How to find the area of a sector - PSAT Math
Card 0 of 42
The radius of the circle above is 
and 
. What is the area of the shaded section of the circle?
The radius of the circle above is
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Area of Circle = πr2 = π42 = 16π
Total degrees in a circle = 360
Therefore 45 degree slice = 45/360 fraction of circle = 1/8
Shaded Area = 1/8 * Total Area = 1/8 * 16π = 2π
Area of Circle = πr2 = π42 = 16π
Total degrees in a circle = 360
Therefore 45 degree slice = 45/360 fraction of circle = 1/8
Shaded Area = 1/8 * Total Area = 1/8 * 16π = 2π
is a square.
The arc from 
to 
is a semicircle with a center at the midpoint of 
.
All units are in feet.
The diagram shows a plot of land.
The cost of summer upkeep is $2.50 per square foot.
In dollars, what is the total upkeep cost for the summer?
The arc from
All units are in feet.
The diagram shows a plot of land.
The cost of summer upkeep is $2.50 per square foot.
In dollars, what is the total upkeep cost for the summer?
Tap to see back →
To solve this, we must begin by finding the area of the diagram, which is the area of the square less the area of the semicircle.
The area of the square is straightforward:
30 * 30 = 900 square feet
Because each side is 30 feet long, AB + BC + CD = 30.
We can substitute BC for AB and CD since all three lengths are the same:
BC + BC + BC = 30
3BC = 30
BC = 10
Therefore the diameter of the semicircle is 10 feet, so the radius is 5 feet.
The area of the semi-circle is half the area of a circle with radius 5. The area of the full circle is 52π = 25π, so the area of the semi-circle is half of that, or 12.5π.
The total area of the plot is the square less the semicircle: 900 - 12.5π square feet
The cost of upkeep is therefore 2.5 * (900 – 12.5π) = $(2250 – 31.25π).
To solve this, we must begin by finding the area of the diagram, which is the area of the square less the area of the semicircle.
The area of the square is straightforward:
30 * 30 = 900 square feet
Because each side is 30 feet long, AB + BC + CD = 30.
We can substitute BC for AB and CD since all three lengths are the same:
BC + BC + BC = 30
3BC = 30
BC = 10
Therefore the diameter of the semicircle is 10 feet, so the radius is 5 feet.
The area of the semi-circle is half the area of a circle with radius 5. The area of the full circle is 52π = 25π, so the area of the semi-circle is half of that, or 12.5π.
The total area of the plot is the square less the semicircle: 900 - 12.5π square feet
The cost of upkeep is therefore 2.5 * (900 – 12.5π) = $(2250 – 31.25π).
A circular, 8-slice pizza is placed in a square box that has dimensions four inches larger than the diameter of the pizza. If the box covers a surface area of 256 in2, what is the surface area of one piece of pizza?
A circular, 8-slice pizza is placed in a square box that has dimensions four inches larger than the diameter of the pizza. If the box covers a surface area of 256 in2, what is the surface area of one piece of pizza?
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The first thing to do is calculate the dimensions of the pizza box. Based on our data, we know 256 = s2. Solving for s (by taking the square root of both sides), we get 16 = s (or s = 16).
Now, we know that the diameter of the pizza is four inches less than 16 inches. That is, it is 12 inches. Be careful! The area of the circle is given in terms of radius, which is half the diameter, or 6 inches. Therefore, the area of the pizza is π * 62 = 36π in2. If the pizza is 8-slices, one slice is equal to 1/8 of the total pizza or (36π)/8 = 4.5π in2.
The first thing to do is calculate the dimensions of the pizza box. Based on our data, we know 256 = s2. Solving for s (by taking the square root of both sides), we get 16 = s (or s = 16).
Now, we know that the diameter of the pizza is four inches less than 16 inches. That is, it is 12 inches. Be careful! The area of the circle is given in terms of radius, which is half the diameter, or 6 inches. Therefore, the area of the pizza is π * 62 = 36π in2. If the pizza is 8-slices, one slice is equal to 1/8 of the total pizza or (36π)/8 = 4.5π in2.
