# GMAT Math: 3D Solids

According to the GMAT OG, there are two 3D solids you should understand in detail are (a) rectangular solids, and (b) cylinders. On any given GMAT, you will see no more than a couple questions on 3D solids, but they definitely could appear, especially if you are already getting many questions right and are moving into harder questions.

## Rectangular Solids

Also known as just a box or a brick, these are pretty straightforward. Let’s say a solid has a length L, a width W, and a height H. The volume, of course, is V = LWH. The surface area is a little trickier: the solid has two faces that are L x W, two that are L x H, and two that are W x H, for a grand total of:

$$Total Surface Area = 2LW + 2LH + 2WH$$

For the special case of a cube with side s, the volume is $$V = s^3$$, and the surface area simplifies to $$S.A. = 6s^2$$

What’s a little more subtle about these: the three-dimensional version of the Pythagorean Theorem.

Let’s say that $$AD = L$$, $$DF = W$$, and $$AB = H$$. Lengths like AC, DE, or BE are called “face diagonals”, and to find them, you would just use the ordinary Pythagorean Theorem with the sides of the relevant rectangle. If we were, though, to construct segment AE, that would go through the empty center of the solid. In geometry, that’s called a “space diagonal.” You don’t need to know that vocabulary, but you need to recognize this when it appears, and you need to know you can apply the three-dimensional version of the Pythagorean Theorem. Let’s say that AE has a length of D. Then:

$$D^2 = L^2 + W^2 +H^2$$

Occasionally, a GMAT math problem will ask you to use this formula to evaluate the length of a space diagonal of a rectangular solid.

## Cylinders

Cylinders are very familiar from everyday life: a circle at the bottom, a congruent circle directly above it, and smooth curving side (the lateral face) joining one circle to the other. The volume is simply $$V = (area of base)*h = \pi r^2*h$$.

The surface area is a little trickier. The top = the bottom = a circle = $$\pi r^2$$, that’s not hard. What’s tricky is the lateral surface area, the sides of the cylinder. Think of a can of soup. In particular, think about the label, which nicely covers the lateral surface of the can. Suppose we make a vertical cut in that label, and unfurl it to lay it flat. We get a rectangle. The height of the rectangle is simply h, the height of the cylinder itself. The width of the rectangle wrapped all the way around the can before we unfurled it. One way to think about it: it ran around the edge of the circles at the top & bottom. Well, that length around the edge of a circle —- that’s called the circumference, and its formula is $$c = 2 \pi r$$. So, the label is rectangle with length $$2 \pi r$$ and height h, so its area is $$A = 2 \pi rh$$. That’s the formula for the lateral surface area of a cylinder. To remember that, remember the soup can.

The total surface area is simply that plus the top & bottom circles.

$$Total ~Surface ~Area = 2 \pi r^2 + 2 \pi rh$$

For more obscure solids, see this blog post on Advanced Geometric Solids.

## Practice Questions

1) K is a rectangular solid. Find the volume of K

(1) a diagonal line across the front face of K has a length of 40

(2) a diagonal line across the bottom face of K has a length of 25

(A) Statement 1 alone is sufficient but statement 2 alone is not sufficient to answer the question asked.
(B) Statement 2 alone is sufficient but statement 1 alone is not sufficient to answer the question asked.
(C) Both statements 1 and 2 together are sufficient to answer the question but neither statement is sufficient alone.
(D) Each statement alone is sufficient to answer the question.

2) M is a rectangular solid. Find the volume of M

(1) The bottom face of M has an area of 28, and the front face, an area of 35.

(2) All three dimensions of M are positive integers greater than one.

3) Q is a cube. Find the volume of the cube.

(1) The total surface area of Q is 150 sq cm

(2) The distance from one vertex of Q to the catty-corner opposite vertex is $$5*\sqrt{3}$$

4) A certain lighthouse is a tall thin cylinder of brick, with a light chamber at the top. The brick extends from the ground to the floor of the light chamber. The brick siding of lighthouse is in need of painting. How many square feet of brick does it have?

