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Question 1 of 5
Find the area of the circle
Round your answer to `1` decimal place
Use `pi=3.14`
Incorrect
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Given Lengths
`\text(radius)=5`
Solve for the area using the formula: `A=pi``r^2`
`\text(Area)` |
`=` |
`pi times``\text(radius)^2` |
Area of a Circle Formula |
|
`=` |
`3.14 times ``5^2` |
Plug in the known values |
|
`=` |
`3.14 times 25` |
Evaluate |
|
`=` |
`78.5 cm^2` |
The given measurements are in centimetres, so the area is measured as square centimetres
`\text(Area)=78.5 cm^2`
The answer will depend on which `pi` you use.
In this solution we used: `pi=3.14`.
`pi=3.14` |
`78.5 cm^2` |
`pi=3.141592654` |
`78.5 cm^2` |
`pi=(22)/(7)` |
`78.6 cm^2` |
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Question 2 of 5
Find the area of the circle
Round your answer to `1` decimal place
Use `pi=3.14`
Incorrect
Given Lengths
`\text(diameter)=21`
First, find the radius of the circle. Note that the radius is half of the diameter
`\text(radius)` |
`=` |
$$\frac{\color{#00880a}{\text{diameter}}}{2}$$ |
|
|
`=` |
$$\frac{\color{#00880a}{\text{21}}}{2}$$ |
|
`\text(radius)` |
`=` |
`10.5` |
Finally, solve for the area using the formula: `A=pi``r^2`
`\text(Area)` |
`=` |
`pi times``\text(radius)^2` |
Area of a Circle Formula |
|
`=` |
`3.14 times ``10.5^2` |
Plug in the known values |
|
`=` |
`3.14 times 110.25` |
Evaluate |
|
`=` |
`346.185` |
|
`=` |
`346.2 m^2` |
Rounded to `1` decimal place |
The given measurements are in metres, so the area is measured as square metres
`\text(Area)=346.2 m^2`
The answer will depend on which `pi` you use.
In this solution we used: `pi=3.14`.
`pi=3.14` |
`346.2 m^2` |
`pi=3.141592654` |
`346.4 m^2` |
`pi=(22)/(7)` |
`346.5 m^2` |
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Question 3 of 5
Find the area of the Circle
Round your answer to `1` decimal place
Use `pi=3.14`
Incorrect
Given Lengths
`\text(diameter)=13.8`
First, find the radius of the circle
Note that the radius is half of the diameter
`\text(radius)` |
`=` |
$$\frac{\color{#00880a}{\text{diameter}}}{2}$$ |
|
|
`=` |
$$\frac{\color{#00880a}{\text{13.8}}}{2}$$ |
|
`\text(radius)` |
`=` |
`6.9` |
Finally, solve for the area using the formula: `A=pi``r^2`
`\text(Area)` |
`=` |
`pi times``\text(radius)^2` |
Area of a Circle Formula |
|
`=` |
`3.14 times ``6.9^2` |
Plug in the known values |
|
`=` |
`3.14 times 47.61` |
Evaluate |
|
`=` |
`149.4954` |
|
`=` |
`149.5 km^2` |
Rounded to `1` decimal place |
The given measurements are in kilometres, so the area is measured as square kilometres
`\text(Area)=149.5 km^2`
The answer will depend on which `pi` you use.
In this solution we used: `pi=3.14`.
`pi=3.14` |
`149.5 km^2` |
`pi=3.141592654` |
`149.6 km^2` |
`pi=(22)/(7)` |
`149.6 km^2` |
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Question 4 of 5
Find the area of the orange-shaded region.
The given measurements are in centimetres.
Round your answer to `1` decimal place.
`pi=3.14`
Incorrect
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Given Lengths
`\text(radius)` (Whole Circle)`=7`
`\text(thickness)` (Shaded Region)`=3`
First, solve for the area of the Whole Circle
`\text(Area)``\text(Whole Circle)` |
`=` |
`pi xx``\text(radius)^2` |
|
`=` |
`3.14 xx``7^2` |
`\text(Area)` |
`=` |
`153.86 cm^2` |
Find the radius of the Inner Circle by subtracting the thickness of the shaded region from the radius of the Whole Circle.
`\text(radius)``\text(Inner Circle)` |
`=` |
`\text(radius)``-``\text(thickness)` |
|
`=` |
`7``-``3` |
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`=` |
`4 cm` |
Next, find the area of the Inner Circle
`\text(Area)``\text(Inner Circle)` |
`=` |
`pi xx``\text(radius)^2` |
|
`=` |
`3.14 times ``4^2` |
`\text(Area)` |
`=` |
`50.24 cm^2` |
Finally, subtract the area of the Inner Circle from the area of the Whole Circle
`\text(Final Area)` |
`=` |
`153.86``-``50.24` |
|
`=` |
`103.62` |
|
`=` |
`103.6 cm^2` |
Rounded to `1` decimal place |
The given measurements are in centimetres, so the area is measured as square centimetres
`\text(Area)=103.6 cm^2`
The answer will depend on which `pi` you use.
In this solution we used: `pi=3.14`.
`pi=3.14` |
`103.6 cm^2` |
`pi=3.141592654` |
`103.7 cm^2` |
`pi=(22)/(7)` |
`103.7 cm^2` |
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Question 5 of 5
Find the area of the yellow-shaded region.
Round your answer to `1` decimal place
Use `pi=3.14`
Incorrect
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Given Lengths
`\text(radius)` (Whole Semicircle)`=9`
`\text(thickness)` (Shaded Region)`=2`
First, solve for the area of the Whole Semicircle
`\text(Area)``\text(Whole Semicircle)` |
`=` |
`1/2 xx pi xx``\text(radius)^2` |
|
`=` |
`1/2 xx 3.14 xx``9^2` |
`\text(Area)` |
`=` |
`127.17 cm^2` |
Find the radius of the Inner Semicircle by subtracting the thickness of the shaded region from the radius of the Whole Semicircle.
`\text(radius)``\text(Inner Semicircle)` |
`=` |
`\text(radius)``-``\text(thickness)` |
|
`=` |
`9``-``2` |
|
`=` |
`7 cm` |
Next, find the area of the Inner Semicircle
`\text(Area)``\text(Inner Semicircle)` |
`=` |
`1/2 xx pi xx``\text(radius)^2` |
|
`=` |
`1/2 xx 3.14 times ``7^2` |
`\text(Area)` |
`=` |
`76.93 cm^2` |
Finally, subtract the area of the Inner Semicircle from the area of the Whole Semicircle
`\text(Final Area)` |
`=` |
`127.17``-``76.93` |
|
`=` |
`50.24` |
|
`=` |
`50.2 cm^2` |
Rounded to `1` decimal place |
The given measurements are in centimetres, so the area is measured as square centimetres
`\text(Area)=50.2 cm^2`
The answer will depend on which `pi` you use.
In this solution we used: `pi=3.14`.
`pi=3.14` |
`50.2 cm^2` |
`pi=3.141592654` |
`50.3 cm^2` |
`pi=(22)/(7)` |
`50.3 cm^2` |