Maximal area

Let $ABC$ be the triangle with $AB = 20 \text{ cm}$ and $AC = 21 \text{ cm}$ .

We know that $\text{Area of triangle} = A = \frac{1}{2} \times AB \times AC \times \sin \angle BAC$ where $0^\circ < \angle BAC < 180^\circ$ .

If we want to maximize the value of $A$ , we have to maximize the value of $\sin \angle BAC$ . That is, $\sin \angle BAC = 1$ , when $\angle BAC =90^\circ$

Therefore

$A_{max} = \frac{1}{2} \times AB \times AC \times \sin \angle BAC$ $A_{max} = \frac{1}{2} \times 20 \text{ cm} \times 21 \text{ cm} \times 1$ $A_{max} = \boxed{210 \text{ cm}^{2}}$

Picture the sides as hinged, and start with them together in a horizontal line (forming a "triangle" with sides of 20, 21 and 1 and an area of 0.) Now leave the side of length 20 horizontal and start opening the hinge. The base of the triangle clearly remains 20, but the height increases as we start opening it out, until the side of length 21 is pointing directly upwards, giving a height of 21. If we open the hinge even more, the triangle becomes obtuse and the height starts decreasing again. So the maximum height is 21 and the maximum area is 1/2 * 20 * 21 = 210.

• Height $= 21$ $cm$

• Base $= 20$ $cm$

Area of triangle $=$ $\dfrac{1}{2}$ $\times$ base $\times$ height

$\Rightarrow 0.5 \times 20 \times 21 = \boxed{210}$ $cm^2$