Question

The medians of a $\Delta ABC$ are 9 cm 12 cm and 15 cm respectively Then the area of the triangle is:

• A. 96 sq cm
• B. 84 sq cm
• C. 72 sq cm
• D. 60 sq cm

Answer 1

The correct option is C 72 sq cm

Given length of medians
$AE=9cm,BF=12cm,CG=15cm$
Point of intersection of medians is centroid and centroid divides median in the ratio 2:1.
So, $AO=6cm,OE=3cm$ , $BO=8cm,OF=4cm$, $CO=10cm,OG=5cm$
Now, draw parallelogram OBDC.
$OB=DC=8cm$ (Opposite sides of parallelogram are equal)
$\Rightarrow OE=ED$ (Diagonals of a parallelogram bisect each other.)
$\Rightarrow ED=3cm$
$\Rightarrow OD=OE+ED=6cm$
Also, $△ODC$ is a right triangle with sides 6,8,10 (forming Pythagorean triplet)
Area of $\Delta ADC=\frac{1}{2}\times 8\times 12=48c{m}^2$
Area of $\Delta EDC=\frac{1}{2}\times 3\times 8\times =12c{m}^2$
Area $\Delta AEC$ = Area $\Delta ADC$ - Area $\Delta EDC$
$\Rightarrow \Delta \left(AEC\right)=48-12=36c{m}^2$
$\Rightarrow \Delta \left(ABC\right)=72c{m}^2$ (Median divides a triangle into equal halves.)