Tag Archive for 'geometry'

Quantitative Comparison Example Problem

The Quantitative Comparison question type on the GRE can be very challenging. Essentially, you are given some information (usually in the form of a sentence, equation, or picture) and then two quantities. You are then asked to determine:

A if the quantity in Column A is greater;

B if the quantity in Column B is greater;

C if the two quantities are equal;

D if the relationship cannot be determined from the information given.

Like most questions on the GRE Quantitative section, the concepts involved in solving this question type are not of themselves very advanced, you just have to be careful in their application. In fact, sometimes there is little or no need at all to do any “real” math; let’s take a look at the following GRE Quantitative Comparison Example Problem:

With geometry problems, it always helps to start with a formula. In this case, the problem is involving the volume of a cylinder. The formula for this is pi times the radius squared times the height.

Volume of a cylinder = πr2h

Therefore, to find the volume of a cylinder you must find both the radius of the cylinder and the height.

In this problem, two cylinders are given. Both cylinders have a different radius, and so it might be assumed that they must have a different volume. However, no information about the height is given. It is possible that the cylinder with the smaller radius has a larger height and might be the larger cylinder. Since there is no way of knowing how these cylinders compare to each other, the answer is D.

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GRE Math: Two More Example Geometry Problems and Solutions

Example Geometry Problem #3

Example Geometry Problem #3

The problem states that the triangle depicted is an isosceles triangle. Remember, there are two qualities that define an isosceles triangle:

1) two identical sides
2) two identical angles

If you have one, you will automatically have the other. The other important fact about isosceles triangles that you need to remember for this question is that the sides that are identical are the two sides that are opposite the two angles that are the same (and vice-versa).

OK, so how does this information help us solve the problem? Well, we know that the two bottom angles are equal, because they are both labeled y. This means that the two sides of the triangle (not the bottom side) are equal because they are opposite the two identical angles. Since they’re equal, we can say

2x – 2 = 3x – 8 [subtract 2x from both sides]
-2 = x – 8 [add 8 to both sides]
6 = x

Now we know that x is 6.

The question is asking about the area of the triangle, which is given by the equation

Atriangle = ½bh

where b is the length of the base, and h is the height. From the diagram, we can see that the height is simply x, which we know to be 6. The base is defined to be

3x – 2

so we simply substitute 6 for x, and evaluate.

3x – 2 = 3(6) – 2
3x – 2 = 16

So the base of the triangle is 16, and the height is 6; thus, the area is

Atriangle = ½(16)(6) = 48

The answer is (C).

Example Geometry Problem #4

Example Geometry Problem #4

The area of a circle is given by the equation

Acircle = πr2

So the area of a semicircle is half of this

Asemicircle = ½πr2

The areas of the three semicircles are given in the problem:

A1 = 8π = ½πr12
A2 = 25π/2 = ½πr22
A3 = 18π = ½πr32

So if we solve for the radius in each of these cases, we get

8π = ½πr12
25π/2 = ½πr22
18π = ½πr32
[multiply all three equations by 2 on both sides]

16π = πr12
25π = πr22
36π = πr32
[divide all three equations by π on both sides]

16 = r12
25 = r22
36 = r32
[take the square root of both sides on all three equations]

4 = r1
5 = r2
6 = r3

Now remember, these are the radii, not the diameters! In order to find the perimeter of the triangle, which is what the question is asking for, we need to multiply each of these by 2 to get the diameters of the semicircles (which are also the sides of the triangle). Thus, the perimeter is

Ptriangle = 2(4) + 2(5) + 2(6) = 30

The answer is (D).

GRE Math: Two Difficult Example Problems and Solutions

Example Difficult Arithmetic Problem

Example Hard Arithmetic Problem

These problems aren’t difficult because they use particularly complicated mathematical principles — they’re difficult because they’re a little tricky. They require you to think a little beyond what’s given directly to you.

This problem looks a lot like an algebra problem because of the variables, but unfortunately it’s not. With three variables, you would need three equations to solve it, which you don’t have.

So how do you approach this problem? Well, once you realize that it’s not an algebra problem, you should be thinking of how you can arrive at values for x, y, and z without using algebra. You know you need these values because eventually you have to add them together. So let’s look at what you do know from the problem.

3x = 4y = 7z

This means that 3x, 4y, and 7z all equal some number, which we will call n; this also means that

3x = n
4y = n
7z = n

From this, we can see that n must be a multiple of 3, a multiple of 4, and a multiple of 7. What is the smallest number that is a multiple of 3, 4, and 7 (LCM of 3, 4, and 7)? It’s 84! (3 * 4 * 7 = 84)

So if n = 84, then we get the following three equations

3x = 84
4y = 84
7z = 84

Solving for each variable, we get

x = 28
y = 21
z = 12

If we add them together, we get

x + y + z = 28 + 21 + 12 = 61

Thus, the answer is (D).


