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Example Questions
Example Question #2 : Diameter And Chords
The perimeter of a circle is 36 π. What is the diameter of the circle?
18
3
36
6
72
36
The perimeter of a circle = 2 πr = πd
Therefore d = 36
Example Question #3 : Diameter And Chords
If the area of the circle touching the square in the picture above is , what is the closest value to the area of the square?
Obtain the radius of the circle from the area.
Split the square up into 4 triangles by connecting opposite corners. These triangles will have a right angle at the center of the square, formed by two radii of the circle, and two 45-degree angles at the square's corners. Because you have a 45-45-90 triangle, you can calculate the sides of the triangles to be , , and . The radii of the circle (from the center to the corners of the square) will be 9. The hypotenuse (side of the square) must be .
The area of the square is then .
Example Question #4 : Diameter And Chords
Two legs of a right triangle measure 3 and 4, respectively. What is the area of the circle that circumscribes the triangle?
For the circle to contain all 3 vertices, the hypotenuse must be the diameter of the circle. The hypotenuse, and therefore the diameter, is 5, since this must be a 3-4-5 right triangle.
The equation for the area of a circle is A = πr2.
Example Question #5 : How To Find The Length Of The Diameter
Note: Figure NOT drawn to scale.
In the above circle, the length of arc is , and . What is the diameter of the circle?
Call the diameter . Since , is of the circle, and is of a circle with circumference .
is in length, so
Example Question #1 : How To Find The Length Of The Diameter
Note: Figure NOT drawn to scale.
In the above circle, the length of arc is 10, and . Give the diameter of the circle. (Nearest tenth).
Insufficient information exists to answer the question.
Call the diameter . Since , is of a circle with circumference . Since it is of length 10, the circumference of the circle is 5 times this, or 50. Therefore, set in the circumference formula:
Example Question #31 : 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 + 36π
12 + 36π
24 + 6π
12 + 6π
24π
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 : Radius
Two concentric circles have circumferences of 4π and 10π. What is the difference of the radii of the two circles?
5
6
3
7
4
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 #2 : How To Find The Length Of A Radius
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 – 18π
204 – 18π
336 – 18π
336 – 36π
96 – 36π
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:
Example Question #1 : How To Find The Length Of A Radius
A circle has an area of 36π inches. What is the radius of the circle, in inches?
36
18
9
6
6
We know that the formula for the area of a circle is πr2. Therefore, we must set 36π equal to this formula to solve for the radius of the circle.
36π = πr2
36 = r2
6 = r
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