## A Case Study of the Area of an Equilateral Triangle

Fact: on the GMAT Math section, you are likely to find questions about the area of an equilateral triangle, and it would be efficient if you knew the formula. (BTW, the formula appears a little further below.)

## Don’t Merely Memorize

I am going to recommend that you work to remember this formula, rather than memorize it. What distinction am I drawing? By memorizing, I mean you write this isolated factoid, area of an equilateral triangle, on an index card, and drill yourself until you can successful regurgitate it. That’s one way to get something into your memory, but it’s hard to retain it, particularly over the long term. One big problem is: on the day of the test, because of increased stress, if you forget the factoid you previously could regurgitate, you are stuck: you have no recourse.

## Memory with Contextual Understanding

By contrast, true learning involves remembering something because you understand it. When you understand every step of an argument and how it fits together, that makes it very easy to remember. This is particularly true in mathematics: if you can remember the series of steps that lead to a conclusion, first of all, such learning makes the conclusion much easier to remember; furthermore, even if you can’t remember the conclusion or want to check it, you will probably remember enough of the argument that you can reconstruct it if need be. This is a much deeper kind of understanding.

## Example: The Area of an Equilateral Triangle

Suppose you have an equilateral triangle with sides of length **s**.

We know all the sides have the same length. It’s not relevant for finding the area, but we also know that all three angles equal 60 degrees.

**Step #1:** We know that the formula for the area of a triangle is . The base is clearly s, but we need a height.

**Step #2:** We draw an altitude, that is, a line from B that is perpendicular to AC. The length of this line is the height needed in A = .

**Step #3**: Point D is the midpoint of AC, so AD = s/2. Also, angle ADB is 90 degrees.

**Step #4:** Call the length of BD h, and apply the Pythagorean Theorem in triangle ADB:

**Step #5**: Now that we have the height in terms of s, we can find the area.

That last formula is, indeed, the area of an equilateral triangle, and remembering it will be a definite time-saver on GMAT Math. Again, I don’t want you to *memorize* it. Rather, I strongly encourage you to remember this five step argument: practice recreating it step-by-step until you can flawlessly recapitulate the entire thing by yourself. Then, you won’t merely remember this formula — you will own it!

I encourage you to do that with each important GMAT math formula. If you don’t know the steps that lead to a particular formula, and can find it online, leave me a comment :).

Try a free Magoosh GMAT Practice Question to practice “remembering” what we just learned!

I think Carly is still right if she sees triangle ABD as half of triangle ABC. in this case 1/2 base *height will be (1/2*S/2* height. ( height is result from step 4). the base of ABD is S/2. At step five she could proceed to eventually obtain an answer which needs to be multiplied by 2 in order to obtain the complete area of ABC because ABD = BCD. .

Hi Emmanuel,

Good eye! That is correct! The issue here, of course, is that Carly was not solving it in this way and I don’t think she saw that solution, but that is definitely another valid way of solving this question. There are often multiple routes to a correct answer–thanks for sharing this! 🙂

Hi Mike,

I have one question from the practice problem which you have explained in co-ordinate geometry part (vertical and horizontal line).

You have used the 30-60-90 triangle formula to solve the problem which i am not able to understand could you please explain the 30-60-90 triangle.

Hi Piyush,

Thanks for your question! Since you’re a premium subscriber, I went ahead and forwarded your question to our remote tutor team. Someone from that team will respond to you directly over email.

Have a great day!

Dani

Hi Mike,

I don’t understand how you get 4 in the denominator in Step 5 when you calculate the final area. Why do you divide the whole thing by 2? The steps below are what I did, and I got 8 for the denominator:

A = (1/2) x bh

A = (1/2) x (s/2) x [ (s x sqrt(3)) /2 ]

A = (s/4) x [ (s x sqrt(3)) /2 ]

A = [ s^2 x sqrt (3) ] / 8

In other words, you multiply the numerators across to get the final numerator,

1 x s x [ s x sqrt (3) ] = s^2 x sqrt (3) –> Fine, no problem here!

But when you multiply the denominators across to get the final denominator,

2 x 2 x 2 = 8 –> How are you getting 4??

Please respond when you can, this is making me crazy!! Thank you in advance!

