Cylinders & Spheres In The GMAT
Posted on
19
Apr 2022

Cylinders & Spheres In The GMAT

Welcome back to our fifth and final article on GMAT circles. Last time we explored the possibilities of treating a circle’s radius as the hypotenuse of a right triangle. This time we will introduce two 3-dimensional shapes built from circles: the cylinder and the sphere.

1. Cylinder

More than likely, you already know what these things are and could describe them. But let’s try to define them in some interesting ways. A cylinder is a “tall circle” or – to use more proper geometric terminology – a circular prism. A prism is the solid shape that results when you take any polygon and “pull it” upward into something like a pillar. The polygon you started with still exists as the “top and bottom” faces of the prism, and the faces around the sides of the prism are rectangles. (Technically they can be parallelograms, which would produce a “leaning” pillar, but this won’t happen on the GMAT.)

Since a circle doesn’t have sides, a cylinder doesn’t have faces – except for the two circles on its top and bottom. In between, there is one smoothly-curving surface. If you need to find the area of this third surface, you can treat it like a rectangle. The length of this rectangle is the height of the cylinder, and the width of this rectangle is the circumference of the circle. The volume of any prism is the area of its base polygon multiplied by the prism’s height. So for a cylinder, the equation is

V = πr²h

2. Sphere

Now for spheres. We all know that a sphere is a perfectly round ball. But think about this: a sphere is like a circle “any way you slice it” – quite literally. If you have some citrus fruits in your kitchen, you can try slicing them in different places at different angles, and the faces of the two resulting pieces will always be circles. Another way to say this is that any cross section taken from a sphere will be a circle. No matter how hard you try, you will never be able to produce an elliptical orange slice. Sorry to disappoint you.

Let’s see how the GMAT employs these shapes in some official problems. Some basic cylinder problems focus on one whole cylinder. More challenging cylinder problems compare one cylinder to another or treat a cylinder as a partially-filled tank. 

3. A Data Sufficiency Problem Featuring Two Cylinders

It costs $2,250 to fill right circular cylindrical Tank R with a certain industrial chemical. If the cost to fill any tank with this chemical is directly proportional to the volume of the chemical needed to fill the tank, how much does it cost to fill right circular cylindrical Tank S with the chemical?

1. The diameter of the interior of Tanks R is twice the diameter of the interior of Tank S.
2. The interiors of Tanks R and S have the same height.

(A) Statement (1) ALONE is sufficient but statement (2) ALONE is not sufficient.
(B) Statement (2) ALONE is sufficient but statement (1) ALONE is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are not sufficient. 

Solution

Since the cost to fill any tank (including tanks R and S) with this chemical is directly proportional to the volume of the cylindrical tank, the only thing we care about here is the ratio of the two tanks’ volumes. Remember that for the volume of a cylinder, we need to know (or be able to derive) both the area of the circle and the height of the cylinder.

Statement 1 gives us the ratio of the tanks’ diameters: 2:1. This means that the ratio of the areas of the tanks’ bases is 4:1 (if lost here, review article 1 on area, circumference, and pi). This is great, but it is still not enough to know the overall ratio of the tanks’ volumes. Statement 1 is insufficient.

Statement 2 tells us that tanks R and S are the same height, specifying “interior” because we are filling up space with a chemical and can’t count whatever volume is taken up by the tank walls. On its own, this information is insufficient.

Combining statements 1 and 2, we have the ratio of the tanks’ diameters (2:1) and the ratio of their heights (1:1). This means that the overall ratio of the tanks’ volumes is fixed. Statements 1 and 2 together are sufficient, and the correct answer is C.

4. Partially Filled Cylinder-as-a-tank Problem

The figures show a sealed container that is a right circular cylinder filled with liquid to 12 its capacity. If the container is placed on its base, the depth of the liquid in the container is 10 centimeters and if the container is placed on its side, the depth of the liquid is 20 centimeters. How many cubic centimeters of liquid are in the container. 

(A) 4,000 π
(B) 2,000 π
(C) 1,000 π
(D) 400 π
(E) 200 π

Solution

This problem is less complex than it might first appear. It all comes together when you realize that the 20cm depth in the second orientation of the tank represents the radius of the circle!  Now you can get the area of the circle in cm² using A = r² and then multiply the result by 10 (the depth in centimeters of the liquid in the upright tank) to get the volume of the liquid in cm³. If you can mentally square 20 and then multiply by 10, you should be just seconds away from selecting correct answer choice A.

5. Final Cylinder-as-a-tank Problem

Solve carefully before reading on.

A tank is filled with gasoline to a depth of exactly 2 feet. The tank is a cylinder resting horizontally on its side, with its circular end oriented vertically. The inside of the tank is exactly 6 feet long. What is the volume of the gasoline in the tank?

1. The inside of the tank is exactly 4 feet in diameter.
2. The top surface of the gasoline forms a rectangle that has an area of 24 square feet.

(A) Statement (1) ALONE is sufficient but statement (2) ALONE is not sufficient.
(B) Statement (2) ALONE is sufficient but statement (1) ALONE is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are not sufficient. 

Solution

Statement 1.

Evaluating statement 1 is fairly straightforward. Combined with the information from the question stem that the depth of the gasoline in the tank is 2 feet, the additional information that the inside of the tank has a 4-foot diameter means that the tank is filled halfway with gasoline. If 4 is the diameter, then 2 is the radius, and the gas fills the tank up to its center line. This looks just like the half-filled tank in the previous problem. The question stem also gave us the length of the tank (called a length rather than a height since this tank is a cylinder “lying down”), so the cylinder’s total and fractional volumes are calculable. Statement 1 is sufficient.

Statement 2.

Statement 2 performs something like a “double flip.” We are told that the top surface of the gasoline is a 24ft² rectangle. Remembering from the question stem that the tank is 6 feet long, you may realize that 24/6 = 4 and think that this tells you the same thing as statement 1: that the tank has a 4-foot diameter. This would be a mistake. The 24ft² rectangle formed by the surface of the gasoline indeed has a length of 6 and a width of 4, but this width of 4 is not necessarily the diameter of the tank. It could just as easily happen in a larger tank that is less than half (or more than half) full. 

Does this make statement 2 insufficient? Well so far, yes. But there’s something we’ve left out that makes it sufficient after all! From the question stem, the depth of the gasoline in the tank is 2 feet. Imagine that the circular end of this tank is transparent. Looking at it this way, the top surface of the gasoline makes a horizontal chord across the circle, and this chord has a length of 4. Simultaneously, this chord is a vertical distance of 2 feet from the bottom of the circle (since the depth of the gasoline in the tank is 2 feet). The only way this can happen is if the 4-foot chord is the diameter of the circle!

