In an isosceles triangle, the base and and the sides are equal to 5 and 20cm respectively. Find the angle bisector at the base of the triangle.
I think by finding the angle bisector, it means to find the length of the bisector. Maybe I'm wrong? Somebody help.... Thanks.
My attempt at answering : Isosceles triangle has the same sides, so 20cm each. Calculating the angle bisector is just like calculating the height in this case, so $\sqrt(20^2-2.5^2)$ that will give 19.8.
Given triangle $ABC$ with the side lengths $a,b$ and $c$, the length of the bisector is given by \begin{align} |AA_b|&= \sqrt{bc\Big(1-\frac{a^2}{(b+c)^2}\Big)} . \end{align}
For $b=a=20,\ c=5$ it is therefore
\begin{align} |AA_b|&= \sqrt{ac\Big(1-\frac{a^2}{(a+c)^2}\Big)} =\sqrt{36}=6 . \end{align}