The area of a rectangle is 66 ft^2 and the length of the rectangle is 7 feet less than three times the width find the dimensions of the rectangle

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14-03-2019
Mathematics

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The area of a rectangle is 66 ft^2 and the length of the rectangle is 7 feet less than three times the width find the dimensions of the rectangle

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laurellee27 laurellee27 · Answer 1 · 2019-03-14T17:26:33+01:00

Length = 11

Width = 6

The area of a rectangle is 66 ft^2:

L * W = 66

Length of the rectangle is 7 feet less than three times the width:

L = 3W-7

Substitute L in terms of W:

(3W-7) * W = 66

Factorise the equation:

3W^2 -7W = 66

3W^2 - 7W - 66 = 0

Factors of 66 =

1 66, 2 33, 3 22, 6 11

(3W + 11) (W - 6) = 0

Solve for W:

3W + 11 = 0

3W = 11

W = 3/11

W - 6 = 0

W = 0 + 6

W = 6

Using the original equation, find L:

L = 3W-7

L = 3(6)-7

L = 18-7

L = 11

L * W = 66

11 * 6 = 66

abidemiokin abidemiokin · Answer 2 · 2021-10-19T10:44:15+02:00

The dimension of the rectangle if the length of the rectangle is 7 feet less than three times the width is 6ft by 11ft

The formula for calculating the area of a rectangle is expressed as:

A = LW

L is the length

W is the width

If the length of the rectangle is 7 feet less than three times the width, this is expressed as L = 3W -7

Substituting the length expression into the area formula;

66 = (3W-7)W

66 = 3W²-7W

3W²-7W-66=0

On factorizing;

W = 6 feet

Since A = LW

L = A/W

L = 66/6

L = 11feeet

Hence the dimension of the rectangle is 6ft by 11ft

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