Question

Triangle $ABC$ is angled at $A$. $AD$ is drawn perpendicular to $BC$. If $AB=10cm,AC=12cm$, find the area of $△ABC$. Also find the length of $AD$.

Answer 1

$BC=\sqrt{{10}^2+{12}^2}=\sqrt{244}$

$cosB=\frac{10}{\sqrt{244}},sinB=\frac{12}{\sqrt{244}}$

also, $sinB=\frac{x}{10}$

$\Rightarrow \frac{x}{10}=\frac{12}{\sqrt{244}}$

$\Rightarrow x=\frac{120}{\sqrt{244}}=\frac{120}{2\sqrt{61}}=\frac{60}{\sqrt{61}}$

and,

Area of $△ABC=\frac{1}{2}\times AC\times AB$

$=\frac{1}{2}\times 12\times 10$

$=6\times 10$

$=60unit{s}^2$