Question

In $△ABC,\angle A={90}^{\circ },AD\perp BC$ and $\angle B={45}^{\circ }$. If $AB=x$, find $AD$ in terms of $x$.

Answer 1

Given that:

In $△ABC,\angle A={90}^{\circ },AD\perp BC,\angle B={45}^{\circ },AB=x$

To find:

$AD=?$

Solution:

If $\angle B={45}^{\circ }$ then $\angle C={45}^{\circ }$

or, $AB=AC=x$ units

$A\left(ABC\right)=\frac{1}{2}\times AB\times AC$

$=\frac{1}{2}\times x\times x$

$=\frac{x^2}{2}$ sq. units. $.................\left(i\right)$

$BC=\sqrt{x^2+x^2}=\sqrt{2}x$ units.

$A\left(ABC\right)=\frac{1}{2}\times AD\times BC$

$=\frac{1}{2}\times AD\times \sqrt{2}x$ sq. units. $.................\left(ii\right)$

From eqn. (i) and (ii) we get,

$\frac{x^2}{2}=\frac{1}{2}\times AD\times \sqrt{2}x$

or, $AD=\frac{x}{\sqrt{2}}$ units.