We are given an isosceles triangle with the following details:
We need to find the area of the triangle.
In an isosceles triangle, the perimeter is the sum of the lengths of all three sides. Let the base of the triangle be \( b \) cm. The perimeter is given by: \[ \text{Perimeter} = 2 \times \text{Equal sides} + \text{Base} \] Substituting the values: \[ 30 = 2 \times 12 + b \] Simplifying: \[ 30 = 24 + b \] Solving for \( b \): \[ b = 30 - 24 = 6 \, \text{cm} \] So, the base of the triangle is \( b = 6 \, \text{cm} \).
To find the area of the triangle, we use Heron's formula: \[ A = \sqrt{s(s - a)(s - b)(s - c)} \] where: - \( a \) and \( c \) are the equal sides, \( a = c = 12 \, \text{cm} \), - \( b \) is the base, \( b = 6 \, \text{cm} \), - \( s \) is the semi-perimeter, given by: \[ s = \frac{\text{Perimeter}}{2} = \frac{30}{2} = 15 \, \text{cm} \] Substituting into Heron's formula: \[ A = \sqrt{15(15 - 12)(15 - 6)(15 - 12)} \] Simplifying: \[ A = \sqrt{15 \times 3 \times 9 \times 3} \] \[ A = \sqrt{15 \times 81} = \sqrt{1215} \] \[ A = 34.86 \, \text{cm}^2 \]
The area of the isosceles triangle is \( \boxed{34.86 \, \text{cm}^2} \).