Question

An isosceles triangle has sides 4a, b and b. Find the length of the perpendicular dropped onto the unequal side from the vertex opposite to this side.

• A. $a\sqrt{b^2-4a^2}$
• B. $\sqrt{b^2-4a^2}$
• C. $\frac{a}{2}\sqrt{b^2-4a^2}$
• D. $\frac{a}{4}\sqrt{b^2-4a^2}$

Answer 1

The correct option is B

$\sqrt{b^2-4a^2}$

Since the perpendicular bisects the unequal side, in the $△$ABC, BD = DC = 2a and BC$\perp$AD, $△$ABD and $△$ADC are right angled triangles. They are also congruent.
So, by Pythagoras theorem, $A{B}^2=B{D}^2+A{D}^2$
$\Rightarrow AD=\sqrt{A{B}^2-B{D}^2}$
$\Rightarrow AD=\sqrt{b^2-(2a{)}^2}$
$\Rightarrow AD=\sqrt{b^2-4a^2}$