a-2b=a+b$
$\Rightarrow 2a-a=b+2b$
$\Rightarrow a=3b$
$\Rightarrow \frac{a}{b}=\frac{3}{1}$

Hence, the required ratio is $3:1$