Question

write the equation you would set up to solve this problem. do not solve.
(image of a right triangle with legs 31mm and 67mm? wait, no, the right angle, one leg 31mm, the other segment 67mm? wait, the triangle has a right angle, one leg 31mm, the hypotenuse? wait, the image shows a right triangle with one leg 31mm, another side 67mm (maybe the other leg or the base), and the hypotenuse x? wait, the ocr text: write the equation you would set up to solve this problem. do not solve.

Explanation:

Step1: Identify the triangle type

The triangle is a right - triangle with one leg \(a = 31\space mm\), the other leg (part of the base) \(b=67 - \text{unknown part}\)? Wait, no, looking at the diagram, the base of the large segment is \(67\space mm\), and the right - triangle has one leg \(31\space mm\), and the hypotenuse? No, wait, the right - triangle has legs \(31\space mm\) and let's say the horizontal leg is \(x\)? Wait, no, maybe the base of the right - triangle is \(x\) and the other leg is \(31\space mm\), and the hypotenuse is related to the \(67\space mm\) segment. Wait, actually, in a right - triangle, we can use the Pythagorean theorem. Wait, maybe the two legs are \(31\space mm\) and \(x\), and the hypotenuse is \(67\space mm\)? No, that can't be because \(31^2+x^2 = 67^2\) would mean \(x\) is less than \(67\), but maybe the base of the right - triangle is \(x\) and the vertical leg is \(31\space mm\), and the hypotenuse is part of the \(67\space mm\) line. Wait, no, the correct approach: the right - triangle has legs \(31\space mm\) and \(x\), and the hypotenuse is the segment from the right - angle to the end of the \(67\space mm\) line? No, maybe the \(67\space mm\) is the length of the base of the whole figure, and the right - triangle has one leg \(31\space mm\) and the other leg \(x\), and the hypotenuse is \(67 - \text{something}\)? No, I think the correct way is to use the Pythagorean theorem. Let's assume that the right - triangle has legs \(31\space mm\) and \(x\), and the hypotenuse is \(67\space mm\)? No, that would not make sense because \(31^2 + x^2=67^2\) is a possible equation. Wait, no, maybe the base of the right - triangle is \(x\) and the vertical leg is \(31\space mm\), and the horizontal leg (the other part of the \(67\space mm\) line) is \(67 - x\)? No, I think the diagram shows a right - triangle with one leg \(31\space mm\), another leg \(x\), and the hypotenuse is equal to the length of the segment which is part of the \(67\space mm\) line. Wait, actually, the Pythagorean theorem states that for a right - triangle with legs \(a\) and \(b\) and hypotenuse \(c\), \(a^{2}+b^{2}=c^{2}\). If we consider the right - triangle with leg \(a = 31\space mm\), leg \(b=x\), and hypotenuse \(c = 67\space mm\) (but that would mean \(31^{2}+x^{2}=67^{2}\)), but maybe the other way: the two legs are \(31\) and \(x\), and the hypotenuse is the segment such that the sum of the squares of the legs equals the square of the hypotenuse. Wait, the correct equation setup is \(31^{2}+x^{2}=67^{2}\)? No, maybe the base of the right - triangle is \(x\) and the vertical leg is \(31\), and the horizontal line is \(67\), so the hypotenuse is \(67 - \text{nothing}\), I think the correct equation is \(31^{2}+x^{2}=67^{2}\) (if \(67\) is the hypotenuse) or \(31^{2}+(67 - x)^{2}=x^{2}\)? No, that's more complicated. Wait, the problem is to set up the equation. Let's assume that the right - triangle has legs \(31\space mm\) and \(x\), and the hypotenuse is \(67\space mm\). Then by Pythagorean theorem, \(31^{2}+x^{2}=67^{2}\). Wait, no, maybe the \(67\space mm\) is the length of the base, and the right - triangle has one leg \(31\space mm\) and the other leg \(x\), and the hypotenuse is the segment from the right - angle to the top of the \(67\space mm\) line. No, I think the correct equation is \(31^{2}+x^{2}=67^{2}\) (using Pythagorean theorem for the right - triangle with legs \(31\) and \(x\) and hypotenuse \(67\)).

Step2: Write the Pythagorean theorem equation

The Pythagorean theorem is \(a^{2}+b^{2}=c^{2}\), where \(a = 31\), \(b=x\), and \(c = 67\). So the equation is \(31^{2}+x^{2}=67^{2}\).

Answer:

\(31^{2}+x^{2}=67^{2}\)