CAT 2001 — QA Question 28

Let the perpendicular on the longest side from the other vertices be h.

$\therefore \frac{1}{2}\times 20\times h$ = 80

∴ h = 8

The perpendicular has two triangles on its two sides. On its left, there is one with a hypotenuse of 10. If the two sides are 10 and 8, the third one must be 6.

∴ The base of the other triangle is 20 − 6 = 14

The two sides being 8 and 14, the hypotenuse must be $\sqrt{{14}^2+8^2}=\sqrt{260}$

Hence, option (a).

Alternatively,

Let the third side of the triangle be x.

$S=\frac{(20+10+x)}{2}=\frac{(30+x)}{2}$

Area of triangle = $\sqrt{s(s-a)(s-b)(s-c)}$

Area of triangle = 80 = $\sqrt{\frac{30+x}{2}\times \frac{x-10}{2}\times \frac{x+10}{2}\times \frac{30-x}{2}}$

$\therefore {80}^2=\frac{({30}^2-x^2)\times (x^2-100)}{4\times 4}$

$(900-x^2)(x^2-100)$ = 6400 × 16 = 102400

Using options, we get, x = $\sqrt{260}$

(900 − 260) × 160 = 102400

Hence, option (a).