The length of a rectangle is two units more than four times the width. What is the length if the perimeter is of 20 units?
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We have to specify that the length of a rectangle is bigger than the width.
We'll put the width of the rectangle to be a units and the length be b inches.
We know, from enunciation, that the width is 2 units more than 4 times it's length and we'll write the constraint mathematically:
a - 2 = 4b
We'll subtract 4b and add 2 both sides:
a - 4b = 2 (1)
The perimeter of the rectangle is 20 units.
We'll write the perimeter of the rectangle:
P = 2(a+b)
20 = 2(a+b)
We'll divide by 2:
10 = a + b
We'll use the symmetric property:
a + b = 10 (2)
We'll add (1) + 4*(2):
a - 4b + 4a + 4b = 2 + 40
We'l eliminate and combine like terms:
5a = 42
We'll divide by 5:
a = 42/5
a = 8.4 units
8.4 + b = 10
b = 10 - 8.4
b = 1.6 units
So, the width of the rectangle is of 8.4 units and the lengths of the rectangle is of 1.6 units.
Since the width cannot be larger than the length, we'll change and we'll put the length of 8.4 units and the width of 1.6 units.
Let the length of the rectangle be L and the width be W.
The length of a rectangle is two units more than four times the width.
=> L = 2 + 4*W
=> W = (L - 2)/4
The perimeter is of 20 units
=> 2*L + 2*W = 20
=> L + W = 10
substitute W = (L - 2)/4
=> L + (L - 2)/4 = 10
=> 4L + L - 2 = 40
=> 5L = 42
=> L = 42/5
The length of the rectangle is 42/5
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