Question

Find the area of a triangle, two of its sides are $8cm$ and $11cm$ and the perimeter is $32cm$.

• A. $8\sqrt{30}c{m}^2$
• B. $15\sqrt{2}c{m}^2$
• C. $4\sqrt{30}c{m}^2$
• D. $5\sqrt{31}c{m}^2$

Answer 1

The correct option is A $8\sqrt{30}c{m}^2$

Since the perimeter of the triangle with sides $a$, $b$ and $c$ is $a+b+c$, therefore, perimeter of triangle with sides $8$ cm, $11$ cm and $c$ cm with perimeter $32$ cm is:

$32=8+11+c$

$c=32-19=13$

We use heron's formula to find the area of the triangle.

Let us find the semi perimeter:

$S=\frac{8+11+13}{2}=\frac{32}{2}=16$ cm

Now area of the triangle is:

$A=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{16(16-8)(16-11)(16-13)}=\sqrt{16\times 8\times 5\times 3}=\sqrt{1920}=8\sqrt{30}$ cm${}^2$

Hence, the area of the triangle is $8\sqrt{30}$ cm${}^2$.