Therefore the tank is still half full, and the volume of the gasoline is half of the (calculable) volume of the cylinder. Statement 2 is also sufficient, and the correct answer choice is D.

 

 

6. Sphere Problem

For the final problem in our circles series, we’ll work with spheres. Spheres are less common on the GMAT than cylinders, and you will never have to memorize any of their formulas. If you need a sphere formula for a problem, it will be supplied with the problem.

For a party, three solid cheese balls with diameters of 2 inches, 4 inches, and 6 inches, respectively, were combined to form a single cheese ball. What was the approximate diameter, in inches, of the new cheese ball? (The volume of a sphere is 433, where r is the radius.)

(A) 12
(B) 16
(C) ∛16
(D) 8
(E) 236

Solution – Long Way

This sounds like a party you don’t want to miss. I don’t know exactly how to combine three solid cheese balls into one, but I do know how to calculate the diameter.

There are two ways to solve this problem: the long way and the best way. The long way is to calculate the volumes of the three original cheese balls, sum your answers into one volume, and then solve for the radius of the combined cheese ball. First you must divide the given diameters of the original cheese balls by 2, since the volume equation uses radius instead.

V = (4π/3)r³
V = (4π/3)1³ + (4π/3)2³ + (4π/3)3³
V = (4π/3)(1³ + 2³ + 3³)
V = (4π/3)(1 + 8 + 27)
V = (4π/3)(36)
V = 48π
V = (4π/3)r³
48π = (4π/3)r³
48 = (4/3)r³
36 = r³
∛36 = r
2(∛36) = D

And the correct answer choice is E.

Solution – Short Way

That was the long way. The best way is to think logically and exploit the answer choices. Since we are effectively adding some cheese onto a ball that already has a diameter of 6 inches, the diameter of the combined cheese ball will be greater than 6 inches. This means that answer choices C and D are nonstarters. (C is somewhere between 2 and 3, and D is exactly 6.) Let’s think next about choices A and B, since they are integers and easier to evaluate than choice E.

Can the diameter of the combined cheese ball be as great as 12 (choice A) or even 16 (choice B)? No, it can’t. Picture a “cheese ball snowman” made of the three original cheese balls – a cooler idea for a party than smashing them into one ball, I argue. His height is 12 inches, but this is not the same as having a single ball with a 12-inch diameter. Three spheres whose diameters sum to 12 cannot combine their volumes to produce a single sphere with a diameter of 12. Therefore choices A and B are also out, leaving us with correct choice E. If we approximate the value of E, it is greater than 6 but less than 8, since the cube root of 36 is greater than 3 but less than 4. A combined cheese ball this size makes logical sense.

 

This concludes our fifth and final article on GMAT circles. Cheers.

 

Contributor: Elijah Mize (Apex GMAT Instructor)

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Radius as Hypotenuse
Posted on
05
Apr 2022

Radius As Hypotenuse – Problems & Solutions

Welcome back to our fourth article on GMAT circles. Last time we considered inscribed angles and learned that where there is a 90-degree inscribed angle, there is a hypotenuse that is also a diameter of the circle. This time we will explore a class of problems where the radius, rather than the diameter, pulls double duty as a hypotenuse. Let’s dive right in with the following official problem.

1. Radius as Hypotenuse  – GMAT Official Problem

Semicircular archway over a flat street problemThe figure above represents a semicircular archway over a flat street. The semicircle has a center at O and a radius of 6 feet. What is the height h, in feet, of the archway 2 feet from its center?

A. √2
B. 2
C. 3
D. 4√2
E. 6

Problem SolutionThis problem is a straightforward application of the Pythagorean theorem. Since we are told that the radius of the semicircle is 6 feet, we can draw a 6-foot radius from center O to the point where height h meets the semicircle. Voila – a right triangle.

h = √(62 – 22)
h = √(36 – 4)
h = √32

This is where you should stop and mark answer choice D since we are taking the square root of a number that is not a perfect square. When we simplify this radical, something will get left inside. Therefore answers B, C, and E are out (Answer A is out because √2 =/= √32), and the correct choice is D.

2. Radius as Hypotenuse Problem 1 

Let’s try something a little different:

In the xy-plane, point (r,s) lies on a circle with center at the origin. What is the value of + s²?

1. The circle has radius 2.
2. The point (2,-2) lies on the circle.

A. Statement (1) ALONE is sufficient but statement (2) ALONE is not sufficient.
B. Statement (2) ALONE is sufficient but statement (1) ALONE is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are not sufficient. 

This is the first problem we’ve seen where a circle is placed on the xy-plane. In such problems, it is usually helpful to remember the basic circle principle that every point on the circle (meaning on its edge or perimeter) is equidistant from its center.

Solution

If you’re unfamiliar with these problems, statement 1 may trip you up. Is the radius of the circle sufficient to determine + ? Yes, it is. If you are concerned about the unknown positivity/negativity of the coordinates r and s, recall that the square of any number (except 0) is positive. This means that for any positive/negative combination of r and s, the sum + will have the same value. 

But what you really need here is to see that the expression + matches the famous + from the Pythagorean theorem, and in fact, it functions the exact same way.

Radius as Hypotenuse ProblemIn this setup, the radius is the hypotenuse of the right triangle with legs r and s. Therefore, applying the Pythagorean theorem, the value + represents the square of the radius. So if we know the value of the radius (2), we know the value r² + s², and statement 1 is sufficient.

Statement 2 offers that the point (√2, -√2) lies on the circle. This statement should be “easier” to evaluate than statement 1. Seeing the radicals in the coordinates ought to help you make the connection to the Pythagorean theorem if you didn’t already while evaluating statement 1. But using the principle that every point on a circle is equidistant from its center, we know that this given point (√2, -√2) is the same distance from the center as the point (r, s) in the question. Therefore if we sum the squares of √2 and -√2, the result (4) will also represent the value r² + s² we were asked about.

3. Problem 2

Let’s try one more:

In cross section, a tunnel that carries one lane of one-way traffic is a semicircle with radius 4.2 m. Is the tunnel large enough to accommodate the truck that is approaching the entrance to the tunnel?

1. The maximum width of the truck is 2.4 m.
2. The maximum height of the truck is 4 m.

A. Statement (1) ALONE is sufficient but statement (2) ALONE is not sufficient.
B. Statement (2) ALONE is sufficient but statement (1) ALONE is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are not sufficient. 

This one is a little more complex. Sometimes on GMAT quant problems, it is helpful to ask why certain details were specified. In this case, we are told that the tunnel “carries one lane of one-way traffic.” This is important because if it were not the case, the truck would have to drive on one side or the other, and there’s no way it would be able to get through the tunnel. Since there is only one lane going through the tunnel, the truck can “center up” to give itself the best chance of fitting through.

