# SAT Tip of the Week: TRYangles! Triangles are one of the first shapes that we learn in elementary school, and yet they are often the source of much consternation on the SAT.  Though there is much to know about trigonometry that can require complex and intricate calculations, the knowledge of triangles required for the SAT is actually quite concise.  Here is a quick review of the basics of triangles and how they might be used on the SAT.

The Basics: A triangle has three sides and three angles.  All the interior angles of a triangle add up to 180 degrees.  In math speak : A + B +  C = 180.  This means if you have two angles of any triangle, you can always find the third (something that comes up frequently on the SAT). The largest side is opposite the largest angle and the smallest side is opposite the smallest angle.

Pythagorean Theorem: This is only useful for right triangles, but right triangles are great on the SAT because they give you all the information needed to find the area of a triangle (which, of course, is ½ A *B, or ½ base * height).  The pythagorean theorem states: A² + B² = C², which means if you have two sides of a right triangle, you can always find the third. Common right triangles that have easy to remember side ratios are triangles with a 3x-4x-5x relationship, and a 5x-12x-13x relationship.  These Pythagorean triples are useful because if two of the sides of a right triangle have this side relationship, the third must follow suit. For example if two sides of a right triangle are 10 and 8, then the third side must be 6 {6-8-10 is the same as 3(2) – 4(2) – 5(2), hence the “x” in the paragraph above}.

Special Triangles: Identifying these special triangles saves a step when doing the work of the Pythagorean theorem. An equilateral triangle, when split in half, becomes a 30 – 60 – 90 triangle, which has the side relationship shown above of X – X √3 – 2X, where X is the side opposite the 30 degree angle. If you cut a square in half you get an isosceles, right triangle or a 45 – 45 – 90 triangle.  This has the side relationship S – S – S√2, where is one of the sides opposite the 45 degree angle.  These special triangles are given on the formula sheet of the SAT but it is very useful to commit them to memory, as it is quite time consuming to constantly refer to the formula sheet when you think you have encountered a special triangle.

An interesting characteristic of the sides of triangles is as follows: If A=5 and B= 8, then 3 < C < 13.  C must be between 3 and 13.In triangle ABC, |B-C| < A < |B+C|. This is to say, any side on a triangle must be between the absolute value of the sum and the difference of the other sides of the triangle.

Here is an example question that will use some triangle knowledge:

“A rectangular pasture has twelve equally spaced poles on its southern border, and sixteen equally spaced poles on its eastern border.  A diagonal pathway from the eastern corner of the pasture to the center of the pasture is 40 ft.  How many feet of fencing would be required to build a fence around the entire pasture?” The first step is always to draw and label what is given.  We are given a rectangular pasture that has twelve equally spaced poles on its southern border, and sixteen equally spaced poles on its eastern border.  We label the distance between poles as X and we notice that we now have two sides of a triangle, one 12x and one 16x.

We remember the rules of Pythagorean triples and deduce that the diagonal of this triangle would have to be 20x.  We then look for what the problem is asking us to find.  We have to find the perimeter of the pasture, but all that is given is the length of a pathway from the eastern corner of the pasture to the center of the pasture.

AHA! We now know the length of HALF of the distance of the diagonal of the rectangular pasture!  We also know that the FULL diagonal is 20x.  We set up a simple equation to solve for X, remembering to double the length given from the center to the corner of the field.

2(40) = 20x

80 = 20x

x = 4

We then use our answer for X to find the length and width of the pasture and add everything together, remembering to multiply the length and width by two, to find the perimeter.

16 (4) = L = 64

12 (4) = W = 48

2W +2L = 2(64) + 2(48) = 224

Voila!  The perimeter of the whole field is 224ft, so that is how much fencing will be needed.

Triangles are a very useful tool that is often used in tandem with other math shapes and concepts on the SAT. Through an understanding of triangles, one can develop a greater understanding of many difficult problems on the SAT.

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David Greenslade is a Veritas Prep SAT instructor based in New York. His passion for education began while tutoring students in underrepresented areas during his time at the University of North Carolina. After receiving a degree in Biology, he studied language in China and then moved to New York where he teaches SAT prep and participates in improv comedy. Read more of his articles here, including How I Scored in the 99th Percentile and How to Effectively Study for the SAT.