Geometry problem 1 by Dhaval Furia

The three medians of a triangle are concurrent. Let the intersection of the three medians be $X$ . So

$AX = \dfrac{2}{3}(12) = 8 cm$

$EX = \dfrac{1}{3}(9) = 3 cm$

$\triangle AXE$ is a right triangle and its area is

$A_{\triangle AXE} = \dfrac{1}{2} \times 3 \times 8 = 12 cm^2$

The three medians together divide a triangle into six equal parts, so the area of $\triangle ABC$ is

$A_{\triangle ABC} = 6 \times 12 = \boxed{\text{72 square centimeters}}$

We have $AD = 12$ and $BE = 9$ . We know that intersections of medians divides the median in the ratio $2 : 1$

So $AG : GD = 2 : 1$ and $AG + GD = 12$

$\Rightarrow AG = 8$ and $GD = 4$

Draw $CF \perp BE$ . Then in $\triangle BCF, D$ is mid point of $BC$ and $DG \parallel CF$ . So $\triangle BDG \sim \triangle BCF$

$\large\frac{BD}{BC}$ $= \large\frac{1}{2}$ $= \large\frac{GD}{CF}$

$\Rightarrow CF = 2GD = 8$

Now, area of $\triangle ABC =$ area of $\triangle ABE \;+$ area of $\triangle BEC$

$\Rightarrow \text{ar}(\triangle ABC) = \large\frac{1}{2}$ $BE\cdot AG + \large\frac{1}{2}$ $BE\cdot CF$

$\Rightarrow \text{ar}(\triangle ABC) = \large\frac{1}{2}$ $\cdot9\cdot 8 + \large\frac{1}{2}$ $\cdot9\cdot 8$

$\Rightarrow \text{ar}(\triangle ABC) = 36 + 36 = 72$