We know that the line DE is parallel to the line BC. Give that the area of triangle ADE is 9 and the area of triangle CDE is 6, find the area of Atriangle BC

Guest Oct 18, 2021

#1**0 **

Angle A = 45º

AD = DE = 3√2

CD = 12 / 3√2 = 2√2

[BCE] = 6 + (2√2)^{2} / 2

**[ABC] = [ADE] + [CDE] + [BCE] **

civonamzuk Oct 19, 2021