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The perimeter of a rhombus is 48 and the sum of the lengths of its diagonals is 26.
What is the area of the rhombus?

Hello Guest!

A diagonal is x.

\((\frac{x}{2})^2+(\frac{26-x}{2}^2)=(\frac{48}{4})^2\)

\(\frac{x^2}{4}+\frac{676-52x+x^2}{4}=\frac{2304}{16}=\frac{576}{4}\\\ x^2+676-52x+x^2=576\\ 2x^2-52x+100=0\\x^2-26x+50=0\)

\(x=13\pm\sqrt{169-50}\\ x=13\pm\sqrt{119}\)

\(x_1=23.91\\x_2=2.091\)

\(A=4\times (\frac{x}{2}\cdot\frac{26-x}{2})/2\\ A=4\times \frac{2.901\cdot (26-2.901)}{2\cdot 2\cdot2}\)

\(A=25\)

\(A=4\times \frac{x(26-x)}{2\cdot 2\cdot 2}\\ A=4\times \frac{23.91\cdot (26-23.91)}{2\cdot 2\cdot 2}\)

\(A=25\)

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