Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

Asked by Pragya Singh | 1 year ago |  76

1 Answer

Solution :-

Ncert solutions class 10 chapter 6-37

Given, ABCD is a square whose one diagonal is AC. ΔAPC and ΔBQC are 

two equilateral triangles described on the diagonals AC and side BC of the square ABCD.

Area(ΔBQC) = \( \dfrac{1}{2}\) Area(ΔAPC)

Since, ΔAPC and ΔBQC are both equilateral triangles, as per given,

 ΔAPC ~ ΔBQC [AAA similarity criterion]

\( \dfrac{area(ΔAPC)}{area(ΔBQC)}\) = (\( \dfrac{AC^2}{BC^2}\)) = \( \dfrac{AC^2}{BC^2}\)

Since, Diagonal = \( \sqrt{2}\) side = \( \sqrt{2}\)BC = AC

Ncert solutions class 10 chapter 6-38

⇒ area(ΔAPC) = 2 × area(ΔBQC)

⇒ area(ΔBQC) = \( \dfrac{1}{2}\)area(ΔAPC)

Hence, proved.

Answered by Abhisek | 1 year ago

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