Mixed Geometry
For questions in the Quantitative Comparison format (“Quantity A” and “Quantity B” given), the answer choices are always as follows:
(A) Quantity A is greater.
(B) Quantity B is greater.
(C) The two quantities are equal.
(D) The relationship cannot be determined from the information given.
For questions followed by a numeric entry box 
, you are to enter your own answer in the box. For questions followed by a fraction-style numeric entry box 
, you are to enter your answer in the form of a fraction. You are not required to reduce fractions. For example, if the answer is 
, you may enter 
or any equivalent fraction.
All numbers used are real numbers. All figures are assumed to lie in a plane unless otherwise indicated. Geometric figures are not necessarily drawn to scale. You should assume, however, that lines that appear to be straight are actually straight, points on a line are in the order shown, and all geometric objects are in the relative positions shown. Coordinate systems, such as xy-planes and number lines, as well as graphical data presentations, such as bar charts, circle graphs, and line graphs, are drawn to scale. A symbol that appears more than once in a question has the same meaning throughout the question.
1.Right triangle ABC and rectangle EFGH have the same perimeter. What is the value of x?
Point O is the center of the circle.
2.If the area of the circle is 36π and the area of the rectangle is 72, what is the length of DC?
3.The center of a circle is (10, –3). The point (10, 9) is outside the circle, and the point (6, –3) is inside the circle; neither point is on the circle. If the radius, r, is an integer, how many possible values are there for r?
(A)Seven
(B)Eight
(C)Nine
(D)Ten
(E)Eleven
A square’s perimeter in inches is equal to its area in square inches.
A circle’s circumference in inches is equal to its area in square inches.
| 4. | Quantity A The side length of the square |
Quantity B The diameter of the circle |
5.In the figure above, point O is the center of the circle, points A and C are located on the circle, and line segment BC is tangent to the circle. If the area of triangle OBC is 24, what is the length of AB?
(A)2
(B)4
(C)6
(D)8
(E)10
In the figure above, the circle is inscribed in the square.
The area of the circle is 9π.
| 6. | Quantity A The area of the square |
Quantity B 30 |
7.In the figure above, the circle is inscribed in a square that has an area of 50. What is the area of the circle?
(A)
(B)
(C)25π
(D)50π
(E)
8.What is the area of the square in the figure above?
9.In the 7-inch square above, another square is inscribed. What fraction of the larger square is shaded?
(A)
(B)
(C)
(D)
(E)
Lines l and m are parallel.
| 10. | Quantity A x + 2y |
Quantity B 180 |
Lines l1 and l2 are parallel.
a > 90
| 11. | Quantity A a + f + g |
Quantity B b + e + h |
12.What is the value of a + b + c + d?
Sector OAB is a quarter-circle.
| 13. | Quantity A x |
Quantity B 15 |
14.What is a in terms of b and y?
(A)b + y + 65
(B)b – y + 65
(C)b + y + 75
(D)b – 2y + 45
(E)b – y + 75
In the figure above, line segments AC and BD are parallel.
| 15. | Quantity A The area of triangle WYX |
Quantity B The area of triangle WZX |
In the figure above, a right triangle is inscribed in a circle with an area of 16π cm2.
| 16. | Quantity A The hypotenuse of the triangle, in centimeters |
Quantity B 8 |
17.A rectangular box has a length of 6 centimeters, a width of 8 centimeters, and a height of 10 centimeters. What is the length of the diagonal of the box, in centimeters?
(A)10
(B)12
(C)
(D)
(E)24
18.Julian takes a 10-inch by 10-inch square piece of paper and cuts it in half along the diagonal. He then takes one of the halves and cuts it in half again from the corner to the midpoint of the opposite side. All cuts are represented in the figure with dotted lines. What is the perimeter of one of the smallest triangles, in inches?
(A)10
(B)
(C)20
(D)10 +
(E)10 + 
Mixed Geometry Answers
1. 4. Triangle ABC is a right triangle, so this must be a 3–4–5 triangle, and the length of side BC is 5. That means the perimeter of triangle ABC is 3 + 4 + 5 = 12.
Thus, the perimeter of rectangle EFGH is also 12. Using that information, find x:
2 × (2 + x) = 12.
4 + 2x = 12
2x = 8
x = 4.
2. 6. The area of this circle is 36π and the area of any circle is πr2, so the radius of this circle is 6. Label both radii (OA and OD) as 6. Because ABCO is a rectangle, its area is equal to base times height, where radius OA is the height.
| Area of a rectangle | = | bh |
| 72 | = | b(6) |
| 12 | = | b |
Since OC is a base of the rectangle, it is equal to 12. Subtract radius OD from base OC to get the length of segment DC: 12 – 6 = 6.
