Question

Find the area of the triangle having medians of length 9 cm , 12 cm and 15 cm.

Answer 1

Task. Find the area of the triangle having medians of length 9 cm, 12 cm and 15 cm.

Solution. Let $a,b,c$ be the sides of the triangle, and

$m_{a}=9,\qquad m_{b}=12,\qquad m_{c}=15$

be the medians to the corresponding sides.

We will use the formula for the area of the triangle:

$S=\frac{1}{2}ab\sin\gamma,$

where $\gamma$ is the angle between $a$ and $b$.

First we obtain a formula expressing the side of the triangle via its medians, and then using cosine theorem will find $\cos\gamma$ and $\sin\gamma$.

Let us recall the formula for the median:

$m_{c}^{2}=\frac{1}{2}\,\sqrt{2(a^{2}+b^{2})-c^{2}},$

whence

$4m_{c}^{2}=2(a^{2}+b^{2})-c^{2}.$

Similarly,

$4m_{a}^{2}=2(b^{2}+c^{2})-a^{2},$

$4m_{b}^{2}=2(a^{2}+c^{2})-b^{2}.$

Adding these equations we will get

$4(m_{a}^{2}+m_{b}^{2}+m_{c}^{2})=2(a^{2}+b^{2})-c^{2}+2(b^{2}+c^{2})-a^{2}+2(a^{2}+c^{2})-b^{2},$

$4(m_{a}^{2}+m_{b}^{2}+m_{c}^{2})=3(a^{2}+b^{2}+c^{2}).$

From

$4m_{a}^{2}=2(b^{2}+c^{2})-a^{2}$

we also obtain that

$b^{2}+c^{2}=2m_{a}^{2}+a^{2}/2,$

whence

$4(m_{a}^{2}+m_{b}^{2}+m_{c}^{2})=3(a^{2}+2m_{a}^{2}+a^{2}/2),$

$4(m_{a}^{2}+m_{b}^{2}+m_{c}^{2})=6m_{a}^{2}+9a^{2}/2,$

$9a^{2}/2=4m_{a}^{2}+4m_{b}^{2}+4m_{c}^{2}-6m_{a}^{2},$

$9a^{2}/2=4m_{b}^{2}+4m_{c}^{2}-2m_{a}^{2},$

and so

$a^{2}=\frac{4}{9}(2m_{b}^{2}+2m_{c}^{2}-m_{a}^{2})=\frac{4}{9}(2*12^{2}+2*15^{2}-9^{2})=292.$

Similarly,

$b^{2}=\frac{4}{9}(2m_{a}^{2}+2m_{c}^{2}-m_{b}^{2})=\frac{4}{9}(2*9^{2}+2*15^{2}-12^{2})=208,$

$c^{2}=\frac{4}{9}(2m_{a}^{2}+2m_{b}^{2}-m_{c}^{2})=\frac{4}{9}(2*9^{2}+2*12^{2}-15^{2})=100.$

Let $\gamma$ be the angle between sides $a$ and $c$. Then by cosine theorem

$c^{2}=a^{2}+b^{2}-2ab\cos\gamma,$

whence

$\cos\gamma$ $=\frac{a^{2}+b^{2}-c^{2}}{2ab}=\frac{292+208-100}{2\sqrt{292*208}}=\frac{400}{2\sqrt{292*208}}=$

$=\frac{200}{\sqrt{64*13*73}}=\frac{200}{8\sqrt{13*73}}=\frac{25}{\sqrt{949}}.$

Therefore

$\sin\gamma=\sqrt{1-\cos^{2}\alpha}=\sqrt{1-\frac{25^{2}}{949}}=\sqrt{\frac{949-625}{949}}=\frac{324}{\sqrt{949}}=\frac{18}{\sqrt{949}},$

and so the area of the triangle is equal to

$S=\frac{1}{2}ab\sin\gamma=\frac{\sqrt{292*208}}{2}\cdot\frac{18}{\sqrt{949}}=\frac{8\sqrt{949}}{2}\cdot\frac{18}{\sqrt{949}}=4*18=72cm^{2}.$