The area of a rhombus is 480 $c m^{2}$ and the length of one of its diagonals is 20 cm. The length of each side of the rhombus is

(a) 24 cm (b) 30 cm (c) 26 cm (d) 28 cm

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Solution

Area of a rhombus $= \frac{1}{2} [d_{1} \times d_{2}]$

$d_{1} = 20 c m A r e a = 480$

i.e, $\frac{1}{2} [d_{1} \times d_{2}] = 480 20 \times d_{2} = 960 d_{2} = \frac{960}{20} = 48 c m$

The diagonals of a rhombus bisect each other and are perpendicular to each other

therefore, In $△ A E D ∠ A E D = 90^{o}$

So by Pythagoras theorem,

$A D^{2} = A E^{2} + E D^{2} A E = \frac{1}{2} \times d_{1} = \frac{1}{2} \times 20 = 10 E D = \frac{1}{2} \times d_{2} = \frac{1}{2} \times 48 = 24 ∴ A D^{2} = A E^{2} + E D^{2} = 10^{2} + 24^{2} = 100 + 576 = 676 A D = \sqrt{676} = 26 c m$

Length of sides of the rhombus = 26 cm

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One side of a rhombus is 20 cm long and one of its diagonals measures 24 cm. The area of the rhombus is

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