Question

The longest rod that can be placed flat on the bottom of a box is $45cm$. The box is $8cm$ longer than it is wide. Find the length and breadth of the box.

Answer 1

Say the breadth $=x$ then Length $=x+8$ now

Using Pythagoras theorem, we get

$x^2+(x+8{)}^2={45}^2,$

$x^2+x^2+64+8x=2025,$

$2x^2+8x-1961=0,$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-8\pm \sqrt{8^2+4\times 2\times 1961}}{2\times 2}$

$x=\frac{-8\pm 126}{2\times 2}$

Taking $+$ sign, we get

$x=29.5$

Taking $-$ sign, we get

$x=-33.5(Non-cosiderable)$

Hence, the length of the box is $37.5cm$ and breadth is $29.5cm.$