Question

in greenpoint, the library is due south of the courthouse and due west of the community swimming pool. if the distance between the library and the courthouse is 15 kilometers and the distance between the courthouse and the city pool is 25 kilometers, how far is the library from the community pool? kilometers

Explanation:

Step1: Identify the triangle type

The library is south of the courthouse, and the pool is west of the courthouse, so their positions form a right triangle with the courthouse at the right angle. The distance between courthouse and library ($a = 15$ km) and between courthouse and pool ($b = 25$ km? Wait, no—wait, the distance between courthouse and city pool (pool) is 25? Wait, no, re-reading: library is south of courthouse (so vertical leg, length 15), pool is west of courthouse (horizontal leg, length let's say $x$? Wait, no, the distance between courthouse and pool is 25? Wait, no, the problem says: library is due south of courthouse (so vertical side, length 15 km), pool is due west of courthouse (horizontal side, length let's say $b$), and we need the distance between library and pool (hypotenuse $c$). Wait, no, wait the numbers: distance between library and courthouse is 15 km (vertical), distance between courthouse and pool is 25 km? Wait, no, maybe I mixed up. Wait, the problem: "distance between the library and the courthouse is 15 kilometers and the distance between the courthouse and the city pool is 25 kilometers"—wait, no, the original: "the distance between the library and the courthouse is 15 kilometers and the distance between the courthouse and the city pool is 25 kilometers"—wait, no, the user's text: "the distance between the library and the courthouse is 15 kilometers and the distance between the courthouse and the city pool is 25 kilometers, how far is the library from the community pool?" Wait, maybe it's a right triangle, so we use Pythagoras: $c = \sqrt{a^2 + b^2}$, but wait, maybe $a = 15$, $b = 25$? Wait, no, maybe I misread. Wait, the library is south (vertical), pool is west (horizontal), so the distance between library and pool is the hypotenuse. So $c = \sqrt{15^2 + 25^2}$? Wait, no, wait, maybe the numbers are 15 and 20? Wait, no, the user's problem: let's parse again.

Wait, the problem: In Greenpoint, the library is due south of the courthouse and due west of the community swimming pool. Wait, no—wait, the original text: "the library is due south of the courthouse and due west of the community swimming pool". Wait, no, maybe: library is south of courthouse, pool is west of courthouse? No, the user's text: "the library is due south of the courthouse and due west of the community swimming pool". Wait, that would mean library is south of courthouse and west of pool, so courthouse is north of library, pool is east of library. So the triangle is courthouse (north of library), pool (east of library), so library to courthouse is south (length 15), library to pool is west? No, maybe the correct interpretation is: courthouse is at the origin, library is at (0, -15) (south), pool is at (-25, 0) (west)? No, that doesn't make sense. Wait, maybe the problem is: library is south of courthouse (so vertical leg: courthouse to library is 15 km, south), pool is west of courthouse (horizontal leg: courthouse to pool is 25 km, west), so library is at (0, -15), pool is at (-25, 0), and the distance between library (0, -15) and pool (-25, 0) is $\sqrt{(25)^2 + (15)^2}$? Wait, no, distance formula: $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(-25 - 0)^2 + (0 - (-15))^2} = \sqrt{625 + 225} = \sqrt{850}$? No, that can't be. Wait, maybe I misread the directions. Maybe the library is south of courthouse, and the pool is east of courthouse? No, the problem says "due west of the community swimming pool"—wait, no, the original: "the library is due south of the courthouse and due west of the community swimming pool". So library is west of pool, and south of courthouse. So courthouse is north of library, pool is east of library. So courthouse (N), library (S of courthouse), pool (E of library). So the triangle is courthouse (0,0), library (0, -15), pool (x, -15), and courthouse to pool is 25 km? Wait, no, courthouse to pool: distance from (0,0) to (x, -15) is 25 km. So $\sqrt{x^2 + 15^2} = 25$. Then $x^2 + 225 = 625$, $x^2 = 400$, $x = 20$. So the distance between library (0, -15) and pool (20, -15) is 20 km? Wait, that makes sense. Wait, maybe the problem is: library is south of courthouse (15 km), pool is east of library, and courthouse is 25 km from pool. So it's a right triangle with legs 15 and 20, hypotenuse 25? Wait, 15-20-25 is a Pythagorean triple (3-4-5 scaled by 5: 15=3*5, 20=4*5, 25=5*5). Ah! So that's the key. So the distance between library and courthouse is 15 (one leg), distance between courthouse and pool is 25 (hypotenuse), so the other leg (library to pool) is 20. Because $15^2 + 20^2 = 225 + 400 = 625 = 25^2$. So that's the Pythagorean theorem.

Step2: Apply Pythagorean theorem

Let $a = 15$ (library to courthouse), $c = 25$ (courthouse to pool, hypotenuse), find $b$ (library to pool).

Pythagorean theorem: $a^2 + b^2 = c^2$

So $15^2 + b^2 = 25^2$

$225 + b^2 = 625$

$b^2 = 625 - 225 = 400$

$b = \sqrt{400} = 20$

Answer:

20