Question

In $\Delta$ABC, a perpendicular AD is brawn to BC. If BD= 8 cm, DC = 2 cm and AD= 4 cm, then ___.

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

Answer 1

The correct option is C

AB=2AC

$\text{Using Pythagoras' theorem in}\Delta ADC,A{C}^2=A{D}^2+D{C}^2.A{C}^2=4^2+2^2=20AC=\sqrt{20}cm----\left(1\right)\text{Similarly, in}\Delta ADBA{B}^2=A{D}^2+B{D}^2=4^2+8^2=80.AB=\sqrt{80}cm=2\sqrt{20}cm----\left(2\right)=2\times AC\left(\text{Using (2)}\right)$