Solution

This is one of those less-common DS problems where each statement on its own is clearly insufficient. If all we know is that the truck is 2.4m wide at its widest point (statement 1), it may still be too tall to fit through the tunnel. If all we know is that the truck is 4m tall at its tallest point, we don’t know whether the truck is narrow enough to make it through the tunnel while being this tall.

Problem 2 - Solution But if we combine statements 1 and 2, we can use the Pythagorean theorem to calculate the max distance of a point on the “centered up” truck from the point at the “center” of the semicircle.

Now here’s the key step: don’t calculate! Running the Pythagorean theorem with our values here would be a waste of time. As long as the value p [from the graphic] is less than 4.2 (the radius of the tunnel), the truck will fit. But for DS, we don’t have to know whether the truck will fit. All we have to know is whether the value p can be calculated, and in this case, it can be. Statements 1 and 2 together are sufficient, and the correct answer choice is C.

 

This concludes our fourth article on the GMAT’s treatment of circles. Next time we will look at circles in two different 3-dimensional shapes: cylinders and spheres.

 

Contributor: Elijah Mize (Apex GMAT Instructor)

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Inscribed Angles & Inscribed Polygons In The GMAT
Posted on
29
Mar 2022

Inscribed Angles & Inscribed Polygons In The GMAT

Welcome back to our third article on GMAT circles. In the second article, we explored central angles, sectors, and arcs. This time we will introduce another kind of angle: the inscribed angle.

An inscribed angle is an angle drawn by using line segments to connect one point on a circle to two other points on the same circle, as in the graphic below:

Inscribed AngleLike a central angle, an inscribed angle creates a “wedge” shape, like a triangle where one side is rounded. The rounded side is an arc of the circle. For a central angle, the measure of the angle corresponds to the measure of the associated arc in a 1:1 relationship. For an inscribed angle, the measure of the angle corresponds to the measure of the associated arc in a 1:2 relationship. A 30 degree inscribed angle creates a 60-degree arc on the other side of the circle. A 60-degree inscribed angle creates a 12- degree arc on the other side of the circle. And, importantly, a 90-degree inscribed angle creates a 180-degree arc (half a circle or a semicircle) on the other side of the circle.

1. Inscribed Angle – GMAT Official Guide Problem

GMAT problems rarely use the term “inscribed angle” or feature an inscribed angle in isolation. Usually, the inscribed angle is part of an inscribed polygon, a polygon drawn inside a circle so that its vertices are points on the circle. Take a look at this official GMAT problem:

Inscribed Angle Official GMAT ProblemIn the figure shown, the triangle is inscribed in the semicircle. If the length of line segment AB is 8 and the length of line segment BC is 6, what is the length of arc ABC?

 

 

A. 15π
B. 12π
C. 10π
D. 7π
E. 5π

The problem refers not to an angle inscribed in a circle but to a triangle inscribed in a semicircle. Still, knowing the “1:2” factor of relationship between an inscribed angle and its associated arc is the key to solving this problem. Your logic might go something like this:

  1. This is a semicircle or a 180-degree arc.
  2. The angle at point B “opens up” to the straight edge of the semicircle, which is like the diameter of a circle. Another semicircle or 180-degree arc could be drawn across from this angle so that it makes a whole circle with the existing piece.
  3. Since the measure of an inscribed angle is 1/2 the measure of the arc it “creates” on the other side of the circle, the angle at point B is a 90-degree angle, and the triangle is a right triangle. 

At this point, your attention should return to the given information about the lengths of line segments AB and BC, which we now know to be the legs of a right triangle. These legs have lengths 6 and 8, which have a 3:4 relationship. Therefore we are looking at a 3-4-5 triangle, and the length of the hypotenuse is 10.

Finally, you must recall that this hypotenuse is the diameter of the circle. Therefore the diameter of the whole circle is 10. However, marking answer choice C would be a mistake, since we were asked for the length of arc ABC, where arc ABC is a semicircle (half a circle). So your final step is to divide your diameter of 10π by 2, leading you to the correct answer choice: E.

2. Inscribed Square – GMAT Official Guide Problem

Let’s try another problem, this time with an inscribed square:

Inscribed Square GMAT Official Guide ProblemThe figure shows a drop-leaf. With all four leaves down the tabletop is a square, and with all four leaves up the tabletop is a circle. What is the radius, in meters, of the tabletop when all four leaves are up?
A. 1/2
B. 
√2/2
C. 1
D. √2
E. 2

Notice that the problem doesn’t mention “a square inscribed in a circle,” but that is nonetheless what we have here. Many GMAT quant problems create scenarios that correspond to some mathematical phenomenon without using the math language. In this case, the fact that we are dealing with a square inscribed in a circle is relatively easy to see.

As in the previous problem, we are asked for a value of the circle (this time it is the radius instead of an arc length) but given only information about the inscribed shape: a square. As in the previous problem, the key is realizing that with any 90-degree inscribed angle, the line segments forming the angle are legs of a right triangle whose hypotenuse is also a diameter of the circle.

Using the Pythagorean theorem, the hypotenuse of this triangle (or the diagonal of the inscribed square) is √2. As before, forgetting to divide this value by 2 (since we were asked for the radius, not the diameter) will lead you to an incorrect answer choice. Don’t trip at the finish line. The value you need is √2 /2, answer choice B.

Here is a related problem:

Square inscribed in a circle problemIf rectangle ABCD is inscribed in the circle above, what is the area of the circular region?

A. 36.00
B. 42.25
C. 64.00
D. 84.50
E. 169.00

Again, we are asked for a value of the circle (its total area) but given only information about the inscribed rectangle. For our purposes, this rectangle is just as good as the square in the previous problem. With the square, we only needed the length of one side, because we know that all four sides are the same length. With a rectangle, we need both the length and the width in order to calculate the diagonal – the diameter of the circle – via the Pythagorean theorem. If you know your Pythagorean triples (like 3-4-5), you may realize immediately that the diagonal of this rectangle is 13.

D = √(5² + 12²)
D = √(25 + 144)
D = √169
D = 13

Now that we have the circle’s diameter, we can solve for its area. The radius of the circle is 13/2 or 6.5, and since Area = r², the square of 13/2 or 6.5 will be the coefficient of in the correct answer choice. It would be a waste of time to fully multiply out 6.5 * 6.5. We know that it will be of form __.25, and the only answer choice that matches this is B.

3. Data Sufficiency – GMAT Official Guide Problem

Let’s transition to data sufficiency for one final problem. Using the diagonal/diameter relationship in the previous problems, it would be possible to construct a variety of DS problems. But some DS inscription problems rely on another property of inscribed polygons.