In the figure, PQ is the arc of a circle with center O. If the area of the sector is 3piwhat is the perimeter of sector?
In the figure, PQ is the arc of a circle with center O. If the area of the sector is 3piwhat is the perimeter of sector?
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First, we figure out what fraction of the circle is contained in sector OPQ: $$\frac{30^{circ}$$$}{360^{circ}$}= $\frac{1}{12}$, so the total area of the circle is 12times 3pi=36 .
Using the formula for the area of a circle, ${pi}r^{2}$, we can see that r=6.
We can use this to solve for the circumference of the circle, 2{pi}r, or 12{pi}.
Now, OP and OQ are both equal to r, and PQ is equal to $\frac{1}{12}$ of the circumference of the circle, or {pi}.
To get the perimeter, we add OP + OQ + PQ, which give us 12+{pi}.
First, we figure out what fraction of the circle is contained in sector OPQ: $$\frac{30^{circ}$$$}{360^{circ}$}= $\frac{1}{12}$, so the total area of the circle is 12times 3pi=36 .
Using the formula for the area of a circle, ${pi}r^{2}$, we can see that r=6.
We can use this to solve for the circumference of the circle, 2{pi}r, or 12{pi}.
Now, OP and OQ are both equal to r, and PQ is equal to $\frac{1}{12}$ of the circumference of the circle, or {pi}.
To get the perimeter, we add OP + OQ + PQ, which give us 12+{pi}.
If B is a circle with line AC = 12 and line BC = 16, then what is the area formed by DBE?
If B is a circle with line AC = 12 and line BC = 16, then what is the area formed by DBE?
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Line AB is a radius of Circle B, which can be found using the Pythagorean Theorem:
$AB^2$$=AC^2$$+BC^2rightarrow$ $AB=$\sqrt{AC^2$$+BC^2$$}$=$\sqrt{16^2$$+12^2$}$=$\sqrt{400}$=20
Since AB is a radius of B, we can find the area of circle B via:
Area=pi $R^2$$=pi(20^2$)=400pi
Angle DBE is a right angle, and therefore 
of the circle so it follows:
Area(DBE)=\frac{400}{4}$pi=100pi
Line AB is a radius of Circle B, which can be found using the Pythagorean Theorem:
$AB^2$$=AC^2$$+BC^2rightarrow$ $AB=$\sqrt{AC^2$$+BC^2$$}$=$\sqrt{16^2$$+12^2$}$=$\sqrt{400}$=20
Since AB is a radius of B, we can find the area of circle B via:
Area=pi $R^2$$=pi(20^2$)=400pi
Angle DBE is a right angle, and therefore
Area(DBE)=\frac{400}{4}$pi=100pi
What is the area of the 
sector above if the radius of circle is 
?
What is the area of the
Tap to see back →
To find the area of a sector, first find the area of the whole circle.
The radius of the circle is 9, so
which can be reduced to 
.
The area of the sector is only a portion of the total area.To find out exactly how large the area is, set up a proportion where one side equals the angle measure over 360
Multiply both sides by 81
and you will solve for x, which equals 
To find the area of a sector, first find the area of the whole circle.
The radius of the circle is 9, so
The area of the sector is only a portion of the total area.To find out exactly how large the area is, set up a proportion where one side equals the angle measure over 360
Multiply both sides by 81
The radius of the circle above is and . What is the area of the shaded section of the circle?
The radius of the circle above is
Tap to see back →
Area of Circle = πr2 = π42 = 16π
Total degrees in a circle = 360
Therefore 45 degree slice = 45/360 fraction of circle = 1/8
Shaded Area = 1/8 * Total Area = 1/8 * 16π = 2π
Area of Circle = πr2 = π42 = 16π
Total degrees in a circle = 360
Therefore 45 degree slice = 45/360 fraction of circle = 1/8
Shaded Area = 1/8 * Total Area = 1/8 * 16π = 2π
is a square.
The arc from to is a semicircle with a center at the midpoint of .
All units are in feet.
The diagram shows a plot of land.
The cost of summer upkeep is $2.50 per square foot.
In dollars, what is the total upkeep cost for the summer?
