Question

problem 6
change one of the values to make this triangle a right triangle.
there are many different solutions. try to calculate at least two.
answers should be exact.

x	y	z

try it
(image of a triangle with y=6, z=7, x=4)

Explanation:

Step1: Check current triangle (Pythagoras)

Check if $4^2 + 6^2 = 7^2$: $\quad 16 + 36 = 52$, $7^2=49$. Not equal.

Step2: Solution 1: Adjust z to right hypotenuse

Set $z = \sqrt{x^2 + y^2}$
$\quad z = \sqrt{4^2 + 6^2} = \sqrt{16+36} = \sqrt{52} = 2\sqrt{13}$

Step3: Solution 2: Adjust y to right leg (z hypotenuse)

Set $y = \sqrt{z^2 - x^2}$
$\quad y = \sqrt{7^2 - 4^2} = \sqrt{49-16} = \sqrt{33}$

Answer:

x	y	z
4	$\sqrt{33}$	7