Question

find the perimeter of a rectangle, whose
area-8whole 2 and 3
length-5 whole 7 and 9
​

Answer 1

Given:

$area (lb)\: = 8 \frac{2}{3} \: {mm} ^{2}\: \: \: \: \: \: \: \: \: \: \: \\ \\ \: \: \: \: \: \: \: \: = \frac{26}{3} \: {mm}^{2}$

$length(l) = 5 \frac{7}{9} \: \: mm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ = \frac{52}{9} \: \: mm$

To find :

Perimeter = 2(l+b) = ?

Solution:

Area = length × breadth

$\frac{26}{3} = l \times b$

By substituting value of $l$ in above equation, we get

$\frac{26}{3} = \frac{52}{9} \times b \\ \\ b = \frac{26}{3} \times \frac{9}{52} \\ \\ b = \frac{1}{1} \times \frac{3}{2} \\ \\ b = \frac{3}{2} \: mm$

Perimeter = 2(l+b)

$= 2( \frac{52}{9} + \frac{3}{2} ) \: \: \: \: \: \: \: \: \: \: \\ \\ = 2( \frac{52(2) + 3(9)}{9 \times 2} ) \\ \\ = \frac{104 + 27}{9} \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ = \frac{131}{9} \: mm \: \: \: \: \: \: \: \: \: \: \: \: \: \:$