Question

ABC is an isosceles triangle with equal sides AB and AC. If each of the equal sides are a units and the base is b units, then the length of altitude drawn from vertex angle to the base is ______ units.

• A. $\sqrt{2}a$
• B. $\sqrt{\frac{3a}{4}}$
• C. $\sqrt{a^2-\frac{b^2}{4}}$
• D. $\sqrt{\frac{2a}{3}}$

Answer 1

The correct option is C

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

It is given that ABC is an isosceles triangle.
AB = AC = a units and base = b units.

Draw $AD\perp BC$.
[Altitudes are perpendiculars drawn from a vertex to its opposite side and they bisect the opposite side in an isosceles triangle]

$\Rightarrow \Delta ADB,\Delta ADC$ are right-angled triangles.

In right angled $\Delta ADB$,

$A{B}^2=A{D}^2+B{D}^2$

$a^2=A{D}^2+(\frac{b}{2}{)}^2$

$\Rightarrow A{D}^2=a^2-\frac{b^2}{4}$

$\Rightarrow AD=\sqrt{a^2-\frac{b^2}{4}}$
$\therefore$ Length of each altitude is
$\sqrt{a^2-\frac{b^2}{4}}units.$