Question

In $\Delta ABC$ given below, AB = 10 cm, AC = 8 cm, and $\angle A={45}^{\circ }$. Find the area of $\Delta ABC$.

Answer 1

Since $\angle A={45}^{\circ }$ and $\angle D={90}^{\circ }$,

$\angle C={90}^{\circ }-{45}^{\circ }={45}^{\circ }$

In $△ADC$,

$cos{45}^{\circ }=\frac{CD}{AC}$

$\Rightarrow \frac{1}{\sqrt{2}}=\frac{CD}{8}$

$\Rightarrow CD=\frac{8}{\sqrt{2}}=4\sqrt{2}cm$

Now, area of triangle $ABC$

$=\frac{1}{2}\times AB\times CD$

$=\frac{1}{2}\times 10\times 4\sqrt{2}$

$=20\sqrt{2}c{m}^2$