All High School Math Resources
Example Questions
Example Question #231 : High School Math
Two equal circles are cut out of a rectangular sheet of paper with the dimensions 10 by 20. The circles were made to have the greatest possible diameter. What is the approximate area of the paper after the two circles have been cut out?
56
23
16
43
43
The length of 20 represents the diameters of both circles. Each circle has a diameter of 10 and since radius is half of the diameter, each circle has a radius of 5. The area of a circle is A = πr2 . The area of one circle is 25π. The area of both circles is 50π. The area of the rectangle is (10)(20) = 200. 200 - 50π gives you the area of the paper after the two circles have been cut out. π is about 3.14, so 200 – 50(3.14) = 43.
Example Question #232 : High School Math
Kate has a ring-shaped lawn which has an inner radius of 10 feet and an outer radius 25 feet. What is the area of her lawn?
525π ft2
275π ft2
125π ft2
175π ft2
325π ft2
525π ft2
The area of an annulus is
where is the radius of the larger circle, and is the radius of the smaller circle.
Example Question #233 : High School Math
A 6 by 8 rectangle is inscribed in a circle. What is the area of the circle?
The image below shows the rectangle inscribed in the circle. Dividing the rectangle into two triangles allows us to find the diameter of the circle, which is equal to the length of the line we drew. Using a2+b2= c2 we get 62 + 82 = c2. c2 = 100, so c = 10. The area of a circle is . Radius is half of the diameter of the circle (which we know is 10), so r = 5.
Example Question #234 : High School Math
A park wants to build a circular fountain with a walkway around it. The fountain will have a radius of 40 feet, and the walkway is to be 4 feet wide. If the walkway is to be poured at a depth of 1.5 feet, how many cubic feet of concrete must be mixed to make the walkway?
None of the other answers are correct.
The following diagram will help to explain the solution:
We are searching for the surface area of the shaded region. We can multiply this by the depth (1.5 feet) to find the total volume of this area.
The radius of the outer circle is 44 feet. Therefore its area is 442π = 1936π. The area of the inner circle is 402π = 1600π. Therefore the area of the shaded area is 1936π – 1600π = 336π. The volume is 1.5 times this, or 504π.
Example Question #1 : Area Of A Circle
How many times greater is the area of a circle with a radius of 4in., compared to a circle with a radius of 2in.?
The area of a circle can be solved using the equation
The area of a circle with radius 4 is while the area of a circle with radius 2 is .
Example Question #235 : High School Math
What is the area of a circle whose diameter is 8?
16π
8π
12π
64π
32π
16π
Example Question #236 : High School Math
In the following diagram, the radius is given. What is area of the shaded region?
This question asks you to apply the concept of area in finding both the area of a circle and square. Since the cirlce is inscribed in the square, we know that its diameter (two times the radius) is the same length as one side of the square. Since we are given the radius, , we can find the area of both the circle and square.
Square:
This gives us the area for the entire square.
The bottom half of the square has area .
Now that we have this value, we must find the area that the circle occupies. The area of a circle is given by .
So the area of this circle will be .
The bottom half of the circle has half that area:
Now that we have both our values, we can subtract the bottom half of the circle from the bottom half of the square to give us the shaded region:
Example Question #237 : High School Math
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
24 + 6π
12 + 36π
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 #352 : Sat Mathematics
Two concentric circles have circumferences of 4π and 10π. What is the difference of the radii of the two circles?
4
5
6
3
7
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.