Question

The perimeter of a triangle is 50 cm. one side of a triangle is 4 cm longer than the smaller side and the third side is 6 cm less than twice the smaller side. Find the area of the triangle.

• A. $10\sqrt{30}{\text{cm}}^2$
• B. $15\sqrt{30}{\text{cm}}^2$
• C. $20\sqrt{30}{\text{cm}}^2$
• D. $25\sqrt{30}{\text{cm}}^2$

Answer 1

The correct option is C $20\sqrt{30}{\text{cm}}^2$

Let the smaller side of a triangle be x cm.
According to the question,
Second side = 4 cm longer than the smaller side = (x + 4) cm
Third side = 6 cm less than twice the smaller side
= (2x – 6) cm
$\therefore$ Perimeter of a triangle = 50 cm
$\Rightarrow$ x + x + 4 + 2x - 6 = 50
$\Rightarrow$ 4x - 2 = 50
$\Rightarrow$ 4x = 52
$\Rightarrow$ x = $\frac{52}{4}$
$\therefore$ Smaller side, x = 13 cm
Second side, x + 4 = 13 + 4 = 17 cm
And third side, 2x - 6 = 2 $\times$ 13 - 6 = 26 - 6 = 20 cm
Now, semi-perimeter, $s=\frac{a+b+c}{2}=\frac{13+17+20}{2}=\frac{50}{2}=25\text{cm}$
$\therefore \text{Area of triangle}=\sqrt{s(s-a)(s-b)(s-c)}$ [by Heron’s formula]
$=\sqrt{25(25-13)(25-17)(25-20)}$
$=\sqrt{25\times 12\times 8\times 5}$
$=\sqrt{5\times 5\times 2\times 2\times 3\times 2\times 2\times 2\times 5}$
$=5\times 2\times 2\sqrt{30}=20\sqrt{30}$
Hence, the area of triangle is $20\sqrt{30}{\text{cm}}^2$