The arc from
All units are in feet.
The diagram shows a plot of land.
The cost of summer upkeep is $2.50 per square foot.
In dollars, what is the total upkeep cost for the summer?
Tap to see back →
To solve this, we must begin by finding the area of the diagram, which is the area of the square less the area of the semicircle.
The area of the square is straightforward:
30 * 30 = 900 square feet
Because each side is 30 feet long, AB + BC + CD = 30.
We can substitute BC for AB and CD since all three lengths are the same:
BC + BC + BC = 30
3BC = 30
BC = 10
Therefore the diameter of the semicircle is 10 feet, so the radius is 5 feet.
The area of the semi-circle is half the area of a circle with radius 5. The area of the full circle is 52π = 25π, so the area of the semi-circle is half of that, or 12.5π.
The total area of the plot is the square less the semicircle: 900 - 12.5π square feet
The cost of upkeep is therefore 2.5 * (900 – 12.5π) = $(2250 – 31.25π).
To solve this, we must begin by finding the area of the diagram, which is the area of the square less the area of the semicircle.
The area of the square is straightforward:
30 * 30 = 900 square feet
Because each side is 30 feet long, AB + BC + CD = 30.
We can substitute BC for AB and CD since all three lengths are the same:
BC + BC + BC = 30
3BC = 30
BC = 10
Therefore the diameter of the semicircle is 10 feet, so the radius is 5 feet.
The area of the semi-circle is half the area of a circle with radius 5. The area of the full circle is 52π = 25π, so the area of the semi-circle is half of that, or 12.5π.
The total area of the plot is the square less the semicircle: 900 - 12.5π square feet
The cost of upkeep is therefore 2.5 * (900 – 12.5π) = $(2250 – 31.25π).
A circular, 8-slice pizza is placed in a square box that has dimensions four inches larger than the diameter of the pizza. If the box covers a surface area of 256 in2, what is the surface area of one piece of pizza?
A circular, 8-slice pizza is placed in a square box that has dimensions four inches larger than the diameter of the pizza. If the box covers a surface area of 256 in2, what is the surface area of one piece of pizza?
Tap to see back →
The first thing to do is calculate the dimensions of the pizza box. Based on our data, we know 256 = s2. Solving for s (by taking the square root of both sides), we get 16 = s (or s = 16).
Now, we know that the diameter of the pizza is four inches less than 16 inches. That is, it is 12 inches. Be careful! The area of the circle is given in terms of radius, which is half the diameter, or 6 inches. Therefore, the area of the pizza is π * 62 = 36π in2. If the pizza is 8-slices, one slice is equal to 1/8 of the total pizza or (36π)/8 = 4.5π in2.
The first thing to do is calculate the dimensions of the pizza box. Based on our data, we know 256 = s2. Solving for s (by taking the square root of both sides), we get 16 = s (or s = 16).
Now, we know that the diameter of the pizza is four inches less than 16 inches. That is, it is 12 inches. Be careful! The area of the circle is given in terms of radius, which is half the diameter, or 6 inches. Therefore, the area of the pizza is π * 62 = 36π in2. If the pizza is 8-slices, one slice is equal to 1/8 of the total pizza or (36π)/8 = 4.5π in2.
In the figure, PQ is the arc of a circle with center O. If the area of the sector is 3piwhat is the perimeter of sector?
In the figure, PQ is the arc of a circle with center O. If the area of the sector is 3piwhat is the perimeter of sector?
Tap to see back →
First, we figure out what fraction of the circle is contained in sector OPQ: $$\frac{30^{circ}$$$}{360^{circ}$}= $\frac{1}{12}$, so the total area of the circle is 12times 3pi=36 .
Using the formula for the area of a circle, ${pi}r^{2}$, we can see that r=6.
We can use this to solve for the circumference of the circle, 2{pi}r, or 12{pi}.
Now, OP and OQ are both equal to r, and PQ is equal to $\frac{1}{12}$ of the circumference of the circle, or {pi}.
To get the perimeter, we add OP + OQ + PQ, which give us 12+{pi}.
