# Express Area as a function of r

• rocomath
In summary, the conversation discusses the calculations for finding the perimeter and area of a shape consisting of a rectangle with semi-circles at each end. There is some confusion over the correct formulas to use, but ultimately it is determined that the perimeter is 2πr + 2l and the area is πr^2 + 2lr. The correct solution is found and it is determined that the book may have an error in its answer.
rocomath
Ok, so I have this picture. A semi-circle at the ends of a rectangle. It tells me that the perimeter is $$\frac 1 4$$. So, isn't the total perimeter just the sum of the circle and rectange? $$P=2\pi r+2(L+W)$$

$$r=radius$$
$$W=2r$$

$$P=2\pi r+2(L+2r)$$

And isn't the total area just the sum of the areas? $$A=A_{circle}+A_{rectange}\rightarrow A=\pi r^2 + 2rL$$

Almost. It's only a semi-circle. So the circumerence of that is just pi*r. And the perimeter of partial rectangle is just 2L+r. Why did you double everything?

Does it look like this:
.._____
(|____|)

or not? If it does look like the above then total perimeter is not the sum of the circle & rectangle.

EnumaElish said:
Does it look like this:
.._____
(|____|)

or not? If it does look like the above then total perimeter is not the sum of the circle & rectangle.
Yes, that is the correct picture! :-]

So it's not? eek.

If by perimeter you mean "edges exposed to the outside" then it is the two half-circles plus the 2 long edges of the rectangle. The short edges are "internalized."

EnumaElish said:
If by perimeter you mean "edges exposed to the outside" then it is the two half-circles plus the 2 long edges of the rectangle. The short edges are "internalized."
AHH! Yes, very true.

Thanks a lot :-]

EnumaElish said:
If by perimeter you mean "edges exposed to the outside" then it is the two half-circles plus the 2 long edges of the rectangle. The short edges are "internalized."
So even after ignoring the width, I still was unable to solve it. The person I was helping me showed me the solution and the internal parts were included. Blah.

Then I guess they tricked you. What was the exact phrasing of the problem?

Dick said:
Then I guess they tricked you. What was the exact phrasing of the problem?
Sorry, I don't have the book, I will post it tomorrow. >:-[

The wording you gave was clear. You have a rectangle with a semicircle at each end of radius r. That end of the rectangle, then, has length 2r. For the moment call the length of the other sides l. Then the area of the two semi-circles totals $\pi r^2$ and the area of the rectangle is lr. The total area of figure is $\pi r^2+ lr$.

The perimeter of the figure is the distance around the two semi-circles, $2\pi r$ and the two lengths, 2l: the perimeter is $2\pi r+ 2l= 1/4$. You can solve that for l as a function of r and replace l by that in the area formula.

HallsofIvy said:
The wording you gave was clear. You have a rectangle with a semicircle at each end of radius r. That end of the rectangle, then, has length 2r. For the moment call the length of the other sides l. Then the area of the two semi-circles totals $\pi r^2$ and the area of the rectangle is lr. The total area of figure is $\pi r^2+ lr$.

The perimeter of the figure is the distance around the two semi-circles, $2\pi r$ and the two lengths, 2l: the perimeter is $2\pi r+ 2l= 1/4$. You can solve that for l as a function of r and replace l by that in the area formula.
$$\frac 1 4=2\pi r+2l \rightarrow l=\frac{1-8\pi r}{8}$$

$$A=\pi r^2+lr$$

$$A=\pi r^2 +\left(\frac{1-8\pi r}{8}\right)r$$

$$A=\frac r 8$$

That's still not the answer in the book! Is the book wrong? I will post the actual problem in a few hours, got to go library.

Last edited:
... and the area of the rectangle is lr

Wait a minute guys! Is not the area of the rectangle = $(2r)l$
?

This would make

$$A=\pi r^2+2lr$$

Your right. I have no idea why I wrote rl!

## 1. What does it mean to "express area as a function of r"?

Expressing area as a function of r means to write the equation for the area of a shape in terms of the radius (r) of that shape. This allows you to calculate the area of the shape for any given radius value.

## 2. How do you express the area of a circle as a function of r?

The equation for the area of a circle is A = πr², where r is the radius of the circle. This means that the area of a circle is a function of r, and you can write it as A(r) = πr².

## 3. Can you express the area of a rectangle as a function of r?

No, the area of a rectangle is not a function of r. The area of a rectangle is given by the formula A = l*w, where l is the length and w is the width of the rectangle. Since the area does not depend on the radius, it cannot be expressed as a function of r.

## 4. Why is it useful to express area as a function of r?

Expressing area as a function of r allows you to easily calculate the area of a shape for any given radius value. It also allows you to compare the areas of different shapes with the same radius, such as circles and annuli.

## 5. Can you express the area of a triangle as a function of r?

No, the area of a triangle is not a function of r. The area of a triangle is given by the formula A = 1/2 * b * h, where b is the base and h is the height of the triangle. Since the area does not depend on the radius, it cannot be expressed as a function of r.

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