prove that the area of an equilateral triangle described one side of the square equal at the half of the area equilateral described on one of its diagonal
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FIGURE IS IN THE ATTACHMENT
Let ABCD be a Square having sides of length ‘a’.
Diagonal of a square (BD)= a√2
Now , construct two equilateral ∆’s.
∆PAB ~ ∆QBD
[Since, equilateral Triangles are similar]
ar (∆PAB ) / ar(∆QBD) = AB² /BD²
[The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides]
ar (∆PAB ) / ar(∆QBD) = a² / (a√2)²
ar (∆PAB ) / ar(∆QBD) = a² / a²×2 = ½
ar (∆PAB ) / ar(∆QBD) = ½
ar (∆PAB ) = ½ ar(∆QBD)
Hence, proved.
HOPE THIS WILL HELP YOU...
Let ABCD be a Square having sides of length ‘a’.
Diagonal of a square (BD)= a√2
Now , construct two equilateral ∆’s.
∆PAB ~ ∆QBD
[Since, equilateral Triangles are similar]
ar (∆PAB ) / ar(∆QBD) = AB² /BD²
[The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides]
ar (∆PAB ) / ar(∆QBD) = a² / (a√2)²
ar (∆PAB ) / ar(∆QBD) = a² / a²×2 = ½
ar (∆PAB ) / ar(∆QBD) = ½
ar (∆PAB ) = ½ ar(∆QBD)
Hence, proved.
HOPE THIS WILL HELP YOU...
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