Question

A point taken on each median of a triangle divides the median in the ratio $1:3$ recking from the vertex. Then the ratio of the area of the triangle with vertices at these points to that of the original triangle is:

• A. $5:13$
• B. $25:64$
• C. $13:32$
• D. None of the above

Answer 1

The correct option is B $25:64$

$LetA(0,0);B(4m,0)andC(4p,4q)$

$M1(2m+2p,2q)$

$M2(2p,2q)andM3(2m,0)$

Let $E,FandG$ be the point on the median.

$\therefore E=(\frac{2m+2p}{4},\frac{2q}{4})=(\frac{m+p}{2},\frac{q}{2})$

$F=(\frac{2p+12m}{4},\frac{2q+0}{4})=(\frac{p+6m}{2},\frac{q}{2})$

$G=(\frac{2m+12p}{4},\frac{0+12q}{4})=(\frac{m+6p}{2},3q)$

$areaof△ABC=\frac{1}{2}$

$areaof△ABC=\frac{1}{2}\left|\begin{array}{ccc}0& 0& 1\\ 4m& 0& 1\\ 4p& 4q& 1\end{array}\right|$

$=\frac{1}{2}\left(16mq\right)=8mqunit$

$areaof△EFG=\frac{1}{2}\left|\begin{array}{ccc}\frac{m+p}{2}& \frac{q}{2}& 1\\ \frac{p+6m}{2}& \frac{q}{2}& 1\\ \frac{m+6p}{2}& 3q& 1\end{array}\right|=\frac{25m^2}{8}$

$⟹\frac{ar\left(△EFG\right)}{ar\left(△ABC\right)}=\frac{\frac{25m^2}{8}}{8m^2}=\frac{25}{64}=25:64$