Welcome to an application of perimeter. In high school,
Frank's basketball coach made the team run fifteen times around the entire court after
every practice. The dimensions of a high school basketball court are fifty feet by eighty-four feet.
This measurement does not include the two-foot edge, or the two-foot boundary line.
Let's begin by labeling the dimensions of the basketball court using the inner
edge and outer edge. So if the court itself is fifty feet by eighty-four feet, which means this
inner edge has a length of fifty feet,
this inner edge has a length of eighty-four feet.
Well now this boundary line, or the boundary edge is two feet wide and therefore, this
distance and this distance are two feet at each corner.
Let's label the lengths of the outer edges.
This outer edge is has a length of fifty feet plus two feet, plus two feet, which equals
fifty-four feet.
This outer edge as the length of eighty-four feet plus two feet, plus two feet, which equals
eighty-eight feet. And now let's look at the questions. If the boys had to stay outside the lines
of the court, what was the least distance they would run? Give the answer in feet and
miles. And b, how much more would someone run if they stayed on the outside edge verses the
inside edge.
Give your final answer in feet and miles. Let's begin by defining the
variables to keep things organized.
Let's let p sub-one equal that perimeter of the inner edge,
p sub-two equal the perimeter of the outer edge and therefore, we'll let r sub-one
equal the distance run using p sub-one, the inner edge. And r sub-two equal
distance run using p sub-two, which is the outer edge.
And then finally, we'll let d equal the difference and distance run.
Let's begin by determining the perimeter using the inner edge and outer edge.
Using the inner edge, p sub-one is equal to the sum of the lengths of the four sides of
the inner rectangle.
Or using the perimeter formula we'd have two times fifty feet, plus two times eighty-four feet,
which is equal to one hundred plus one hundred sixty-eight, which equals two hundred sixty-eight feet.
And now we'll determine p sub-two, the perimeter around the outer edge.
This would be equal to two times fifty-four feet, plus two times eighty-eight feet,
which is equal to one hundred eight plus one hundred fifty-six, which equals two hundred eighty-four feet.
And now let's determine r sub-one and r sub-two. The total distance run if the boys have to
run fifteen times around the entire court.
So r sub-one is equal to fifteen times p sub-one,
the perimeter around the inner edge. So fifteen times two hundred sixty-eight feet, which is equal to four
thousand twenty feet. And r sub-two is a distance run fifteen times around the outer
edge, which is fifteen times two hundred eighty-four feet,
which is equal to four thousand two hundred sixty feet.
Let's continue on the next slide. Part a asks, if the boys had to stay outside
the lines of the court, what was the least distance they would run? Give the answer
in feet and miles.
And we now know that the boys ran around the outside edge,
they would run at least four thousand, two hundred sixty feet. But we're asked
to express this distance also in miles.
So now let's convert four thousand, two hundred sixty feet to miles. To do this let's first write this distance
as a fraction with the denominator of one. So we'd have four thousand, two hundred
sixty feet over one. And now we will multiply by a unit fraction, or a
fraction equal to one to convert feet to miles. To do this we'll use the
conversion one mile equals five thousand, two hundred eighty feet. Because we want feet to simplify out,
we want a unit fraction where the units of feet are in the denominator
and we'll have miles in the numerator. And the conversion is one mile is equal
to five thousand, two hundred eighty feet.
Notice this fraction is equal to one and in this form, the units of feet simplify out.
So multiplying we have four thousand, two hundred sixty divided by five thousand, two hundred eighty. And now the units are miles.
And now we simplify the fraction. Four thousand, two hundred sixty and five thousand, two hundred eighty share a common factor of sixty.
So to simplify, we divide the numerator and denominator by sixty, which gives us the
simplified fraction of seventy-one eighty-eighths, and the units are now miles.
So now we know if the boys stay outside the lines of the court,
they will run at least four thousand two hundred sixty feet, which equals seventy-one, eighty-eighths of a mile.
Now for part b
we're asked, how much more would someone run that stayed on the outside edge verses
the inside edge? Give the answer in feet and miles. To answer this question we
first need to find the difference in distance run, running along the outside
edge verses the inside edge. So we first determine the value of d. D is equal to
the difference in the distance run, which would be four thousand, two hundred
sixty feet
minus four thousand twenty feet, which is equal to two hundred forty feet.
So this is how much more someone would run, running along the outside edge verses
the inside edge. But we also need to express this distance in miles.
So starting with two hundred forty feet, as a fraction, with the denominator of one we multiply it by
a unit fraction, to convert the units from feet to miles.
We will use the same unit fraction we did above, which is one mile per five thousand, two hundred eighty feet.
Again notice how the units of feet simplify out leaving us with the
fraction of two hundred forty divided by five thousand, two hundred eighty. And now the units are miles.
To simplify this fraction, two hundred forty
and five thousand, two hundred eighty share a common factor of two hundred forty. And therefore, we divide two hundred forty by two hundred forty.
And divide five thousand, two hundred eighty by two hundred forty,
which gives us a simplified fraction of one over twenty-two miles.
So now we know if someone runs along the outer edge
he runs at least two hundred forty feet, or one twenty-second of a mile longer.
I hope you found this helpful.