Find the area to the nearest cm 2 of the triang... - JAMB Mathematics 2023 Question

Let the sides of the triangle be \(2x, 3x, \) and \(4x\), where \(x\) is a positive constant.

The perimeter of the triangle is given as:

\[2x + 3x + 4x = 180\]

Combine like terms:

\[9x = 180\]

Solve for \(x\):

\[x = \frac{180}{9} = 20\]

Now, the sides of the triangle are:

\[2x = 2 \times 20 = 40\]

\[3x = 3 \times 20 = 60\]

\[4x = 4 \times 20 = 80\]

The semi-perimeter (\(s\)) is half of the perimeter:

\[s = \frac{40 + 60 + 80}{2} = \frac{180}{2} = 90\]

Now, use Heron's formula to find the area (\(A\)):

\[A = \sqrt{s(s-a)(s-b)(s-c)}\]

Substitute the values:

\[A = \sqrt{90 \cdot (90-40) \cdot (90-60) \cdot (90-80)}\]

\[A = \sqrt{90 \cdot 50 \cdot 30 \cdot 10} = \sqrt{1350000}\]

Therefore, \(A \approx 1162\) cm\(^2\) (to the nearest cm\(^2\)).

So, the correct answer is 1162 cm\(^2\).