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

## Question

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 . The hypotenuse squared must be less than , so 48 is a good place to start. If the hypotenuse squared is 48, then we can do the following:

Area of 45:45:90 triangle is , which equals .

*How’d you do? *

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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

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

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

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!

Got it. Thanks Chris.