How to find the length of a radius - GRE Quantitative Reasoning

A radius of 5 means we need a distance of 5 from the center to any points on the circle. We need 52 = (1 – _x_2)2 + (2 – _y_2)2. Let's start with the first point, (3,4). (1 – 3)2 + (2 – 4)2 ≠ 25. Next let's try (4,6). (1 – 4)2 + (2 – 6)2 = 25, so (4,6) is our answer. The same can be done for the other three points to prove they are incorrect answers, but this is something to do ONLY if you have enough time.

This question is easy on the whole, though you must not be intimidated by one small fact that we will soon see. Set up your standard circumference equation:

The circumference is feet, so we can say:

Solving for , we get:

Some students may be intimidated by having in the denominator; however, there is no need for such intimidation. This is simply the answer!

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:

We know the triangle inscribed within the circle must be isosceles, as it contains a 90-degree angle and fixed radii. As such, the opposite angles must be equal. Therefore we can use a simplified version of the Pythagorean Theorem,

a2 + a2 = c2 → 2r2 = 16 → r2 = 8; r = √8 < 3. (since we know √9 = 3, we know √8 must be less); therefore, Quantity B is greater.

A radius of 5 means we need a distance of 5 from the center to any points on the circle. We need 52 = (1 – _x_2)2 + (2 – _y_2)2. Let's start with the first point, (3,4). (1 – 3)2 + (2 – 4)2 ≠ 25. Next let's try (4,6). (1 – 4)2 + (2 – 6)2 = 25, so (4,6) is our answer. The same can be done for the other three points to prove they are incorrect answers, but this is something to do ONLY if you have enough time.

This question is easy on the whole, though you must not be intimidated by one small fact that we will soon see. Set up your standard circumference equation:

The circumference is feet, so we can say:

Solving for , we get:

Some students may be intimidated by having in the denominator; however, there is no need for such intimidation. This is simply the answer!

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:

We know the triangle inscribed within the circle must be isosceles, as it contains a 90-degree angle and fixed radii. As such, the opposite angles must be equal. Therefore we can use a simplified version of the Pythagorean Theorem,

a2 + a2 = c2 → 2r2 = 16 → r2 = 8; r = √8 < 3. (since we know √9 = 3, we know √8 must be less); therefore, Quantity B is greater.

A radius of 5 means we need a distance of 5 from the center to any points on the circle. We need 52 = (1 – _x_2)2 + (2 – _y_2)2. Let's start with the first point, (3,4). (1 – 3)2 + (2 – 4)2 ≠ 25. Next let's try (4,6). (1 – 4)2 + (2 – 6)2 = 25, so (4,6) is our answer. The same can be done for the other three points to prove they are incorrect answers, but this is something to do ONLY if you have enough time.

This question is easy on the whole, though you must not be intimidated by one small fact that we will soon see. Set up your standard circumference equation:

The circumference is feet, so we can say:

Solving for , we get:

Some students may be intimidated by having in the denominator; however, there is no need for such intimidation. This is simply the answer!

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:

We know the triangle inscribed within the circle must be isosceles, as it contains a 90-degree angle and fixed radii. As such, the opposite angles must be equal. Therefore we can use a simplified version of the Pythagorean Theorem,

a2 + a2 = c2 → 2r2 = 16 → r2 = 8; r = √8 < 3. (since we know √9 = 3, we know √8 must be less); therefore, Quantity B is greater.

A radius of 5 means we need a distance of 5 from the center to any points on the circle. We need 52 = (1 – _x_2)2 + (2 – _y_2)2. Let's start with the first point, (3,4). (1 – 3)2 + (2 – 4)2 ≠ 25. Next let's try (4,6). (1 – 4)2 + (2 – 6)2 = 25, so (4,6) is our answer. The same can be done for the other three points to prove they are incorrect answers, but this is something to do ONLY if you have enough time.

This question is easy on the whole, though you must not be intimidated by one small fact that we will soon see. Set up your standard circumference equation:

The circumference is feet, so we can say:

Solving for , we get:

Some students may be intimidated by having in the denominator; however, there is no need for such intimidation. This is simply the answer!

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:

We know the triangle inscribed within the circle must be isosceles, as it contains a 90-degree angle and fixed radii. As such, the opposite angles must be equal. Therefore we can use a simplified version of the Pythagorean Theorem,

a2 + a2 = c2 → 2r2 = 16 → r2 = 8; r = √8 < 3. (since we know √9 = 3, we know √8 must be less); therefore, Quantity B is greater.

A radius of 5 means we need a distance of 5 from the center to any points on the circle. We need 52 = (1 – _x_2)2 + (2 – _y_2)2. Let's start with the first point, (3,4). (1 – 3)2 + (2 – 4)2 ≠ 25. Next let's try (4,6). (1 – 4)2 + (2 – 6)2 = 25, so (4,6) is our answer. The same can be done for the other three points to prove they are incorrect answers, but this is something to do ONLY if you have enough time.

This question is easy on the whole, though you must not be intimidated by one small fact that we will soon see. Set up your standard circumference equation:

The circumference is feet, so we can say:

Solving for , we get:

Some students may be intimidated by having in the denominator; however, there is no need for such intimidation. This is simply the answer!

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: