In triangle ABC, medians AD and CE intersect at P , PE=1.5, PD=2 and .DE= 2.5 What is the area of AEDC?
+128474 Area of triangle DPE can be found as
Semiperimemter, S, = [2.5 + 2 + 1.5 ] / 2 = 3
Area = sqrt [ 3 ( 3 - 2) (3 - 2.5) ( 3 -1.5) ] = 1.5
By the property of medians
PC = 2 * PE = 2 * 1.5 = 3
PA = 2 * PD = 2 * 2 = 4
And triangles DPE and APC are similar with each side of APC twice as long as that of DPC
So scale factor of triangle APC to triangle DPE = 2
Area of triangle APC = area of DPE * scale factor^2 = 1.5 * 2^2 = 6
And by another median property areas of triangle AEP plus triangle CDP = area of triangle APC
So....area of AEDC = 1.5 + 6 + 6 = 13.5
