Question

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

Answer 1

Given :- ∆ABC & ∆ADE both are equilateral triangle , AD is a altitude of ∆ABC & base for ∆ADE

To Prove :- area(∆ADE) : area(∆ ABC) = 3:4

Solution :-

As angles of equilateral triangle is 60°

hence ,

∆ ABC ~ ∆ ADE

As per the properties of Similar triangle "AAA"

Altitude AD

Let all the side of ∆ ABC be "a"

In ∆ ADC apply Pythagorean theorem

AD² = AC² - DC²

AD² = $a^{2} \: - \: \Big[\dfrac{a}{2}\Big]^{2}$

AD² = $\dfrac{4a^{2} \: -\: a^{2}}{4}$

AD² = $\dfrac{3a^{2}}{4}$

AD = $\sqrt{\dfrac{3a^{2}}{4}}$

AD = $\sqrt{3} \dfrac{a}{2}$

By properties of similar triangles

The ratio of area of similar triangles are equal to the ratio of square of its corresponding sides.

$\dfrac{area(\bigtriangleup ADE)}{ area(\bigtriangleup ABC)}\: = \: \dfrac{AD^{2}}{AB^{2}}$

$\dfrac{area(\bigtriangleup ADE)}{ area(\bigtriangleup ABC)}\: = \: \dfrac{\Big[\sqrt{3} \dfrac{a}{2}\Big]^{2}}{a^{2}}$

$\dfrac{area(\bigtriangleup ADE)}{ area(\bigtriangleup ABC)}\: = \: \dfrac{3\dfrac{a^{2}}{4}}{a^{2}}$

$\dfrac{area(\bigtriangleup ADE)}{ area(\bigtriangleup ABC)}\: = \: \dfrac{3}{4}$

Hence Prove

™Brainly