Square ABCD is inscribed in circle O. What is the area of square region ABCD?

  1. The area of circular region O is 64π.
  2. The circumference of circle O  is 16π.

Solution

To answer this problem, all you need to know is that there is only one way to inscribe a square in a circle. The vertices of the square must lie on the circle. The perimeters and areas of the square and the circle will scale together. This means that if we know any value for either shape, we can calculate every value for both of them. Therefore each statement on its own is sufficient, and the answer to this problem is D.

As long as any polygon can be established as regular (having sides of equal length and angles of equal measure), there is only one way to inscribe it in a circle. A square is a regular quadrilateral, so this works for squares every time. But this same problem could have used a regular pentagon, a regular hexagon, or any regular polygon you like, and the correct answer would still be D. The regularity of the polygon is sufficient – and necessary – for this to work. If the regularity of the polygon cannot be established, then there are an infinite number of ways to inscribe it in a circle, each with its own unique area and perimeter.

It is also possible to flip the relationship and inscribe a circle inside a polygon. A related term is circumscription. The shape on the inside is inscribed in the shape on the outside. The shape on the outside is circumscribed around the shape on the inside. GMAT problems where the circle is on the inside usually use a square, so that the diameter of the circle is equal to the length of each side of the square. Such problems tend to be of lower difficulty level.

 

This concludes our third article on the GMAT’s treatment of circles. Next time we will look at what happens when the radius – rather than the diameter – pulls double duty as the hypotenuse of a right triangle.

 

Contributor: Elijah Mize (Apex GMAT Instructor)

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Pieces of Pi: Sectors, Arcs, and Central Angles
Posted on
22
Mar 2022

Pieces of Pi: Sectors, Arcs, and Central Angles

Welcome back to our series on GMAT circles. In the first article, we introduced the properties of radius/diameter, circumference, and area, discussing the relationships between all of these. This time we will introduce something called a central angle, which creates portions of a circle’s area and circumference called sectors and arcs, respectively.

The best way to define these things is probably with a simple visual.

Pieces of Pi

A central angle is an angle created by using line segments to connect a circle’s center to two points on its edge. A sector is the part of a circle’s area bounded by this central angle, and an arc is the part of a circle’s circumference between the two points used to draw the angle. A 90-degree central angle creates both a 90-degree sector and a 90-degree arc. An important note is that the lines used to form the central angle are radii of the circle.

As a further illustration, think about a pizza (something I do regularly). The pizza is a circle, the pieces are sectors separated by central angles, the crust is the circle’s circumference, and each piece’s section of crust is an arc. From this, you can see that any central angle creates both a sector and an arc that correspond to one another. When you pull a piece from a pizza or cut out a piece from a pie, you use a central angle to create a sector with an arc on its rounded edge.

To represent these things mathematically, we consider a circle to be like a 360-degree central angle. In this setup, the fractional relationship of a central angle to 360 corresponds to two things:

  1. The fractional relationship of the resulting sector to the circle’s total area
  2. The fractional relationship of the resulting arc to the circle’s total circumference

Since 90 is ¼ of 360, the area of a 90-degree sector is ¼ of its circle’s total area, and the length of a 90-degree arc is ¼ of its circle’s circumference.

To show all of this algebraically, let’s use the variable x for the degree measure of a central angle:

x / 360 = arc length / circumference
x / 360 = sector area / circle area 

Most pizzas are divided into 8 slices. This means that each slice has a central angle of 360/8 = 45° and that each slice is ⅛ of the area of the entire pizza.

Examples:

1. What is the central angle for three slices of pizza?
The central angle formed by 3 slices of pizza is 3 * 360 / 8 = 135 degrees.

2. What’s the area of a slice if the diameter is 20cm and there are six slices?
The area of a slice of pizza is 1/6 of its pizza’s total area. So, the area of a pizza can be found by using this formula A= π*r2 = 3.14*102 = 314cm
The area of a slice of pizza is  314/ 6 = 52.33cm


Keep in mind that you may have to consider this relationship in either direction. You may be given some info about the whole circle and then tasked with concluding something about a sector or an arc. Or you may be given some info about a sector or an arc and then tasked with concluding something about the whole circle. You may even be given info about both the whole circle (its area or circumference) and a sector or arc and then tasked with calculating the central angle. Each of these represents a perspective shift, and when doing a problem form, you can rewrite the problem from each of these perspectives to make sure you can fully navigate problems of this sort.

Pieces of Pi: Official GMAT Problems

Now for some official GMAT problems. Let’s start with two straightforward sector problems, one problem solving and one data sufficiency.

Problem-Solving Problem 

The annual budget of a certain college is to be shown on a circle graph. If the size of each sector of the graph is to be proportional to the amount of the budget it represents, how many degrees of the circle should be used to represent an item that is 15 percent of the budget? 

A. 15°
B. 36°
C. 54°
D. 90°
E. 150°

From the question, we can tell that the “circle graph” mentioned here is what we usually call a “pie chart,” a handy way to show the breakdown of a whole (like a budget) into its various parts. If we want to represent 15% of the budget, we need a sector with a central angle using 15% of the (360) available degrees in the circle. 0.15 * 360 = 54, so the correct answer is C. Piece of cake. Or piece of pie.

Now for a DS pie chart problem

TOTAL EXPENSES FOR THE FIVE DIVERSIONS OF COMPANY H

DS Pie Chart ProblemThe figure represents a circle graph of Company’s H total expenses broken down by the expenses for each of its five diversions. If O is the center of the circle and if Company H’s total expenses are $5,400,000, what are the expenses for Division R? 

1. x = 94
2. The total expenses for Division S and T are twice as much as the expenses for Division R.

Once again, this pie chart (which the GMAT apparently prefers to call a “circle graph”) is being used to represent a budget breakdown. Here we are told that the value represented by the whole circle is $5,400,000. We can think of this value as the area of the circle. We are asked for the expenses for division R, or in circle terms, the area of sector R.

Statement 1: x = 94

This is the measure of the central angle bounding the sector whose area we need to know (sector R). Since we already know the area of the whole circle, the measure of this central angle is the final piece of the puzzle. (Area of sector R = 94/360 * $5,400,000) Statement 1 is sufficient.

Statement 2: The total expenses for Divisions S and T are twice as much as the expenses for Division R.

This statement relates the total of two unknown sectors to another unknown sector. Given this statement alone, we don’t know the relationship of any of these sectors to the whole circle, so we can’t solve for any of their areas. Statement 2 is insufficient.

The answer:
Statement 1 is sufficient.
Statement 2 is insufficient.
The correct answer is A.

