Question

In a $△ABC,AB=CA=BC=2a$ and $AD\perp BC$. Prove that
(i) $AD=a\sqrt{3}$
(ii) Area $\left(△ABC\right)=\sqrt{3}{a}^2$

Answer 1

In $△ABC$, $AB=CA=BC=2a$ [ Given ]

$AD\perp BC$ [ Given ]

$△ABD\cong △ACD$

Using Phythagores theorem in $△ABD$ to find $AD.$

$\therefore$ $A{B}^2=B{D}^2+A{D}^2$

$\Rightarrow$ $(2a{)}^2=(a{)}^2+A{D}^2$

$\Rightarrow$ $4a^2=a^2+A{D}^2$

$\Rightarrow$ $A{D}^2=4a^2-a^2$

$\therefore$ $AD=3a^2$

$\therefore$ $AD=\sqrt{3}a$ --- [ Proved ]

Now, $Area\left(△ABC\right)=\frac{1}{2}\times BC\times AD$

$\therefore$ $Area\left(△ABC\right)=\frac{1}{2}\times 2a\times \sqrt{3}a$

$\therefore$ $Area\left(△ABC\right)=\sqrt{3}{a}^2$ --- [ Proved ]