# Parts of Areas and Circumferences of Circles

Rate this video

Ø 5.0 / 2 ratings

The author
Eugene Lee

## Information about the videoParts of Areas and Circumferences of Circles

When dealing with a fractional part of a circle, called a sector, multiply the fraction by πr² to get its area or by 2πr to get its circumference. A knowledge of parts of areas and circumferences of circles plays a major role in an architect’s design of a dome that is essential in terms of determining material requirement and ensuring safety. This also comes in handy in creating graphic designs and is always put into consideration when constructing curves and arcs in amusement parks. Learn how to compute for areas and circumferences of sectors by helping the event coordinators of the 5th Annual Polar Games find suitable ice floe locations for the Penguin Sumo Wrestling, Polar Bear Figure Skating, and Walrus Swimming events. Common Core Reference: CCSS.MATH.CONTENT.7.G.B.4

## Parts of Areas and Circumferences of Circles Exercise

Would you like to practice what you’ve just learned? Practice problems for this video Parts of Areas and Circumferences of Circles help you practice and recap your knowledge.
• ### Find the area of the sectors.

Hints

$A_{\text{circle}} =${$\pi~r^2$}, where $\pi\approx 3.14$ and $r$ the radius of the circle.

Don't forget about the units. Areas are $m^2$ or ...

Solution

For the polar bear figure skating the coordinators need the largest area. The other area will be used for the penguin sumo wrestling.

So the coordinators have to determine the area of a part of a circle two times.

They need the formula for the area of circles:

$A_{\text{circle}}=\pi~r^2$.

with $\pi\approx3.14$ and $r$ the radius of the circle.

Because each ice floe is just a part of the circle, they have to multiply this formula by the corresponding fraction.

Let's start with ice floe number $1$:

$A_{Floe1}=\frac13~\pi~r^2$.

Plugging in the given values ($\pi\approx 3.14$ and $r=30~m$) we get

$\begin{array}{rcl} A_{Floe1}&=&\frac13(3.14)(30)^2\\ &=&\frac13(3.14)(900)\\ &\approx&942 \end{array}$

The area of ice floe number $1$, which is what we were looking for, is $942~m^2$.

In a similar way we can determine the area of ice floe number $2$. Here we have to multiply the area formula by $\frac34$:

$A_{Floe2}=\frac34~\pi~r^2$.

Again we plug in the given values for $\pi$ and $r=10~m$

$\begin{array}{rcl} A_{Floe2}&=&\frac34(3.14)(10)^2\\ &=&\frac34(3.14)(100)\\ &\approx&235.5 \end{array}$

The area of ice floe number 2 is $235.5~m^2$.

Thus the coordinators decide to take ice floe number $1$ for the polar bear figure skating and number $2$ for the penguin sumo wrestling.

• ### Find the circumferences of the sectors.

Hints

$C_{\text{circle}}=${$2~\pi~r$}, where $\pi\approx 3.14$ and $r$ is the radius of the circle.

Don't forget about the units. The circumference of a circle is a length, which in this case will be given as $m$, and in other cases could be $ft$, $km$, or $mi$.

Solution

The coordinators of the games have a lot of work to do. Now, for the walrus swimming event, they have to find out which sector has the longest circumference.

Therefore, they have to determine the circumferences for both ice floes.

Fortunately, they know the formula for the circumferences of circles:

$C_{\text{circle}}=2~\pi~r$,

with $\pi\approx3.14$ and $r$ the radius of the circle.

Because each ice floe isn't a whole circle (it's just a part) they have to multiply this formula with the regarding fraction.

They start with ice floe number $1$:

$C_{Floe1}=\frac13~2~\pi~r^2$.

Plugging in the given values ($\pi\approx 3.14$ and $r=30~m$) they get

$\begin{array}{rcl} C_{Floe1}&=&\frac13(2)(3.14)(30)\\ &=&\frac13(3.14)(60)\\ &\approx&62.8 \end{array}$

The circumference of ice floe number $1$ is $62.8~m$.

They just have to determine the circumference of ice floe number $2$. For this they multiply the circumference formula by $\frac34$:

$C_{Floe2}=\frac34~2~\pi~r^2$.

Finally they plug in the given values for $\pi$ and $r=10~m$:

$\begin{array}{rcl} C_{Floe2}&=&\frac34(2)(3.14)(10)\\ &=&\frac34(3.14)(20)\\ &\approx&47.1 \end{array}$

This gives them the circumference of ice floe number 2 as approximately $47.1~m$.

Wow, they got it. They decide to use ice floe number 1 for the walrus swimming event because it has a longer circumference.

The games will start.

• ### Decide which ice floe has enough area to house the Polar Games Village.

Hints

Use the approximate value $\pi\approx 3.14$.

Each time you have to multiply the area formula

$A_{\text{circle}}=\pi~r^2$

by the given fraction.