First, we figure out what fraction of the circle is contained in sector OPQ: $$\frac{30^{circ}$$$}{360^{circ}$}= $\frac{1}{12}$, so the total area of the circle is 12times 3pi=36 .
Using the formula for the area of a circle, ${pi}r^{2}$, we can see that r=6.
We can use this to solve for the circumference of the circle, 2{pi}r, or 12{pi}.
Now, OP and OQ are both equal to r, and PQ is equal to $\frac{1}{12}$ of the circumference of the circle, or {pi}.
To get the perimeter, we add OP + OQ + PQ, which give us 12+{pi}.
If B is a circle with line AC = 12 and line BC = 16, then what is the area formed by DBE?
If B is a circle with line AC = 12 and line BC = 16, then what is the area formed by DBE?
Tap to see back →
Line AB is a radius of Circle B, which can be found using the Pythagorean Theorem:
$AB^2$$=AC^2$$+BC^2rightarrow$ $AB=$\sqrt{AC^2$$+BC^2$$}$=$\sqrt{16^2$$+12^2$}$=$\sqrt{400}$=20
Since AB is a radius of B, we can find the area of circle B via:
Area=pi $R^2$$=pi(20^2$)=400pi
Angle DBE is a right angle, and therefore of the circle so it follows:
Area(DBE)=\frac{400}{4}$pi=100pi
Line AB is a radius of Circle B, which can be found using the Pythagorean Theorem:
$AB^2$$=AC^2$$+BC^2rightarrow$ $AB=$\sqrt{AC^2$$+BC^2$$}$=$\sqrt{16^2$$+12^2$}$=$\sqrt{400}$=20
Since AB is a radius of B, we can find the area of circle B via:
Area=pi $R^2$$=pi(20^2$)=400pi
Angle DBE is a right angle, and therefore
Area(DBE)=\frac{400}{4}$pi=100pi
What is the area of the sector above if the radius of circle is ?
What is the area of the
Tap to see back →
To find the area of a sector, first find the area of the whole circle.
The radius of the circle is 9, so
which can be reduced to .
The area of the sector is only a portion of the total area.To find out exactly how large the area is, set up a proportion where one side equals the angle measure over 360
Multiply both sides by 81 and you will solve for x, which equals
To find the area of a sector, first find the area of the whole circle.
The radius of the circle is 9, so
The area of the sector is only a portion of the total area.To find out exactly how large the area is, set up a proportion where one side equals the angle measure over 360
Multiply both sides by 81
The radius of the circle above is and . What is the area of the shaded section of the circle?
The radius of the circle above is
Tap to see back →
Area of Circle = πr2 = π42 = 16π
Total degrees in a circle = 360
Therefore 45 degree slice = 45/360 fraction of circle = 1/8
Shaded Area = 1/8 * Total Area = 1/8 * 16π = 2π
Area of Circle = πr2 = π42 = 16π
Total degrees in a circle = 360
Therefore 45 degree slice = 45/360 fraction of circle = 1/8
Shaded Area = 1/8 * Total Area = 1/8 * 16π = 2π
is a square.
The arc from to is a semicircle with a center at the midpoint of .
All units are in feet.
The diagram shows a plot of land.
The cost of summer upkeep is $2.50 per square foot.
In dollars, what is the total upkeep cost for the summer?
The arc from
All units are in feet.
The diagram shows a plot of land.
The cost of summer upkeep is $2.50 per square foot.
In dollars, what is the total upkeep cost for the summer?
Tap to see back →
To solve this, we must begin by finding the area of the diagram, which is the area of the square less the area of the semicircle.
The area of the square is straightforward:
30 * 30 = 900 square feet
Because each side is 30 feet long, AB + BC + CD = 30.
We can substitute BC for AB and CD since all three lengths are the same:
BC + BC + BC = 30
3BC = 30
BC = 10
Therefore the diameter of the semicircle is 10 feet, so the radius is 5 feet.
The area of the semi-circle is half the area of a circle with radius 5. The area of the full circle is 52π = 25π, so the area of the semi-circle is half of that, or 12.5π.
