The floor of a shed given on the right has an area of 55 square feet. the floor is in the shape of a rectangle whose length is 1foot less than twice the width. find the length and the width of the floor of the shed.

The length is 10 feet and the width is 11/2 feet of the floor of the shed.

Given that,

The floor of a shed given on the right has an area of 55 square feet.

The floor is in the shape of a rectangle whose length is 1 foot less than twice the width.

We have to determine,

The length and the width of the floor of the shed.

According to the question,

Let, The length of the floor is L,

And the width of the floor is W.

Area of the floor = 55 square feet.

The floor is in the shape of a rectangle whose length is 1 foot less than twice the width.

L = 2W -1

Then, The area of the floor is given by the formula;

Substitute all the values in the formula,

The value w = -5 was rejected because width can not be negative.

Then, The value of w is 11/2 feet.

Substitute the value of w in the equation,

The length of the floor is 10 feet.

Hence, The required value of the length is 10 feet and the width is 11/2 feet of the floor of the shed.

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The answer to your question is: l = 10 ft; w = 11/2 ft

Step-by-step explanation:

Data

Area = 55 ft²

length = l

width = w

l = 2w - 1

Process

Area = l x w

Area = (2w - 1) w

Area =2w² - w

2w² -w - 55 = 0                  Solve the quadratic equation

2w² + 10w - 11w - 55 = 0

2w(w + 5) - 11(w + 5) = 0

(w + 5) (2w -11) = 0

w1 = -5     w2 = 11/2   There is not negative width, so the answer is 11/2 ft

l = 2(11/2) - 1

l = 11 - 1

l = 10 ft

Area = (10)(11/2) = 110/ 2 = 55 ft²

sss (side, side, side)

sas (side, angle, side)

asa (angle, side, angle)

aas (angle, angle, side)

hl (hypotenuse, leg)