Question

In fig . 10.26 , AB = BC =CA =2a and segment AD $\perp$ side BC . Show that
$i)AD=a\sqrt{3}$
ii)area of $\Delta ABC$ = $a^2\sqrt{3}$

Answer 1

Using Pythagorus theorem,

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

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

$A{D}^2=4a^2-a^2=3a^2$

$\therefore AD=a\sqrt{3}$

Area of $△ABC=\frac{BC\times AD}{2}$

$=\frac{2a\times a\sqrt{3}}{2}$

$=a^2\sqrt{3}$