Identify given parameters

From the given question, the equal sides of the isosceles triangle are $a = 100\text{ cm}$ and the base is $b = 56\text{ cm}$.

$a = 100\text{ cm}, \quad b = 56\text{ cm}$

Calculate altitude to the base

An altitude dropped from the vertex angle to the base of an isosceles triangle bisects the base into two equal halves of length $\frac{b}{2} = \frac{56}{2} = 28\text{ cm}$. Using the Pythagorean theorem on one of the right-angled triangles formed:

$h = \sqrt{a^2 - \left(\frac{b}{2}\right)^2} = \sqrt{100^2 - 28^2}$

Simplify the square root

Apply the algebraic identity $x^2 - y^2 = (x - y)(x + y)$ to make computation quick and simple:

$h = \sqrt{(100 - 28)(100 + 28)} = \sqrt{72 \times 128} = \sqrt{9216} = 96\text{ cm}$

Calculate the area of the triangle

Use the standard area formula $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$:

$\text{Area} = \frac{1}{2} \times 56 \times 96 = 28 \times 96 = 2688\text{ cm}^2$