Isosceles Triangle Area: Height 20% Greater than Base with DC=10

Triangle ABC is an isosceles triangle AB=AC

AD is the height of the BC

Given DC=10

The length of the height AD is greater by 20% than the length of the side BC.

Calculate the area of the triangle ABC

Given that it is an isosceles triangle if $DC=10$ then $BC=20$.

We are told that the height $AD$ is greater by$20$ percent

than the length of the side$BC$.

That is:

$AD=1.2\cdot BC$

$\frac{100}{100}+\frac{20}{100}=\frac{120}{100}=1.2$

$AD=1.2\cdot20=24$

From here we can calculate the area of the triangle $ΔABC$:

$AΔ\text{ABC}=\frac{AD\cdot BC}{2}=\frac{24\cdot20}{2}=\frac{480}{2}=240$