Question

The sum of length, breadth and height of a cuboid is 19 cm and its diagonal is 5$\sqrt{5}$ cm. Its surface area is

(a) 361 $c{m}^2$ (b) 125 $c{m}^2$ (c) 236 $c{m}^2$ (d) 486 $c{m}^2$

Answer 1

Let the length, breadth, height be l, b, h respectively.

Now, l + b + h = 19 cm

Given it's diagonal = $5\sqrt{5}$ cm

Diagonal = $\sqrt{l^2+b^2+h^2}$ =$5\sqrt{5}$
Squaring on both sides we get

125 = $l^2+b^2+h^2$

Also, T. S. A = 2(lb+bh+hl)

Now, Take l + b + h = 19

Squaring on both sides

$(l+b+h{)}^2$ = 19²

$l^2+b^2+h^2$ + 2 ( lb+ bh+hl) = 361

125 + TSA = 361

TSA = $236c{m}^2$

Therefore, T. S. A = $236c{m}^2$

(c) $236c{m}^2$