Question

The ratio of the length and breadth of a rectangular field in 3 :2 the area of filed is 3 4 5 6 m2 the costing finishing and fild at the rate of rs 3 -50 per metre length and breadth of a rectangle find

Answer 1

$\sf\blue{Correct \ Question:}$

$\sf{The \ ratio \ of \ the \ length \ and \ breadth}$

$\sf{of \ a \ rectangular \ field \ is \ 3:2. \ The}$

$\sf{area \ of \ field \ is \ 3456 \ m^{2}. \ The}$

$\sf{cost \ of \ fencing \ the \ field \ is \ Rs \ 3.50}$

$\sf{per \ metre. \ Find \ length \ and \ breadth \ of}$

$\sf{rectangle \ and \ cost \ for \ fencing \ the \ land.}$

______________________________________

$\sf\red{\underline{\underline{Answer:}}}$

$\sf{Length \ and \ breadth \ of \ rectangle \ are}$

$\sf{72 \ m \ and \ 48 \ m \ respectively,}$

$\sf{cost \ of \ fencing \ is \ Rs \ 840.}$

$\sf\orange{Given:}$

$\sf{\implies{Length \ and \ breadth \ of \ the}}$

$\sf{rectangle \ are \ in \ ratio \ of \ 3:2}$

$\sf{\implies{Area \ of \ rectangle=3456}}$

$\sf{\implies{Cost \ of \ fencing =Rs \ 3.50 \ per \ metre.}}$

$\sf\pink{To \ find:}$

$\sf{1. \ Length \ and \ b}$

$\sf{2. \ Cost \ of \ fencing \ the \ field.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{Let \ the \ common \ factor \ be \ x.}$

$\sf{According \ to \ first \ condition.}$

$\sf{\implies{Length=3x}}$

$\sf{\implies{Breadth=2x}}$

$\sf{According \ to \ the \ second \ condition.}$

$\sf{Area \ of \ rectangle=Length\times \ Breadth}$

$\sf{\therefore{(3x)(2x)=3456}}$

$\sf{\therefore{6x^{2}=3456}}$

$\sf{x^{2}=\frac{3456}{6}}$

$\sf{\therefore{x^{2}=576}}$

$\sf{On \ taking \ square \ root \ of \ both \ sides}$

$\sf{x=24 \ or \ -24}$

$\sf{But, \ x \ can't \ be \ negative.}$

$\sf{\therefore{x=24}}$

$\sf{\therefore{Length=3(24)=72 \ m}}$

$\sf{\therefore{Breadth=2(24)=48 \ m}}$

$\sf{Perimeter=2(length+breadth)...formula}$

$\sf{=2(72+48)}$

$\sf{=2\times120}$

$\sf{=240 \ m}$

$\sf{Cost \ of \ fencing=240\times3.50}$

$\sf{\therefore{Cost \ of \ fencing=Rs \ 840}}$

$\sf\purple{\tt{\therefore{Length \ and \ breadth \ of \ rectangle \ are}}}$

$\sf\purple{\tt{72 \ m \ and \ 48 \ m \ respectively,}}$

$\sf\purple{\tt{cost \ of \ fencing \ is \ Rs \ 840.}}$

Answer 2

$\large\bf\underline{Correct\: Question:-}$

The ratio of the length and breadth of a rectangular field is 3 :2 and the area of filed is 3456m² If the cost of fencing the fild at the rate of Rs 3.50 per metre.

Then find the length, breadth and cost of fencing whole field.

$\large\bf\underline \pink{Given:-}$

• Ratio of Length and breadth = 3:2
• Area of field = 3456m²

$\large\bf\underline \pink{To \: find:-}$

• length and breadth.
• cost of fencing the field.

$\huge\bf\underline \green{Solution:-}$

Let length and breadth of rectangle be 3x and 2x meters .

$\large\bf \: we \: know \: that \:$

$\bf \pink{area \: of \: rectangle = length \times breadth}$

$: \implies \rm \:3456 = 3x \times 2x \\ \\ : \implies \rm \:3456 = 6 {x}^{2} \\ \\ : \implies \rm \: \frac{3456}{6} = {x}^{2} \\ \\ : \implies \rm \:576 = {x}^{2} \\ \\ : \implies \rm\:x = \sqrt{576} \\ \\ : \implies \rm \:x = \pm24 \\ \\ : \implies \bf \red{ x = 24}$

$\therefore \rm \: length \: = 3 \times 24 = \bf72\\ \: \: \: \: \: \rm \: breadth = 2 \times 24 = \bf \: 48$

As, fencing can be done around the field

hence,

$\bf\pink{perimeter = 2(length + breadth)} \\ \\ \rm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2(72 + 48) \\ \\ \rm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2(120) \\ \\ \bf\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 240m$

Now,

Rate of fencing = Rs 3.50 per meter

$\therefore \rm \: total \: cost \: of \: fencing \: = 240 \times\:Rs 3.50 \\ \\ \bf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \pink{ = Rs\:840}$