Each side of a rhombus shaped field is 30 m and its one of the diagonals is 48 m. What is the area of the field?
864 m2
The diagonals of a rhombus bisect each other at right angles, so the point of intersection divides each diagonal into two equal halves.
Half of the given diagonal is \(\frac{48}{2}=24\) m.
Each side of the rhombus is the hypotenuse of a right triangle formed by the two half-diagonals, so the other half-diagonal is \(\sqrt{30^2-24^2}=\sqrt{900-576}=\sqrt{324}=18\) m, making the full second diagonal \(2\times18=36\) m.
The area of a rhombus is half the product of its diagonals: \(\frac{1}{2}\times48\times36=864\) sq m.
Hence, the area of the field is 864 sq m.
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