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## Polynomials word problems

Current time:0:00Total duration:4:37

# Polynomial word problem: rectangle and circle area

CCSS Math: HSA.APR.A.1, HSA.CED.A.1

## Video transcript

Write a binomial to express
the difference between the area of a rectangle with
length p and width 2r and the area of a
circle with diameter 4r. And they tell us that
p is greater than 7r. So let's first
think about the area of a rectangle with
length p and width 2r. So this is our
rectangle right here. It has a length of p and
it has a width of 2r. So what's its area? Well, it's just going to be
the length times the width. So the area here is going to be
p-- or maybe I should say 2rp. This is the length
times the width, or the width times the length. So area is equal to
2rp for the rectangle. Now, we also want to find the
difference between this area and the area of a
circle with diameter 4r. So what's the area of
the circle going to be? So let me draw our
circle over here. So our circle looks like that. Its diameter is 4r. How do we figure out
the area of a circle? Well, area is equal
to pi r squared for a circle, where
r is a radius. They gave us the diameter. The radius is half of that. So the radius here is going to
be half this distance, or 2r. So the area of our circle
is going to be pi times 2r, the whole thing squared. This is the radius, right? So we're squaring
the entire radius. So this is going to be equal
to pi times 4 times r squared. I'm just squaring
each of these terms. Or if we were to change the
order, the area of the circle is equal to 4 pi r squared. And we want to find
the difference. So to find a difference,
It's helpful-- just so we don't end up
with a negative number-- to figure out which of
these two is larger. So they're telling us that
p is greater than 7 r. So let's think about this. If p is greater than 7r, then
2-- let me write it this way. We know that p is
greater than 7r. So if we're going to multiply
both sides of this equation by 2rr-- and 2r
is positive, we're dealing with positive
distances, positive lengths-- so if we multiply both sides
of this equation by 2r, it shouldn't change
the equation. So multiply that by 2r, and
then multiply this by 2r. And then our equation
becomes 2rp is greater than 14r squared. Now, why is this interesting? Actually, why did I even
multiply this by 2r? Well, that's so that
this becomes the same as the area of the rectangle. So this is the area
of the rectangle. And what's 14r squared? Well, 4 times pi, is going to
get us something less than 14. This is less than 14. So this is 4 pi is less than 14. 14 is 4 times 3 and 2--
let me put it this way. 4 times 3.5 is equal to 14. Right? So 4 times pi, which
is less than 3.5, is going to be less than 14. So we know that
this over here is larger than this
quantity over here. It's larger than 4 pi r squared. And so we know
that this rectangle has a larger area
than the circle. So we can just subtract
the circle's area from the rectangle's area
to find the difference. So the difference is going to
be the area of the rectangle, which we already
figured out is 2rp. And we're going to subtract from
that the area of the circle. The area of the circle
is 4 pi r squared. So hopefully that made sense. And one point I want to clarify. I gave the equation of the area
of a circle to be pi r squared. And then we said that the radius
is actually 2r in this case. So I substituted 2r for r. Hopefully that
doesn't confuse you. This r is the general
term for any radius. They later told us
that the actual radius is 2 times some letter r. So I substitute that
into the formula. Anyway, hopefully you
found that useful.