Pieces of Pi: More Difficult Problems

Let’s ratchet up the difficulty a bit with another sector problem that involves more smoke and mirrors.

Three identical circles problem

The figure consists of three identical circles that are tangent to each other. If the area of the shaded region is 64√3 – 32*π, what’s the radius of each circle?

A. 4
B. 8
C. 16
D. 24
E. 32

(tangent means just touching and not overlapping)

The problem mentions only three circles and a shaded region, but the graphic includes something more: an equilateral triangle drawn by connecting the centers of the three circles. You can solve this problem without knowing anything about the formula for the area of an equilateral triangle (although you should know this formula).  It should occur to you that the area of the shaded region could be expressed as the area of the triangle minus the combined area of those three sectors, which matches the given expression 64√3 – 32*π. So the (irrelevant) area of the triangle is 64√3, and the area of the three sectors is 32*π. 

You might start by trying to get the area of a single sector by dividing 32*π by 3. But 32 won’t divide nicely by 3, which should signal you to try something else. If you can’t go from the combined area of the three sectors to the area of one sector, maybe you can go from the combined area of the three sectors to the area of one circle. You might use your spatial reasoning and conclude that the three sectors together look like they make up half a circle. Or you might recall that each interior angle of an equilateral triangle measures 60 degrees. Therefore each of these sectors is ⅙ (60/360) of a whole circle, and the three of them together do indeed make up half a circle (3 * ⅙ = ½).

Now, if the sectors’ combined area is 32*π, and this is half a circle, then the area of the whole circle is 2 * 32*π = 64. Having found the area of a circle, we can now solve for the radius. 

A = 64 = π*r2
64 = r2
r = 8

And the correct answer choice is B.

Arc Length Problem

Let’s try one more problem, this time focusing on arc length.

The points R, T, and U lie on a circle that has radius 4. If the length of arc RTU is 4*π/3 what is the length of line segment RU?

A. 4/3
B. 8/3
C. 3
D. 4
E.  6

Points that “lie on a circle”

A note about points that “lie on a circle.” This always means that the points are on the edge or perimeter of the circle.

It may be helpful to visualize or even draw out what has been described here.

This is a good opportunity to introduce some terminology. We see that line segment RU connects two points on the circle. Such a line segment is called a chord. If the line continues on to either side of the circle so that the circle is “skewered,” the line is called a secant (the GMAT does not expect you to know this term). When a chord or secant passes through the circle’s center, it creates a diameter. A line outside a circle that just touches the circle at one point is called a tangent.

If you aren’t sure how to calculate the length of a chord like RU, start with what you know. We are given the length of arc RTU (4*π/3) and the radius of the circle. A good step is to calculate the circumference of the circle so that we can see how it relates to arc RTU. 

C = 2*π*r
C = 2*π*4
C = 8*π

The circumference is 8*π, and arc RTU is 4*π/3. 8*π/6 = 4*π/3. Therefore arc RTU represents ⅙ of the circumference of the circle, and its corresponding central angle is 60 degrees (360/6). Drawing out this information helps us to see its relevance.

The length of line segment RUThe central angle and chord RU form an equilateral triangle. Since the radius of the circle is 4, chord RU also has length 4, and the correct answer is D.

This concludes our second article on the GMAT’s treatment of circles. Next time we will look at another kind of angle inside a circle: an inscribed angle, and at the related topic of inscribed polygons.

 

Contributor: Elijah Mize (Apex GMAT Instructor)

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Circles On The GMAT 101 Area, Pi, and Circumference
Posted on
15
Mar 2022

Circles On The GMAT 101 – Area, Circumference, and Pi

Circles on the GMAT function like any other GMAT quant topic: the list of “knowledge bits” you need is short, but the questions creatively combine and/or disguise these few “knowledge bits” to create complex problems. 

In this first article, we will discuss the most basic properties of circles and the formulas that relate to these properties. The properties in question are area, circumference, and radius/diameter.

Circle Properties: Radius & Diameter 

I treat radius (r) and diameter (D) together because they essentially express the same thing, and because the relationship between them is so simple.

Radius (r) and diameter (D) of the circle

Diameter tells you how “wide” a circle is at its widest point. If you draw a straight line all the way across a circle, hitting the center on the way, that line is a diameter, and the length of that line is the diameter of the circle. The radius of the circle is half of this line – or any straight line drawn from the center of the circle to a point on its edge.

This is the most basic property of a circle. The two next most basic properties involve bringing in a specific number, one of the most famous numbers in mathematics: pi.

Circle Properties: Pi

Pi is the name of a Greek letter that looks (in its lowercase form) like this: π.

The value represented by this character is irrational (the numbers after the decimal never stop), but for most purposes and for the GMAT, 3.14 is enough. If you happen to be working with fractions or if you prefer fractions to decimals, 22/7 is a rather precise way to express pi. Rounded to the thousandths place, pi is 3.142, and 22/7 is 3.143. That’s close enough for jazz and certainly close enough for GMAT quant.

Interestingly – and for reasons relating to math beyond what is required for the GMAT – this same value can be combined with the radius/diameter to calculate both the circumference (C)  and area (A) of the circle. Circumference is the distance around the circle (its perimeter), and area is the space inside the circle.

C = π*D     or     C = 2*π*r
A = π*r²

The two options for the circumference equation are, in fact, equivalent, since D = 2r. You may be wondering why mathematicians split the 2 and the r around the π, instead of just saying π2r. One reason is simply the conventions that have developed for algebraic notation; it “looks wrong” to have the 2 in the middle after the π. But another reason is a (non-GMAT math) connection between the circumference and area equations. Both equations have a π, an r, and a 2. In the area equation, the 2 functions as an exponent on the r. In the circumference equations, it functions as a coefficient multiplying the expression. Area is pi times the square of the radius. Circumference is pi times the radius doubled.

Examples:

1. If a radius is 3, what’s the area of the circle?

A = π*r²= π*3² ≈ 28.27433

2. If the area of a circle is 25π, what is the diameter of the circle?

A = π*
r² = 25*π÷π
r = √25 = 5

3. If a radius is 4, what’s the circumference of the circle?

C = 2*r*π
C = 2*4*π
C = 8*π = 25.13274 

Official GMAT Problem

In all likelihood, none of this looks new. But like we said at the beginning, GMAT quant can get creative with common mathematical knowledge. Take a look at this official GMAT problem, and try to answer it before reading on:

Official GMAT Circle ProblemIn the figure shown, if the area of the shaded region is 3 times the area of the smaller circular region, then the circumference of the larger circle is how many times the circumference of the smaller circle?

