Question

Find the area of the rhombus having perimeter 40cm and the length of one of
its diagonal is 16cm.

Answer 1

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Perimeter of rhombus = 40cm

Side = 1/4 × 40 = 10cm

One diagonal = 16cm [ given ]

We know, diagonal of rhombus bisect each other at right angles.

*refer the attachment*

$\rm{OB = \sqrt{{AB}^{2} - {OA}^{2}}} \\ = \rm{ \sqrt{ {(10)}^{2} - {(8)}^{2}}} \\ = \: \: \: \: \: \rm{\sqrt{100 \: - \: 64 }} \\ \rm{ = \: \: \: \: \: \sqrt{36} = 6cm } \\ \\ \rm{\therefore \: diagonal \: BD = 6 \times 2 = 12cm} \\ \\ \rm{Now, area \: of \: rhombus =} \\ \rm{\frac{1}{2} \times product \: of \: diagonals} \\ \rm{ \implies \: \frac{1}{2} \times 12 \times 16 } \\ \rm{ \implies \: \: \: \: \: \: \: \: \: \: \: \: 96c {m}^{2} }$

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