3. (A). This problem does not actually require any special formulas regarding circles. Calculate the distance between the center point (10, –3) and point (10, 9). Since the x-coordinates are the same and 9 – (–3) = 12, the two points are 12 apart. Because (10, 9) is outside of the circle, the radius must be less than 12. Similarly, calculate the distance between the center point (10, –3) and point (6, –3). Since the y-coordinates are the same, the distance is 10 – 6 = 4. Because (6, –3) is inside the circle, the radius must be more than 4. The radius must be an integer that is greater than 4 and less than 12, so it can only be 5, 6, 7, 8, 9, 10, or 11. Thus, there are seven possible values for r.
4. (C). The perimeter of a square is 4s and the area of a square is s2 (where s is a side length). If the square’s perimeter equals its area, set the two expressions equal to each other and solve:
| 4s | = | s2 |
| 0 | = | s2 – 4s |
| 0 | = | s(s – 4) |
| 4 or 0 | = | s |
Only s = 4 would result in an actual square, so s = 0 is not a valid solution.
The circumference of a circle is 2πr and the area of a circle is πr2 (where r is the radius). If the circle’s circumference equals its area, set the two expressions equal to each other and solve:
| 2πr | = | πr2 |
| 2r | = | r2 |
| 0 | = | r2 – 2r |
| 0 | = | r(r – 2) |
| 2 or 0 | = | r |
Only r = 2 would result in an actual circle, so r = 0 is not a valid solution.
If the radius of the circle is 2, then the diameter is 4. Thus, the two quantities are equal.
5. (B). Because BC is tangent to the circle, angle OCB is a right angle. Thus, radius OC is the height of the triangle. If the area of the triangle is 24, use the area formula for a triangle (and 8 as the base, from the figure) to determine the height:
| A | = |  |
| 24 | = |  |
| 48 | = | 8 × OC |
| 6 | = | OC |
Thus, the radius of the circle is 6 (note that there are two radii on the diagram, OC and OA). Since two sides of the right triangle are known, use the Pythagorean theorem to find the third:
| 62 + 82 | = | (OB)2 |
| 36 + 64 | = | (OB)2 |
| 100 | = | (OB)2 |
| 10 | = | OB |
(The 6–8–10 triangle is one of the special right triangles you should memorize for the GRE!)
Since the hypotenuse OB is equal to 10 and the radius OA is equal to 6, subtract to get the length of AB. The answer is 10 – 6 = 4.
6. (A). If the area of the circle is 9π = πr2, the radius must be 3. The radius represents half the side of the square, so the square is 6 on each side. The area of the square is thus A = s2 = 36, which is greater than 30. Quantity A is greater.
7. (B). If the area of the square is 50, the sides of the square are 
.
If the square is 
“tall,” so is the circle. That is, the side of the square is equal to the diameter of the circle. Since the diameter of the circle is , the radius is 
. Using the formula for the area of a circle, A = πr2:
A = 
A = 
A = 
Note that even if you got a bit lost in the math, you could estimate quite reliably! The square is a bit larger than the circle, so the circle area should be a bit less than 50. Put all the answers in the calculator, using 3.14 as an approximate value for π, and you will quickly see that choice (A) is equal to 19.625, which is too small, and choice (B) is equal to 39.25, while the other three choices are much too large (larger than the square).
8. 50. One way to solve this problem is by using the Pythagorean theorem. All sides of a square are equal to s, so:
| s2 + s2 | = | 102 |
| 2s2 | = | 100 |
| s2 | = | 50 |
Note that you could solve for s (s = 
), but the area of the square is s2, which is already calculated above. The area of the square is 50.
9. (B). Each of the shaded triangles is a 3–4–5 Pythagorean triple. (Or, just note that each shaded triangle has legs of 3 and 4; the Pythagorean theorem will tell you that each hypotenuse is equal to 5).
Since each hypotenuse is also a side of the square, the square has an area of 5 × 5 = 25.
The larger square (the overall figure) has an area of 7 × 7 = 49.
Subtract to find the area of the shaded region: 49 – 25 = 24.
The fraction of the larger square that is shaded is therefore 
.
10. (A). When two parallel lines are cut by a transversal, same-side interior angles are supplementary. Thus, x + y = 180. Since y is not 0 (the transversal and line m do not “overlap”), x + 2y is greater than x + y, which also means that x + 2y is greater than 180.
11. (A). While the exact measures of any of the angles are not given, when parallel lines are cut by a transversal, only two angle measures are created: all the “big” angles are the same and all the “small” angles are the same. Further, the sum of a “small” and a “big” angle is 180° (These angles are said to “form a line.”) Use this fact to simplify both quantities.
