{"id":16241,"date":"2013-01-14T09:00:19","date_gmt":"2013-01-14T16:00:19","guid":{"rendered":"https:\/\/gmatclub.com\/blog\/?p=16241"},"modified":"2013-01-04T07:32:20","modified_gmt":"2013-01-04T14:32:20","slug":"re-thinking-pythagoras-is-a-triangle-obtuse","status":"publish","type":"post","link":"https:\/\/gmatclub.com\/blog\/re-thinking-pythagoras-is-a-triangle-obtuse\/","title":{"rendered":"Re-thinking Pythagoras: Is a triangle obtuse?"},"content":{"rendered":"

\"\"The\u00a0Pythagorean Theorem<\/a>\u00a0is one of the most remarkable theorems in all of mathematics.\u00a0 It has a treasure trove of ramification up its sleeve, any one of which could provide you with invaluable help on the GMAT Quantitative section.\u00a0 For example, consider this practice problem.<\/p>\n

1) Consider the following three triangles<\/p>\n

I. a triangle with sides 6-9-10
\nII.\u00a0 a triangle with sides 8-14-17
\nIII. a triangle with sides 5-12-14<\/p>\n

Which of the following gives a complete set of the triangles that have at least one obtuse angle, that is, an angle greater than 90\u00b0?<\/p>\n

    \n
  1. I<\/li>\n
  2. II<\/li>\n
  3. III<\/li>\n
  4. I & II<\/li>\n
  5. II & III<\/li>\n<\/ol>\n

     <\/p>\n

    The basic theorem<\/h2>\n

    Everyone knows the basic formula:<\/p>\n

    \"\"<\/a><\/p>\n

    To distinguish this formula from the theorem itself (something not often done), I will refer to that as the \"Pythagorean formula<\/strong>.\"\u00a0 Many folks don't realize that the Theorem is something different from this single formula.\u00a0 Some people realize that, in order for this formula to work, the triangle must be a right triangle, and some people even remember that \"c\" has to be the hypotenuse of a right triangle.\u00a0 This leads to the basic statement of the theorem:<\/p>\n

    If a triangle is a right triangle with sides a < b < c, then the Pythagorean formula is true of the sides.\u00a0<\/strong><\/p>\n

    One important thing to appreciate about Mr. Pythagoras' famous theorem is that it goes both ways: in logical parlance, it is \"biconditional.\" In other words<\/p>\n

    A. If you know the triangle is a right triangle, if you are given that fact, then you can conclude that the Pythagorean formula works for its sides.<\/p>\n

    and<\/p>\n

    (B) If you are given the three sides of a triangle, and you know (or can verify) that these three sides satisfy the Pythagorean formula, then that triangle absolutely must be a right triangle.<\/p>\n

    In other words, if you consider these two qualities (i) being a right triangle, and (ii) sides satisfying the Pythagorean formula, then, those two qualities always come together, and either one necessitates the other.\u00a0 To use highly dramatic language, God Himself\u00a0could not create<\/a>\u00a0a triangle that has one of those qualities and not the other.<\/p>\n

    There are common three-number sets know as\u00a0Pythagorean triplets<\/a>:\u00a0 these sets, such as (3, 4, 5), are sets of numbers that satisfy the Pythagorean formula, which necessarily means they would also be the sides of a right triangle.\u00a0 The reader familiar with the common Pythagorean triplets discussed in that post will find the numbers in the above problem evocative of, but not equal to, these sets of triplets.<\/p>\n

     <\/p>\n

    Other triangles<\/h2>\n

    Most triangles in the world are not right triangles.\u00a0 The Pythagorean Theorem applies only to right triangles, but with a little reconfiguring, we can also use it to deduce facts about other triangles.\u00a0\u00a0 We know that if there's an equal sign in the Pythagorean formula, it means the triangle is a right triangle.\u00a0\u00a0 What if there's either a greater-than or a less-than sign instead of equals sign?<\/p>\n

     <\/p>\n

    Case one: Bigger big side or shorter legs<\/h2>\n

    For this case, we will consider this inequality:<\/p>\n

    \"\"<\/a><\/p>\n

    This would be true if we started with a right triangle, and then made the hypotenuse bigger while leaving the two legs the same size.\u00a0\u00a0 That pushes the two legs apart.<\/p>\n

