Question

AD is an altitude of an equilateral triangle ABC. On AD as base another equilateral triangle ADE is constructed. Prove that
Area( triangle ADE): Area (triangle 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/2ΔABC  ~ 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.

Answer 2

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/2ΔABC  ~ 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.