Question

In a $△ABC,$ perpendicular $AD$ from $A$ on $BC$ meets $BC$ at $D.$ If $BD=8$ cm, $DC=2$ cm and $AD=4$ cm, then

• A. $△ABC$ is isosceles
• B. $△ABC$ is equilateral
• C. $AC=2AB$
• D. $△ABC$ is right angled at $A$

Answer 1

The correct option is D $△ABC$ is right angled at $A$

In $△ABD$, Using Pythagoras Theorem,
$A{D}^2+B{D}^2=A{B}^2$
$4^2+8^2=A{B}^2$
$AB=\sqrt{80}$
In $△ACD$, Using Pythagoras Theorem,
$A{D}^2+C{D}^2=A{C}^2$
$4^2+2^2=A{C}^2$
$AC=\sqrt{20}$
Now, $A{B}^2+A{C}^2=80+20=100$
and, $B{C}^2=100$
Since, $A{B}^2+A{C}^2=B{C}^2$
Thus, $△ABC$ is a right angled triangle at $A$