# New SAT Math Section: Pythagorean Theorem on Soccer Fields

So I attempted this problem and I'm very sure I'm doing it right but I keep getting it wrong as my answer choice is not even one of the answer choices listed. There is a picture that goes with the problem that I have attached.

Question: The picture below shows the dimensions of a soccer field. Let x be the distance from the northwest corner to the center of the eastern side and let y be the distance from the northwest corner to the southeast corner. To the nearest meter, what is y – x?

So what I did was I drew out the triangles, the smaller one would have a horizontal side length of 90 m and a vertical side length of 40 m so by the pythagorean theorem, the x value (the hypotenuse) would be 98.49 m. I did the same with the larger triangle that had side lengths of 120 m and 80 m giving me a y of 144.22 m. Calculating y - x results in 45.73 m rounded to 46 m which is not an answer choice. The correct answer choice is actually A: 18 m. I'm not sure if the answer key is wrong or if I'm doing the problem wrong.

• $120$ and $40$, not $90$ and $40$. – André Nicolas Jan 28 '16 at 18:38
• @AndréNicolas but it says Northwest corner to the center of the eastern side, so wouldn't the center of the eastern half of the field be at 90 not 120? or does it mean the center of the all the way east line? – Dude1310 Jan 28 '16 at 18:40
• A matter of interpretation. They should have said Eastern edge, then it would be unambiguous. But my preferred interpretation of centre of the Eastern side makes it $120$. – André Nicolas Jan 28 '16 at 18:43
• Yup, makes sense now, thanks! – Dude1310 Jan 28 '16 at 18:45
• You are welcome. The question as worded should not have survived proof-reading, one always checks again and again for potential ambiguity. Not much fun, but got to be done. – André Nicolas Jan 28 '16 at 18:49

First, we find $x$ by Pythagoras. The legs of the right triangle are $120$ and $40.$ Therefore, $x = \sqrt{120^{2} + 40^{2}} = 40\sqrt{10}.$
Next, we find $y,$ also using Pythagoras. The legs of this triangle are $120$ and $80.$ We have that $y = \sqrt{120^{2} + 80^{2}} = 40\sqrt{13}.$
We solve and get $y - x = 40\sqrt{10} - 40\sqrt{13},$ which is closest to $\boxed{\text{(A)}18\text{m}}.$