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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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