If length and breadth of a rectangle are in ratio $3 : 2$ . If the area of the field is $3456 m^{2}$ . Find the cost of fencing the field at $23.50 p e r m e t e r$ .

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Solution

$\begin{matrix} L e t l e n g t h a n d b r e a d t h o f r e c t a n g l e a r e 3 x a n d 2 x r e s p e c t i v e l y . \Rightarrow A r e a o f r e c t a n g l e = ℓ \times b \Rightarrow 3456 c m^{2} = 3 x \times 2 x \Rightarrow 3456 = 6 x^{2} \Rightarrow 576 = x^{2} \Rightarrow \sqrt{576} = x x = 24 c m L e n g t h = 3 x = 72 c m B r e a d t h = 2 x = 48 m \Rightarrow p e r i m e t e r o f r e c t a n g l e = 2 (ℓ + b) = 2 (72 + 48) = 240 m \Rightarrow f a n c i n g t h e f i e l d a t 23.50 p e r i m e t e r . a n d f a n c i n g f o r 240 m i s = 23.5 \times 240 = 5640 R s . \end{matrix}$

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