    \"\"<\/a><\/p>\n

    This would also be true if we made either one of the smaller, while leave the other two sides the same size.\u00a0 This pulls the vertex that previously had a right angle closer to the former hypotenuse, which has the effect of pushing the legs apart.<\/p>\n

    \"\"<\/a><\/p>\n

    In either case, notice that the right angle becomes an\u00a0obtuse<\/strong>\u00a0angle.\u00a0\u00a0 This allows us to state variation #1 on the Pythagorean Theorem:<\/p>\n

    \"\"<\/a><\/p>\n

    Notice: an obtuse triangle is a triangle in which one angle is obtuse.\u00a0 It is impossible for more than one angle to be obtuse, because then the angles in the triangle would add up to more than 180\u00b0.<\/p>\n

     <\/p>\n

    Case two: Smaller big side or bigger legs<\/h2>\n

    For this case, we will consider this inequality:<\/p>\n

    \"\"<\/a><\/p>\n

    This would be true if we started with a right triangle, and then made the hypotenuse smaller while leaving the two legs the same size.\u00a0\u00a0 That pulls the two legs toward each other.<\/p>\n

    \"\"<\/a><\/p>\n

    This would also be true if we made either one of the bigger, while leave the other two sides the same size.\u00a0 This pushes the vertex that previously had a right angle further from to the former hypotenuse, which has the effect of pulling the legs toward each other.<\/p>\n

    \"\"<\/a><\/p>\n

    In either case, notice that the right angle becomes an\u00a0acute<\/strong>\u00a0angle.\u00a0\u00a0 This allows us to state variation #2 on the Pythagorean Theorem:<\/p>\n

    \"\"<\/a><\/p>\n

    Notice: an acute triangle is a triangle in which all three angles are acute.\u00a0 In any triangle, at least two of the angles must be acute. In an acute triangle, even the largest angle is acute.<\/p>\n

     <\/p>\n

    Summary<\/h2>\n

    We can combine all this information in one place.\u00a0 Suppose a triangle has three sides such that a < b < c.\u00a0 Then:<\/p>\n

    \"\"<\/a><\/p>\n

    Remembering how to morph the Pythagorean Theorem could help you in a rare GMAT problem such as the one at the top.\u00a0 More importantly, perhaps this discussion gave you some insight into the interrelationship between equations and spatial reasoning, and insights along those lines certain could help you in challenging GMAT questions.\u00a0\u00a0 Now that you have seen all of this, take another look at the practice problem before reading the solutions below.<\/p>\n

     <\/p>\n

    Practice problem explanation<\/h2>\n

    1) Remembering the common Pythagorean Triplets is very helpful in this problem.\u00a0\u00a0 Triangle I is very close to the (6, 8, 10) right triangle, but we made one of the legs longer.\u00a0\u00a0 This makes the sum of the squares of the legs greater than the longest side squared, so Triangle I is acute.\u00a0\u00a0 Triangle II is very close to the (8, 15, 17) right triangle, but we made one of the legs shorter.\u00a0\u00a0 This makes the sum of the squares of the legs less than the longest side squared, so Triangle II is obtuse.\u00a0 Triangle III is very close to the (5, 12, 13) right triangle, but we made the longest side bigger.\u00a0\u00a0 This makes the sum of the squares of the legs less than the longest side squared, so Triangle III is obtuse.\u00a0 Answer =\u00a0E<\/strong><\/p>\n

    This post was written by Mike McGarry, GMAT expert at Magoosh<\/a>, and originally posted here<\/a>.<\/p>\n","protected":false},"excerpt":{"rendered":"

    The\u00a0Pythagorean Theorem\u00a0is one of the most remarkable theorems in all of mathematics.\u00a0 It has a treasure trove of ramification up its sleeve, any one of which could provide you with…<\/p>\n","protected":false},"author":133,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[9,783,243,736],"tags":[],"class_list":["post-16241","post","type-post","status-publish","format-standard","hentry","category-gmat","category-magoosh-blog","category-blog","category-quant-gmat","entry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/posts\/16241","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/users\/133"}],"replies":[{"embeddable":true,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/comments?post=16241"}],"version-history":[{"count":1,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/posts\/16241\/revisions"}],"predecessor-version":[{"id":16243,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/posts\/16241\/revisions\/16243"}],"wp:attachment":[{"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/media?parent=16241"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/categories?post=16241"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/tags?post=16241"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}