Find the area, to the nearest cm\(^2\), of the triangle whose sides are in the...

Find the area, to the nearest cm\(^2\), of the triangle whose sides are in the ratio 2 : 3 : 4 and whose perimeter is 180 cm.

Let the length of the sides of triangle be 2x, 3x and 4x.

Perimeter of triangle = 180cm

⇒ \(2x +3x + 4x = 180\)

⇒ \(9x = 180\)

⇒ \(x = \frac{180}{9}\) = 20cm

Then the sides of the triangle are:

\(2x = 2\times20 = 40cm; 3x = 3\times20\) = 60cm and \(4x\) = 4\(\times20 \)= 80cm

Using Heron's formula

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

Where s = \(\frac{a + b + c}{2}\)

Let a = 40cm, b = 60cm, c = 80cm and s = \(\frac{40 + 60 + 80}{2} = \frac{180}{2}\) = 90cm

⇒ A = \(\sqrt90 (90 - 40) (90 - 60) (90 - 80) = \sqrt90 \times 50 \times 30 \times 10 = \sqrt1350000\)

∴ A =1162cm\(^2\) (to the nearest cm\(^2)\)

There is an explanation video available below.

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