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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 

 Oct 18, 2021
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Angle A = 45º       

AD = DE = 3√2

CD = 12 / 3√2 = 2√2

[BCE] = 6 + (2√2)2 / 2

[ABC] = [ADE] + [CDE] + [BCE] 

 Oct 19, 2021
edited by civonamzuk  Oct 19, 2021

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