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What is the area of triangle in which two of its medians are 9 cm and 12 cm long intersect at the right angles?

I tried this but could not get to the answer.

Does the triangle will also be right angle if the median intersect at right triangle?

Thanks in advance.

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You may want to prove that the triangle is right angled with sides $9 \operatorname{cm}$ and $12\operatorname{cm}$. –  user21436 Mar 30 '12 at 18:10

2 Answers 2

up vote 5 down vote accepted

Let our triangle be $ABC$, and let the two medians be $AM$ (of length $12$) and $BN$ (of length $9$), meeting at the centroid $X$. Recall that the centroid of a triangle divides each median in the proportions $2:1$. It follows that $\triangle ABX$ is right-angled at $X$ and has "legs" $8$ and $6$. Thus $\triangle ABX$ has area $24$.

But the area of $\triangle ABX$ is one-third of the area of $\triangle ABC$, since the two triangles share a base $AB$, and the median from $C$ is split by $X$ in the proportion $2:1$, which implies that that the height of $\triangle ABC$ is $3$ times the height of $\triangle ABX$. So $\triangle ABC$ has area $72$.

Another way: Here is a simpler but less symmetric way. Let $AM=12$ and $BN=9$. Then $BX=6$, so $\triangle ABM$ has area $(12)(6)/2=36$. It follows that $\triangle ABC$ has area $72$.

Another way: Draw all the medians. They divide the triangle into $6$ triangles of equal area. But one of them, $\triangle MXB$, is right-angled at $X$ and has legs $4$ and $6$, so area $12$.

Remark: Part of your question asked whether the original triangle is right-angled. It isn't. Calculation shows that $\triangle ABC$ with $AB=10$, $BC=4\sqrt{13}$, and $CA=2\sqrt{73}$ is the only triangle that has medians of length $12$ and $9$ meeting at right angles. This triangle is not right-angled.

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Can you please explain? By which property you got area 24.? –  vikiiii Mar 30 '12 at 18:19
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@vikiiii: I have added detail –  André Nicolas Mar 30 '12 at 18:23
    
The centroid divides each median into parts in the ratio 2:1, with the centroid being twice as close to the midpoint of a side as it is to the opposite vertex. –  Salech Alhasov Mar 30 '12 at 18:27

Consider the triangle below: $|\overline{AC}|=2|\overline{AE}|$ and $|\overline{AB}|=2|\overline{AD}|$.

By SAS (side-angle-side), $\triangle ADE\sim\triangle ABC$. Therefore, $\overline{DE}||\overline{BC}$ and $|\overline{BC}|=2|\overline{DE}|$.

Since $\angle DEB=\angle EBC$ and $\angle EDC=\angle DCB$, ASA (angle-side-angle) yields $\triangle FED\sim\triangle FBC$.

Thus, $|\overline{FB}|=2|\overline{EF}|$ and $|\overline{FC}|=2|\overline{DF}|$. Therefore, the lengths are as listed. Since all the angles at $F$ are right, we get that $|\triangle BEC|=\frac128\cdot9=36$. The altitude of $\triangle ABC$ is twice that of $\triangle BEC$, so $|\triangle ABC|=72$.

$\hspace{3.5cm}$enter image description here

Now that I've posted this, I see that it is the same as André Nicolas' second method (on the other internal triangle). I will leave it because of the diagram.

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Do you know about the formula $\text{Area}=\frac23 m_1m_2\sinθ$, where θ is the angle between the medians? Using it we can do it in notime. –  Sawarnik Mar 6 at 18:32
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Yes. The area of $\triangle BFC$ is $\frac12\left(\frac23\overline{CD}\right) \left(\frac23\overline{BE}\right) \sin(\angle BFC)$. Since $\triangle BFC$ has the same base and $\frac13$ the altitude of $\triangle ABC$, we get the area of $\triangle ABC$ to be $$\frac23\overline{BE}\cdot\overline{CD}\sin(\angle BFC)$$ –  robjohn Mar 6 at 19:29

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