(1) The outer diameter of the brick portion of the lighthouse is 28 ft

(2) There are 274 steps from ground level to the floor of the light chamber, and each one is 8 inches high.

Here’s another free practice question with a video explanation.

(1) E; (2) C; (3) D; (4) C

## Practice Questions Solutions

(1) So, the prompt gives us relatively little information. We are looking for the volume. Statement #1 tells us a diagonal line across the front face has a length of 40. If L is the length and H is the height, then this means: $$L^2 + H^2 = 40^2 = 1600$$. That’s one equation with two unknowns, and no way to solve for the volume. Statement #1 by itself is insufficient.

Statement #2 tells us a diagonal line across the bottom face has a length of 25. If L is the length and W is the width, then this means: $$L^2 + W^2 = 25^2 = 625$$. Again, that’s one equation with two unknowns, and no way to solve for the volume. Statement #2 by itself is insufficient.

Combine the statements: we now know both $$L^2 + H^2 = 40^2 = 1600$$ and $$L^2 + W^2 = 25^2 = 625$$. We now have two equations but three unknowns: still not enough information to solve for the individual values, nor can we somehow rearrange to get the product of LWH. Even combined, the statements are insufficient.

2) Again, a sparse prompt, and we need to find the volume.

Statement #1 tells us the area of the bottom — so $$L*W = 28$$, and the area of the front, so $$L*H=35$$. With this statement alone, we have no idea what the individual dimensions could be. There’s no guarantee that they are integers: for example, $$L=3$$, $$W=28/3$$, and $$H = 35/3$$ satisfy these conditions. The individual dimensions could be almost anything, which means the volume could be almost anything. Statement #1by itself is insufficient.

Statement #2 tells us each individual dimension is an integer greater than one, but other than that, we have zero information. This could be $$3*4*5$$, or $$76*84*137$$. The volume will be a positive integer, but beyond that, we can determine nothing about it. Statement #2by itself is insufficient.

Combine the statements. Now we know $$L*W=28$$ and $$L*H=35$$, and that L, W, and H are all integers greater than one. Well, 28 and 35 have only one common factor greater than one, and that’s 7. That must mean, L = 7, which means W = 4 and H = 5, which means we could calculate the volume. Combined, the statements are sufficient.

3) Another simple prompt, and again, we need the volume.

Statement #1 tells us surface area is 150. Well, for a cube, $$SA = 6s^2$$. If we know $$150 = 6s^2$$, we can solve for s, which will allow us to calculate the volume. Statement #1 is by itself sufficient.

Statement #2 tell us the space diagonal has a length of $$5*\sqrt{3}$$. Well, the three-dimensional version of the Pythagorean Theorem tell us that:

$$(5*\sqrt{3})^2 = s^2 + s^2 + s^2$$

That would allow us to solve for s, which would allow us to calculate the volume. Statement #2 is by itself sufficient.

Both statements sufficient by themselves. Answer = D.

4) The lighthouse is essentially a cylinder, and the area that needs to be painted is the lateral surface area. We would need both r and h to calculate this.

Statement #1 tells us diameter = 28, which means radius = 14. We know the radius, but not the height. Statement #1 by itself is insufficient.

In Statement #2, the 274 steps go from the very bottom to the very top of the cylinder in question. We know each step is 8 inches, so the total height traversed by the steps is $$274*8 in$$. We would have to change that result from inches to feet, but essentially, with this piece of information, we have the height. We know the height, but not the radius. Statement #2 by itself is insufficient.

Combined. Statement #1 gives us the radius, statement #2 gives us the height, and we’re in business! We now have enough information to calculate the lateral surface area. Combined, the statements are sufficient.

