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प्रश्न
A square and an equilateral triangle have equal perimeters. If the diagonal of the square is \[12\sqrt{2}\] cm, then area of the triangle is
विकल्प
- \[24\sqrt{2} c m^2\]
- \[24\sqrt{3} c m^2\]
- \[48\sqrt{3} c m^2\]
- \[64\sqrt{3} c m^2\]
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उत्तर
It is given the perimeter of a square ABCD is equal to the perimeter of triangle PQR.
The measure of the diagonal of the square is given `12 sqrt(2)` cm.We are asked to find the area of the triangle
In square ABCD, we assume that the adjacent sides of square be a.
Since, it is a square then a= b
By using Pythagorean Theorem
`a^2 +b^2 = (12sqrt(2))^2`
`a^2 +a^2 = 288`
`2a^2 = 288`
`a^2 = 288/2`
` a = sqrt(144)`
a = 12 cm
Therefore, side of the square is 12 cm.
Perimeter of the square ABCD say P is given by
p = 4 × side
Side = 12 cm
p = 4 × 12
p = 48 cm
Perimeter of the equilateral triangle PQR say P1 is given by
p1= 3 × side
p = p1
p = 3 × side
48 = 3 × side
side = `48/3`
side = 16 cm
The side of equilateral triangle PQR is equal to 16 cm.
Area of an equilateral triangle say A, having each side a cm is given by
`A = sqrt(3)/4 a^2`
Area of the given equilateral triangle having each equal side equal to 4 cm is given by
a = 16 cm
`A = sqrt(3)/4 (16)^2 `
`A = sqrt(3)/4 xx 256`
`A=64 sqrt(3) cm^2`
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