Question

In a $\Delta$ $ABC,AC=4,BC=2,AB=6$, find the value of $AD$. (Use Apollonius theorem).

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

The correct option is D 5

According to the Apollonius theorem,
$A{B}^2+A{C}^2=2[A{D}^2+{\frac{BC}{2}}^2]$
$6^2+4^2=2[A{D}^2+{\frac{2}{2}}^2]$
$36+16=2[A{D}^2+1]$
$2A{D}^2=52-2$
$2A{D}^2=50$
$A{D}^2=\frac{50}{2}$
$A{D}^2=25$
$AD=5$