Question

AD is an altitude of an equilateral triangle ABC.on AD as base,another triangle ADE is constructed.prove that area(ADE):(ABC)=3:4

Answer 1

Given ABC and ADE are equilateral triangles. Let AB=BC=CA = a Recall that the altitude of an equilateral triangle is √3/2 times its side Hence AD = √3a/2 ΔABC  ~ ΔADE [By AAA similarity criterion] [Since ratio of areas of two similar triangles is equal to the ratio of their                                                   corresponding sides] Hence area(ADE):area(ABC)=3:4.