Question

Area of an Isosceles Triangle Show that the area $A$of an isosceles triangle whose equal sides are of length $s$and $\theta$is the angle between them is

$A=\frac{1}{2}{s}^2\sin \theta$

[Hint: See the illustration. The height $h$bisects the angle $\theta$and is the perpendicular bisector of the base.]

Short answer

The area is$A=\frac{1}{2}{s}^2\sin \theta$

Step-by-step solution

Given information

Given length of isosceles triangle as$s$and$\theta$is angle between them

Use the formula for area of a triangle and calculate

The area of a triangle is $A=\frac{1}{2}bh$so we can write as follows;

$h=s\cos \left(\frac{\theta }{2}\right)$

Base will be$$\begin{gathered} \frac{b}{2}=\left(\frac{\theta }{2}\right) \\ b=2s\sin \left(\frac{\theta }{2}\right) \end{gathered}$$

Substitute into formula for area of a triangle

Substituting, we get

$$\begin{gathered} A=\frac{1}{2}bh \\ A=\frac{1}{2}·2s\sin \left(\frac{\theta }{2}\right)·S\cos \left(\frac{\theta }{2}\right) \\ A=\frac{1}{2}·2s\sin \left(\frac{\theta }{2}\right)·S\cos \left(\frac{\theta }{2}\right) \\ A=s^2\sin \left(\frac{\theta }{2}\right)\cos \left(\frac{\theta }{2}\right) \\ A=s^2·\frac{1}{2}\sin \theta \\ A=\frac{1}{2}{s}^2\sin \theta \end{gathered}$$