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प्रश्न
AD is an altitude of an equilateral triangle ABC. On AD as base, another equilateral triangle ADE is constructed. Prove that Area (ΔADE): Area (ΔABC) = 3: 4
उत्तर
We have,
ΔABC is an equilateral triangle
Then, AB = BC = AC
Let, AB = BC = AC = 2x
Since, AD ⊥ BC then BD = DC = x
In ΔADB, by Pythagoras theorem
𝐴𝐵2 = (2𝑥)2 − (𝑥)2
⇒ 𝐴𝐷2 = 4𝑥2 − 𝑥2 = 3𝑥2
⇒ 𝐴𝐷 = `sqrt3`𝑥 cm
Since, ΔABC and ΔADE both are equilateral triangles then they are equiangular
∴ ΔABC ~ ΔADE [By AA similarity]
By area of similar triangle theorem
`("area"(triangleADE))/("area"(triangleABC))="AD"^2/"AB"^2`
`=(sqrt3x)^2/(2x)^2`
`=(3x^2)/(4x^2)`
`=3/4`
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