#### Special Note:

To find out where 3D geometry sits in the “big picture” of GMAT Quant, and what other Quant concepts you should study, check out our post entitled:

What Kind of Math is on the GMAT? Breakdown of Quant Concepts by Frequency

### 7 Responses to GMAT Math: 3D Solids

1. Michael October 21, 2015 at 8:49 pm #

Hi Mike,

For question #1 wouldn’t you know since it’s a rectangle that the corners have 90* angles and as such if the hypotenuse is 40 that you could determine that it’s a special right triangle (using 3-4-5 the sides would be 24-32-40)? If so that gives you L & W and then only H would be left to solve for in part 2?

If I have this right, wouldn’t the answer be C?

• Mike McGarry October 22, 2015 at 11:30 am #

Dear Michael,
I’m happy to respond. 🙂 My friend, not all right triangles are 3-4-5 triangles. Of all possible right triangles, 3-4-5 constitute only a tiny little percentage. From the first statement, all we know is that we have a right triangle with a hypotenuse of c = 40. This means the other sides could be:
(a) a = 24 and b = 32 (the 3-4-5 case)
(b) a = 11.2 and b = 38.4 (the 7-24-25 case)
(c) a = 20 and b = 20sqrt(3) (the 30-60-90 case)
(d) a = b = 20sqrt(2) (the 45-45-90 case)
(e) a = 1 and b = sqrt(1599)
(f) a = 8sqrt(5) and b = 16sqrt(5)
(g) a = 3 and b = sqrt(1591)
(h) a = sqrt(3) and b = sqrt(1597)
These are just 8 cases, but the possibilities are infinite. There is absolutely no reason to assume that the length and width of the rectangle are integers. That is one of the biggest math traps on the test, assuming the requirement of integers when that requirement is not present in the problem.
Does all this make sense?
Mike 🙂

• Michael October 22, 2015 at 7:56 pm #

It does 🙂

For clarify, if it said “K is a regular rectangle solid” then the x-x-x[sqrt(2)] technique would work, right? My logic is that all angles then must be equal (90*) and that all sides would also be equal, so it would be a 45-45-90. Or am i off here too?

Also, I appreciate your timely response above. You guys have a phenomenal blog and it’s been very helpful so far – so thank you!

Michael

• Mike McGarry October 23, 2015 at 12:10 am #

Dear Michael,
I’m happy to respond. 🙂 I am very glad you appreciate the blog and that you have found it helpful. Unfortunately, you are off again. The x-x-xsqrt(2) works only if one of the sides of the rectangular solid is a square! Think about boxes in the real world — a shoe box, a cereal box, a box of envelops, etc. Think of office buildings, which often are big rectangular solids (ignoring windows and such). It’s exceedingly rare that a pair of faces on a rectangular solid would be squares. Yes, it could happen in an exceptional case, but you are making this exceptional case the norm. Also, in any ordinary 2D rectangles, all four angles are equal, but this does not guarantee that all four sides are equal. Equiangular implies equilateral in triangles, but that rule does not generalize to anything other than triangles.
Does all this make sense?
Mike 🙂

• Michael October 24, 2015 at 10:57 am #

Ahhhh yes, I see – thank you! I’m trying to get into the habit of drawing things to the extreme (as you guys do in your videos!) and from that standpoint it’s very obvious.

Many thanks! 😀

Michael

2. Mountain14 July 2, 2013 at 5:07 pm #

Hi Mike,

I really thought as much I can and then realized there can be typo in the below, i.e W=4 . :)…

i hope , I am right else, i need to know this funda as well 🙁

Combine the statements. Now we know and , and that L, W, and H are all integers greater than one. Well, 28 and 35 have only one common factor greater than one, and that’s 7. That must mean, L = 7, which means L = 4 and H = 5, which means we could calculate the volume. Combined, the statements are sufficient.

• Mike July 2, 2013 at 5:54 pm #

Dear Mountain,
Yes, quite right. Thank you for catching this typo. I corrected it on the blog.
Mike 🙂

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