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A classroom will be assembled so that the desks fit within an area that is 8 m by 12 m. The desks will be surrounded by a border that will be the same width on all sides of the rectangular room, to

allow for walking space. The area of the border will equal 20% of the area of the space that the desks
occupy. What is the width of the border, to the nearest hundredth of a meter?

User Joshualan
by
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1 Answer

5 votes

Answer:

0.46 m

Explanation:

area of desks: 8 m by 12 m = 8 m * 12 m = 96 m^2

20% of this area is 20% * 96 m^2 = 19.2 m^2

The area of the border is 19.2 m^2.

Let the path around the desks have width x.

The area of desks plus path is a rectangle 2x + 8 by 2x + 12.

area of desks plus path = (2x + 8)(2x + 12)

= 4x^2 + 24x + 16x + 96 = 4x^2 + 40x + 96

The area of the border is the area of the rectangle that includes the border minus the rectangle that has just the desks.

area of border = (4x^2 + 40x + 96) - (96) =

= 4x^2 + 40x

Above, we have the area of border = 19.2, so we get this equation:

4x^2 + 40x = 19.2

4x^2 + 40x - 19.2 = 0

x^2 + 10x - 4.8 = 0

We now use the quadratic formula to solve the equation for x.


x = (-b \pm √(b^2 - 4ac))/(2a)


x = \frac{-10 \pm √(10^2 - 4(1)(-4.8)){2(1)}


x = \frac{-10 \pm √(100 + 19.2 )){2}


x = \frac{-10 \pm √(119.2 )){2}


x = 0.46 or
x = -10.46

We discard the negative solution.

Answer: the border is 0.46 m wide.

User Nick Clarke
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