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

##### Solution :-

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

⇒ area(ΔAPC) = 2 × area(ΔBQC)

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

Hence, proved.

Answered by Abhisek | 1 year ago

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