Dear Carly,

I’m happy to respond! 🙂 I don’t want you to go crazy! 🙂

The problem lies in how you copied from what I had printed above to what you typed in your comment. To you, I am sure it looks as if you copied faithfully, but you actually made a big mistake in copying.

Look above at the formula, the line following A = (1/2)*bh

In the big numerator, there’s a (1/2) time s times (s * sqrt(3)). So far, so good. In the denominator of that huge fraction, there’s a two. We are dividing that entire denominator by a single factor of 2.

Your mistake was to quasi-distribute the division over the different factors. This is a gross misapplication of the Distributive Law. You see, we can distribute division over addition or subtraction

(a + b)/c = a/c + b/c

(a – b)/c = a/c – b/c

What you did was the super-illegal move of distributing division over multiplication

(a*b)/c does NOT equal (a/c)*(b/c).

This illegal move would produce an extra factor of c in the denominator, an extra factor out of thin air. That is precisely how you produced an extra factor of 2 out of thin air, and then wondered why your answer was a factor of 2 off.

Here’s a better way to think about the whole thing. When we have [big numerator]/2, that is entirely equivalent to

(1/2)*[big numerator]

In other words, dividing something by 2 is the same as multiplying by 1/2. We are dividing by only one factor of two, so the entire expression should be multiplied by one factor of (1/2). If you write that line as

(1/2)*[(1/2)*s*(s*sqrt(3)]

Then it become very clear why we have only 4 in the denominator.

Does all this make sense?

Mike 🙂

Yes, thank you! Now that I’m looking at it again I realize what a silly mistake that is. I always mess up on the small stuff!!

Thanks again! I LOVE this blog, you are a lifesaver 🙂

Hi Mike!

Just came along with a shorter explanation for the area:

1- triangle ADB can be seen as a form of 1, sqrt(3), 2 (because the hypotenuse is the double of the smaller base), h = (s x sqrt(3))/2

2- now applying the normal triangle area formula:

(s x h) / 2 = (s x (s x sqrt(3))/2)) / 2 = (s^2 x sqrt(3)) / 4

love your reasoning, understand the hole process is to master the rules, an not know them by hard.

Dear Juan,

Yes, if one thinks in terms of the 30-60-90 triangle, one can get this formula very quickly. I was showing everything in a slow, step-by-step way, just to make everything clear to folks who do not have your facility with math.

Mike 🙂

Having an issue in Step 4. How do you go from

H^2 = s^2 – (s^2)/4 (3rd equation)

to

H^2 = 4 (s^2)/4 – (s^2)/4 (4th equation)

Thanks,

Ryan

Dear Ryan,

I’m happy to respond. 🙂 I simply multiplied the first term, the s^2, by 4/4 to get a common denominator. We can always multiply by 4/4, because that’s a quantity equal to 1, so it doesn’t change the value of the expression.

Does all this make sense?

Mike 🙂

Thanks, Mike.

This makes sense.

Ryan

Ryan,

You are more than welcome! I’m very glad this made sense to you. Best of luck to you in your studies!

Mike 🙂

While this seems like the most thorough way to learn the formulas, I think it is an unrealistic approach. When will I find the time to locate and fully understand the derivations of all of the needed math formulas, while also plugging away at lesson videos and practice problems? Just recalling this five-step argument would eat up the entire ninety seconds needed to solve the problem.

Dear Caroline,

I’m happy to respond. 🙂 First of all, you will notice that, almost always in the Magoosh lesson videos, we explain the derivation of the formula as part and parcel of the lesson. We never simply say: here’s the formula, just memorize it. It’s already part of the lesson. I’m suggesting that you take notes on it, remember it, and include in your studying, rather than studying just the formula in isolation. A very good way to study the formula is to look at the formula itself and explain to yourself: from whence does this formula come? Why is this true? If you study & practice that way consistently, it will become ingrained, and by the time you get to the test, you will simply know it inside-out. That’s my hope.

Does this make sense?

Mike 🙂

Isn’t the sum of all three angles inside a triangle 180, instead of 60?

Dear Cami,

Absolutely! The sum of the three angles inside

anytriangle is 180. In an equilateral triangle,eachangle is 60, so the sum is still 180. Does all this make sense?Mike 🙂