Question

The perimeter of the top of rectangular table is $28$m,whereas its area is $48$sq.m.What is the length of its diagonal?

Answer 1

Let $x$ and $y$ be the length and breadth of the rectangle.

$xy=48,2x+2y=28$

$\Rightarrow x+y=14\dots \left(1\right)$

${(x-y)}^2={(x+y)}^2-4xy$

$={14}^2-4\times 48$

$=196-192=4$

$\therefore x-y=\sqrt{4}=2\dots \left(2\right)$

Now,$x+y=14$ and $x-y=2$

$x+y+x-y=14+2$

$\Rightarrow 2x=16$

$\Rightarrow x=\frac{16}{2}=8$

Put $x=8$ in $x+y=14$

$y=14-x=14-8=6$

$\therefore x=8,y=6$

$\therefore$ Length of diagonal$=\sqrt{x^2+y^2}$

$=\sqrt{8^2+6^2}=\sqrt{64+36}=\sqrt{100}$

$\therefore$ Length of diagonal$=10$m