A uniform-width boardwalk is built around the inside edge of a rectangular parkland that is 10m by 5m. If the boardwalk takes up 20% of the lot, how wide is the boardwalk to the nearest centimetre?
I have no idea what to do after figuring out that:
Total area= 150m^2
Boardwalk area= 30m^2
Parkland area= 120m^2
Draw a figure. Now look at the area of the boardwalk: 2(Lenght*w ) + 2(Width-2w)w
where w is the width of the boardwarlk, and Lenght, Width is the Lot dimensions, which are given, as is area Length*width>
solve for w
To find the width of the boardwalk, we need to use the information given and set up an equation.
Let's denote the width of the boardwalk as 'x'. The length and width of the parkland, including the boardwalk, would then be (10+2x) and (5+2x) meters, respectively.
We know that the total area of the parkland, including the boardwalk, is 150 square meters. Therefore, we can set up the following equation:
(10+2x)(5+2x) = 150
Now, we need to solve this equation to find the value of 'x' to the nearest centimeter.
Expanding the left side of the equation:
(10+2x)(5+2x) = 150
50 + 20x + 4x + 10x^2 = 150
10x^2 + 24x + 50 = 150
Rearranging the equation:
10x^2 + 24x + 50 - 150 = 0
10x^2 + 24x - 100 = 0
Dividing both sides by 2 to simplify the equation:
5x^2 + 12x - 50 = 0
Now, you can solve this quadratic equation either by factoring, completing the square, or using the quadratic formula. Once you find the two solutions (possible values for 'x'), select the positive value since the width cannot be negative.
Finally, you can round your answer to the nearest centimeter to find the width of the boardwalk.