Question

Find the area of a triangle two whose sides are are 18 cm and 10 cm perimeter is 42 cm.

Answer 1

Perimeter of a triangle = 42 cm

Two sides are 18 cm and 10 cm.

$\therefore$ Third side =42-28= 14 cm

$\therefore s=\frac{Perimeter}{2}=\frac{42}{2}=21\therefore Area=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{21\times (21-10)(21-14)(21-18)}=\sqrt{21\times 11\times 7\times 3}=\sqrt{3\times 7\times 7\times 3\times 11}=3\times 7\times \sqrt{11}=21\sqrt{11}=21\times 3.316=69.65c{m}^2$