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Magoosh Brain Twister: Looking for Mr. Right Triangle – Explanation


Before we get started, make sure you remember Monday’s Brain Twister!


Right triangle DEF has a hypotenuse of less than 7. What is the maximum possible area of DEF, if the area of DEF is an integer?

Numeric Entry: [_____________]

Answer and Explanation

To maximize the area of a right triangle, we must assume that the triangle is a 45:45:90. Now, we don’t know if this will lead to an integer area, but it is a good first line of attack.

Using the Pythagorean theorem, we know that x^2 + x^2 = hypotenuse^2. The hypotenuse squared must be less than 49 (7^2 = 49), so 48 is a good place to start. If the hypotenuse squared is 48, then we can do the following:

x^2 + x^2 = 48

2x^2 = 48

x^2 = 24

x = 2{sqrt{6}}

Area of 45:45:90 triangle is x^2/2, which equals (2{sqrt {6}})^2/2 = 24/2 = 12.


How’d you do?


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5 Responses to Magoosh Brain Twister: Looking for Mr. Right Triangle – Explanation

  1. Jenil November 14, 2014 at 2:24 pm #

    a^2 + b^2 = c^2

    So, if we consider the hypotenuse as 7 => Sqrt. of 49 (we are looking for a value slightly less than 7)

    The least greatest value less than 7 (Sqrt 49) => Sqrt 48

    Also since the area has to be an integer, we should find two such values that add to the hypotenuse and when divided by 2 gives an integer value

    A – Sqrt 24 : B – Sqrt 24 (one will be the base and other the height)

    (Sqrt 24)^2 + (Sqrt 24)^2 = C^2
    24 + 24 = C^2
    48 = C^2
    C = Sqrt 48 = 6.9

    Area of triangle = 1/2 * b * h = (Sqrt 24 * Sqrt 24)/2 = 24/2 = 12

    • Chris Lele
      Chris Lele November 17, 2014 at 2:18 pm #

      That’s definitely the right approach, and is very similar to the one above 🙂

  2. Asaad November 11, 2014 at 2:02 am #

    Hi Chris,

    I’m not sure about “of less than 7”.. But you know the language 🙂

    Would that work?

    3:4:5 — 16+9=25

    1/2 b*h

    1/2 3*4=6

    • Chris Lele
      Chris Lele November 11, 2014 at 11:27 am #

      Hi Asaad,

      Yes, that would work…however, it wouldn’t maximize the area. You would want a triangle with a hypotenuse much closer to 7, and that triangle (to maximize area) would have to be a 45:45:90.

      Hope that helps!

      • Asaad November 12, 2014 at 12:25 pm #

        Got it. Thanks Chris.

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