The total area of the plot is the square less the semicircle: 900 - 12.5π square feet
The cost of upkeep is therefore 2.5 * (900 – 12.5π) = $(2250 – 31.25π).
To solve this, we must begin by finding the area of the diagram, which is the area of the square less the area of the semicircle.
The area of the square is straightforward:
30 * 30 = 900 square feet
Because each side is 30 feet long, AB + BC + CD = 30.
We can substitute BC for AB and CD since all three lengths are the same:
BC + BC + BC = 30
3BC = 30
BC = 10
Therefore the diameter of the semicircle is 10 feet, so the radius is 5 feet.
The area of the semi-circle is half the area of a circle with radius 5. The area of the full circle is 52π = 25π, so the area of the semi-circle is half of that, or 12.5π.
The total area of the plot is the square less the semicircle: 900 - 12.5π square feet
The cost of upkeep is therefore 2.5 * (900 – 12.5π) = $(2250 – 31.25π).
A circular, 8-slice pizza is placed in a square box that has dimensions four inches larger than the diameter of the pizza. If the box covers a surface area of 256 in2, what is the surface area of one piece of pizza?
A circular, 8-slice pizza is placed in a square box that has dimensions four inches larger than the diameter of the pizza. If the box covers a surface area of 256 in2, what is the surface area of one piece of pizza?
Tap to see back →
The first thing to do is calculate the dimensions of the pizza box. Based on our data, we know 256 = s2. Solving for s (by taking the square root of both sides), we get 16 = s (or s = 16).
Now, we know that the diameter of the pizza is four inches less than 16 inches. That is, it is 12 inches. Be careful! The area of the circle is given in terms of radius, which is half the diameter, or 6 inches. Therefore, the area of the pizza is π * 62 = 36π in2. If the pizza is 8-slices, one slice is equal to 1/8 of the total pizza or (36π)/8 = 4.5π in2.
The first thing to do is calculate the dimensions of the pizza box. Based on our data, we know 256 = s2. Solving for s (by taking the square root of both sides), we get 16 = s (or s = 16).
Now, we know that the diameter of the pizza is four inches less than 16 inches. That is, it is 12 inches. Be careful! The area of the circle is given in terms of radius, which is half the diameter, or 6 inches. Therefore, the area of the pizza is π * 62 = 36π in2. If the pizza is 8-slices, one slice is equal to 1/8 of the total pizza or (36π)/8 = 4.5π in2.
In the figure, PQ is the arc of a circle with center O. If the area of the sector is 3piwhat is the perimeter of sector?
In the figure, PQ is the arc of a circle with center O. If the area of the sector is 3piwhat is the perimeter of sector?
Tap to see back →
First, we figure out what fraction of the circle is contained in sector OPQ: $$\frac{30^{circ}$$$}{360^{circ}$}= $\frac{1}{12}$, so the total area of the circle is 12times 3pi=36 .
Using the formula for the area of a circle, ${pi}r^{2}$, we can see that r=6.
We can use this to solve for the circumference of the circle, 2{pi}r, or 12{pi}.
Now, OP and OQ are both equal to r, and PQ is equal to $\frac{1}{12}$ of the circumference of the circle, or {pi}.
To get the perimeter, we add OP + OQ + PQ, which give us 12+{pi}.
First, we figure out what fraction of the circle is contained in sector OPQ: $$\frac{30^{circ}$$$}{360^{circ}$}= $\frac{1}{12}$, so the total area of the circle is 12times 3pi=36 .
Using the formula for the area of a circle, ${pi}r^{2}$, we can see that r=6.
We can use this to solve for the circumference of the circle, 2{pi}r, or 12{pi}.
Now, OP and OQ are both equal to r, and PQ is equal to $\frac{1}{12}$ of the circumference of the circle, or {pi}.
To get the perimeter, we add OP + OQ + PQ, which give us 12+{pi}.
If B is a circle with line AC = 12 and line BC = 16, then what is the area formed by DBE?
If B is a circle with line AC = 12 and line BC = 16, then what is the area formed by DBE?
