Question

In an acute trianGle ABC, an altitude AD is drawn. Find the area of triangle ABC if AB=15 in, AC=18 in, and BD= 10 in.

Answer 1

Consider the acute triangle as shown in the figure below:

Use Pythagoras theorem in triangle $ABD$ as,

$AB^2=BD^2+AD^2$

$15^2=10^2+AD^2$

$225=100+AD^2$

Subtract 100 from both sides to isolate $AD$ as,

$225-100=100+AD^2-100$

$125=AD^2$

$AD=\sqrt{125}$

$AD=5\sqrt{5}$

Again, use Pythagoras theorem in triangle $ADC$ as,

$AD^2+DC^2=AC^2$

$(5\sqrt{5})^2+DC^2=18^2$

$125+DC^2=324$

Subtract $125$ from both sides to isolate $DC$ as,

$125+DC^2-125=324-125$

$DC^2=199$

$DC=\sqrt{199}$

So, the length of base of the acute triangle $BC$ is,

$BC=BD+DC$

$BC=10+\sqrt{199}$

Now, the area of he triangle is evaluated as,

Area$(A)$ $=\frac{1}{2}(BC)(AD)$

$=\frac{1}{2}(10+\sqrt{199})(5\sqrt{5})$

$\approx134.76$

Therefore, the area of the acute triangle is Area$(A)$ $=\frac{1}{2}(10+\sqrt{199})(5\sqrt{5})\approx134.76$ $in.^2$