Just two ice floes are suitable.

Solution

We use the formula for the area of a circle

$A_{\text{circle}}=\pi~r^2$

Each time we have to multiply this formula by the given fraction.

$~$

Ice Floe #1

$A_{\text{Floe1}}=\frac12~\pi~r^2=\frac12(3.14)(20)^2=628$

Great, this ice floe is suitable because the area $628~m^2$ is larger than $527~m2$.

$~$

Ice Floe #2

$A_{\text{Floe2}}=\frac23~\pi~r^2=\frac23(3.14)(15)^2=471$

Too bad, this area $471~m^2$ is less than they require.

$~$

Ice Floe #3

$A_{\text{Floe2}}=\frac25~\pi~r^2=\frac25(3.14)(20)^2=502.4$

This floe isn't suitable either. The area $502.4~m^2$ is to small.

$~$

Ice Floe #4

$A_{\text{Floe4}}=\frac34~\pi~r^2=\frac34(3.14)(15)^2\approx 530$

Awesome. This ice floe, with an area of $530~m^2$, is suitable for the village.

$~$

Ice Floe #5

$A_{\text{Floe5}}=\frac14~\pi~r^2=\frac14(3.14)(25)^2\approx 491$

Sorry, that's not large enough. The village can't be build on this ice floe.

• ### Calculate the area of the stadium seating.

Hints

The circumferences of the sectors are already given.

Take care of the measures: The circumference is a length and thus has the measure $m$ or ...

Just multiply the given circumferences by the corresponding fraction.

Solution

We already know the circumferences of the ice floes:

• $C_{\text{Floe1}}=62.8~m$
• $C_{\text{Floe2}}=47.1~m$
To determine the needed length for the seating we still have to multiply those values with the given fraction.

• Regarding to ice floe #1 we get $\frac16\times 62.8~m=10.47~m$.
• The length corresponding to ice floe #2 is given by $\frac38\times 47.1~m=17.67~m$.
• ### Recall the formulas for the area and circumference of both a circle and a sector of a circle.

Hints

$A$ is the symbol for the area while $C$ stands for the circumference.

Pay attention to the meaning of the values

• $r$ is the radius of the circle.
• $d=2r$ is the diameter of the circle.
• $\pi\approx 3.14$

For area we can use units such as $m^2$ and for circumference we can use units such as $m$.

$r^2$ leads to the unit $m^2$.

Solution

Here are all the formulas you need.

• $r$ is the radius of the circle.
• $d=2r$ is the diameter of the circle.
• $\pi\approx 3.14$

• The area is given by $A_{\text{circle}}=\pi~r^2$ The units for the area are $m^2$ or ...
• The circumference is given by $C_{\text{circle}}=2~\pi~r=\pi~d$ The units for the circumference are $m$ or ...
If we want to determine the area or circumference of part of a circle, we have to multiply those formulas by the corresponding fraction.

For the given examples we get:

• The area of a third of a circle $A_{\text{sector of a circle}}=\frac13~\pi~r^2$
• The circumference of three quarters of a circle $C_{\text{sector of a circle}}=\frac34~2~\pi~r$
• ### Solve for the circumference of the ice floe for the Dolphin sprint.

Hints

Use the formula for the circumference of a circle.

$C_{\text{circle}}=2~\pi~r$.

Multiply the circumference formula by the corresponding fraction.

The smallest value is $26.17~m$ and the largest one $47.1~m$.

Solution

We have to use formula for the circumference of a circle

$C_{\text{circle}}=2~\pi~r$

For each given ice floe we multiply this formula with the corresponding fraction. The ice floe are already in the right order:

Ice Floe #5

$C_{\text{Floe5}}=\frac16~2~\pi~r=\frac16(2)(3.14)(25)\approx 26.17$

The circumference of this floe is given by $26.17~m$.

Ice Floe #4

$C_{\text{Floe4}}=\frac14~2~\pi~r=\frac14(2)(3.14)(18)=28.26$

The circumference of this floe is given by $28.26~m$.

Ice Floe #1

$C_{\text{Floe1}}=\frac12~2~\pi~r=\frac12(2)(3.14)(10)=31.4$

The circumference of this floe is given by $31.4~m$.

Ice Floe #2

$C_{\text{Floe2}}=\frac25~2~\pi~r=\frac25(2)(3.14)(15)=37.68$

The circumference of this floe is given by $37.68~m$.

Ice Floe #3

$C_{\text{Floe3}}=\frac35~2~\pi~r=\frac35(2)(3.14)(12)=45.216$

The circumference of this floe is given by $45.216~m$.

Ice Floe #6

$C_{\text{Floe6}}=\frac38~2~\pi~r=\frac38(2)(3.14)(20)=47.1$

The circumference of this floe is given by $41.7~m$.

So best the organizers should choose the ice floe#6 if it's still available.