How to find circumference - GRE Quantitative Reasoning

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.

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

A = pi $r^2$

25pi = pi $r^2$

25 = $r^2$

Then calculate circumference:

C = 2pi r

C = 2pi times 5 = 10pi approx 31.4 inches (approximating pi as 3.14)

To find the perimeter of the square, we can use P = 4s , where P is the perimeter and s is the side length:

31.4>28, so the circle's circumference is greater.

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.)

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

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.

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

A = pi $r^2$

25pi = pi $r^2$

25 = $r^2$

Then calculate circumference:

C = 2pi r

C = 2pi times 5 = 10pi approx 31.4 inches (approximating pi as 3.14)

To find the perimeter of the square, we can use P = 4s , where P is the perimeter and s is the side length:

31.4>28, so the circle's circumference is greater.

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.)

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

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.

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

A = pi $r^2$

25pi = pi $r^2$

25 = $r^2$

Then calculate circumference:

C = 2pi r

C = 2pi times 5 = 10pi approx 31.4 inches (approximating pi as 3.14)

To find the perimeter of the square, we can use P = 4s , where P is the perimeter and s is the side length:

31.4>28, so the circle's circumference is greater.

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.)

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

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.

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

A = pi $r^2$

25pi = pi $r^2$

25 = $r^2$

Then calculate circumference:

C = 2pi r

C = 2pi times 5 = 10pi approx 31.4 inches (approximating pi as 3.14)

To find the perimeter of the square, we can use P = 4s , where P is the perimeter and s is the side length:

31.4>28, so the circle's circumference is greater.

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.)