CAT 1997 — QA Question 36

Let L and B denote the length and the breadth of the rectangle. So the diagonal will be $\sqrt{(L^2+B^2)}$. Hence, from the condition given, (L + B) – $\sqrt{(L^2+B^2)}=\frac{1}{2}L$

$\Rightarrow \sqrt{(L^2+B^2)}=\frac{L}{2}+B$

Squaring both sides, we get

(L²+ B²) = ${\left(\frac{L}{2}+B\right)}^2$

⇒ L² = $\frac{L^2}{4}+LB$

⇒ $\frac{3L^2}{4}=LB$

⇒ $\frac{B}{L}=\frac{3}{4}$

Shortcut:
First write the relation (L + B) – $\sqrt{(L^2+B^2)}=\frac{1}{2}L.$

or $\frac{L}{2}+B=\sqrt{(L^2+B^2)}$

Put the values of options. Option (d) satisfies. So the answer is (d).