Radius - GMAT Quantitative

Let be the sidelength of the square. Then its area is .

From Statement 1, it follows that the radius of the circle is .

From Statement 2, it follows that , since the perimeter of the square is , the circumference of the circle is , and the radius is - the same fact given in Statement 1.

Either way, it follows that the area of the circle in terms of is

,

so all we have to do is compare to 1 in order to determine whether the square or the circle is larger in area.

From Statement 2, we can calculate the area of the triangle, but we are given no clues about the area of the circle, actual or relative.

From Statement 1, we know that if we call the radius of the circle , we know the sidelength of the triangle is .

The area of the circle is .

The area of the triangle is .

All we have to do is compare to to determine whether the circle or the triangle has the greater area.

The radius of the larger circle is equal to the diameter of the smaller, and, subsequently, twice the radius of the smaller. If one radius is known, then the other can be calculated, and the areas of the two circles can be as well; the difference between their areas is the area of the gray region. Statement 1 tells us the diameter of the large circle, from which its radius can be determined by dividing by 2; Statement 2 tells us the radius of the radius of the smaller circle. From either, the radius of the other circle can be calculated.

Examine the diagram below, which shows the hexagon, segments from its center to two vertices and the midpoint of a side, and the two circles.

From Statement 1 alone, the radius of the circumscribed circle can be calculated using the area formula. From Statement 2 alone, the perimeter can be divided by 6 to obtain ; since two consecutive radii and one side of a hexagon can be proved to form an equilateral triangle, . As a consequence, from either statement alone, can be calculated.

can be proved a 30-60-90 triangle, so the 30-60-90 Theorem can be used to calculate , the radius of the inscribed circle. From this, the area of the inscribed circle can be calculated.

The area of a circle can be calcuated using the equation:

and the circumference calculated using:

The radius is the only information required for calculating the area of a circle and that can be obtained from the circumference, therefore, either statement is sufficient.

Let be the sidelength of the square. Then its area is .

From Statement 1, it follows that the radius of the circle is .

From Statement 2, it follows that , since the perimeter of the square is , the circumference of the circle is , and the radius is - the same fact given in Statement 1.

Either way, it follows that the area of the circle in terms of is

,

so all we have to do is compare to 1 in order to determine whether the square or the circle is larger in area.

From Statement 2, we can calculate the area of the triangle, but we are given no clues about the area of the circle, actual or relative.

From Statement 1, we know that if we call the radius of the circle , we know the sidelength of the triangle is .

The area of the circle is .

The area of the triangle is .

All we have to do is compare to to determine whether the circle or the triangle has the greater area.

The radius of the larger circle is equal to the diameter of the smaller, and, subsequently, twice the radius of the smaller. If one radius is known, then the other can be calculated, and the areas of the two circles can be as well; the difference between their areas is the area of the gray region. Statement 1 tells us the diameter of the large circle, from which its radius can be determined by dividing by 2; Statement 2 tells us the radius of the radius of the smaller circle. From either, the radius of the other circle can be calculated.

Examine the diagram below, which shows the hexagon, segments from its center to two vertices and the midpoint of a side, and the two circles.

From Statement 1 alone, the radius of the circumscribed circle can be calculated using the area formula. From Statement 2 alone, the perimeter can be divided by 6 to obtain ; since two consecutive radii and one side of a hexagon can be proved to form an equilateral triangle, . As a consequence, from either statement alone, can be calculated.

can be proved a 30-60-90 triangle, so the 30-60-90 Theorem can be used to calculate , the radius of the inscribed circle. From this, the area of the inscribed circle can be calculated.

The standard form of the equation of a circle is

,

where the radius is and the center is .

The equation given is this same form, with replacing , replacing , and replacing , so to find the radius, we need to find .

Statement 1 alone tells us that the center is but it tells us nothing about the radius. Statement 2 alone tells us only that the circle passes through .

The two together, however, reveal enough information to give the radius. The radius is the distance from the center to a point on the circle, so we can use the distance formula to find the distance between and . This is the radius.

We are given the area and circumference of a circle and asked to find the radius.

Given the following equations:

We can use either equation to work backwards and find our radius, therefore; Each statement alone is enough to solve the question.

To find the radius of a cricle, we need its circumference, diameter, or area.

I) Seems to be helpful, but it is really just giving us pi, so it is not sufficient.

II) Gives us the area of the circle, which we can use to work backward to find the radius.

So II is sufficient to answer the question, but I is not.

Statement 1) gives the circumference of a circle. The formula for finding the circumference of a circle is . The radius can be solved by using this formula.

Statement 2) gives the standard form of a circle, where is the center of the circle:

The radius is also given in the equation.

Therefore, either statement alone is enough to solve for the radius of a circle.

First, locate the other midpoints of the sides of the triangle and construct the segments from each vertex to the opposite midpoint.

Since is equilateral, , , and are all altitudes that insersect at the center of the circumscribed circle, , so that . is the radius of the circumscribed circle.

Assume Statement 1 alone. The length of one side of an equilateral triangle can be calculated using the formula

, or, equivalently,

Once is calculated, then, since is also a perpendicular bisector of and a bisector of , making a 30-60-90 triangle, can be calculated to be one half of ; can be multiplied by to yield , and, since the three altitudes of an equilateral triangle divide one another into segments whose lengths have ratio 2:1, can be multiplied by to obtain radius .

Statement 2 gives us explicitly, so we can take two thirds of this to get the radius

.

If a right triangle is inscribed inside a circle, the hypotenuse of the triangle is a diameter of the circle; therefore, its length is the diameter, and half this is the radius. However, from the two statements, it cannot be determined which segment is the hypotenuse.

If is the hypotenuse, then the radius is half its length; since , the radius is 10.

If is the hypotenuse, then, since the hypotenuse is the longest side of a right triangle, - that is, . The radius is greater than 10.

Therefore, the radius depends on which side is the hypotenuse; since that is not clear, the radius cannot be determined.

The diameter of a circle with an inscribed rectangle is equal to the length of the diagonal of the rectangle; once this diameter is found, it can be divided by 2 to yield the radius.

Statement 1 alone gives this length, from which the radius can be found to be . Statement 2 alone gives only the length of one set of opposite sides, from which the length of the diagonal cannot be determined.

Statement 1 alone yields insufficient information, since, as seen in the diagram below, the circles that circumscribe a square and a triangle with the same sidelength have different sizes:

Statement 2 is insufficient since it gives no hints about the size of the hexagon.

Assume both statements. The six radii of a regular hexagon divide it into six equilateral triangles, by symmetry; therefore, the radius of a regular hexagon is equal to its sidelength, which is given in Statement 1 as 10. Since the radius of a regular hexagon is equal to that of the circle in which it is inscribed, the circle has radius 10.

This can be seen by examining the figure below:

From Statement 2 alone, , a diagonal of the square, measures . The diameter of the circle is equal to the length of a diagonal of an inscribed square, so the radius of the circle is equal to half this, or .

From Statement 1 alone, since the area of the square is 100, its sidelength is the square root of this, or 10. By the 45-45-90 Theorem, a diagonal of the square measures times this, or , which makes Statement 2 a consequence of Statement 1. Therefore, it follows again that the circle has radius .

The figure referenced is below:

By symmetry, is one third of a circle. Therefore, its length is one third of the circumference, so, if Statement 1 alone is assumed, the circumference can be determined to be ; this can be divided by to yield radius .

Statement 2 yields no helpful information; from the body of the problem, can already be deduced to be two thirds of a circle, or, equivalently, an arc of measure

.