Geometry - GRE Quantitative Reasoning
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What is the circumference of a circle with an area of 36π?
What is the circumference of a circle with an area of 36π?
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
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
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What is the area of a circle, one-quarter of the circumference of which is 5.5 inches?
What is the area of a circle, one-quarter of the circumference of which is 5.5 inches?
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.
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.
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Quantitative Comparison
Quantity A: Area of a right triangle with sides 7, 24, 25
Quantity B: Area of a circle with radius 5
Quantitative Comparison
Quantity A: Area of a right triangle with sides 7, 24, 25
Quantity B: Area of a circle with radius 5
Quantity A: area = base * height / 2 = 7 * 24/2 = 84
Quantity B: area = πr_2 = 25_π
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.
Quantity A: area = base * height / 2 = 7 * 24/2 = 84
Quantity B: area = πr_2 = 25_π
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.
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If the outer arc of 1/12th of a circular pie is 7π, what is the area of 1/4th of the pie?
If the outer arc of 1/12th of a circular pie is 7π, what is the area of 1/4th of the pie?
Our initial data tells us that (1/12)c = 7π or (1/12)πd = 7π. This simplifies to (1/12)d = 7 or d = 84. Furthermore, we know that r is 42. Given this, we can ascertain the area of a quarter of the whole pie by taking one fourth of the whole area or 0.25 * π * 422 = 0.25 * 1764 * π = 441π
Our initial data tells us that (1/12)c = 7π or (1/12)πd = 7π. This simplifies to (1/12)d = 7 or d = 84. Furthermore, we know that r is 42. Given this, we can ascertain the area of a quarter of the whole pie by taking one fourth of the whole area or 0.25 * π * 422 = 0.25 * 1764 * π = 441π
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What is the perimeter of an isosceles triangle given that the sides 5 units long and half of the base measures to 4 units?
What is the perimeter of an isosceles triangle given that the sides 5 units long and half of the base measures to 4 units?
The base of the triangle is 4 + 4 = 8 so the total perimeter is 5 + 5 + 8 = 18.
The base of the triangle is 4 + 4 = 8 so the total perimeter is 5 + 5 + 8 = 18.
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A triangle has sides 3, 5, and x. What can side x not be equal to?
A triangle has sides 3, 5, and x. What can side x not be equal to?
This question draws from the Third Side Rule of triangles. The length of any side of a triangle must be greater than the difference between the other sides and less than the sum of the other two sides.
This means that side x must be between 2 and 8 since the difference between 5 – 3 = 2 and the sum of 3 + 5 = 8.
Choices 3, 4, and 6 all fall within the range of 2 to 8, but choice 9 does not. The answer is 9.
This question draws from the Third Side Rule of triangles. The length of any side of a triangle must be greater than the difference between the other sides and less than the sum of the other two sides.
This means that side x must be between 2 and 8 since the difference between 5 – 3 = 2 and the sum of 3 + 5 = 8.
Choices 3, 4, and 6 all fall within the range of 2 to 8, but choice 9 does not. The answer is 9.
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Find the area of an equilateral triangle when one of its sides equals 4.
Find the area of an equilateral triangle when one of its sides equals 4.
All sides of an equilateral triangle are equal, so all sides of this triangle equal 4.
Area = 1/2 base * height, so we need to calculate the height: this is easy for an equilateral triangle, since you can bisect any such triangle into two identical 30:60:90 triangles.
The ratio of lengths of a 30:60:90 triangle is 1:√3:2. The side of the equilateral triangle is 4, and we divided the base in half when we bisected the triangle, so that give us a length of 2, so our triangle must have sides of 2, 4, and 2√3; thus we have our height.
One of our 30:60:90 triangles will have a base of 2 and a height of 2√3. Half the base is 1, so 1 * 2√3 = 2√3.
We have two of these triangles, since we divided the original triangle, so the total area is 2 * 2√3 = 4√3.
You can also solve for the area of any equilateral triangle by applying the formula (s2√3)/4, where s = the length of any side.
All sides of an equilateral triangle are equal, so all sides of this triangle equal 4.
Area = 1/2 base * height, so we need to calculate the height: this is easy for an equilateral triangle, since you can bisect any such triangle into two identical 30:60:90 triangles.
The ratio of lengths of a 30:60:90 triangle is 1:√3:2. The side of the equilateral triangle is 4, and we divided the base in half when we bisected the triangle, so that give us a length of 2, so our triangle must have sides of 2, 4, and 2√3; thus we have our height.
One of our 30:60:90 triangles will have a base of 2 and a height of 2√3. Half the base is 1, so 1 * 2√3 = 2√3.
We have two of these triangles, since we divided the original triangle, so the total area is 2 * 2√3 = 4√3.
You can also solve for the area of any equilateral triangle by applying the formula (s2√3)/4, where s = the length of any side.
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Which set of side lengths CANNOT correspond to a right triangle?
Which set of side lengths CANNOT correspond to a right triangle?
Because we are told this is a right triangle, we can use the Pythagorean Theorem, _a_2 + _b_2 = _c_2. You may also remember some of these as special right triangles that are good to memorize, such as 3, 4, 5.
Here, 6, 8, 11 will not be the sides to a right triangle because 62 + 82 = 102.
Because we are told this is a right triangle, we can use the Pythagorean Theorem, _a_2 + _b_2 = _c_2. You may also remember some of these as special right triangles that are good to memorize, such as 3, 4, 5.
Here, 6, 8, 11 will not be the sides to a right triangle because 62 + 82 = 102.
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A parallelogram has a base of
and a height measurement that is
the base length. Find the area of the parallelogram.
A parallelogram has a base of and a height measurement that is
the base length. Find the area of the parallelogram.
By definition a parallelogram has two sets of opposite sides that are congruent/parallel. However, to find the area of a paralleogram, you need to know the base and height lengths. Since this problem provides both the base and height measurements, apply the formula:

