Question

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

• A. $83.67$ cm${}^2$
• B. $46.12$ cm ${}^2$
• C. $77.15$ cm${}^2$
• D. $69.69$ cm${}^2$

Answer 1

The correct option is

D

$69.69$ cm${}^2$

Let $a$, $b$ and $c$ be the sides of triangle.

Perimeter $=a+b+c$=42 \text{ cm}$

$a=18$, $b=10$

As $a+b+c=42$

$\therefore$ $c=42-b-a=42-18-10$

$\therefore c=14$

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

where, $s=\frac{a+b+c}{2}=\frac{18+10+14}{2}=\frac{42}{2}=21\text{cm}$

$\therefore$ Area of triangle $=\sqrt{21(21-18)(21-10)(21-14)}$

$=\sqrt{21\left(3\right)\left(11\right)\left(7\right)}$

$=\sqrt{21(3\times 7)\left(11\right)}$

$=\sqrt{21\times 21\times 11}$

$=21\sqrt{11}$

$=21\times 3.3166$

$=69.64$

$\therefore$ Area of triangle $=69.69{\text{cm}}^2$