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# the perimeter of a rectangle is 48 meters, and its area is 135 m2. the sides of the rectangle are

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### Mohammed

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## A rectangle has a perimeter of 48m and area 135 . Find the dimensions of the rectangle.

A rectangle has a perimeter of 48m and area 135 . Find the dimensions of the rectangle.

ALGEBRA 1 Natalie N. asked • 08/30/20

## A rectangle has a perimeter of 48m and area 135 . Find the dimensions of the rectangle.

I don't understand at all, I need an explanation.

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## 3 Answers By Expert Tutors

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Dylan B. answered • 08/30/20

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Rectangles have two dimensions, a length which we will call L, and a width that we will call W. For problems like this, it might be useful to draw a rectangle and label all its sides, like below:

W

_________________________________

| |

| |

L | | L

| |

| |

|________________________________|

W

Because these values are unknown right now, we have two variables, L and W. To solve for any number of variables, you'll need at least the same number of independent equations, which means we need to write two equations to be able to find L and W. Your question gives us two important pieces of info: the perimeter is equal to 48, and the area is equal to 135. Let's use this info to write our two equations.

The perimeter of a shape is the length of its outline. To get the perimeter of a rectangle, you'll add up the length of all its sides. To write our perimeter, P in math terms, it could look like this:

P = W + L + W + L = 2W + 2L

Because our perimeter is equal to 48, we can say:

P = 2W + 2L 48 = 2W + 2L

Now let's use our area equation. The area of a shape is the amount of space contained in its outline. To get our area, A you need to multiply a rectangle's length by its width, like below:

A = LW

Because we know our area is 135, we can also say:

A = LW 135 = LW

Now we have two equations with two variables, L and W. To be able to solve for a variable, you need to use one equation to solve what a variable is equal to in terms of the other, then use the second equation to rewrite everything in terms of one variable. To write this in math terms, let's start with the second equation and solve for W:

135 = LW W = 135/L

Now we know what W is equal to in terms of L. We can use this in the first equation to get:

48 = 2W + 2L = 2(135/L) + 2L

= 270/L + 2L. Because we're dividing 270/L, let's multiply everything by L so there are no fractions

48L = 270 + 2L2. Now because we have L2, L, and a number without L, we have what's called a quadratic. You can solve these by moving everything to one side. I prefer to keep the L2 part positive, so let's subtract both sides by 48L.

48L = 270 + 2L2 0 = 270 + 2L2 - 48L

2L2 - 48L + 270 = 0. This is the most common way to write a quadratic, with the variable squared part first, the variable part, then the number without the variable.

Now you can solve a quadratic two ways, either by factoring or using the quadratic formula. The most direct way would be to use the quadratic formula. The quadratic formula is given below:

L = [-b ± √(b2 - 4ac)] / 2a

Here, our variable is L instead of x, and our equation is 2L2 - 48L + 270, so

a = 2 b = -48 c = 270

Plug these numbers into the quadratic formula to get:

L = [-(-48) ± √((-48)2 - 4(2)(270))] / 2(2)

= [48 ± √(2304 - 2160)] / 4

= [48 ± √(144)] / 4 = (48 ± 12) / 4 = 48/4 ± 12/4 = 12 ± 3

For our last step, we have a ±, which tells us we have two answers: 12 + 3, and 12 - 3. Knowing this, we have L = 15, and L = 9.

Now we have take both of these values, and plug them into our area equation to get a corresponding W. For L = 15,

A = LW 135 = (15)W W = 135/15 = 9 For L = 9, A = LW 135 = (9)W W = 135/9 = 15

So if L = 15, W = 9, and if L = 9, W = 15. Notice these make the same rectangle, so you could probably just choose one of these for your answer.

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Arjun R. answered • 08/30/20

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the perimeter of a rectangle I given by P = 2*L + 2*W where P is perimeter, L is length, and W is width

the area of a rectangle is given by A = L*W

here we have 48m = 2L + 2W and 135 = L*W ...

48 = 2*(L+W) 24 = (L + W) L = 24 - W

now we can plugin L in terms of W into our are equation and solve for W...

135 = L*W 135 = (24 - W)*W 135 = 24W -W2 W2 - 24W + 135 = 0

use the quadratic formula to solve for W...

W = [24 ± sqrt((-24)2 - 4(1)(135)] / 2

W = 9 or 15

since L = 24 - W, L = 24-9 = 15 or L = 24-15 = 9

the side lengths of the rectangle measure 9m and 15m.

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Tom K. answered • 08/30/20

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The perimeter of a rectangle is 2(l + w). The area is lw.

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## If perimeter of a rectangle is 48 metres and its area is \( \Large 135 m^{2} \), then sides of the rectangle are

If perimeter of a rectangle is 48 metres and its area is \( \Large 135 m^{2} \), then sides of the rectangle are

Quadrilateral and parallelogram

If perimeter of a rectangle is 48 metres and its area is

135 m 2 135m2

, then sides of the rectangle are

A) 15 m, 9 m B) 19 m, 5 m C) 45 m, 3 m D) 27 m, 5 m Correct Answer: A) 15 m, 9 m

Description for Correct answer:

Let sides of the rectangle are x and y meters

By hypothesis, 2(x+y)=48 2(x+y)=48 => x+y=24 x+y=24 => x=24−y x=24−y ...(i) and xy=135 xy=135 ...(ii)

From equations (i) and (ii), we get

(24−y)y=135 (24−y)y=135 => 24y− y 2 =135 24y−y2=135 => y 2 −24y+135=0 y2−24y+135=0 => y 2 −15y−9y+135=0 y2−15y−9y+135=0 => y(y−15)−9(y−15)=0 y(y−15)−9(y−15)=0 => (y−15)(y−9)=0 (y−15)(y−9)=0 => y=15, or y=9 From equation (i), x = 24 - 15 = 9 or x = 24 - 9 = 15

Hence, sides are 15 m, 9 m.

Part of solved Quadrilateral and parallelogram questions and answers : >> Elementary Mathematics >> Quadrilateral and parallelogram

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## If the perimeter of a rectangle is 48 metres and its area is 135m2, then what are the sides of the rectangle?

Answer (1 of 3): LET LENGTH= L BREADTH= B PERIMETER= 2(L+B)= 48 L+B= 24 — —(1) AREA= LB= 135 (L-B)²= (L+B)²-4LB= 24²-4×135= 576–540 (L-B)²= 36 L-B= 6 — —(2) FROM (1) AND (2) 2L=30 L= 15 B= 9 ANSWER LENGTH = 15 M BREADTH = 9 M

If the perimeter of a rectangle is 48 metres and its area is 135m2, then what are the sides of the rectangle?

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The perimeter of the rectangle=2(l+w)=48 m

l+w=48/2=24 m

The area of the rectangle=l*w=135 sq m

l=135/w 135/w+w=24 multiply with w 135+w^2=24w w^2–24w+135=0 (w-15)(w-9)=0 w=15 or 9

The length of the rectangle=15 m

The width of the rectangle=9 m

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