We know that the area of a circle can be expressed: a = πr²

If we know that the area is 36π, we can substitute this into said equation and get: 36π = πr²

Solving for r, we get: 36 = r²; (after taking the square root of both sides:) 6 = r

Now, we know that the circuference of a circle is expressed: c = πd.  Since we know that d = 2r (two radii, placed one after the other, make a diameter), we can rewrite the circumference equation to be: c = 2πr

Since we have r, we can rewrite this to be: c = 2π*6 = 12π

Starting with the circle, we need to find the radius in order to get the circumference. Find  by plugging our given area into the equation for the area of a circle:

Then calculate circumference:

 (approximating  as 3.14)

To find the perimeter of the square, we can use  , where  is the perimeter and  is the side length:

, so the circle's circumference is greater.

Based on our information, we know that the 121π = πr²; 121 = r²; r = 11.

Our other circle with half the radius of A has a diameter equal to the radius of A.  Therefore, the circumference of this circle is 11π.  Half of this is 5.5π.  However, since this is a semi circle, it is enclosed and looks like this:

Therefore, we have to include the diameter in the perimeter.  Therefore, the total perimeter of the semi-circle is 5.5π + 11.

Let's compute each value separately. We know that the radii are positive numbers that are greater than or equal to . This means that we do not need to worry about the fact that the area could represent a square of a decimal value like .

Quantity A

Since , we know:

Quantity B

If the diameter is one-fourth the radius of A, we know:

Thus, the radius must be half of that, or .

Now, we need to compute the area of this circle. We know:

Therefore, 

Now, notice that if , Quantity A is larger.

However, if we choose a value like , we have:

Quantity A: 

Quantity B: 

Therefore, the relation cannot be determined!

Since we know that the area of Square  is , we know , where  is the length of one of its sides. From this, we can solve for  by taking the square root of both sides. You will have to do this by estimating upward. Therefore, you know that  is . By careful guessing, you can quickly see that  is . From this, you know that the diameter of your circle must be half of , or  (because it is circumscribed). Therefore, you can draw:

The circumference of this circle is defined as:

 or, for your values:

(You could also compute this from the diameter, but many students just memorize the formula above.)

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.

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

Try different values for the radius to see if a pattern emerges.  The formulas needed are Area = πr² and Perimeter = 2πr.

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

If r = 4, then the area = 16π and the perimeter = 8π, 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 = πr² = 4π.  The circumference = 2πr = 4π.  

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 = πr² = 25π 

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.