12. Each of equal sides of an isosceles triangle is 4 cm greater than its height. If the base of the triangle is 24 cm ; calculate the perimeter and the area of the triangle.

To solve for the height of the isosceles triangle, let's denote the height as \( h \). According to the problem, each of the equal sides is \( h + 4 \) cm. The base of the triangle is given as 24 cm. By dropping a perpendicular from the apex to the base, you create two right triangles. Each half of the base is \( 12 \) cm (half of 24 cm). Using the Pythagorean theorem, we have: \[ (h + 4)^2 = h^2 + 12^2 \] Expanding and simplifying gives: \[ h^2 + 8h + 16 = h^2 + 144 \] This simplifies to: \[ 8h + 16 = 144 \] \[ 8h = 128 \quad \Rightarrow \quad h = 16 \, \text{cm} \] Now, substitute \( h \) back to find the lengths of the equal sides: \[ \text{Equal sides} = h + 4 = 16 + 4 = 20 \, \text{cm} \] Now, we can calculate the perimeter: \[ \text{Perimeter} = \text{base} + 2 \times \text{equal sides} = 24 + 2 \times 20 = 64 \, \text{cm} \] Next, let's calculate the area. The area of a triangle can be calculated using the formula: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 24 \times 16 = 192 \, \text{cm}^2 \] Thus, the perimeter of the isosceles triangle is **64 cm** and the area is **192 cm²**.