Question

$ABC$ is ab isosceles triangle with AB=AC=12 and BC=8cm. Find the altitude on BC and hence calculate its area.

Answer 1

Let $AD$ be the altitude of $ABC.$
Given $AB=AC=12cm$
$BC=8cm$
The altitude to the base of an isosceles triangle bisects the base.
So $BD=DC$
$BD=8/2=4cm$
$DC=4cm$
$ADC$ is a right triangle.
$A{B}^2=B{D}^2+A{D}^2$[Pythagoras theorem]
$A{D}^2=A{B}^2-B{D}^2$
$A{D}^2=122-42$
$A{D}^2=144-16$
$A{D}^2=128$
Taking square root on both sides,
$AD=\surd 128=\surd (2\times 64)=8\surd 2cm$
Area of $ABC=1/2\times base\times height$
$=1/2\times 8\times 8\surd 2$
$=4\times 8\surd 2$
$=32\surd 2c{m}^2$