Median problem

Geometry Level 3

In $\triangle$ $ABC$ $AD$ is the median to $BC$ .If $AC=1$ and $\angle BAC=30$ . If area of $\triangle BAD=4$ Find AB?

The answer is 32.

3 solutions

Mehul Chaturvedi
Jan 10, 2015

Please upvote if you like it

$\Rightarrow\angle BAC= 30^\circ$

Area of $\triangle ABD =\triangle ADC$

$\therefore$ total area of $\triangle ABC=8$

Consruct $BE \perp AC$ now as $AC=1 \therefore \dfrac{1\times BE}{2}=8$

$\therefore BE=16$ Now $\sin\angle BAC=\dfrac{BE}{AB}$

$\Rightarrow\sin 30^\circ=\dfrac{16}{AB}$

$\Rightarrow\dfrac{1}{2}=\dfrac{16}{AB} \therefore AB=32$

The answer is $\Huge\Rightarrow\color{royalblue}{\boxed{32}}$

Maybe you could also get the answer directly by the sine area formula. It seems like you took a few more steps to get to the answer with the extra height and stuff but that's my opinion. :P

Sudeshna Pontula - 6 years, 5 months ago

Paola Ramírez
Feb 28, 2015

As $BD$ is a median $\triangle BAD=\triangle DCA=4$ thus $ABC=8$

We know that $8=\frac{1}{2}×sin 30°×1×AB$ so $\boxed{AB=32}$

Rifath Rahman
Jan 12, 2015

Median divides a triangle into 2 equal halves,so area of the triangle is 2 * BAD=2 * 4=8,so (1/2) * 1 *AB * AC * sin 30=8 or AB/4=8 or AB=32(ans)

Median problem

The answer is 32.

3 solutions

0 pending reports