Question

In a $\Delta$ $ABC,AC=8,BC=2,AB=6$, find the value of $AD$. (Use Apollonius theorem).

• A. 3
• B. 5
• C. 7
• D. 9

Answer 1

The correct option is C 7

According to the Apollonius theorem,
$A{B}^2+A{C}^2=2[A{D}^2+{\frac{BC}{2}}^2]$
$6^2+8^2=2[A{D}^2+{\frac{2}{2}}^2]$
$36+64=2[A{D}^2+1]$
$2A{D}^2=100-2$
$2A{D}^2=98$
$A{D}^2=\frac{98}{2}$
$A{D}^2=49$
$AD=7$