Question

17. Prove that the area of an equilateral triangle de-
scribed on one side of the square is equal to half
the area of the equilateral triangle described on
one of its diagonal.
[3]
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Answer 1

Step-by-step explanation:

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Answer 2

Answer:

$\Large{\underline{\underline{\sf{Given:}}}}$

• A square ABCD
• Equilateral triangle FCD
• Equilateral triangle ACE

$\Large{\underline{\underline{\sf{To\:Prove:}}}}$

• Area of ΔFCD = 1/2 × Area of ΔACE

$\Large{\underline{\underline{\sf{Solution:}}}}$

→ Let CF = FD = DC = a units

→ Hence area of ΔFCD = √3/4 × a² units -----equation 1 (∵ It is an equilateral triangle)

→ The diagonal of the square ABCD would be √2 a units which is a side of ΔACE.

→ Area of ΔACE = (√3/4) × (√2 a )²

  Area of Δ ACE = (√3/4 ) × 2 a²

  Area of ΔACE = √3/2 × a² ---------equation 2

→ Divide equation 1 by 2

  Area of ΔFCD/ Area of ΔACE = ( √3/4 × a² ) / (√3/2 × a²)

  Area of ΔFCD/ Area of ΔACE = (√3/4 × a² ) × ( 2/√3 × a²)

→ Cancelling the common terms,

  Area of ΔFCD/ Area of ΔACE = 1/2

  Area of ΔFCD = 1/2 × Area of ΔACE

→ Hence proved.

$\Large{\underline{\underline{\sf{Notes:}}}}$

→ The area of an equilateral triangle is given by the formula,

   Area = √3/4 × a²

  where a is a side of the triangle

→ The diagonal of a square is given by

   Diagonal = √2 a

   where a is a side of the square.