In a triangle $$ABC, \angle{A}=\dfrac{\pi}{3},b=40,c=30,AD$$ is the median through $$A,$$ then $$4{\left(AD\right)}^{2}$$ must be:
  • A
    $$3007$$
  • B
    $$3070$$
  • C
    $$3700$$
  • D
    $$7003$$

Solution


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