Question

The base $BC$ of an isosceles triangle $ABC$ is $21cm$ and $AB+BC=29cm$. Find the altitude from the apex vertex $A$.

Answer 1

Let $ABC$ be an isosceles triangle with $AB=AC$. We are given $AB+AC=29cm$ and $BC=21cm$. Hence $AB=AC=29/2=14.5cm$. The altitude $AD$ is drawn to the base bisects the base and is perpendicular to the base. Hence
$CD=21/2=10.5cm$. In $△ADC,\angle ADC={90}^o$. Hence
${AC}^2={AD}^2+{CD}^2$ [by pythagoras' theorem]
This gives ${AC}^2={AD}^2+{CD}^2={\left(14.5\right)}^2-{\left(10.5\right)}^2=(14.5+10.5)(14.5-10.5)=25\times 4=100$
We obtain $AD=10cm$.