Before applying the formula you must find
of
.
The solution is:

Note: when working with multiples of ten remove zeros and then tack back onto the product.

There were two total zeros in the factors, so tack on two zeros to the product:

By definition a parallelogram has two sets of opposite sides that are congruent/parallel. However, to find the area of a paralleogram, you need to know the base and height lengths. Since this problem provides both the base and height measurements, apply the formula:
Before applying the formula you must find of
.
The solution is:
Note: when working with multiples of ten remove zeros and then tack back onto the product.
There were two total zeros in the factors, so tack on two zeros to the product:
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Find the area of a square with a side length of 4.
Find the area of a square with a side length of 4.
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.
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.
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If a square has a side length of √2, how long is the diagonal of the square?
If a square has a side length of √2, how long is the diagonal of the square?
A diagonal divides a square into two 45-45-90 triangles, which have lengths adhering to the ratio of x: x: x√2. Therefore, 2 is the correct answer as the diagonal represents the hypotenuse of the triangle. the Pythagorean theorem can also be used: √22+ √22 = c2.
A diagonal divides a square into two 45-45-90 triangles, which have lengths adhering to the ratio of x: x: x√2. Therefore, 2 is the correct answer as the diagonal represents the hypotenuse of the triangle. the Pythagorean theorem can also be used: √22+ √22 = c2.
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What is one possible equation for a line parallel to the one passing through the points (4,2) and (15,-4)?
What is one possible equation for a line parallel to the one passing through the points (4,2) and (15,-4)?
(4,2) and (15,-4)
All that we really need to ascertain is the slope of our line. So long as a given answer has this slope, it will not matter what its y-intercept is (given the openness of our question). To find the slope, use the formula: m = rise / run = (y1 - y2) / (x1 - x2):
(2 - (-4)) / (4 - 15) = (2 + 4) / -11 = -6/11
Given this slope, our answer is: y = -6/11x + 57.4
(4,2) and (15,-4)
All that we really need to ascertain is the slope of our line. So long as a given answer has this slope, it will not matter what its y-intercept is (given the openness of our question). To find the slope, use the formula: m = rise / run = (y1 - y2) / (x1 - x2):
(2 - (-4)) / (4 - 15) = (2 + 4) / -11 = -6/11
Given this slope, our answer is: y = -6/11x + 57.4
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For the line

