Question

In a triangle $ABC,\angle A=\frac{\pi }{3},b=40,c=30,AD$ is the median through $A,$ then $4{\left(AD\right)}^2$ must be:

• A. $3007$
• B. $3070$
• C. $3700$
• D. $7003$

Answer 1

The correct option is C $3700$

$\cos A=\frac{b^2+c^2-a^2}{2bc}$
Given $A=\frac{\pi }{3}$
$\Rightarrow \cos \frac{\pi }{3}=\frac{1}{2}=\frac{b^2+c^2-a^2}{2bc}$
$\Rightarrow bc=b^2+c^2-a^2$ ..............$\left(1\right)$
In $△ABD,$
${AD}^2=c^2+\frac{a^2}{4}-2c\times \frac{a}{2}\cos B$
$\Rightarrow 4{AD}^2=4c^2+a^2-4ca\left(\frac{c^2+a^2-b^2}{2ca}\right)$
$\Rightarrow 4{AD}^2=4c^2+a^2-2(c^2+a^2-b^2)$
$\Rightarrow 4{AD}^2=4c^2+a^2-2c^2-2a^2+2b^2$
$\Rightarrow 4{AD}^2=2c^2-a^2+2b^2$
$\Rightarrow 4{AD}^2=c^2+b^2+(c^2+b^2-a^2)$
$\Rightarrow 4{AD}^2=c^2+b^2+bc$ from $\left(1\right)$
Substituting for $b=40,c=30$ we get
$\Rightarrow 4{AD}^2={30}^2+{40}^2+40\times 30$
$=900+1600+1200=3700$