Question

Base BC of an isosceles $\Delta ABC$ is $20$cm and equal sides are of $26$cm, then length of altitude through the vertex A is

• A. $23$cm
• B. $24$cm
• C. $25$cm
• D. $26$cm

Answer 1

The correct option is B $24$cm

Given $BC=20AB=AC=26$

The Altitude from A will divide BC into 2 equal parts

$BD=CD=\frac{1}{2}20=10$

Now, in $△ADC$ which is a right Angled Triangle

$A{D}^2=A{C}^2-C{D}^2\Rightarrow A{D}^2={26}^2-{10}^2=676-100=576\Rightarrow AD=\sqrt{576}=24$