If its length is x, then its width is 2x-8
x(2x-8) = 93
solve for x, and then use the Pythagorean Theorem to find the diagonal.
The width of a rectangle is 8 less than twice its length. If the area of the rectangle is 93
cm
2
, what is the length of the diagonal?
2 answers
L = Length
W = Width
A = Area
The width of a rectangle is 8 less than twice its length means:
W = 2 L - 8
A = L ∙ W
93 = L ∙ ( 2 L - 8 )
93 = 2 L² - 8 L
Subtract 93 to both sides
0 = 2 L² - 8 L - 93
Now solve equation:
2 L² - 8 L - 93 = 0
The solutions are:
L = 2 - √ ( 101 / 2 ) = - 5.1063 cm
and
L = 2 + √ ( 101 / 2 ) = 9.1063 cm
Length cannot be negative, so:
L = 2 + √ ( 101 / 2 )
W = 2 L - 8
W = 2 [ 2 + √ ( 101 / 2 ) ] - 8
W = 4 + 2 √ ( 101 / 2 ) - 8
W = 2 √ ( 101 / 2 ) - 4
W = √4 ∙ √ ( 101 / 2 ) - 4
W = (√ 4 ∙ 101 / 2 ) - 4
W = (√ 404 / 2 ) - 4
W = √202 - 4
W = 14.21267 - 4 = 10.21267 cm
Length of the diagonal:
d = √ ( L² + W² )
d = √ ( 9.1063² + 10.21267² )
d = 13.633 cm
W = Width
A = Area
The width of a rectangle is 8 less than twice its length means:
W = 2 L - 8
A = L ∙ W
93 = L ∙ ( 2 L - 8 )
93 = 2 L² - 8 L
Subtract 93 to both sides
0 = 2 L² - 8 L - 93
Now solve equation:
2 L² - 8 L - 93 = 0
The solutions are:
L = 2 - √ ( 101 / 2 ) = - 5.1063 cm
and
L = 2 + √ ( 101 / 2 ) = 9.1063 cm
Length cannot be negative, so:
L = 2 + √ ( 101 / 2 )
W = 2 L - 8
W = 2 [ 2 + √ ( 101 / 2 ) ] - 8
W = 4 + 2 √ ( 101 / 2 ) - 8
W = 2 √ ( 101 / 2 ) - 4
W = √4 ∙ √ ( 101 / 2 ) - 4
W = (√ 4 ∙ 101 / 2 ) - 4
W = (√ 404 / 2 ) - 4
W = √202 - 4
W = 14.21267 - 4 = 10.21267 cm
Length of the diagonal:
d = √ ( L² + W² )
d = √ ( 9.1063² + 10.21267² )
d = 13.633 cm