All SAT Math Resources
Example Questions
Example Question #51 : Plane Geometry
A circle has the equation below. What is the circumference of the circle?
(x – 2)2 + (y + 3)2 = 9
The radius is 3. Yielding a circumference of .
Example Question #6 : How To Find Circumference
Find the circumferencce fo a circle given radius of 7.
To solve, simply use the formula for the circumference of a circle. Thus,
Like the prior question, it is important to think about dimensions if you don't remember the exact formula. Circumference is 1 dimensional, so it makes sense that the variable is not squared as cubed. If you rather, you can use the following formula, but realize by defining diameter, it equals the prior one.
Thus,
Example Question #7 : How To Find Circumference
The area of a circle is . What is its circumference?
None of the given answers.
First, let's find the radius r of the circle by using the given area.
Now, plug this radius into the formula for a circle's circumference.
.
Example Question #5 : How To Find Circumference
The surface area of a sphere is .
is a point on the surface of the sphere; is the point on the sphere farthest from . A curve is drawn from to entirely on the surface of the sphere. Give the length of the shortest possible curve fitting this description.
Below is a sphere with its center and with points and as described.
For and to be on the surface of the sphere and to be a maximum distance apart, they must be endpoints of a diameter of the sphere. The shortest curve connecting them that is entirely on the surface is a semicircle whose radius coincides with that of the sphere. Therefore, first find the radius of the sphere using the surface area formula
Setting and solving for :
The length of the curve is half the circumference of a circle with radius 10, or
Substituting 10 for , this is
.
Example Question #101 : Circles
Give the circumference of a circle with the following diameter:
The circumference of a circle is its diameter multiplied by , so, since the diameter is 100, the circumference is simply .
Example Question #991 : Grade 7
Ashley has a square room in her apartment that measures 81 square feet. What is the circumference of the largest circular area rug that she can fit in the space?
In order to solve this question, first calculate the length of each side of the room.
The length of each side of the room is also equal to the length of the diameter of the largest circular rug that can fit in the room. Since , the circumference is simply
Example Question #82 : Circles
In a large field, a circle with an area of 144π square meters is drawn out. Starting at the center of the circle, a groundskeeper mows in a straight line to the circle's edge. He then turns and mows ¼ of the way around the circle before turning again and mowing another straight line back to the center. What is the length, in meters, of the path the groundskeeper mowed?
24 + 6π
12 + 6π
24π
24 + 36π
12 + 36π
24 + 6π
Circles have an area of πr2, where r is the radius. If this circle has an area of 144π, then you can solve for the radius:
πr2 = 144π
r 2 = 144
r =12
When the groundskeeper goes from the center of the circle to the edge, he's creating a radius, which is 12 meters.
When he travels ¼ of the way around the circle, he's traveling ¼ of the circle's circumference. A circumference is 2πr. For this circle, that's 24π meters. One-fourth of that is 6π meters.
Finally, when he goes back to the center, he's creating another radius, which is 12 meters.
In all, that's 12 meters + 6π meters + 12 meters, for a total of 24 + 6π meters.
Example Question #2 : How To Find The Length Of A Radius
Two concentric circles have circumferences of 4π and 10π. What is the difference of the radii of the two circles?
5
4
6
7
3
3
The circumference of any circle is 2πr, where r is the radius.
Therefore:
The radius of the smaller circle with a circumference of 4π is 2 (from 2πr = 4π).
The radius of the larger circle with a circumference of 10π is 5 (from 2πr = 10π).
The difference of the two radii is 5-2 = 3.
Example Question #109 : Plane Geometry
In the figure above, rectangle ABCD has a perimeter of 40. If the shaded region is a semicircle with an area of 18π, then what is the area of the unshaded region?
96 – 36π
336 – 18π
336 – 36π
96 – 18π
204 – 18π
96 – 18π
In order to find the area of the unshaded region, we will need to find the area of the rectangle and then subtract the area of the semicircle. However, to find the area of the rectangle, we will need to find both its length and its width. We can use the circle to find the length of the rectangle, because the length of the rectangle is equal to the diameter of the circle.
First, we can use the formula for the area of a circle in order to find the circle's radius. When we double the radius, we will have the diameter of the circle and, thus, the length of the rectangle. Then, once we have the rectangle's length, we can find its width because we know the rectangle's perimeter.
Area of a circle = πr2
Area of a semicircle = (1/2)πr2 = 18π
Divide both sides by π, then multiply both sides by 2.
r2 = 36
Take the square root.
r = 6.
The radius of the circle is 6, and therefore the diameter is 12. Keep in mind that the diameter of the circle is also equal to the length of the rectangle.
If we call the length of the rectangle l, and we call the width w, we can write the formula for the perimeter as 2l + 2w.
perimeter of rectangle = 2l + 2w
40 = 2(12) + 2w
Subtract 24 from both sides.
16 = 2w
w = 8.
Since the length of the rectangle is 12 and the width is 8, we can now find the area of the rectangle.
Area = l x w = 12(8) = 96.
Finally, to find the area of just the unshaded region, we must subtract the area of the circle, which is 18π, from the area of the rectangle.
area of unshaded region = 96 – 18π.
The answer is 96 – 18π.
Example Question #31 : Circles
Consider a circle centered at the origin with a circumference of . What is the x value when y = 3? Round your answer to the hundreths place.
5.77
10.00
5.8
None of the available answers
5.778
5.77
The formula for circumference of a circle is , so we can solve for r:
We now know that the hypotenuse of the right triangle's length is 13.5. We can form a right triangle from the unit circle that fits the Pythagorean theorem as such:
Or, in this case: