Question

gimme the answer of this question​

Answer 1

Answer:

ABC is an equilateral triangle with AD as altitude

(altitude of any equilateral triangle bisects the side)

so, AD bisects BC and also AD is perpendicular to BC

By pythagoras theorem,

AD² + BD²= AB²

AD²=AB²-BD²

AD²= AB²- (1/2BC)²         (BD= 1/2BC)

AD²=AB²-(1/2AB)²           (BC=AB in an equi. triangle)

AD²=AB²- (AB²/4)

AD=√3 AB/2

Another equi. triangle with base AD is triangle ADE

side of triangle ADE is √3 AB/2

AREA OF TRIANGLE ABC = √3/4 × AB²

AREA OF TRIANGLE ADE = √3/4 × (√3 AB/2)²= 3√3 AB²/16

so, area of ΔABC/ area of ΔADE = √3/4 AB²/ (3√3 AB² /16)

= 4:3

HENCE, AR(ΔADE) : AR(ΔABC) = 3:4

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