From the figure
Sketch the third median CF. It is noted that:
The point of intersection of medians, known as the centroid (G), divides each median into a 2:1 ratio.
All three medians collectively partition the triangle into six equal areas.
\(GD=\frac{1}{3}×AD=\frac{1}{3}×12=4\)
\(GB=\frac{2}{3}×BE=\frac{2}{3}×9=6\)
Area of triangle BGD =\(\frac{1}{2}\times GB\times GD=\frac{1}{2}\times6\times4=12\)
Hence, area of triangle \(ABC = 6×12=72\)
ABCD is a trapezoid where BC is parallel to AD and perpendicular to AB . Kindly note that BC<AD . P is a point on AD such that CPD is an equilateral triangle. Q is a point on BC such that AQ is parallel to PC . If the area of the triangle CPD is 4√3. Find the area of the triangle ABQ.