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A rectangle has a length 10 more than its width. If the width is increased by 8

and the length by 4, the resulting rectangle has an area of 135 square units.
Part A
Write an equation to model the above scenario. Use the model to find the length
of the original rectangle?
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esmos
Part B
What is the perimeter of the expanded rectangle?

User GoGonzo
by
4.6k points

1 Answer

11 votes

Explanation:

length = the original length

width = the original width

length = width + 10 or

width = length - 10

(length + 4) × (width + 8) = 135

A

now we are using the first equation in the second to solve then for length :

(length + 4) × (length - 10 + 8) = 135

(length + 4) × (length - 2) = 135

length² - 2×length + 4×length - 8 = 135

length² + 2×length = 143 or

length² + 2×length - 143 = 0

this is the equation to model the scenario.

solving in general such a quadratic equation :

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

x = length

a = 1

b = 2

c = -143

length = (-2 ± sqrt(2² - 4×1×-143))/(2×1) =

= (-2 ± sqrt(4 + 572))/2 =

= (-2 ± sqrt(576))/2 =

= (-2 ± 24)/2 = (-1 ± 12)

length1 = -1 + 12 = 11

length2 = -1 - 12 = -13

a negative solution for a length is not applicable, so,

length = 11

width = length - 10 = 11 - 10 = 1

B

the expanded length = 11 + 4 = 15

the expanded width = 1 + 8 = 9

the expanded perimeter is then

2×15 + 2×9 = 30 + 18 = 48

User Prabhat G
by
4.4k points