Area of $\triangle ABC$, given $AB=7$, $AC=15$, and median $AM=10$

Question

I've been working on this interesting problem for a while already, and here it is:

In $\triangle ABC$, $AB = 7$, $AC = 15$, and median $AM = 10$. Find the area of $\triangle ABC$.

I have figured out that $BM$ and $CM$ are both $4\sqrt2$ using Stewart's Theorem. Now, I tried to use Heron's Formula to calculate the area, which was a mess.

Any help is appreciated. Thanks.

Answer 1

Are you sure? $p = \dfrac{7+15+8\sqrt{2}}{2} = 11+4\sqrt{2}\implies S^2 = p(p-15)(p-7)(p-8\sqrt{2}) = (11+4\sqrt{2})(11-4\sqrt{2})(-4+4\sqrt{2})(4+4\sqrt{2}) = (121-32)(32-16) = 89\cdot 16.$

So $S = 4\sqrt{89}.$

Answer 2

You can avoid Heron's formula by also computing the measure of $∠AMB$. Let $\theta=m∠AMB$, $x=BM$. Then (as in the derivation of Stewart's/Apollonius's theorem) we have $$ 49=100+x^2-20x \cos \theta\\ 225=100+x^2+20x\cos \theta $$ and so $20x \cos \theta=\frac{225-49}{2}=88$, and $x^2=\frac{49+225}{2}-100=37$.

Then the area of the triangle can be computed as $$ 10x\sin \theta = 10\sqrt{x^2-x^2\cos^2 \theta} $$

Answer 3

Using Stewart's theorem, you can find that $BM$ and $CM$ are both $\sqrt{37}$ and not $4\sqrt2$, which means our third side ($BC$) is of length $2\sqrt{37}$ (= $\sqrt{148}$)

Then, using the less messier form of the Heron's formula given by: $$16 \;|\triangle ABC|^2 = 2a^2 b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4$$ you can calculate the area to be equal to exactly $42$!

Answer 4

Let $|AB|=7=c$, $|AC|=15=b$, $|AM|=10=m_a$, $|BM|=|MC|=\tfrac{a}2=x$.

Then by Stewart’s theorem for $\triangle ABC$ \begin{align} b^2x+c^2x&=2x(m_a^2+x^2) ,\\ x&=\sqrt{\tfrac12(b^2+c^2)-m_a^2} =\sqrt{\tfrac12(49+225)-100} =\sqrt{37} ,\\ a&=2x=2\sqrt{37} . \end{align}

And the area

\begin{align} S&=\tfrac14\sqrt{4a^2b^2-(a^2+b^2-c^2)^2} \\ &=\tfrac14\sqrt{4\cdot4\cdot37 \cdot 225 -(4\cdot37+225-49)^2} . \end{align}

So, the answer is$\dots$ still 42.

Answer 5

Extend $AM$ to $D$ such that $MA=MD$. Then triangles $AMB$ and $DMC$ are congruent, so triangles $ABC$ and $ACD$ have equal area. $\text{Area of }ABC=\text{Area of }ACD$, $AC=15$, $AD=20$, and $CD=7$.

Now use the Heron’s formula and you're done!