Question

In an isosceles $△$ ABC, AB = AC = 25 cm, BC = 14 cm. Calculate the value of AD

Answer 1

Here altitude from $A$ to $BC$ is $AD$, we know in isosceles altitude on non equal sides also median,

so, $BD=CD=\frac{BC}{2}=\frac{14}{2}=7cm$

Now, we applying Pythagoras theorem in $△ABD$,

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

$(25{)}^2=(7{)}^2+A{D}^2$

$A{D}^2=625-49$

$A{D}^2=576$

$AD=24cm$

$\therefore$ Altitude from $A$ to $BC=AD=24cm$