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Question 1 of 5
Find the area of the circle
Round your answer to 1 decimal place
Use π=3.14
Incorrect
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Solve for the area using the formula: A=πr2
Area |
= |
π×radius2 |
Area of a Circle Formula |
|
= |
3.14×52 |
Plug in the known values |
|
= |
3.14×25 |
Evaluate |
|
= |
78.5 cm2 |
The given measurements are in centimetres, so the area is measured as square centimetres
Area=78.5 cm2
The answer will depend on which π you use.
In this solution we used: π=3.14.
π=3.14 |
78.5 cm2 |
π=3.141592654 |
78.5 cm2 |
π=227 |
78.6 cm2 |
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Question 2 of 5
Find the area of the circle
Round your answer to 1 decimal place
Use π=3.14
Incorrect
Given Lengths
diameter=21
First, find the radius of the circle. Note that the radius is half of the diameter
radius |
= |
diameter2 |
|
|
= |
212 |
|
radius |
= |
10.5 |
Finally, solve for the area using the formula: A=πr2
Area |
= |
π×radius2 |
Area of a Circle Formula |
|
= |
3.14×10.52 |
Plug in the known values |
|
= |
3.14×110.25 |
Evaluate |
|
= |
346.185 |
|
= |
346.2 m2 |
Rounded to 1 decimal place |
The given measurements are in metres, so the area is measured as square metres
Area=346.2 m2
The answer will depend on which π you use.
In this solution we used: π=3.14.
π=3.14 |
346.2 m2 |
π=3.141592654 |
346.4 m2 |
π=227 |
346.5 m2 |
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Question 3 of 5
Find the area of the Circle
Round your answer to 1 decimal place
Use π=3.14
Incorrect
Given Lengths
diameter=13.8
First, find the radius of the circle
Note that the radius is half of the diameter
radius |
= |
diameter2 |
|
|
= |
13.82 |
|
radius |
= |
6.9 |
Finally, solve for the area using the formula: A=πr2
Area |
= |
π×radius2 |
Area of a Circle Formula |
|
= |
3.14×6.92 |
Plug in the known values |
|
= |
3.14×47.61 |
Evaluate |
|
= |
149.4954 |
|
= |
149.5 km2 |
Rounded to 1 decimal place |
The given measurements are in kilometres, so the area is measured as square kilometres
Area=149.5 km2
The answer will depend on which π you use.
In this solution we used: π=3.14.
π=3.14 |
149.5 km2 |
π=3.141592654 |
149.6 km2 |
π=227 |
149.6 km2 |
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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.
π=3.14
Incorrect
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Given Lengths
radius (Whole Circle)=7
thickness (Shaded Region)=3
First, solve for the area of the Whole Circle
AreaWhole Circle |
= |
π×radius2 |
|
= |
3.14×72 |
Area |
= |
153.86 cm2 |
Find the radius of the Inner Circle by subtracting the thickness of the shaded region from the radius of the Whole Circle.
radiusInner Circle |
= |
radius-thickness |
|
= |
7-3 |
|
= |
4 cm |
Next, find the area of the Inner Circle
AreaInner Circle |
= |
π×radius2 |
|
= |
3.14×42 |
Area |
= |
50.24 cm2 |
Finally, subtract the area of the Inner Circle from the area of the Whole Circle
Final Area |
= |
153.86-50.24 |
|
= |
103.62 |
|
= |
103.6 cm2 |
Rounded to 1 decimal place |
The given measurements are in centimetres, so the area is measured as square centimetres
Area=103.6 cm2
The answer will depend on which π you use.
In this solution we used: π=3.14.
π=3.14 |
103.6 cm2 |
π=3.141592654 |
103.7 cm2 |
π=227 |
103.7 cm2 |
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Question 5 of 5
Find the area of the yellow-shaded region.
Round your answer to 1 decimal place
Use π=3.14
Incorrect
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Given Lengths
radius (Whole Semicircle)=9
thickness (Shaded Region)=2
First, solve for the area of the Whole Semicircle
AreaWhole Semicircle |
= |
12×π×radius2 |
|
= |
12×3.14×92 |
Area |
= |
127.17 cm2 |
Find the radius of the Inner Semicircle by subtracting the thickness of the shaded region from the radius of the Whole Semicircle.
radiusInner Semicircle |
= |
radius-thickness |
|
= |
9-2 |
|
= |
7 cm |
Next, find the area of the Inner Semicircle
AreaInner Semicircle |
= |
12×π×radius2 |
|
= |
12×3.14×72 |
Area |
= |
76.93 cm2 |
Finally, subtract the area of the Inner Semicircle from the area of the Whole Semicircle
Final Area |
= |
127.17-76.93 |
|
= |
50.24 |
|
= |
50.2 cm2 |
Rounded to 1 decimal place |
The given measurements are in centimetres, so the area is measured as square centimetres
Area=50.2 cm2
The answer will depend on which π you use.
In this solution we used: π=3.14.
π=3.14 |
50.2 cm2 |
π=3.141592654 |
50.3 cm2 |
π=227 |
50.3 cm2 |