Area of a rectangle with missing dimensions

[rachelmaddie]

I need my work to be checked please.
Area of a rectangle = L * W
Given: area = 144 square inches, since the length is 10 inches longer than the width
length = w + 10, thus the area = x(x + 10)
area = x^2 + 10x
144 = x^2 + 10x
0 = x^2 + 10x - 144
Factoring the expression:
(x + 18)(x - 8) = 0

clearing for x
x + 18 = 0 —> x = -18 (drop negative sign)
x - 8 = 0 —> x = 8

18in x 8in = 144in

Length: 18 inches
Width: 8 inches

 

[Deleted member 4993]

I need my work to be checked please.
Area of a rectangle = L * W
Given: area = 144 square inches, since the length is 10 inches longer than the width
length = w + 10, thus the area = x(x + 10)
area = x^2 + 10x
144 = x^2 + 10x
0 = x^2 + 10x - 144
Factoring the expression:
(x + 18)(x - 8) = 0

clearing for x
x + 18 = 0 —> x = -18 (drop negative sign)
x - 8 = 0 —> x = 8

18in x 8in = 144in

Length: 18 inches
Width: 8 inches
View attachment 20323

Your numerical work is correct - but there are lot of deficiencies in your written work See below:

Area of a rectangle = L * W [first define your variables. What is 'L'? What is 'W'?]

Given: area = 144 square inches, since the length is 10 inches longer than the width

length = w + 10, [ Now you switched from capital W to small w - not allowed]

thus the area = x(x + 10)........[What is 'x'? ]

area = x^2 + 10x

144 = x^2 + 10x

0 = x^2 + 10x - 144

Factoring the expression:

(x + 18)(x - 8) = 0

clearing for x

x + 18 = 0 —> x = -18 (drop negative sign value of width)

x - 8 = 0 —> x = 8

18in x 8in = 144in² ...........[unit of area is in² ]

Length: 18 inches
Width: 8 inches

If I missed something - other members of this board will catch those......

 

[rachelmaddie]

Your numerical work is correct - but there are lot of deficiencies in your written work See below:

Area of a rectangle = L * W [first define your variables. What is 'L'? What is 'W'?]

Given: area = 144 square inches, since the length is 10 inches longer than the width

length = w + 10, [ Now you switched from capital W to small w - not allowed]

thus the area = x(x + 10)........[What is 'x'? ]

area = x^2 + 10x

144 = x^2 + 10x

0 = x^2 + 10x - 144

Factoring the expression:

(x + 18)(x - 8) = 0

clearing for x

x + 18 = 0 —> x = -18 (drop negative sign value of width)

x - 8 = 0 —> x = 8

18in x 8in = 144in² ...........[unit of area is in² ]

Length: 18 inches
Width: 8 inches

If I missed something - other members of this board will catch those......

How is this?
Area of a rectangle = L * W where L = Length, W = Width
Given: area = 144 square inches, since the length is 10 inches longer than the width
length = W + 10, let the width be x thus the area = x(x + 10)
area = x^2 + 10x
144 = x^2 + 10x
0 = x^2 + 10x - 144
Factoring the expression:
(x + 18)(x - 8) = 0

clearing for x
x + 18 = 0 —> x = -18 (drop the negative value of width)
x - 8 = 0 —> x = 8

18in x 8in = 144in^2

Length: 18 inches
Width: 8 inches

 

[Deleted member 4993]

How is this?
Area of a rectangle = L * W where L = Length, W = Width
Given: area = 144 square inches, since the length is 10 inches longer than the width
length = W + 10, let the width be x thus the area = x(x + 10)
area = x^2 + 10x
144 = x^2 + 10x
0 = x^2 + 10x - 144
Factoring the expression:
(x + 18)(x - 8) = 0

clearing for x
x + 18 = 0 —> x = -18 (drop the negative value of width)
x - 8 = 0 —> x = 8

18in x 8in = 144in^2

Length: 18 inches
Width: 8 inches

Correct

 

[Steven G]

Your work is fine--good job!

I just don't understand why you defined W to be the width and then you defined the width to be x.