If the lengths of the three sides of a trapezium other than the base are $10 \mathrm{~cm}$ each, then the maximum area of the trapezium is:

We have a trapezium where three sides (two legs and the top parallel side) are each $10 \text{ cm}$, and we need to find the maximum possible area.

Since both legs are equal, this is an isosceles trapezium. Drop perpendiculars from the top corners to the base. Let each horizontal extension beyond the top side be $x$ cm, and let $h$ be the height.

Then the base $= 10 + 2x$ and by the Pythagorean theorem on the right triangle formed at either side:

$x^2 + h^2 = 100$

$h = \sqrt{100 - x^2}$

---

The area of a trapezium is:

$A = \dfrac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$

$A = \dfrac{1}{2} \times (10 + 10 + 2x) \times \sqrt{100 - x^2}$

$A = (10 + x)\sqrt{100 - x^2}$

---

To maximize, differentiate with respect to $x$ and set $\dfrac{dA}{dx} = 0$:

$\dfrac{dA}{dx} = \sqrt{100 - x^2} - \dfrac{x(10 + x)}{\sqrt{100 - x^2}} = 0$

Multiplying both sides by $\sqrt{100 - x^2}$:

$100 - x^2 = 10x + x^2$

$2x^2 + 10x - 100 = 0$

$x^2 + 5x - 50 = 0$

---

Using the quadratic formula:

$x = \dfrac{-5 \pm \sqrt{25 + 200}}{2} = \dfrac{-5 \pm 15}{2}$

Since $x$ must be positive, $x = 5$.

---

Substituting $x = 5$:

$h = \sqrt{100 - 25} = 5\sqrt{3} \text{ cm}$

$\text{Base} = 10 + 2(5) = 20 \text{ cm}$

$A = \dfrac{1}{2} \times (10 + 20) \times 5\sqrt{3}$

$A = \dfrac{1}{2} \times 30 \times 5\sqrt{3}$

$A = 75\sqrt{3} \text{ cm}^2$