A. 4
B. 3
C. 2
D. √3
E. √2

Understanding the Problem 

The only properties at play here are area and circumference, perhaps with radius/diameter as a “bridge” between the two, but the answer to the problem may not be immediately obvious. Part of this is due to a common GMAT technique – the removal of numbers to make the problem more abstract.

This problem disguises the relationship between the two circles by referring not to the area of each circle, but to the area of the shaded region. A helpful preliminary deduction is that if, as the problem says, the area of the shaded region is 3 times the area of the smaller circular region, then the area of the larger circle is 4 times the area of the smaller circle (comprising the 3 “parts” in the shaded region and the 1 “part” in the smaller circle).

So we have the factor relating the circles’ areas: 4. But we were asked about circumference, not area. How do we get from area to circumference? Well, both the area and circumference equations have the radius in them, so one option is to pick two values for area, with one being 4 times the other, solve for the radius of each circle, and then plug each of these radius values into the circumference equation.

Solving the Problem 

Let’s say the area of the larger circle is 16, and the area of the smaller circle is 4. Since π is common to all the circle equations, in this case it is irrelevant, and we should just remove it instead of keeping it as “dead weight” to move around algebraically.

Large Circle                             Small Circle
A = π*r²                                     A = π*r² 
16 = π*r²                                   4 = π*r² 
r = 4                                         r = 2

So when the areas of two circles are related by a factor of 4, their radii (plural for radius) are related by a factor of 2: that is, the square root of 4. This makes sense, since radius is a linear value and area is a square value. This is just like how when you double the length of the side of a square, you quadruple its area (since 2² = 4). When you triple the length of the side of a square, you make its area nine times what it was before (since 3² = 9). The factor of increase for area is the square of the factor of increase for the linear measure.

C = 2*π*r
C = 2*π*4
C = 2*π*2

Since we are only concerned about the factor relating the circumferences of the circles, we can ignore whatever is common to both (2*π), leaving us with 4 and 2. Since 4 is twice as much as 2, the circumference of the larger circle is twice the circumference of the smaller circle, and the correct answer choice is C.

The Final Step

That last step (and any real calculation) is avoided with two insights:

  1. There is a square relationship between area and radius. Since the area of the larger circle is 4 times that of the smaller circle, the radius of the larger circle is the square root of 4 (2) times the radius of the smaller circle.
  2. There is a linear relationship between circumference and radius. Since the radius of the larger circle is twice that of the smaller circle, the circumference of the larger circle is also twice that of the smaller circle.

When you move beyond rote memorization and understand the circle equations, these insights happen naturally and lead you to the correct answer choice quickly, with virtually no calculation.

 

This concludes the first article in our series on the GMAT’s treatment of circles. Next, we will dive into what happens when you use something called a central angle to mark off only a part of the area (a sector) and circumference (an arc) of a circle.

 

Contributor: Elijah Mize (Apex GMAT Instructor) 

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21
Jul 2021

GMAT 3D Geometry Problem – GMAT Math – Quant Section

GMAT 3D Geometry Problem 

In this problem we’re going to take a look at 3D objects and in particular a special problem type on the GMAT that measures the longest distance within a three-dimensional object. Typically, they give you rectangular solids, but they can also give you cylinders and other such objects. The key thing to remember about problems like this one is that effectively we’re stacking Pythagorean theorems to solve it – we’re finding triangles and then triangles within triangles that define the longest distance.

This type of problem is testing your spatial skills and a graphic or visual aid is often helpful though strictly not necessary. Let’s take a look at how to solve this problem and because it’s testing these skills the approach is generally mathematical that is there is some processing because it’s secondary to what they’re actually testing.

gmat 3d geometry question

GMAT 3D Geometry Problem Introduction

So, we have this rectangular solid and it doesn’t matter which way we turn it – the longest distance is going to be between any two opposite corners and you can take that to the bank as a rule: On a rectangular solid the opposite corners will always be the longest distance. Here we don’t have any way to process this central distance so, what we need to do is make a triangle out of it.

Notice that the distance that we’re looking for along with the height of 5 and the hypotenuse of the 10 by 10 base will give us a right triangle. We can apply Pythagoras here if we have the hypotenuse of the base. We’re working backwards from what we need to what we can make rather than building up. Once you’re comfortable with this you can do it in either direction.

Solving the Problem

In this case we’ve got a 10 by 10 base. It’s a 45-45-90 because any square cut in half is a 45-45-90 which means we can immediately engage the identity of times root two. So, 10, 10, 10 root 2. 10 root 2 and 5 makes the two sides. We apply Pythagoras again. Here it’s a little more complicated mathematically and because you’re going in and out of taking square roots and adding and multiplying, you want to be very careful not to make a processing error here.

Careless errors abound particularly when we’re distracted from the math and yet we need to do some processing. So, this is a point where you just want to say “Okay, I’ve got all the pieces, let me make sure I do this right.” 10 root 2 squared is 200 (10 times 10 is 100, root 2 times root 2 is 2, 2 times 100 is 200). 5 squared is 25. Add them together 225. And then take the square root and that’s going to give us our answer. The square root of 225 is one of those numbers we should know. It’s 15, answer choice A.

Okay guys for another 3D and Geometry problem check out GMAT 680 Level Geometry Problem – No Math Needed! We will see you next time.

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14
Jul 2021

GMAT 680 Level Geometry Problem – No Math Needed!

GMAT Geometry Problem

Hey guys, top level geometry problems are characterized typically by stringing a whole bunch of different rules together and understanding how one thing relates to the next thing, to the next thing. Until you get from the piece of information you started with to the conclusion. We’re going to start out by taking a look at this problem using the z equals 50° and seeing how that information goes down the line.

top level geometry problem

But afterwards we’re going to see a super simple logical pathway utilizing a graphic scenario that makes the z equals 50° irrelevant. To begin with we’re being asked for the sum of x and y and this will come into play on the logical side. We need the sum not the individual amounts but let’s begin with the y. We have a quadrilateral and it has parallel sides which means the two angles z and y must equal 180°. That’s one of our geometric rules. If z is 50° that means y is 130° and we’re halfway there.

Next we need to figure out how x relates and there are several pathways to this. One way we can do it is drop. By visualizing or dropping a third parallel line down, intersecting x, so on the one hand we’ll have 90 degrees. We’ll have that right angle and on the other we’ll have that piece. Notice that the parallel line we dropped and the parallel line next to z are both being intersected by the diagonal line going through which means that that part of x equals z. So we have 50° plus 90° is 140°. 130° from the y, 140° from the x, gives us 270°.

