Learn the most common solutions to the Pythagorean Theorem
There aren’t many numbers you need to memorize for success on the GMAT Quantitative section, but knowing a few key Pythagorean triplets will save you a ton of time. First, try these GMAT practice question: remember: no calculator!
1) In right triangle ABC, BC = 48 and AB = 60. Find AC

(A) 32
(B) 36
(C) 40
(D) 42
(E) 45
2) In the xy plane, what is the distance between (4, 2) and (11, 6)?

(A) 16
(B) 17
(C) 18
(D) 19
(E) 20
3) In the diagram above, ∠L = ∠M = 90°, KL = 4, LM = 8, MN = 10, and JN = JK = 13. What is the area of JKLMN?

(A) 92
(B) 96
(C) 100
(D) 108
(E) 116
Good numbers to memorize
In almost all cases, I will recommend that GMAT student learn to remember mathematical facts without memorizing them. This is one of the few cases in which I will unapologetically recommend memorizing. There are certain sets of positive integers that satisfy the Pythagorean Theorem: these sets of three integers are called Pythagorean triplets. Some of them are very obscure, but some of them are extremely common. The most common by far is the triplet (3, 4, 5). In all of these, I am listing a set (a, b, c) that satisfies , so the largest number would be the hypotenuse of the triangle. Two other common sets of Pythagorean triplets are (5, 12, 13) and (8, 15, 17). Right there, BAM! Memorize those three sets, and you will spare yourself many stressful moments of lengthy calculations on the GMAT Quantitative section, when you have no calculator.
First of all, if you encounter a right triangle with legs 8 & 15, you won’t have to square and add things up: rather, you will just know that the hypotenuse is 17. The benefits, though, of that wee bit of memorizing expend wildly when you realize: you can multiply any of those three sets by any integer and get a new set of Pythagorean Triplets. The first set, (3, 4, 5), is the most common to see in multiplies — (6, 8, 10), (9, 12, 15), (12, 16, 20), etc. — but once in a while you may see one of the other two multiplied by something small, like 2 or 3.
Imagine on the GMAT, you see this:
That’s the diagram, and you are asked to find the length of YZ — you are given five answer choices, any of which looks like it could be plausible at first blush. Well, the unskilled GMAT taker will consume a great deal of time squaring 24, then squaring 45, then adding those together and — Gadzooks! — trying to figure out the square root of the fourdigit number that results. 🙁 It’s MUCH easier simply to recognize that 24 = 3*8 and 45 = 3*15, so we are clearly dealing with 3 times the (8, 15, 17) triplet, which we conveniently have memorized. That means the answer must be YZ = 3*17 = 51. 🙂
Armed with these tricks, take another look at the practice problems again, before reading the explanations below. Also, here’s a Magoosh practice GMAT question that uses one of these Pythagorean triplets:
4) http://gmat.magoosh.com/questions/100
Practice question explanations
1) Clearly, the wrong approach would be to square 48 and 60, subtract, and without calculator try to take the square root of the resultant fourdigit number. Not fun! A little GCF exploration reveals: 48 and 60 have a GCF = 12. More to the point, 48 = 4*12, and 60 = 5*12, so clearly we have the (3, 4, 5) triplet multiplied by 12. That means the missing side must be AC = 3*12 = 36. Answer = B
2) For those who would like the visual, here’s a diagram:
The purple line is the actual distance we want to find. One very important trick to know: when you need to find a diagonal distance in the xy plane, always use the Pythagorean Theorem. We draw the connecting horizontal and vertical lines, shown in green here, to create a right triangle, the hypotenuse of which is the distance we want. Horizontal & vertical distances are very easy in the xy plane: we simply count, or subtract the corresponding coordinates. These x and ydistances are:
These are the lengths of the two green lines. Lo and behold: our old friend, the (8, 15, 17) triplet! Without any further calculations, we see immediately that the distance between these two points must be 17 units. Answer =
3) We will subdivide the figure as shown.
Notice, we constructed KR, such that KR is perpendicular to MN. This makes KLMR a rectangle and KRN a right triangle. S is the midpoint of KN, so that JS is the median & altitude of isosceles triangle JKN. (Since it’s easy to find the area of rectangles and right triangles, those make particularly good choices for subdivision units when you have to find area.)
Now, we will figure things out piecebypiece. First of all, a particularly easy piece is rectangle KLMR: Area = bh = 4*8 = 32.
The problem gives that MN = 10, and we know MR = 4, so NR must equal 6, and KR must equal 8. These are two legs of the Pythagorean triplet (6, 8, 10), which is the (3, 4, 5) triplet times two. This means we know KN = 10. We also know the area of triangle KRN is: Area = 0.5*bh = 0.5*6*8 = 24.
Since we know KN = 10, we know the midpoint divides that in half, so KS = SN = 5. Notice, we now have two right triangles, KJS and NJS, each with a leg of 5 and a hypotenuse of 13. This is another one of the triplets, (5, 12, 13)! Immediately, without further calculation, we know JS = 12. Now we can find the area of the big isosceles triangle JKN: Area = 0.5*bh = 0.5*12*10 = 60.
Total area = (area of rect. KLMR) + (area of triangle KRN) + (area of triangle JKN)
= 32 + 24 + 60 = 116
Answer = E
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The sound quality in this video is poor, but the technique is super useful and more intuitive IMO.
Pythagorean triplets can be generated from the first number:
For odds: n , n^2/2.5, n^2/2+.5
For evens: n, (n/2)^21, (n/2)^2+1
https://www.youtube.com/watch?v=86YAPbZmsRI
Great explanation, thanks for the blog.
Quick question, How did we conclude that S is the midpoint of KN?
I’m glad you liked the explanations. And great question about S and its place on KN. A triangle proerty that’s very important on the GMAT is in play here. Note that in triangle JKN, JN and JK have the same length. this means you’re dealing with an isosceles triangle. As such, if you bisect triangle JKN directly down the middle between the two sides, you’ll create a 90 degree angle in the center of the triangle’s base, cutting the base exactly in half. This is true of any triangle with two equal sides.
Because JK and JN have the same length so if you cut the base at 90 degrees connecting the top, that can only happen at the midpoint. If you cut the base at 90 degrees at any other point it would also intersect one of the sides.
Using units digit calculations you can quickly figure out question #1.
If you understand that 60 > 48 then use units digit multiplication and subtraction = 0 (0x0) and 4 (8×8)
then 0 – 4 = 6 (remember 60×60 > 48×48)
the only answer choice that can have a units digit of 6 is 36 or B.
Dear Vitor,
That’s a great observation! There is often more than one valid method for approaching difficult GMAT problems. Thanks for sharing!
Mike 🙂
Thanks for the great blog. Under Good numbers to memorise: 24 = 3*8 and “45 = 3*45”
Seems there’s a typo there as 45 = 3*15 (not 3*45)
I have just memorised the common triplets you suggested. Thanks again for the helpful tips.
– JC
JC
Thanks for pointing out the typo — I just corrected it. I’m glad you found the tips helpful. Best of luck to you.
Mike 🙂
Thanks for the great blog. Under Good numbers to memorise: 24 = 3*8 and “45 = 3*45″
Seems there’s a typo there as 45 = 3*15 (not 3*45)
Dear Yo Yo,
I’m glad you found this blog helpful. 🙂 I believe I fixed all the typos in the blog already. I think the only place 3*45 appears is in the previous comment, JC pointing out the typo over a year ago.
Mike 🙂