Question

The area of rectangle is 26/3 cm2. if it's length is 52/9 cm, find it's perimeter​

Answer 1

Question :

The area of rectangle is $\sf{8 \frac{2}{3} \: {cm}^{2} }$. If its length is is $\sf{5 \frac{7}{9} \: cm }$,find its perimeter.

Given :

• Area of rectangle $\sf{ \frac{26}{3} \:cm}$
• Length of rectangle$\sf{ \frac{52}{9}\:cm }$

To find :

• Perimeter of rectangle

According to the question,

Area of rectangle = length (l) × breadth (b)

$\implies \sf{ \frac{26}{3} = \frac{52}{9} \times breadth}$

$\sf \implies{breadth = \frac{ \cancel{26}}{ \cancel 3} \times \frac{ \cancel{9} {}^{ \: \: 3} } { \cancel{52} {}^{ \: \: 2} }}$

$\sf \implies{breadth = \frac{3}{2} } \green \bigstar$

Now we will find perimeter of rectangle,

$\star \boxed{ \sf \purple{Perimeter \: of _{ \: rectangle} = 2(length \times breadth) }}$

$\sf{ \implies2( \frac{52}{9} + \frac{3}{2} )}$

$\sf{ \implies2( \frac{104 + 27}{18} })$

$\sf{ \implies \cancel{2} \times \frac{131}{ \cancel{18} {}^{ \: \: 9} } }$

$\sf{ \implies \frac{131}{9} } \: cm$

$\sf \implies{ 14\frac{5}{9} } \: cm \green\bigstar$

So,the correct option is (b)

Answer 2

Given,

• The area of rectangle=$8\frac{2}{3}$
• Length of rectangle=$5\frac{7}{9}$

To find,

• The perimeter of the rectangle

Solution:

Let the breadth be$x$

We know that,

Area of rectangle=$</strong><strong>l×</strong><strong>b</strong><strong>$

A/Q

$8\frac{2}{3}$=$5\frac{7}{9}×x$

$\frac{26}{3}$=$\frac{52}{9}×x$

$\frac{26}{3}$=$\frac{52}{9}x$

$\frac{26}{3}×\frac{9}{52}x$

$x=\frac{3}{2}$

Hence,We got the breadth that is $\frac{3}{2}$.

Now,

We know that,

Perimeter of rectangle:-$\boxed{\fcolorbox{white}{pink}{2×(l+b)}}$

$\leadsto2×\frac{52}{9}+\frac{3}{2}$

$\leadsto2×\frac{104+27}{18}$

$\leadsto2×\frac{131}{18}$

$\leadsto\frac{131}{9}$

$\leadsto14\frac{5}{9}$

Therefore,the perimeter of the rectangle=$\leadsto14\frac{5}{9}$.

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