Pythagorean Theorem 2

Directions: Using the digits 0 to 9, at most one time each, fill in the boxes to find the lengths of the missing sides such that the missing leg’s length is as long as possible.


How do we figure out which side length would be easiest to assume?
How do we figure out the length of the remaining side?
Which side of a right triangle is the longest?


The longest possible length for the hypotenuse is 10. If it was 11, you would not be able to make the leg long enough and only two digits. It also can’t be 4 or less as the hypotenuse has to be the longest side.

So, the possible side lengths for the triangles are (a, b, and c being the leg, leg, and hypotenuse):

a, b, c
4, √84, 10
4, √65, 09
4, √48, 08
4, √33, 07
4, √20, 06
4, 03, 05

After removing the solutions that duplicate digits, you are left with only two options:

a, b, c
4, √65, 09 (from Karl S.)
4, √84, 10

Therefore, 4, √84, 10 is the optimal solution. Be on the lookout though for students who make the leg longer than the hypotenuse. While we want the missing leg to be as long as possible, it still has to be less than the length of the hypotenuse. So, this is a good chance to check for misunderstandings.

Source: Robert Kaplinsky

Print Friendly, PDF & Email

Check Also

Exponential Powers

Directions: Using the digits 0 to 9 at most one time each, place a digit …


  1. Love it!

  2. 4 and 84 share a digit of 4. Shouldn’t that response be eliminated, too?

  3. Couldn’t 4, sqrt(65), and 09 also be a possible solution, albeit not as long as your given optimal solution?

  4. You are totally right Karl. I’ve updated the answer accordingly.

  5. I don’t really understand

Leave a Reply

Your email address will not be published. Required fields are marked *