Question

∆ABC is an isosceles triangle with AB = AC = 13 cm. The length of the altitude from A on BC is 5 cm. Find BC.

Answer 1

It is given that $△ABC$ is an isosceles triangle.
Also, AB = AC = 13 cm
Suppose the altitude from A on BC meets BC at D. Therefore, D is the midpoint of BC.
AD = 5 cm
$△ADBand△ADC$are right-angled triangles.
Applying Pythagoras theorem, we have:

$$\begin{gathered} A{B}^2=A{D}^2+B{D}^2 \\ B{D}^2=A{B}^2-A{D}^2={13}^2-5^2 \\ B{D}^2=169-25=144 \\ BD\hspace{0.17em}=\sqrt{144}=12 \end{gathered}$$

Hence,
BC = 2(BD) = 2 $\times 12$ = 24 cm