Working Backward and Measuring Parts of Circles | Grade 7 Math | LevelKite

You can already measure a circle when its radius or diameter is stated. Here is a quick reminder of the names and letters:

The radius $r$ is the distance from the center to the edge of the circle.
The diameter $d$ is the distance straight across the circle through its center, so $d = 2r$ .
The circumference $C$ is the distance around the circle, and $C = \pi d$ .
The area $A$ is the amount of flat space inside the circle, and $A = \pi r^2$ .

The radius, diameter and circumference are lengths, measured in units such as centimeters (cm), and each is greater than $0$ . The area is measured in square units such as square centimeters ( $\text{cm}^2$ ).

This lesson adds two new skills. You will work backward from a stated circumference or area to the radius or diameter. You will also find the area of, and the distance around, half of a circle and one-fourth of a circle.

Working backward from circumference. The relationship $C = \pi d$ says: multiply the diameter by $\pi$ and you get the distance around. Now suppose a problem states the distance around, a positive length, and asks for the diameter. You need to undo the multiplication by $\pi$ .

Division does that. Dividing both sides of a true equation by the same number keeps it true, as long as that number is not zero. Since $\pi$ is not zero, you may divide both sides of $C = \pi d$ by $\pi$ . On the right side, $\pi d \div \pi$ is just $d$ . So for a stated circumference $C > 0$ , the diameter is $d = C \div \pi$ , with $\pi$ in the form the problem states.

For example, a diameter of $5$ cm gives a circumference of $5\pi$ cm. Going backward, $5\pi \div \pi = 5$ , and you are back at $5$ cm.

Once you have the diameter, halve it to get the radius, because $d = 2r$ .

Worked example

From circumference to diameter and radius

A circle has a circumference of $25.12$ cm. Use $3.14$ for $\pi$ . Find the diameter and the radius of the circle.

Start from $C = \pi d$ , where $d$ is the diameter in centimeters. To undo the multiplication by $\pi$ , divide both sides by $\pi$ . The equation stays true because both sides are divided by the same number, and that number is not zero.
$d = C \div \pi$
Put in the stated circumference, $25.12$ , and use $3.14$ for $\pi$ . To divide by a decimal, multiply both numbers by $100$ , which does not change the result of the division. Then $2512 \div 314 = 8$ , because $314 \times 8 = 2512$ .
$d \approx 25.12 \div 3.14 = 2512 \div 314 = 8$
The radius is half of the diameter, because $d = 2r$ . Divide the diameter by $2$ .
$r \approx 8 \div 2 = 4$
Check by working forward with the same $3.14$ : multiply your diameter by $3.14$ . You get the stated circumference back, so the answers fit.
$3.14 \times 8 = 25.12$

Answer

Using $3.14$ for $\pi$ , the diameter is about $8$ cm and the radius is about $4$ cm.

Working backward from area. The relationship $A = \pi r^2$ does two things to the radius. First it squares the radius ( $r^2$ means $r \times r$ ). Then it multiplies the result by $\pi$ . To get back to the radius from a stated positive area, undo those two things in reverse order:

Divide the area by $\pi$ , in the form the problem states. You may do this because $\pi$ is not zero. It leaves $r \times r = A \div \pi$ .
Find the positive number that, multiplied by itself, gives that value. That number is the radius.

For example, a radius of $5$ cm gives an area of $\pi \times 5 \times 5 = 25\pi\text{ cm}^2$ . Going backward, $25\pi \div \pi = 25$ . Both $5 \times 5$ and $(-5) \times (-5)$ equal $25$ , but a radius is a length, so $r > 0$ and only $5$ fits. You are back at $5$ cm.

In this lesson, $r \times r$ is always a perfect square you can recognize: a whole number multiplied by itself, such as $25 = 5 \times 5$ or $100 = 10 \times 10$ . After you find the radius, double it to get the diameter, because $d = 2r$ .

Worked example

From area to radius and diameter

A circle has an area of $49\pi\text{ cm}^2$ , stated in terms of $\pi$ . Find the radius and the diameter of the circle.

Write the area relationship with the stated area in place of $A$ . Let $r$ be the radius in centimeters. It is a length, so $r > 0$ .
$49\pi = \pi r^2$
Undo the multiplication by $\pi$ : divide both sides by $\pi$ . The equation stays true because $\pi$ is not zero and both sides are divided by it. On the left, $49\pi \div \pi = 49$ . On the right, $\pi r^2 \div \pi = r^2$ .
$49\pi \div \pi = \pi r^2 \div \pi \quad \text{so} \quad 49 = r^2$
Now $r \times r = 49$ . Find the positive number that, multiplied by itself, gives $49$ . That number is $7$ , because $7 \times 7 = 49$ and the radius must be positive.
$r = 7 \quad \text{because } 7 \times 7 = 49 \text{ and } r > 0$
The diameter is twice the radius, because $d = 2r$ .
$d = 2 \times 7 = 14$
Check by working forward from your radius with $A = \pi r^2$ . You get the stated area back, so the answers fit.
$\pi \times 7^2 = \pi \times 49 = 49\pi$

Answer

The radius is $7$ cm and the diameter is $14$ cm.

