Question

If the base of an isosceles triangle is of length $2p$ and the length of the altitude dropped to the base is $q$, then the distance from the mid point of the base to the side of the triangle is

• A. $\frac{pq}{\sqrt{p^2+q^2}}$
• B. $\frac{2pq}{\sqrt{p^2+q^2}}$
• C. $\frac{3pq}{\sqrt{p^2+q^2}}$
• D. $\frac{4pq}{\sqrt{p^2+q^2}}$

Answer 1

The correct option is A $\frac{pq}{\sqrt{p^2+q^2}}$

We know that altitude dropped on the base $BC$ is also a median from $A$ to $BC$ in isosceles triangle
$AB=\sqrt{q^2+p^2}$
So, area of $\Delta ABC=\frac{1}{2}2q\times p=pq$ ---(1)
And also, area of $\Delta =$ area of $\Delta ACE+$ area of $\Delta ABE$
$=\frac{1}{2}\times AB\times l+\frac{1}{2}AC\times l$
$=\frac{2\times AB\times l}{2}$
$=AB\times l$
$=\sqrt{p^2+q^2}\times l$ ---(2)
From 1 & 2,
$l=\frac{pq}{\sqrt{p^2+q^2}}$