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

In order to solve this, we must first find the area of the containing square and then remove the inscribed circle. Once this is done, we need to divide our result by 4 in order to get the one-forth that is the one shaded region.

One side of the square will be equal to the circle's diameter (2r). Since r = 5, d = 10. Therefore, the area of the square is d2 = 102 = 100. The area of the circle is πr2 = 52π = 25π.

Therefore, the area of the four "corner regions" is equal to 100 - 25π. One of these is equal to (100 - 25π) / 4. Simplified, this is 25 - (25/4)π.

To compare, first calculate the area of figure 1. Since it shares dimensions with the semi-circle, we will put all our variables in terms of the radius of that semi-circle:

A1 = (2r)2 + πr2/2 = 4r2 + πr2/2 = r2(4 + π)/2

If we double r, we get:

A2 = (2 * 2r)2 + π(2r)2/2 = 16r2 + π4r2/2 = 4r2(4 + π)/2

This means that the new figure is 4x the size of the original. This is an increase of 300%.

The obvious answer for this problem is that they are equal. Remember quantitative comparisons are often tricky and require you to check your initial inclination. If 1 m equals 100 cm then a square with side 1 m (100 cm) has an area of 100 cm x 100 cm or 10,000 cm2. The area of a square with a 1 cm side is 1 cm x 1 cm or 1 cm2. One hundred times 1 cm2 is 100 cm2. 10,000 cm2 is larger than 100 cm2 so Quantity A is greater.

Since the diameter of the circle is 10, we know the radius of the circle is 5 feet. We can then draw radii that go from the center to two consecutive corners of the square. These radii are both 5 feet, and form a 90 degree angle (since they are diagonals of a square). Thus, with the enclosed side of the square they form a 45-45-90 triangle. Thus, the side of the square must be 5$\sqrt{2}$ (this can also be found with the Pythagorean theorem). Then, the area of the square is

5$\sqrt{2}$times $5$\sqrt{2}$=50ft^{2}$

The diameter of the circle and the sides of the square form a 45-45-90 triangle. Since the hypotenuse is 5 cm, then a leg of the triangle (a side of the square) is .

The area of the square is then .

If you draw points and on the coordinate plane, you know that they are opposite ends of the square, since the sides are parallel to the axes. This means that the squares have sides with lengths of 5, making the area 25.

To compare, first calculate the area of figure 1. Since it shares dimensions with the semi-circle, we will put all our variables in terms of the radius of that semi-circle:

A1 = (2r)2 + πr2/2 = 4r2 + πr2/2 = r2(4 + π)/2

If we double r, we get:

A2 = (2 * 2r)2 + π(2r)2/2 = 16r2 + π4r2/2 = 4r2(4 + π)/2

This means that the new figure is 4x the size of the original. This is an increase of 300%.

The obvious answer for this problem is that they are equal. Remember quantitative comparisons are often tricky and require you to check your initial inclination. If 1 m equals 100 cm then a square with side 1 m (100 cm) has an area of 100 cm x 100 cm or 10,000 cm2. The area of a square with a 1 cm side is 1 cm x 1 cm or 1 cm2. One hundred times 1 cm2 is 100 cm2. 10,000 cm2 is larger than 100 cm2 so Quantity A is greater.

In order to solve this, we must first find the area of the containing square and then remove the inscribed circle. Once this is done, we need to divide our result by 4 in order to get the one-forth that is the one shaded region.

One side of the square will be equal to the circle's diameter (2r). Since r = 5, d = 10. Therefore, the area of the square is d2 = 102 = 100. The area of the circle is πr2 = 52π = 25π.

Therefore, the area of the four "corner regions" is equal to 100 - 25π. One of these is equal to (100 - 25π) / 4. Simplified, this is 25 - (25/4)π.

All sides are equal in a square. To find the area of a square, multiply length times width. We know length = 4 but since all sides are equal, the width is also 4. 4 * 4 = 16.

Since the diameter of the circle is 10, we know the radius of the circle is 5 feet. We can then draw radii that go from the center to two consecutive corners of the square. These radii are both 5 feet, and form a 90 degree angle (since they are diagonals of a square). Thus, with the enclosed side of the square they form a 45-45-90 triangle. Thus, the side of the square must be 5$\sqrt{2}$ (this can also be found with the Pythagorean theorem). Then, the area of the square is

5$\sqrt{2}$times $5$\sqrt{2}$=50ft^{2}$

The diameter of the circle and the sides of the square form a 45-45-90 triangle. Since the hypotenuse is 5 cm, then a leg of the triangle (a side of the square) is .

The area of the square is then .

If you draw points and on the coordinate plane, you know that they are opposite ends of the square, since the sides are parallel to the axes. This means that the squares have sides with lengths of 5, making the area 25.

All sides are equal in a square. To find the area of a square, multiply length times width. We know length = 4 but since all sides are equal, the width is also 4. 4 * 4 = 16.

In order to solve this, we must first find the area of the containing square and then remove the inscribed circle. Once this is done, we need to divide our result by 4 in order to get the one-forth that is the one shaded region.

One side of the square will be equal to the circle's diameter (2r). Since r = 5, d = 10. Therefore, the area of the square is d2 = 102 = 100. The area of the circle is πr2 = 52π = 25π.

Therefore, the area of the four "corner regions" is equal to 100 - 25π. One of these is equal to (100 - 25π) / 4. Simplified, this is 25 - (25/4)π.

To compare, first calculate the area of figure 1. Since it shares dimensions with the semi-circle, we will put all our variables in terms of the radius of that semi-circle:

A1 = (2r)2 + πr2/2 = 4r2 + πr2/2 = r2(4 + π)/2

If we double r, we get:

A2 = (2 * 2r)2 + π(2r)2/2 = 16r2 + π4r2/2 = 4r2(4 + π)/2

This means that the new figure is 4x the size of the original. This is an increase of 300%.

The obvious answer for this problem is that they are equal. Remember quantitative comparisons are often tricky and require you to check your initial inclination. If 1 m equals 100 cm then a square with side 1 m (100 cm) has an area of 100 cm x 100 cm or 10,000 cm2. The area of a square with a 1 cm side is 1 cm x 1 cm or 1 cm2. One hundred times 1 cm2 is 100 cm2. 10,000 cm2 is larger than 100 cm2 so Quantity A is greater.

Since the diameter of the circle is 10, we know the radius of the circle is 5 feet. We can then draw radii that go from the center to two consecutive corners of the square. These radii are both 5 feet, and form a 90 degree angle (since they are diagonals of a square). Thus, with the enclosed side of the square they form a 45-45-90 triangle. Thus, the side of the square must be 5$\sqrt{2}$ (this can also be found with the Pythagorean theorem). Then, the area of the square is

5$\sqrt{2}$times $5$\sqrt{2}$=50ft^{2}$

The diameter of the circle and the sides of the square form a 45-45-90 triangle. Since the hypotenuse is 5 cm, then a leg of the triangle (a side of the square) is .

The area of the square is then .