Question

AD is an altitude of an equilateral triangle ABC. On AD as base, another equilateral triangle ADE is constructed.Prove that Area $Area (\trinagle ADE) : Area (\trinagle ABC)$ = 3 : 4.

Answer 1

SOLUTION :  

Given : ∆ABC and ∆ADE are equilateral triangles. AD is an altitude of an equilateral triangle ABC.

Let the sides of equilateral ∆ be  AB = BC = CA = a

altitude of an equilateral triangle,h (AD) =  √3/2 × a

AD = √3a/2

Since , ΔABC  &  ΔADE are equilateral triangles, then  

ΔABC  ~ ΔADE  

[By AAA similarity criterion]

By Using the Theorem, the ratio of the areas of two triangles is equal to the square of the ratio of their corresponding sides.

ar(ΔADE) / ar(ΔABC) = (AD/AB)²

ar(ΔADE) / ar(ΔABC) = AD²/AB²

ar(ΔADE) / ar(ΔABC) = (√3a/2)² / a²

ar(ΔADE) / ar(ΔABC) = (3a²/4) /a² = ¾

ar(ΔADE) / ar(ΔABC) = ¾  

ar(ΔADE) :  ar(ΔABC) = 3 : 4  

Hence, proved that the ratio of the areas of ar(ΔADE) :  ar(ΔABC) = 3 : 4  

HOPE THIS ANSWER WILL HELP YOU ..

Answer 2

ANSWER in attachment........................