Question

The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3:2.
Find the area of the triangle.

• A. $\sqrt{32}c{m}^2$
• B. $32c{m}^2$
• C. $32\sqrt{4}c{m}^2$
• D. $32\sqrt{2}c{m}^2$

Answer 1

The correct option is D $32\sqrt{2}c{m}^2$

Perimeter of an isosceles triangle = 32 cm
Ratio = 3 : 2
From the given ratio we get, a = 3x,
b = 3x and c = 2x
Now we can write,
3x + 3x + 2x = 32
Therefore, 8x = 32
$\Rightarrow$x = 4
Sides of triangle are a = 12 cm, b = 12 cm, c = 8 cm
Semi perimeter (s) = $\frac{12+12+8}{2}=16$
s = 16 cm
Area of the triangle $=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{16(16-12)(16-12)(16-8)}$
$=\sqrt{16\times 4\times 4\times 8}$
$=\sqrt{8\times 2\times 4\times 4\times 8}$
$=32\sqrt{2}c{m}^2$