Question

10. The sides of a rectangular park are in the ratio 4 : 3. If its area is 2028 sq.m, find the cost of fencing it at ru.3 per metre.​

Answer 1

Answer:

let any no. be x

therefore , sides = 4x , 3x

area of rectangle = l×b

therefore , 2028 = 4x × 3x

12x²= 2028

x²= 2028/12 = 169

x = √169 = 13

length = 4×13 = 52 m

breadth = 3 × 13 = 39 m

perimeter of rectangle = 2(l+b)

therefore , perimeter = 2(52+39)

perimeter = 182 m

cost of fencing per metre = ₹3

therefore , cost of fencing the park = 182 × 3 = ₹546 .

Answer 2

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✦ Required Answer:

♦️ GiveN:

• The park is rectangular in shape
• Sides are in ratio 4:3
• Area of the rectangular park = 2038 m^2

♦️ To FinD:

• Find cost of fencing Rs.3 /metre.

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✦ How to solve?

The above question is based on area and perimeter of a rectangle. We are provided with ratio of sides, area and we have to find the perimeter of the rectangle.

As the sides are in ratio, we can write them in terms of single variable. This will be easier to find the sides as area is given. Then we can find the perimeter, finally the cost of fencing.

[Note: Fencing is done around the boundary of park]

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✦ Solution:

Let the sides be 4x and 3x

Area of park given = 2028 m^2

We know,

$\large{\boxed{\rm{\purple{Area \:of\: rectangle = length × breadth}}}}$

By using formula,

$\large{ \rm{ \longrightarrow \: 4x \times 3x = 2028 {m}^{2}}} \\ \\ \large{ \rm{ \longrightarrow \: 12 {x}^{2} = 2028 {m}^{2}}} \\ \\<strong> </strong>\large{ \rm{ \longrightarrow \: {x}^{2} = \frac{2028}{12} \: {m}^{2} }} \\ \\ \large{ \rm{ \longrightarrow \: {x}^{2} = 169 \: {m}^{2} }} \\ \\ \large{ \rm{ \longrightarrow \: x = \sqrt{169} \: m}} \\ \\ <strong> </strong>\large{ \rm{ \longrightarrow \: x = 13 \: m}}$

Then the sides of the rectangular park:

$\large{ \rm{ \longrightarrow \: 4x = 4 \times 13 = 52 \: m}} \\ \\\large{ \rm{ \longrightarrow \: 3x = 3 \times 13 = 39 \: m}}$

Now we know that,

$\large{\boxed{\rm{\purple{Perimeter \:of \:rectangle = 2(l+b)}}}}$

By using formula,

$\large{ \rm{ \longrightarrow \: Perimeter = 2(52 + 39) \: m}} \\ \\ \large{ \rm{ \longrightarrow \: Perimeter = 2(91) \: m}} \\ \\ \large{ \rm{ \longrightarrow \: Perimeter = 182 \: m}}$

We have, Cost of fencing

• 1 m = Rs. 3

Then,

$\large{ \rm{ \longrightarrow \: Total \: cost \: of \: fencing = Rs. 182 \times 3 \: }} \\ \\ \large{ \rm{ \longrightarrow \: Total \: cost \: of \: fencing = \boxed{ \rm{ \red{Rs. 546 }}}}} \\ \\ \large{ \therefore{ \underline{\underline{ \rm{ \purple{Hence, \: solved \: \dag}}}}}}$

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