1. **State the problem:** We have an isosceles triangle with a base length of 24 cm and a perimeter of 54 cm. We need to find the area of this triangle. 2. **Identify known values:** - Base, $b = 24$ cm - Perimeter, $P = 54$ cm - The triangle is isosceles, so the two equal sides have the same length, call each side $s$. 3. **Find the length of the equal sides:** The perimeter is the sum of all sides: $$P = b + 2s$$ Substitute known values: $$54 = 24 + 2s$$ Solve for $s$: $$2s = 54 - 24 = 30$$ $$s = \frac{30}{2} = 15 \text{ cm}$$ 4. **Find the height of the triangle:** Draw a perpendicular from the apex to the base, splitting the base into two equal segments of length $\frac{b}{2} = 12$ cm. Use the Pythagorean theorem in one of the right triangles formed: $$s^2 = h^2 + \left(\frac{b}{2}\right)^2$$ Substitute known values: $$15^2 = h^2 + 12^2$$ $$225 = h^2 + 144$$ Solve for $h^2$: $$h^2 = 225 - 144 = 81$$ $$h = \sqrt{81} = 9 \text{ cm}$$ 5. **Calculate the area of the triangle:** Area formula for a triangle: $$\text{Area} = \frac{1}{2} \times b \times h$$ Substitute values: $$\text{Area} = \frac{1}{2} \times 24 \times 9 = 12 \times 9 = 108 \text{ cm}^2$$ **Final answer:** The area of the triangle is $108$ cm$^2$.