Parts of a circle. A semicircle is exactly half of a circle. Imagine cutting a circle straight across through its center. The cut runs along a diameter and makes two matching halves, so a semicircle has half of the circle's area. Its outline has two parts: a curved edge that is half of the circumference, and one straight edge that is the diameter.

A quarter circle is exactly one-fourth of a circle. Imagine two straight cuts from the center to the edge, along two radii (the plural of radius) that meet in a right angle. Four pieces like this fill the whole circle, because four right angles make a full turn. So a quarter circle has one-fourth of the circle's area. Its outline has three parts: a curved edge that is one-fourth of the circumference, and two straight edges that are radii.

The perimeter of a shape is the total distance around its outline. For a semicircle or a quarter circle, that means the curved edge plus every straight edge.

The table works this out for one circle with a radius of $6$ cm and a diameter of $12$ cm, keeping $\pi$ as a symbol.

One circle with radius 6 cm and diameter 12 cm, and two parts cut from it

Measure	Full circle	Semicircle	Quarter circle
Share of the circle	all of it	$\tfrac{1}{2}$	$\tfrac{1}{4}$
Area	$\pi \times 6^2 = \pi \times 36 = 36\pi\text{ cm}^2$	$\tfrac{1}{2} \times 36\pi = 18\pi\text{ cm}^2$	$\tfrac{1}{4} \times 36\pi = 9\pi\text{ cm}^2$
Curved edge	$\pi \times 12 = 12\pi$ cm	$\tfrac{1}{2} \times 12\pi = 6\pi$ cm	$\tfrac{1}{4} \times 12\pi = 3\pi$ cm
Straight edges	none	one diameter: $12$ cm	two radii: $6 + 6 = 12$ cm
Perimeter	$12\pi$ cm	$(6\pi + 12)$ cm	$(3\pi + 12)$ cm

Key concept

Two relationships, run backward and cut into parts

For a circle with radius $r > 0$ and diameter $d = 2r$ , the relationships $C = \pi d$ and $A = \pi r^2$ also work backward and in parts. Use $\pi$ in the form the problem states.

Backward: a stated positive circumference gives $d = C \div \pi$ . A stated positive area gives $r \times r = A \div \pi$ , and $r$ is the positive number whose square is that value (a perfect square you can recognize, in this lesson).
In parts: a semicircle is $\tfrac{1}{2}$ of a circle and a quarter circle is $\tfrac{1}{4}$ . Take that fraction of the area, and of the circumference for the curved edge. The perimeter is the curved edge plus the straight edges: $\tfrac{1}{2}\pi d + d$ for a semicircle and $\tfrac{1}{4}\pi d + 2r$ for a quarter circle.

C = \pi d \qquad A = \pi r^2 \qquad (d = 2r > 0)

Worked example

Area and perimeter of a quarter circle

A quarter circle is one-fourth of a circle with a radius of $10$ cm. Its two straight edges are radii, each $10$ cm long, and they meet in a right angle. Use $3.14$ for $\pi$ . Find the area and the perimeter of the quarter circle.

Find the area of the full circle from the stated radius with $A = \pi r^2$ , using $3.14$ for $\pi$ . Square the radius first: $10^2 = 10 \times 10 = 100$ . The area is in square centimeters.
$A \approx 3.14 \times 10^2 = 3.14 \times 100 = 314$
The quarter circle is one-fourth of the full circle, so take one-fourth of the full area by dividing by $4$ .
$\tfrac{1}{4} \times 314 = 314 \div 4 = 78.5$
Find the full circumference with $C = \pi d$ . The diameter is twice the radius, because $d = 2r$ . Lengths are in centimeters.
$d = 2 \times 10 = 20, \qquad C \approx 3.14 \times 20 = 62.8$
The curved edge is one-fourth of the circumference, so divide by $4$ .
$\tfrac{1}{4} \times 62.8 = 62.8 \div 4 = 15.7$
The perimeter goes around the whole outline, so add the two straight edges to the curved edge. Each straight edge is a radius of $10$ cm.
$15.7 + 10 + 10 = 35.7$

Answer

Using $3.14$ for $\pi$ , the quarter circle has an area of about $78.5\text{ cm}^2$ and a perimeter of about $35.7$ cm.

Takeaway. For a circle with $d = 2r > 0$ , the relationships $C = \pi d$ and $A = \pi r^2$ work in more than one direction. To go backward, divide a stated positive circumference by $\pi$ to get the diameter. Or divide a stated positive area by $\pi$ , then find the positive number whose square is the result to get the radius. For a semicircle take $\tfrac{1}{2}$ , and for a quarter circle take $\tfrac{1}{4}$ , of the full area and of the full circumference. That share of the circumference is only the curved edge, so add the straight edges to get the perimeter. Keep $\pi$ in the form the problem states. Next, you will use these skills to solve circumference and area word problems.