Another way we can do this is by taking a look at the right triangle that’s already built in z is 50° so y is 1 30°. now the top angle in the triangle must then be 180° minus the 130° that is 50°. it must match the z again we have the parallel lines with the diagonal coming through then the other angle the one opposite x is the 180° degrees that are in the triangle minus the 90° from the right triangle brings us to 90° minus the 50° from the angle we just figured out means that it’s 40° which means angle x is 180° flat line supplementary angles minus the 40° gives us 140° plus the 130° we have from y again we get to 270°.

Graphic Solution Path

Now here’s where it gets really fun and really interesting. We can run a graphic scenario here by noticing that as long as we keep all the lines oriented in the same way we can actually shift the angle x up. We can take the line that extends from this big triangle and just shift it right up the line until it matches with the y. What’s going to happen there, is we’re going to see that we have 270° degrees in that combination of x and y and that it leaves a right triangle of 90°, that we can take away from 360° again to reach the 270°.

Here the 50° is irrelevant and watch these two graphic scenarios to understand why no matter how steep or how flat this picture becomes we can always move that x right up and get to the 270°. That is the x and y change in conjunction with one another as z changes. You can’t change one without the other while maintaining all these parallel lines and right angles. Seeing this is challenging to say the least, it requires a very deep understanding of the rules and this is one of those circumstances that really points to weaknesses in understanding most of what we learn in math class in middle school, in high school. Even when we’re prepping only scratches the surface of some of the more subtle things that we’re either allowed to do or the subtle characteristics of rules and how they work with one another and so a true understanding yields this very rapid graphic solution path.

Logical Solution Path

The logical solution path where immediately we say x and y has to be 270° no matter what z is and as you progress into the 80th, 90th percentile into the 700 level on the quant side this is what you want to look for during your self prep. You want to notice when there’s a clever solution path that you’ll overlook because of the rules. Understand why it works and then backtrack to understand how that new mechanism that you discovered fits into the framework of the rules that we all know and love. Maybe? I don’t know if we love them! But they’re there, we know them, we’re familiar with them, we want to become intimate. So get intimate with your geometry guys put on some al green light some candles and I’ll see you next time.

If you enjoyed this problem, try other geometry problems here: GMAT Geometry.

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17
Feb 2021

Data Sufficiency: Area of a Triangle Problem

Hey guys! Today we’re checking out a geometry Data Sufficiency problem asking for the area of a triangle, and while the ask might seem straightforward, it’s very easy to get caught up in the introduced information on this problem. And this is a great example of a way that the GMAT can really dictate your thought processes via suggestion if you’re not really really clear on what it is you’re looking for on DS. So here we’re looking for area but area specifically is a discrete measurement; that is we’re going to need some sort of number to anchor this down: whether it’s the length of sides, or the area of a smaller piece, we need some value!

Begin with Statement #2

Jumping into the introduced information, if we look at number 2, because it seems simpler, we have x = 45 degrees. Now you might be jumping in and saying, well, if x = 45 and we got the 90 degree then we have 45, STOP. Because if you’re doing that you missed what I just said. We need a discrete anchor point. The number of degrees is both relative in the sense that the triangle could be really huge or really small, and doesn’t give us what we need. So immediately we want to say: number 2 is insufficient. Rather than dive in deeply and try and figure out how we can use it, let’s begin just by recognizing its insufficiency. Know that we can go deeper if we need to but not get ourselves worked up and not invest the time until it’s appropriate, until number 1 isn’t sufficient and we need to look at them together.

Consider Statement #1

Number 1 gives us this product BD x AC = 20. Well here, we’re given a discrete value, which is a step in the right direction. Now, what do we need for area? You might say we need a base and a height but that’s not entirely accurate. Our equation, area is 1/2 x base x height, means that we don’t need to know the base and the height individually but rather their product. The key to this problem is noticing in number 1 that they give us this B x H product of 20, which means if we want to plug it into our equation, 1/2 x 20 is 10. Area is 10. Number 1 alone is sufficient. Answer choice A.

Don’t Get Caught Up With “X”

If we don’t recognize this then we get caught up with taking a look at x and what that means and trying to slice and dice this, which is complicated to say the least. And I want you to observe that if we get ourselves worked up about x, then immediately when we look at this BD x AC product, our minds are already in the framework of how to incorporate these two together. Whereas, if we dismiss the x is insufficient and look at this solo, the BD times AC, then we’re much more likely to strike upon that identity. Ideally though, of course, before we jump into the introduced information, we want to say, well, the area of a triangle is 1/2 x base x height. So, if I have not B and H individually, although that will be useful, B x H is enough. And then it’s a question of just saying, well, one’s got what we need – check. One is sufficient. Two doesn’t have what we need – isn’t sufficient, and we’re there. So,

I hope this helped. Look for links to other geometry and fun DS problems below and I’ll see you guys soon. Read this article about Data sufficiency problems and triangles next to get more familiar with this type of GMAT question.

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Posted on
16
Feb 2021

Triangles With Other Shapes

As discussed before, now that we’ve talked about the basic triangles, we can start looking at how the GMAT can make problems difficult by embedding triangles in other figures, or vice versa. 

Here are just a few examples, which include triangles within and outside of squares, rectangles, and circles:

triangles in other shapes GMAT article

Today, we’ll talk about some crucial connections that are often made between triangles and other figures, starting with the 45-45-90 triangle, also known as the isosceles right triangle.

You’ve probably seen a rectangle split in two along one of its diagonals to produce two right triangles:

triangles in other shapes gmat article gmat problem

But one of the oft-overlooked basic geometric truths is that when that rectangle is a square (and yes, remember a square is a type of rectangle), the diagonal splits the square into two isosceles right triangles. This makes sense when you think about it, because the diagonal bisects two 90-degree angles to give you two 45-degree angles:

triangles in other shapes gmat article, 45 45 90 degree angle

(For clarification, the diagonal of a rectangle is a bisector when the rectangle is a square, but it is not a bisector in any other case.)

Another very common combination of shapes in more difficult GMAT Geometry problems is triangles with circles. This can manifest itself in three common ways:

Triangles Created Using the Central Angle of a Circle

triangle in a circle, gmat geometry article

In this case, notice that two of the sides of the triangle are radii (remember, a radius is any line segment from the center of the circle to its circumference). What does that guarantee about the triangle?

Since two side are of equal length, the triangle is automatically isosceles. Remember that the two angles opposite those two sides are also of equal measure. So any triangle with the center of the circle as one vertex and points along the circumference as the other two vertices will automatically be an isosceles triangle.

Inscribed Triangles

triangle inscribed in circle, gmat problem

An inscribed triangle is any triangle with a circle’s diameter as one of its sides and a vertex along the circumference. And a key thing to note: an inscribed triangle will ALWAYS be a right triangle. So even if you don’t see the right angle marked, you can rest assured the inscribed angle at that third vertex is 90 degrees.

