Question

AD is an altitude of 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

Answer:

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Answer 2

SOLUTION

Let the side of ∆ABC be a units

$= > </u></strong><strong><u>AD</u></strong><strong><u> = \frac{ \sqrt{ 3} }{2} a$

Now, since ∆ABC & ∆ADE are equilateral ∆

=) ∆ABC ∼ ∆ADE

$= > \frac{Ar( \triangle \: ADE)}{Ar( \triangle \: ABC)} = \frac{(side \: of \triangle \: ADE) {}^{2} }{(side \: of \triangle \: ABC) {}^{2} } \\ \\ = > \frac{ (\frac{ \sqrt{3} }{2}a) {}^{2} }{ {a}^{2} } \\ \\ = > \frac{3}{4}$

Hence, Ar (∆ADE): Ar (∆ABC)= 3:4

[Proved]

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