Which one of these coordinates can be found on the line?
For the line
Which one of these coordinates can be found on the line?
To test the coordinates, plug the x-coordinate into the line equation and solve for y.
y = 1/3x -7
Test (3,-6)
y = 1/3(3) – 7 = 1 – 7 = -6 YES!
Test (3,7)
y = 1/3(3) – 7 = 1 – 7 = -6 NO
Test (6,-12)
y = 1/3(6) – 7 = 2 – 7 = -5 NO
Test (6,5)
y = 1/3(6) – 7 = 2 – 7 = -5 NO
Test (9,5)
y = 1/3(9) – 7 = 3 – 7 = -4 NO
To test the coordinates, plug the x-coordinate into the line equation and solve for y.
y = 1/3x -7
Test (3,-6)
y = 1/3(3) – 7 = 1 – 7 = -6 YES!
Test (3,7)
y = 1/3(3) – 7 = 1 – 7 = -6 NO
Test (6,-12)
y = 1/3(6) – 7 = 2 – 7 = -5 NO
Test (6,5)
y = 1/3(6) – 7 = 2 – 7 = -5 NO
Test (9,5)
y = 1/3(9) – 7 = 3 – 7 = -4 NO
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What is the slope of line 3 = 8y - 4x?
What is the slope of line 3 = 8y - 4x?
Solve equation for y. y=mx+b, where m is the slope
Solve equation for y. y=mx+b, where m is the slope
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What is the slope of the equation
?
What is the slope of the equation ?
To find the slope of a line, you should convert an equation to the slope-intercept form. In this case, the equation would be
, which means the slope is
.
To find the slope of a line, you should convert an equation to the slope-intercept form. In this case, the equation would be , which means the slope is
.
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what would be the slope of a line perpendicular to
4x+3y = 6
what would be the slope of a line perpendicular to
4x+3y = 6
switch 4x+ 3y = 6 to "y=mx+b" form
3y= -4x + 6
y = -4/3 x + 2
m = -4/3; the perpendicular line will have the negative reciprocal of this line so it would be 3/4
switch 4x+ 3y = 6 to "y=mx+b" form
3y= -4x + 6
y = -4/3 x + 2
m = -4/3; the perpendicular line will have the negative reciprocal of this line so it would be 3/4
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A cylinder has a height of 4 and a circumference of 16π. What is its volume
A cylinder has a height of 4 and a circumference of 16π. What is its volume
circumference = πd
d = 2r
volume of cylinder = πr2h
r = 8, h = 4
volume = 256π
circumference = πd
d = 2r
volume of cylinder = πr2h
r = 8, h = 4
volume = 256π
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A cube with a surface area of 216 square units has a side length that is equal to the diameter of a certain sphere. What is the surface area of the sphere?
A cube with a surface area of 216 square units has a side length that is equal to the diameter of a certain sphere. What is the surface area of the sphere?
Begin by solving for the length of one side of the cube. Use the formula for surface area to do this:

s= length of one side of the cube



The length of the side of the cube is equal to the diameter of the sphere. Therefore, the radius of the sphere is 3. Now use the formula for the surface area of a sphere:






The surface area of the sphere is
.
Begin by solving for the length of one side of the cube. Use the formula for surface area to do this:
s= length of one side of the cube
The length of the side of the cube is equal to the diameter of the sphere. Therefore, the radius of the sphere is 3. Now use the formula for the surface area of a sphere:
The surface area of the sphere is .
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What is the circumference of a circle with an area of 36π?
What is the circumference of a circle with an area of 36π?
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
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
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What is the area of a circle, one-quarter of the circumference of which is 5.5 inches?
What is the area of a circle, one-quarter of the circumference of which is 5.5 inches?
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.
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.
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