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3 years ago

Find the length and width of a rectangle whose perimeter is 26 feet and whose area is 42 square feet

Mathematics

1 answer:

likoan [24]3 years ago

8 0

Answer: length = 7 feet's

Width = 6 feets

Step-by-step explanation:

Perimeter of a rectangle is the distance round the rectangle.

Perimeter of rectangle is expressed as

2(length + width)

The perimeter of the given rectangle is 26 feet. Therefore

2(L + W) = 26

Dividing by 2,

L+W = 13

The area of a rectangle is expressed as length × width

The given area is 42 square feet. Therefore,

L×W = 42

Substituting L = 13 - W into LW = 42, it becomes

W(13-W) = 42

13W - W^2 = 42

W^2 -13W + 42 = 0

W^2 -7W - 6W + 42 = 0

W(W - 7) - 6(W - 7) = 0

(W-6)(w-7)=0

W-6 = 0 or W-7=0

W=6 or W = 7

L= 13-6 or L= 13-7

L= 7 or L = 6

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Step-by-step explanation:

two line are perpendicular to each other if product of their slope equal to -1

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∠SUT=∠RUS

3x+6=5x-4

x=5

∠SUT=3x+6=15+6=21°

∠SUT=21°

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3 years ago

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2 years ago

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3 years ago

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hjlf

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Step-by-step explanation:

The cross section AFGD is a rectangle. Then, its area can be calculated with this formula:

$A=l*w$

Where:

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"w" is the width of the rectangle.

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