The perimeter of triangle is 50 cm. One side of the triangle is 4 cm longer than the smallest side and the third side is 6 cm less than twice the smallest side.

Question:

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

Solution:

Let ABC be any triangle with perimeter 50 cm.
Let the smallest side of the triangle be x.
Then the other sides be x + 4 and 2x − 6.

Now,
x + x + 4 + 2x − 6 = 50        (∵ perimeter is 50 cm)
⇒ 4x − 2 = 50
⇒ 4x = 50 + 2
⇒ 4x = 52
⇒ x = 13

∴ The sides of the triangle are of length 13 cm, 17 cm and 20 cm.
∴ Semi-perimeter of the triangle is

$s=\frac{13+17+20}{2}=\frac{50}{2}=25 \mathrm{~cm}$

∴ By Heron's formula,

Area of $\Delta A B C=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{25(25-13)(25-17)(25-20)}$

$=\sqrt{25(12)(8)(5)}$

 

$=20 \sqrt{30} \mathrm{~cm}^{2}$

Hence, the area of the triangle is $20 \sqrt{30} \mathrm{~cm}^{2}$.

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