Question

Please help, this one had me upset all day!

Answer 1

check the picture below.

let's do the red ones, to see if they check out.


`\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=√(a^2+b^2) \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}\\\\ -------------------------------\\\\ √(24^2+69^2)~~~~\approx~~~~ 73.054774~~~~\\e 74 \\\\\\ √(8^2+15^2)~~~~=~~~~17\\e 19 \\\\\\ √(30^2+60^2)~~~~\approx~~~~67.08204\\e 68`

and now, you can check any of the others, to see if the root of that sum is the last or longest side, namely the hypotenuse.

Answer 2

7, 24, 25 makes one. 9, 40, 41 as well. 39, 52, 65 works. 11, 60, 61 does too. Lastly, 40, 42, 58 is a right triangle as well.

How I found out was plugging each into the Pythagorean Theorem for proving/solving right triangles of a² + b² = c². The last number in each combo is your C. The other two numbers are interchangeable in regards to position.