Question

find the area of Rhombus whose diagonals are of length 10 cm and 8.2 CM

Answer 1

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$diagonal = area = diagonal \times$

$length \div 2$

$= 10 \: cm \: \times \: 8.2 \: cm \: \div \: 2$

$82 \: {cm}^{2} \: \div \: 2$

$= 41 \: {cm}^{2}.$





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Answer 2

To find  : area rhombus

Given : length = $10\ cm$   , breadth = $8.2\ cm$

Solution :

• Let , $d_{1}$ and $d_{2}$ be two diagonals measuring as , $d_{1} = 10 \ cm$ and $d_{2} = 8.2\ cm$
• Area of Rhombus is given by , product of diagonals divided by $2$ .
• Formula is given as :
• Area of Rhombus =  $\frac{1}{2} \times d_{1} d_{2}$  

                                      $= \frac{10\times 8.2 }{2} \\= 5 \times 8.2 \\= 41\ cm^{2}$

Hence , area of Rhombus is $41\ cm^{2}$ .