Question

ABC is an isosceles triangle with AB and AC as the equal sides. Find the length of the altitude drawn from the vertex angle to the base if each of the equal sides is 'a' units and the base is 'b' units.

Answer 1

It is given that ABC is an isosceles triangle.
AB = AC = 'a' units and base BC = '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 the altitude drawn from the vertex angle to the base is
$\sqrt{a^2-\frac{b^2}{4}}units.$