The length of three medians of a triangle are $9$,$12$ and $15$cm.The area (in sq. cm) of the triangle is
a) $48$
b) $144$
c) $24$
d) $72$
I don't want whole solution just give me the hint how can I solve it.Thanks.
[Math] How to find area of triangle from its medians
geometrytrianglestrigonometry
Related Solutions
Let $X$, $Y$ and $Z$ be the side midpoints. Construct the parallelogram PYBZ and connect PC.
By construction, PY = BZ = AZ, PY || AZ and then PYZA is also a parallelogram, which leads to AP || ZY || XC and AP = XC, and in turn the parallelogram APCX. Thus, PC = AX and the sides of the triangle PCZ are the medians of ABC.
The parallelogram PYZA also yields AQ = QY = $\frac14$AC and then
$$\frac{Area_{PCZ}}{Area_{ABC}}=\frac{Area_{QCZ}}{Area_{AZC}}=\frac{QC}{AC} = \frac34 $$
From JimmyK4542's formulas it follows that $a={2\over3}\sqrt{2s^2-m_a^2}\>$, where $s:=\sqrt{m_b^2+m_c^2}$. From this one derives the following construction of ${3\over2}a$:
Construct $s$ as hypotenuse of a right triangle with legs $m_b$, $m_c$; then draw a square with side $s$ and find $\sqrt{2}\>s$ as length of the diagonal $d$. Draw a Thales semicircle with diameter $d$; then construct a right triangle with hypotenuse $d$ and one leg $m_a$. The other leg then is ${3\over2}a$. The rest should be easy.
Best Answer
Area of a Triangle from the Medians
A triangle is divided in to $6$ equal areas by its medians:
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In the case where the two blue triangles share a common side of the triangle, it is pretty simple to see they share a common altitude (dotted) and equal bases; therefore, equal areas.
In the case where the two red triangles share a common $\frac23$ of a median, the altitudes (dotted) are equal since they are corresponding sides to two right triangles with equal hypotenuses and equal vertically opposite angles, and they share a common base; therefore, equal areas.
Now duplicate the original triangle (dark outline) by rotating it one-half a revolution on the middle of one of its sides:
$\hspace{3cm}$
The triangle in green has sides $\frac23a$, $\frac23b$, and $\frac23c$, and by Heron's formula has area $$ \frac49\sqrt{s(s-a)(s-b)(s-c)}\tag{1} $$ where $s=(a+b+c)/2$. Thus, each of the $6$ small, equal-area triangles in the original triangle has an area of half of that. Therefore, the area of the original triangle is $3$ times that given in $(1)$: $$ \frac43\sqrt{s(s-a)(s-b)(s-c)}\tag{2} $$