Question

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

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

Answer 1

The correct option is C

AC = 2AB

In $△ADC,A{C}^2=A{D}^2+D{C}^2$

$\Rightarrow A{C}^2=4^2+2^2=20$

$\Rightarrow AC=\sqrt{20}\text{cm}$ --------(1)

In $△ADB,A{B}^2=4^2+8^2=80$

$\Rightarrow AB=\sqrt{80}=2\sqrt{20}\text{cm}$ -------(2)

From (1) and (2), we get $AC=2AB$