Quantity A: a + f + g = a + (f + g) = a + (180)
Quantity B: b + e + h = b + (e + h) = b + (180)
Subtract 180 from each quantity, and the question is really asking for the comparison of a and b. Since a > 90 and a + b = 180, b = 180 – a = 180 – (greater than 90) = less than 90. If a is greater than 90 and b is less than 90, Quantity A is greater.
12. 290. Angles that “go around in a circle” sum to 360°. It may be tempting to just subtract 35 from 360 and answer 325, but don’t overlook the unlabeled angle, which is opposite and therefore equal to 35°. Therefore, subtract 35 + 35 = 70 from 360 to get the answer, 290.
13. (C). If sector OAB is a quarter-circle, then the angle inside the quarter-circle at O measures 90°. Since angles that make up a straight line must sum to 180, 2x + 3x + x must sum to 90:
2x + 3x + x = 90
6x = 90
x = 15
The two quantities are equal.
14. (E). An exterior angle of a triangle is equal to the sum of the two opposite interior angles. From the left triangle, a = x + 90. From the right triangle, b = (x + 10) + (y + 5) = x + y + 15.
Alternatively, you could use the facts that the interior angles of a triangle sum to 180, as do angles that form a straight line. From the left triangle, x + y + 90 and y + a both equal 180, so x + y + 90 = y + a, or x + 90 = a. From the right triangle, 180 = (x + 10) + (y + 5) + (180 – b), or b = x + y + 15.
The question asks for a in terms of b and y, so x is the variable that needs to be eliminated. Eliminate the variable by solving one equation for x, and substituting this expression for x in the other equation. From the right triangle:
b = x + y + 15
x = b – y – 15
From the left triangle:
a = x + 90
a = (b – y – 15) + 90
a = b – y + 75
15. (C). Both triangles, WYX and WZX, share a common base of segment WX. Consider the formula for the area of a triangle:
Area = 
(base)(height)
If two triangles have equal bases, the triangle with the greater area is the one with the greater height. The height is a perpendicular line drawn from the highest point on the triangle to the base. In this case, the heights would be the gray lines below:
By the definition of parallel lines, AC and BD are uniform distance apart. Therefore, the heights shown are the same. Because these triangles have equal bases and heights, they must have equal areas.
16. (C). To solve this problem, recall that a triangle inscribed in a semicircle will be a right triangle if and only if one side of the triangle is the diameter (i.e., the center of the circle must lie on one side of the triangle). Because this is a right triangle, the hypotenuse must be the diameter of the circle.
To find the diameter of the circle, recall the formula for area, Area = πr2, and set up an equation:
| 16π cm2 | = | πr2 |
| 16 cm2 | = | r2 |
| r | = | 4 cm |
Given that diameter is twice the radius, the diameter (i.e., the hypotenuse of the triangle) is 8 cm. Quantity A is 8, so the two quantities are equal.
17. (C). A fast approach to solving this problem is to use the “Super Pythagorean theorem,” which states that the diagonal of any rectangular box is d in the following formula:
d2 = l2 + w2 + h2
where l, w, and h are the length, width, and height of the box, respectively. Plugging in the given information yields:
d2 = 62 + 82 + 102
d2 = 36 + 64 + 100
d2 = 200
d = 
Alternatively, if you don’t remember the Super Pythagorean, apply the normal Pythagorean theorem twice. To find the diagonal of the box, first find the diagonal of one of the sides. Use the bottom side of the figure below as the base:
where the dashed line represents the diagonal of the base. Applying the Pythagorean theorem:
c2 = a2 + b2
c2 = 62 + 82
c2 = 36 + 64
c2 = 100
c = 10
From here, draw the diagonal of the box and apply the Pythagorean theorem again to the vertically oriented triangle with legs 10 and 10 again as shown:
d2 = 102 + 102
d2 = 100 + 100
d2 = 200
d = 
18. (D). In order to compute the perimeter of one of the smaller triangles, first compute the length of the diagonal. For a square with side length 10 inches, the length of the diagonal can be computed by the Pythagorean theorem, (diagonal)2 = (side)2 + (side)2:
d2 = 102 + 102
d2 = 200
d =
Alternatively, recognize that the diagonal of a square is always 
times the side length.
The second cut goes from the corner to the midpoint of the diagonal, so that slice is half as long as the diagonal of the square: 
= . This can be seen as
Similarly, because the remaining line in each of the smaller triangles is half of a diagonal, each is of length inches:
Summing the lengths of the sides, the perimeter of the smallest triangle is:
Perimeter = 10 + +
Perimeter = 10 +