Example Difficult Geometry Problem

Example Hard Geometry Problem

The goal of this problem is to find the area of the shaded region. There are two ways to do this problem — I’ll go over both.

Method 1: Find the area of the shaded trapezoid
A trapezoid is defined as a quadrilateral with two parallel sides and two non-parallel sides. The shaded region is a trapezoid.

(Brief aside: We know that the shaded region is a trapezoid because both the top line segment (y) and the bottom line segment (x) form right angles with the left line segment (also labeled x). Because both y and x intersect the left line segment at the same angle, they must be parallel.)

So what is the formula for the area of a trapezoid?

Atrapezoid = ½(b1 + b2)h

b1 and b2 are the bases of the trapezoid, which are the two parallel line segments; h is the height of the trapezoid, which is the perpendicular distance from one base to the other base.

What do we already know?

b1 = x
b2 = y

So all we need to know is the height of the trapezoid. Based on the drawing, it sort of looks like the height is simply half of the left segment — but that’s not good enough (and it will lead you to the wrong answer, in this problem)! We should never, ever have to “guess” at a value on the GRE! There must be a way to figure it out.

Notice that both the bottom segment and the left segment have length x. There’s a name for triangles that have two identical sides — isosceles triangles. Isosceles triangles also have two identical angles as well. The angles that are opposite of the identical sides are also identical. Let’s call it a.

We know that the sum of the angles in a triangle is 180º. So the sum of the angles in the whole triangle is

a + a + 90 = 180
2a + 90 = 180
2a = 90
a = 45

Both angles are 45º.

Now let’s examine the smaller triangle (the unshaded portion). The top angle for that triangle is the same — 45º. It also has a right angle, which is 90º, so that means that the remaining angle is also 45º.

Hey, look at that! This triangle is also an isosceles triangle, and the sides that are the same are the bottom (y) and the left segment — which is also y (this concept is called similar triangles).

We know that the entire left segment has length x, and the top part of that segment has length y — this means that the bottom part has length x – y. This bottom part also happens to be the height of our trapezoid!

Let’s plug it all into the area equation now:

Atrapezoid = ½(b1 + b2)h
Atrapezoid = ½(x + y)(x – y)

Note that

(x + y)(x – y) = x2 – y2

Substituting,

Atrapezoid = ½(x2 – y2)

which is answer choice (A) (multiplying by ½ is the same as dividing by 2).

Method 2: Whole Minus Part
The shaded region of the triangle is part of the whole triangle. In order to find the shaded region, we could find the area of the whole triangle and subtract away the area of the smaller triangle (the unshaded region).

The formula for area of a triangle is

Atriangle = ½(bh)

where b is the base of the triangle, and h is the height.

For the large triangle,

Abig_triangle = ½(x*x) = ½x2

For the small triangle, we know the base is y, and from the previous explanation above, we know that the height (the left segment) is also y (because the triangle is isosceles). Thus, the area of the small triangle is

Asmall_triangle = ½(y*y) = ½y2

Subtracting the two, we get

Abig_triangle – Asmall_triangle = ½x2 – ½y2 = ½(x2 – y2)

which is answer choice (A).


Questions? Concerns? Suggestions? Criticisms? Leave a comment below!

GRE Math: Two Example Math Problems

Example Problem #1

Example Geometry Problem

This is a fun problem.  The problem tells us that ∆ABC and ∆DEF have the same area.  It also tells us that AD > CF.  The problem is asking us about altitude, which is height.

From this second piece of information, we know that the base of ∆ABC is longer than the base of ∆DEF.  How do we know this?  From the picture, we see that the base of ∆ABC is composed of two segments: AD + DC.  The base of ∆DEF is also composed of two segments: DC + CF.  Both triangles share DC — this is the same for both triangles.  But we know that AD is bigger than CF, which means that AD + DC > DC + CF.  Thus, the base of triangle ∆ABC is larger.

Now, since we know that both triangles have the same area, but the base of ∆ABC is bigger, this means that the height of ∆DEF has to be bigger in order to compensate.  The answer should be B, but let’s prove it.

In math terms:

Area∆ABC = Area∆DEF
½b1h1 = ½b2h2

We know that b1 (base of ∆ABC) is bigger than b2 (base of ∆DEF), so let’s just choose some arbitrary numbers to make it a bit easier to see the relationship.  Let’s let b1 = 6 and b2 = 4.

½(6)(h1) = ½(4)(h2)
3h1 = 2h2
h1 = ⅔h2

Therefore, we know that the height of ∆ABC is smaller than the height of ∆DEF.  Thus, the answer is B.

Continue reading “GRE Math: Two Example Math Problems” »

GRE Math: Two Example Geometry Problems and Solutions

I’m still trying to hammer out a coherent and apposite answer to the nightmare prompt from yesterday’s post, so I haven’t made much other progress on my applications.  Since there’s nothing to report, today we’ll just do a couple geometry problems together!

Continue reading “GRE Math: Two Example Geometry Problems and Solutions” »