Squares and Rectangles Inscribed in Circles

rectangle in circle, gmat geometry

What’s important to note here is that the diagonal of the rectangle (or square) is equivalent to the diameter of the circle.

Now that we’ve seen a few common relationships between triangles and other figures, let’s take a look at an example Official Guide problem:

A small, rectangular park has a perimeter of 560 feet and a diagonal measurement of 200 feet. What is its area, in square feet?

A) 19,200
B) 19,600
C) 20,000
D) 20,400
E) 20,800

Explanation

The diagonal splits the rectangular park into two similar triangles:

triangle in other shapes gmat problem

Use Signals to Avoid Algebra

It can be tempting to then jump straight into algebra. The formulas for perimeter and diagonal are P = 2L + 2W an D2 = L2 + W2, respectively, where L and W are the length and width of the rectangle. The second formula, you’ll notice, arises out of the Pythagorean Theorem, since we now have two right triangles. We are trying to find area, which is LW, so we could set out on a cumbersome algebraic journey.

However, let’s try to use some SIGNALS the problem gives us and our knowledge of how the GMAT operates to see if we can short-circuit this problem.

We know the GMAT is fond of both clean numerical solutions and common Pythagorean triples. The large numbers of 200 for the diagonal and 560 for the perimeter don’t change that we now have a very specific rectangle (and pair of triangles). Thus, we should suspect that one of our basic Pythagorean triples (3-4-5, 5-12-13, 7-24-25) is involved.

Could it be that our diagonal of 200 is the hypotenuse of a 3-4-5 triangle multiple? If so, the 200 would correspond to the 5, and the multiplying factor would be 40. That would also mean that the legs would be 3*40 and 4*40, or 120 and 160.

Does this check out? Well, we’re already told the perimeter is 560. Adding 160 and 120 gives us 280, which is one length and one width, or half the perimeter of the rectangle. We can then just double the 280 to get 560 and confirm that we do indeed have the correct numbers. The length and width of the park must be 120 and 160. No algebra necessary.

Now, to get the area, we just multiply 120 by 160 to get 19,200 and the final answer of A.

Check out the following links for our other articles on triangles and their properties:

A Short Meditation on Triangles
The 30-60-90 Right Triangle
The 45-45-90 Right Triangle
The Area of an Equilateral Triangle
Isosceles Triangles and Data Sufficiency
Similar Triangles
3-4-5 Right Triangle
5-12-13 and 7-24-25 Right Triangles

 

By: Rich Zwelling, Apex GMAT Instructor
Date: 16th February, 2021

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Posted on
09
Feb 2021

Triangle Inequality Rule

One of the less-common but still need-to-know rules tested on the GMAT is the “triangle inequality” rule, which allows you to draw conclusions about the length of the third side of a triangle given information about the lengths of the other two sides.

Often times, this rule is presented in two parts, but I find it is easiest to condense it into one, simple part that concerns a sum and a difference. Here’s what I mean, and we’ll use a SCENARIO:

Suppose we have a triangle that has two sides of length 3 and 5:

triangles inequalities 1

What can we say about the length of the third side? Of course, we can’t nail down a single definitive value for that length, but we can actually put a limit on its range. That range is simply the difference and the sum of the lengths of the other two sides, non-inclusive.

So, in this case, since the difference between the lengths of the other two sides is 2, and their sum is 8, we can say for sure that the third side of this triangle must have a length between 2 and 8, non-inclusive. [Algebraically, this reads as (5-3) < x < (5+3) OR 2 < x < 8.]

If you’d like to see that put into words:

**The length of any side of a triangle must be shorter than the sum of the other two side lengths and longer than the difference of the other two side lengths.**

It’s important to note that this works for any triangle. But why did we say non-inclusive? Well, let’s look at what would happen if we included the 8 in the above example. Imagine a “triangle” with lengths 3, 5, and 8. Can you see the problem? (Think about it before reading the next paragraph.)

Imagine a twig of length 3 inches and another of length 5 inches. How would you form a geometric figure of length 8 inches? You’d simply join the two twigs in a straight line to form a longer, single twig of 8 inches. It would be impossible to form a triangle with a side of 8 inches with the original two twigs.

triangle inequalities 2

 

If you wanted to form a triangle with the twigs of 3 and 5, you’d have to “break” the longer twig of 8 inches and bend the two twigs at an angle for an opportunity to have a third side, guaranteed to be shorter than 8 inches:

triangle inequalities 3

The same logic would hold for the other end of the range (we couldn’t have a triangle of 3, 5, and 2, as the only way to form a length of 5 from lengths of 2 and 3 would be to form a longer line segment of 5.)

Now that we’ve covered the basics, let’s dive into a few problems, starting with this Official Guide problem:

If k is an integer and 2 < k < 7, for how many different values of k is there a triangle with sides of lengths 2, 7, and k?
(A) one
(B) two
(C) three
(D) four
(E) five

Strategy: Eliminate Answers

As usual with the GMAT, it’s one thing to know the rule, but it’s another when you’re presented with a carefully worded question that tests your ability to pay close attention to detail. First, we are told that two of the lengths of the triangle are 2 and 7. What does that mean for the third side, given the triangle inequality rule? We know the third side must have a length between 5 (the difference between the two sides) and 9 (the sum of the two sides).

Here, you can actually use the answer choices to your advantage, at least to eliminate some answers. Notice that k is specified as an integer. How many integers do we know now are possible? Well, if k must be between 5 and 9 (and remember, it’s non-inclusive), the only options possibly available to us are 6, 7, and 8. That means a maximum of three possible values of k, thus eliminating answers D and E.

Since the GMAT is a time-intensive test, you might have to end up guessing now and then, so if you can strategically eliminate answers, it increases your chances of guessing correctly.

Now for this problem, there’s another condition given, namely that 2 < k < 7. We already determined that k must be 6, 7, or 8. However, of those numbers, only 6 fits in the given range 2 < k < 7. This means that 6 is the only legal value that fits for k. The correct answer is A.

Note

It’s important to emphasize that the eliminate answers strategy is not a mandate. We’re simply presenting it as an option that works here because it is useful on many GMAT problems and should be explored and practiced as often as possible.

Check out the following links for our other articles on triangles and their properties:

A Short Meditation on Triangles
The 30-60-90 Right Triangle
The 45-45-90 Right Triangle
The Area of an Equilateral Triangle
Triangles with Other Shapes
Isosceles Triangles and Data Sufficiency
Similar Triangles
3-4-5 Right Triangle
5-12-13 and 7-24-25 Right Triangles

 

By: Rich Zwelling, Apex GMAT Instructor
Date: 9th February, 2021

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