Tap to see back →
Line AB is a radius of Circle B, which can be found using the Pythagorean Theorem:
$AB^2$$=AC^2$$+BC^2rightarrow$ $AB=$\sqrt{AC^2$$+BC^2$$}$=$\sqrt{16^2$$+12^2$}$=$\sqrt{400}$=20
Since AB is a radius of B, we can find the area of circle B via:
Area=pi $R^2$$=pi(20^2$)=400pi
Angle DBE is a right angle, and therefore of the circle so it follows:
Area(DBE)=\frac{400}{4}$pi=100pi
Line AB is a radius of Circle B, which can be found using the Pythagorean Theorem:
$AB^2$$=AC^2$$+BC^2rightarrow$ $AB=$\sqrt{AC^2$$+BC^2$$}$=$\sqrt{16^2$$+12^2$}$=$\sqrt{400}$=20
Since AB is a radius of B, we can find the area of circle B via:
Area=pi $R^2$$=pi(20^2$)=400pi
Angle DBE is a right angle, and therefore
Area(DBE)=\frac{400}{4}$pi=100pi
What is the area of the sector above if the radius of circle is ?
What is the area of the
Tap to see back →
To find the area of a sector, first find the area of the whole circle.
The radius of the circle is 9, so
which can be reduced to .
The area of the sector is only a portion of the total area.To find out exactly how large the area is, set up a proportion where one side equals the angle measure over 360
Multiply both sides by 81 and you will solve for x, which equals
To find the area of a sector, first find the area of the whole circle.
The radius of the circle is 9, so
The area of the sector is only a portion of the total area.To find out exactly how large the area is, set up a proportion where one side equals the angle measure over 360
Multiply both sides by 81
The radius of the circle above is and . What is the area of the shaded section of the circle?
The radius of the circle above is
Tap to see back →
Area of Circle = πr2 = π42 = 16π
Total degrees in a circle = 360
Therefore 45 degree slice = 45/360 fraction of circle = 1/8
Shaded Area = 1/8 * Total Area = 1/8 * 16π = 2π
Area of Circle = πr2 = π42 = 16π
Total degrees in a circle = 360
Therefore 45 degree slice = 45/360 fraction of circle = 1/8
Shaded Area = 1/8 * Total Area = 1/8 * 16π = 2π
is a square.
The arc from to is a semicircle with a center at the midpoint of .
All units are in feet.
The diagram shows a plot of land.
The cost of summer upkeep is $2.50 per square foot.
In dollars, what is the total upkeep cost for the summer?
The arc from
All units are in feet.
The diagram shows a plot of land.
The cost of summer upkeep is $2.50 per square foot.
In dollars, what is the total upkeep cost for the summer?
Tap to see back →
To solve this, we must begin by finding the area of the diagram, which is the area of the square less the area of the semicircle.
The area of the square is straightforward:
30 * 30 = 900 square feet
Because each side is 30 feet long, AB + BC + CD = 30.
We can substitute BC for AB and CD since all three lengths are the same:
BC + BC + BC = 30
3BC = 30
BC = 10
Therefore the diameter of the semicircle is 10 feet, so the radius is 5 feet.
The area of the semi-circle is half the area of a circle with radius 5. The area of the full circle is 52π = 25π, so the area of the semi-circle is half of that, or 12.5π.
The total area of the plot is the square less the semicircle: 900 - 12.5π square feet
The cost of upkeep is therefore 2.5 * (900 – 12.5π) = $(2250 – 31.25π).
To solve this, we must begin by finding the area of the diagram, which is the area of the square less the area of the semicircle.
The area of the square is straightforward:
30 * 30 = 900 square feet
Because each side is 30 feet long, AB + BC + CD = 30.
We can substitute BC for AB and CD since all three lengths are the same:
BC + BC + BC = 30
3BC = 30
BC = 10
Therefore the diameter of the semicircle is 10 feet, so the radius is 5 feet.
The area of the semi-circle is half the area of a circle with radius 5. The area of the full circle is 52π = 25π, so the area of the semi-circle is half of that, or 12.5π.
The total area of the plot is the square less the semicircle: 900 - 12.5π square feet
The cost of upkeep is therefore 2.5 * (900 – 12.5π) = $(2250 – 31.25π).