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(D) Let the rhombus be $ABCD$ with side $AB = 26 \ cm$ and diagonal $AC = 48 \ cm$. The diagonals of a rhombus bisect each other at right angles at po

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$480$

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(D) Let the rhombus be $ABCD$ with side $AB = 26 \ cm$ and diagonal $AC = 48 \ cm$. The diagonals of a rhombus bisect each other at right angles at point $O$.Therefore,$OA = OC = \frac{48}{2} = 24 \ cm$.In the right-angled triangle $\Delta OAB$,by the Pythagorean theorem:$(OB)^2 = (AB)^2 - (OA)^2$$(OB)^2 = (26)^2 - (24)^2$$(OB)^2 = (26 - 24)(26 + 24) = 2 	imes 50 = 100$$OB = \sqrt{100} = 10 \ cm$.The other diagonal $BD = 2 	imes OB = 2 	imes 10 = 20 \ cm$.The area of the rhombus is given by $\frac{1}{2} 	imes d_1 	imes d_2$.Area $= \frac{1}{2} 	imes 48 	imes 20 = 480 \ cm^2$.

Each side of a rhombus is $26 \ cm$ and one of its diagonals is $48 \ cm$ long. The area of the rhombus is (in $cm^2$)

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