Question

sports the distance between each base on a baseball infield is 90 feet. the third baseman throws a ball from third base to point p. to the nearest foot, how far did the player throw the ball? blank ft

Explanation:

Step1: Identify the triangle type

The baseball infield is a square (each side 90 ft), so the distance from 3rd base to 1st base (infield diagonal) and the path from 3rd base to P forms a right triangle? Wait, no. Wait, point P is 30 ft from 1st base along the extension? Wait, the diagram: 3rd base, 2nd base, 1st base, home plate form a square (each side 90 ft). The infield diagonal (3rd to 1st) is a right triangle with legs 90 ft. Then point P is 30 ft beyond 1st base? Wait, the horizontal distance from 3rd base to P's horizontal component: from 3rd to 1st is 90 ft (vertical? Wait, no, the square: 3rd to 2nd is 90 ft (right angle), 2nd to 1st is 90 ft (right angle), 1st to home is 90 ft, home to 3rd is 90 ft. The infield diagonal (3rd to 1st) is a right triangle with legs 90 ft, so length $\sqrt{90^2 + 90^2} = 90\sqrt{2}$? Wait, no, maybe the horizontal and vertical distances from 3rd base to P: 3rd base to 2nd base is 90 ft (let's say along y-axis), 2nd base to 1st base is 90 ft (x-axis), 1st base to P is 30 ft (x-axis). So 3rd base is at (0,90), 2nd at (90,90), 1st at (90,0), P at (90 + 30, 0) = (120, 0). Wait, no, maybe 3rd base is at (0,0), 2nd at (90,0), 1st at (90,90), home at (0,90). Then infield diagonal is from (0,0) to (90,90). Then point P is 30 ft from 1st base: 1st base is (90,90), so P is (90, 90 - 30) = (90,60)? No, the diagram shows "30 ft" from 1st base towards P. Wait, maybe the right triangle has legs: one leg is 90 ft (3rd to 2nd), and the other leg is 90 + 30 = 120 ft? Wait, let's re-examine. The problem: distance from 3rd base to P. Let's model coordinates: let 3rd base be (0,0), 2nd base (90,0), 1st base (90,90), home (0,90). Then point P is 30 ft from 1st base along the line extending from 1st base towards... Wait, the diagram has a dashed line (infield) from 3rd to 1st? No, the dashed line is infield, and P is 30 ft from 1st base. So the horizontal distance from 3rd base to P: x-coordinate: from 3rd (0,0) to 2nd (90,0) to 1st (90,90) to P (90, 90 - 30)? No, the "30 ft" is marked from 1st base towards P, so vertical? Wait, maybe the right triangle is with legs 90 ft (3rd to 2nd) and 90 + 30 = 120 ft (2nd to P)? Wait, no, the angle at 3rd base: 3rd base, 2nd base, and P: 3rd to 2nd is 90 ft, 2nd to 1st is 90 ft, 1st to P is 30 ft, so 2nd to P is 90 + 30 = 120 ft? Wait, no, 2nd base to 1st base is 90 ft (horizontal), 1st base to P is 30 ft (horizontal), so 2nd to P is 90 + 30 = 120 ft (horizontal). 3rd base to 2nd base is 90 ft (vertical). So the distance from 3rd base (0,0) to P (120, 90)? No, 3rd base (0,0), 2nd (90,0), 1st (90,90), P (90 + 30, 90) = (120,90)? No, the diagram's "30 ft" is from 1st base, so maybe 1st base to P is 30 ft downward (vertical). Wait, I think the correct approach is: the distance from 3rd base to P is the hypotenuse of a right triangle with one leg 90 ft (3rd to 2nd) and the other leg 90 + 30 = 120 ft (2nd to P). Wait, 3rd to 2nd is 90 ft (let's say vertical), 2nd to 1st is 90 ft (horizontal), 1st to P is 30 ft (horizontal), so total horizontal from 2nd to P is 90 + 30 = 120 ft. So the two legs are 90 ft (vertical) and 120 ft (horizontal). Then distance is $\sqrt{90^2 + 120^2}$.

Step2: Calculate the distance

Compute $90^2 = 8100$, $120^2 = 14400$. Sum: $8100 + 14400 = 22500$. Square root of 22500 is 150. Wait, that's nice. So $\sqrt{90^2 + 120^2} = \sqrt{8100 + 14400} = \sqrt{22500} = 150$.

Answer:

150