How to find the area of a circle - GRE Quantitative Reasoning

Here, you need to “solve backward” from the data you have been given. We know that 0.25C = 5.5; therefore, C = 22. In order to solve for the area, we will need the radius of the circle. This can be obtained by recalling that C = 2πr. Replacing 22 for C, we get 22 = 2πr.

Solve for r: r = 22 / 2π = 11 / π.

Now, we solve for the area: A = πr2. Replacing 11 / π for r: A = π (11 / π)2 = (121π) / (π2) = 121 / π.

There are two steps to this problem: determining the area of the circle and determining the area of the square. The area of the circle is πr2 which is π(2/1)2 or π. AD is a diameter of circle O and creates two isosceles right triangles with ACD and ABD. The relationship between sides of an isosceles right triangle is 1 : 1 : √2. Thus the sides of square ABCD are √2 and the area is 2. The area of the shaded region is the area of the circle minus the area of the square, or π – 2.

Therefore the 16 inch pizza is the better deal.

To find the radius of circle B, use the circumference formula (c = πd = 2πr):

2πr = 36π

Divide each side by 2π: r = 18

Now, if circle A has a radius half the length of that of B, A's radius is 18 / 2 = 9.

The area of a circle is πr2. Therefore, for A, it is π*92 = 81π.

Since the answers are in terms of pi, simply find the area of each circle in terms of x and ∏:

Smaller: ∏(5)2 = 25∏

Larger: ∏x2

We must subtract the inner circle from the outer circle; this translates to ∏x2-25∏.

Try different values for the radius to see if a pattern emerges. The formulas needed are Area = π r_2 and Perimeter = 2_πr.

If r = 1, then the Area = π and the Perimeter = 2_π_, so the perimeter is larger.

If r = 4, then the area = 16_π_ and the perimeter = 8_π_, so the area is larger.

Therefore the relationship cannot be determined from the information given.

This is one of the only special cases where the area equals the circumference of the circle. The Area = πr_2 = 4_π. The circumference = 2_πr_ = 4_π_.

Note: For a quantitative comparison such as this one where the columns have numeric values instead of variables, the answer will rarely be "cannot be determined".

Quantity A: area = base * height / 2 = 7 * 24/2 = 84

Quantity B: area = πr_2 = 25_π

Now we have to remember what π is. Using π = 3, the area is approximately 75. Using π = 3.14, the area increases a little bit, but no matter how exact an approximation for π, this area will never be larger than Quantity A.

To solve this problem, you must find the area of the entire circle (garden and sidewalk) and subtract it by the area of the inner garden. The entire area has a radius of 5 ft. (3 ft. radius of the garden plus the 2 ft. wide sidewalk), giving it an area of . The inner garden has a radius of 3 ft. and an area of . The difference is , which is the area of the sidewalk.

To find the area of the garden, you need to find the entire area and subtract that by the area of the inner circle, or the monument. The radius of the larger circle is 50, which makes its area . The radius of the inner circle is 30, which makes its area . The difference is .

If the area of the square is 9, then s2 = 9 and s = 3. If the sides thus equal 3, we can calculate the diagonals (either CB or AD) by using the 45-45-90 triangle ratio. For a side of 3, the diagonal will be 3√(2). Note that since the square is inscribed in the circle, this diagonal is also the diameter of the circle. If it is such, the radius is one half of that or 1.5√(2).

Based on that value, we can computer the circle’s area:

A = πr2 = π(1.5√(2))2 = (2.25 * 2)π = 4.5π

Here, you need to “solve backward” from the data you have been given. We know that 0.25C = 5.5; therefore, C = 22. In order to solve for the area, we will need the radius of the circle. This can be obtained by recalling that C = 2πr. Replacing 22 for C, we get 22 = 2πr.

Solve for r: r = 22 / 2π = 11 / π.

Now, we solve for the area: A = πr2. Replacing 11 / π for r: A = π (11 / π)2 = (121π) / (π2) = 121 / π.

There are two steps to this problem: determining the area of the circle and determining the area of the square. The area of the circle is πr2 which is π(2/1)2 or π. AD is a diameter of circle O and creates two isosceles right triangles with ACD and ABD. The relationship between sides of an isosceles right triangle is 1 : 1 : √2. Thus the sides of square ABCD are √2 and the area is 2. The area of the shaded region is the area of the circle minus the area of the square, or π – 2.

Therefore the 16 inch pizza is the better deal.

To find the radius of circle B, use the circumference formula (c = πd = 2πr):

2πr = 36π

Divide each side by 2π: r = 18

Now, if circle A has a radius half the length of that of B, A's radius is 18 / 2 = 9.

The area of a circle is πr2. Therefore, for A, it is π*92 = 81π.

Since the answers are in terms of pi, simply find the area of each circle in terms of x and ∏:

Smaller: ∏(5)2 = 25∏

Larger: ∏x2

We must subtract the inner circle from the outer circle; this translates to ∏x2-25∏.

Try different values for the radius to see if a pattern emerges. The formulas needed are Area = π r_2 and Perimeter = 2_πr.

If r = 1, then the Area = π and the Perimeter = 2_π_, so the perimeter is larger.

If r = 4, then the area = 16_π_ and the perimeter = 8_π_, so the area is larger.

Therefore the relationship cannot be determined from the information given.

This is one of the only special cases where the area equals the circumference of the circle. The Area = πr_2 = 4_π. The circumference = 2_πr_ = 4_π_.

Note: For a quantitative comparison such as this one where the columns have numeric values instead of variables, the answer will rarely be "cannot be determined".

Quantity A: area = base * height / 2 = 7 * 24/2 = 84

Quantity B: area = πr_2 = 25_π

Now we have to remember what π is. Using π = 3, the area is approximately 75. Using π = 3.14, the area increases a little bit, but no matter how exact an approximation for π, this area will never be larger than Quantity A.