Question

The area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm is $21\sqrt{11}c{m}^2$.

• A. True
• B. False

Answer 1

The correct option is A True

Let the sides of the triangle be $a=18cm,b=10cm$

Perimeter (2s) $=42$ cm

$s=\frac{42}{2}=21cm$

$c=42-(18+10)cm=14cm$

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

$=\sqrt{21(21-18)(21-10)(21-14)}$

$=21\sqrt{11}c{m}^2$