The perimeter of a rhombus is 40 centimeters. The length of one diagonal is 12 centimeters. How do you find the area of the rhombus?

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Answer 1 · 2017-03-27T03:06:45

$96 {cm}^{2}$ $96 "cm"^2$

A rhombus has 4 edges of equal length. If it has perimeter of 40cm, then each length would be

$\frac{40 cm}{4} = 10 cm$ ${40"cm"}/4 = 10"cm"$

Divide the rhombus into 4 right-angle triangle by cutting along the 2 diagonals.

Each right-angle triangle will have a hypotenuse of 10cm and one other edge with length

$\frac{12 cm}{2} = 6 cm$ ${12"cm"}/2 = 6"cm"$

Use Pythagoras Theorem to find the length of the remaining edge.

$\sqrt{{(10 cm)}^{2} - {(6 cm)}^{2}} = 8 cm$ $sqrt{(10"cm")^2 - (6"cm")^2} = 8 "cm"$

Now find the area of the triangle.

$\frac{1}{2} \times 6 cm \times 8 cm = 24 {cm}^{2}$ $1/2 xx 6 "cm" xx 8"cm" = 24 "cm"^2$

Since the original rhombus was made up of four such triangles, the area of the rhombus would be

$24 {cm}^{2} \times 4 = 96 {cm}^{2}$ $24 "cm"^2 xx 4 = 96 "cm"^2$