Question

Question:-

The sides of a rectangle are in the ratio 3:2. If the area is 486 sqm, find the cost of fencing it at 40 per metre.

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Answer 1

$\Large{\bf{\green{\mathfrak{\dag{\underline{\underline{Given:-}}}}}}}$

• Sides of a rectangle are in the ratio 3:2. Area of the rectangle = 486 m². Cost of fencing for 1 meter is Rs. 40.

$\Large{\bf{\orange{\mathfrak{\dag{\underline{\underline{To \: Find:-}}}}}}}$

• Cost of fencing the rectangle.

$\Large{\bf{\red{\mathfrak{\dag{\underline{\underline{Solution:-}}}}}}}$

• Let the sides of the rectangle be x.

Then,

• The sides of the rectangle are 3x and 2x.

• Area of the rectangle = 486 m².

We know that,

$\boxed{\pink{\sf Area \: = \: l \: * \: b \: units^2}}$

Substituting the values,

• 3x * 2x = 486

• 6x² = 486

• x² = $\sf \dfrac{486}{6}$

• x² = $\sf \cancel{\dfrac{486}{6}}$

• x² = 81

• x = $\sf \sqrt{81}$

• x = ± 9

Since, length cannot be negative.

• x = 9

Therefore,

• Length of the rectangle =

• 3x

• 3 * 9

• 27 m

• Breadth of the rectangle =

• 2x

• 2 * 9

• 18 m

Now,

• Perimeter = ?

We know that,

$\boxed{\pink{\sf Perimeter \: = \: 2(l \: + \: b) \: units}}$

Substituting the values,

• 2 (27 + 18)

• 2 (45)

• 90 m

Now,

• Cost of fencing per metre = Rs.40

Then,

• Cost of fencing for 90m =

• Rs. 40 * 90

• Rs. 3,600

Therefore,

• $\boxed{\purple{\textsf{Cost of fencing = Rs. 3,600.}}}$

Answer 2

${\large{\underline{\pmb{\frak{Given...}}}}}$

→ The sides of a rectangle are in the ratio 3:2

→ The area of the rectangle is 486 cm²

${\large{\underline{\pmb{\frak{To \; Find...}}}}}$

→ The cost of fencing it at Rs.40 per metre.

${\large{\underline{\pmb{\frak{Understanding \; the \; Concept...}}}}}$

❍ Concept : Here we have been said that the sides of the rectangle are in the ratio of 2 : 3 and the area of the rectangle is 486 cm ² and said that we'll have to find the cost of fencing the rectangle at the rate of Rs. 40 per m

✰ Now let's find the measure of the sides of the rectangle and then find thee perimeter of the rectangle and later find the cost of fencing it

${\large{\underline{\pmb{\frak{Using \; Concepts...}}}}}$

✪ Formula to find the area of a rectangle:

$\tt Area \; of \; a \; rectangle = Length \times Breadth$

✪ Formula to find the perimeter of a rectangle:

$\tt Perimeter \; of \; a \; rectangle = 2 ( L + B )$

${\large{\underline{\pmb{\sf{RequirEd \; Solution...}}}}}$

★ The cost of fencing the rectangle is Rs.3,600

${\large{\underline{\pmb{\frak{Full \; Solution...}}}}}$

⋆ Let's Assume That,

⠀⠀⠀» The length of the rectangle is 3x

⠀⠀⠀» The breadth of the rectangle is 2x

~ As per statement 1 we know that the area of the rectangle is 486m² so, let's find the dimensions

$:\implies \sf Length \times Breadth = 486 cm^2$

$:\implies \sf 3x \times 2x = 486 cm^2$

$:\implies \sf 6x^2 = 486cm^2$

$:\implies \sf x^2 = \dfrac{486cm^2}{6}$

$:\implies \sf x^2 = 81cm^2$

$:\implies \sf x = \sqrt{81 cm^2}$

$:\implies \sf x = 9cm$

~ Now let's find the dimensions

${:\implies} \sf Length = 3x = 3(9cm) = 21cm$

${:\implies} \sf Breadth = 2x = 2(9cm) = 18cm$

• Henceforth the length and breadth are 21cm and 18cm

~ Now let's find the perimeter of thee rectangle

$:\implies \sf Perimeter_{(rectangle) } = 2 ( l+ b )$

${:\implies} \sf Perimeter_{(rectangle) } = 2 ( 21cm + 18cm )$

${:\implies} \sf Perimeter_{(rectangle) } = 2( 45cm)$

${:\implies} \sf Perimeter_{(rectangle) } = 90cm$

• Henceforth the perimeter of the rectangle is 90cm

~ Now let's find the cost of fencing it

${:\implies} \sf Cost \; of \; fencing = Rate \times Perimeter$

${:\implies} \sf Cost \; of \; fencing = 40 \times 90$

${:\implies} \sf Cost \; of \; fencing = Rs.3600$

• Henceforth the cost of fencing the field is Rs.3600

${\large{\underline{\pmb{\frak{Additional \; Information...}}}}}$

⋆ Related Formulae

$\;\;\;\;\;\;\;\; {\leadsto} \tt \; Area \; of \; a \; square = Side ^2$

$\;\;\;\;\;\;\;\; {\leadsto} \tt \; Perimeter \; of \; a \; square = 4\times side$

$\;\;\;\;\;\;\;\; {\leadsto} \tt \; Area \; of \; circle =\pi r^2$

$\;\;\;\;\;\;\;\; {\leadsto} \tt \; Perimeter \; of \; a\; circle = 2\pi r$

$\;\;\;\;\;\;\;\; {\leadsto} \tt \; Area \; of \; a \; triangle = \dfrac{1}{2} \times b \times h$

⋆ Diagrams,

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\multiput(2.1,-0.7)(0,4.2){2}{\sf\large 21 cm}\multiput(-1.4,1.4)(6.8,0){2}{\sf\large 18 cm}\end{picture}$