How to find the area of a circle - SAT Math

The area of a circle is πr2.

The diameter of the first circle = 6” so radius of the first circle = 3” so the area = π * 32 = 9π in2

The radius of the second circle = 8” so the area = π * 82 = 64π in2

The area of the interstitial space = area of the first circle – area of the second circle.

Area = 64π in2 - 9π in2 = 55π in2

We know that the equation for the area of a circle is π r2. To solve this problem, we pick radii for Circles A and B, making sure that Circle A’s radius is three times Circle B’s radius, as the problem specifies. Then we will divide the resulting areas of the two circles. For example, if we say that Circle A has radius 6 and Circle B has radius 2, then the ratio of the area of Circle A to B is: (π 62)/(π 22) = 36π/4π. From here, the π's cancel out, leaving 36/4 = 9.

Area of a circle = A = πr2

R = 1/2d = ½(10) = 5

A = 52π = 25π

Explanation: Visualizing the rectangle inside the circle (corners touching the circumference of the circle and the center of the rectangle is the center of the circle) you will see that the rectangle can be divided into 8 congruent right triangles, with the hypotenuse as the radius of the circle. Calculating the radius you divide each side of the rectangle by two for the sides of each right triangle (giving 6 and 8). The hypotenuse (by pythagorean theorem or just knowing right triangle sets) the hypotenuse is give as 10. Area of a circle is given by πr2. 102 is 100, so 100π is the area.

The area of a circle is A=πr2. Since the radius was tripled 144π =π(3r)2. Divide by π and then take the square root of both sides of the equal sign to get 12=3r, and then r=4. The diameter (d) is equal to twice the radius so d= 2(4) = 8.

The equation is already in a circle equation, and the right side of the equation stands for r2 → r2 = 81 and r = 9

The area of a circle is πr2, so the area of this circle is 81π.

The circumference of a circle = πd where d = diameter. Therefore, this circle’s diameter must equal 16. Knowing that diameter = 2 times the radius, we can determine that the radius of this circle = 8. Halving the radius would give us a new radius of 4. To find the area of this new circle, use the formula A=πr² where r = radius. Plug in 4 for r. Area will equal 16π.

Using the Pythagorean Theorem, the diameter of the circle (also the diagonal of the square) can be found to be 4√2. Thus, the radius of the circle is half of the diameter, or 2√2. The area of the circle is then π(2√2)2, which equals 8π. Next, the area of the square must be subtracted from the entire circle, yielding an area of 8π-16 square inches.

The area of an annulus is

where is the radius of the larger circle, and is the radius of the smaller circle.

The area of the circle is (30/2)2*3.14 (π) = 706.5, since the shaded region is simply the area difference between the circle and the star, it’s 706.5-345 = 361.5

If the circumference is , then since we know . We further know that , so

The original radius would be 4, making the new radius 8 and by the area of a circle (A=π(r)2) the new area would be 64π.

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.

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.

If you answered that the whirlpool’s area is 18.84 feet and therefore fits, you are incorrect because 18.84 is the circumference of the whirlpool, not the area.

If you answered that the area of the whirlpool is 56.52 feet, you multiplied the area of the whirlpool by 1.5 and assumed that that was the correct area, not the legal limit.

If you answered that the area of the backyard was smaller than the area of the whirlpool, you did not calculate area correctly.

And if you thought the area of the backyard was 30 feet, you found the perimeter of the backyard, not the area.

The correct answer is that the area of the whirlpool is 28.26 feet and, when multiplied by 1.5 = 42.39, which is smaller than the area of the backyard, which is 56 square feet.

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.

The diameter of the circle is the length of a side of the square. Therefore, first solve for the length of the square's sides. The area of the square is:

A = s2 or 1089 = s2. Taking the square root of both sides, we get: s = 33.

Now, based on this, we know that 2r = 33 or r = 16.5. The area of the circle is πr2 or π16.52 = 272.25π.

The diameter of the circle is the length of a side of the square. Therefore, first solve for the length of the square's sides. The area of the square is:

A = s2 or 32 = s2. Taking the square root of both sides, we get: s = √32 = √(25) = 4√2.

Now, based on this, we know that 2r = 4√2 or r = 2√2. The area of the circle is πr2 or π(2√2)2 = 4 * 2π = 8π.

The smallest block from which a circle could be made would be a square that perfectly matches the diameter of the given circle. (This is presuming we have perfectly calibrated equipment.) Such a square would have dimensions equal to the diameter of the circle, meaning it would have sides of 88 inches for our problem. Its total area would be 88 * 88 or 7744 in2.

Now, the waste amount would be the "corners" remaining after the circle was cut. The area of the circle is πr2 or π * 442 = 1936π in2. Therefore, the area remaining would be 7744 – 1936π. The cost of the waste would be 0.25 * (7744 – 1936π). This is not an option for our answers, so let us simplify a bit. We can factor out a common 4 from our subtraction. This would give us: 0.25 * 4 * (1936 – 484π). Since 0.25 is equal to 1/4, 0.25 * 4 = 1. Therefore, our final answer is: 1936 – 484π dollars.

The smallest block from which a circle could be made would be a square that perfectly matches the diameter of the given circle. (This is presuming we have perfectly calibrated equipment.) Such a square would have dimensions equal to the diameter of the circle, meaning it would have sides of 50 inches for our problem. Its total area would be 50 * 50 or 2500 in2.

Now, the waste amount would be the "corners" remaining after the circle was cut. The area of the circle is πr2 or π * 252 = 625π in2. Therefore, the area remaining would be 2500 - 625π. The cost of the waste would be 0.2 * (2500 – 625π). This is not an option for our answers, so let us simplify a bit. We can factor out a common 5 from our subtraction. This would give us: 0.2 * 5 * (500 – 125π). Since 0.2 is equal to 1/5, 0.2 * 625 = 125. Therefore, our final answer is: 500 – 125π dollars.