A shortstop is standing in the base path between second base and third base when she fields the ball she is 34 feet from the second base and needs to make the throw to first base. Assuming the field is square, how far does she need to throw the ball round your answer to the nearest tenth I tried 96.1 not working and 96.15 not working so helpppp
To find the distance the shortstop needs to throw the ball, we can use the Pythagorean theorem. Let's consider the square field with the shortstop at point A, second base at point B, and first base at point C. The distance from A to B is given as 34 feet, and we need to find the distance from A to C (the distance the shortstop needs to throw the ball).
Using the Pythagorean theorem, we have:
AC² = AB² + BC²
AC² = 34² + 34²
AC² = 2 * (34)²
AC = √(2 * 34²)
AC ≈ 48.2 feet (rounded to the nearest tenth)
Therefore, the shortstop needs to throw the ball approximately 48.2 feet to reach first base.
Not working
Apologies for the confusion. Let's try solving the problem again.
The shortstop is standing at point A, 34 feet from second base (point B). We need to find the distance from point A to first base (point C).
Since the field is square, the distance from second base to first base is the same as the distance from second base to third base. Let's consider the distance from point B to point D as the distance from second base to third base.
By using the Pythagorean theorem, we have:
AB² = AD² + BD²
34² = AD² + BD²
AD² = 34² - BD²
AD² = 34² - 34²
AD² = 0
Therefore, the distance AD is zero. This means that point D, which is third base, coincides with point B, second base.
Thus, the distance from point A to point C (the distance the shortstop needs to throw the ball) is also 34 feet like the distance from point A to B.
Therefore, the shortstop needs to throw the ball approximately 34 feet to reach first base.
To find the distance the shortstop needs to throw the ball, we can use the Pythagorean theorem, which states that for a right triangle, the square of the hypotenuse (longest side) is equal to the sum of the squares of the other two sides.
In this case, we can consider the distance from second base to first base as the hypotenuse, and the distance from second base to the shortstop as one of the legs of the triangle.
Let's define:
- Distance from second base to the shortstop = 34 feet (leg 1 of the triangle)
- Distance from the shortstop to first base = unknown (leg 2 of the triangle)
- Distance from second base to first base = unknown (hypotenuse)
Using the Pythagorean theorem, we can set up this equation:
(34 feet)² + (leg 2)² = (hypotenuse)²
We need to solve for the hypotenuse, which is the distance the shortstop needs to throw the ball (rounded to the nearest tenth). Let's calculate:
(34 feet)² + (leg 2)² = (hypotenuse)²
1156 + (leg 2)² = (hypotenuse)²
Next, we can rearrange the equation to solve for leg 2:
(leg 2)² = (hypotenuse)² - 1156
(leg 2)² = (hypotenuse)² - 1156
Now, we can substitute the values into the equation:
(leg 2)² = (hypotenuse)² - 1156
(leg 2)² = (96.1)² - 1156
(leg 2)² = 9235.21 - 1156
(leg 2)² = 8079.21
To find the square root of 8079.21, we get:
leg 2 = √8079.21
leg 2 ≈ 89.9 (rounded to the nearest tenth)
Therefore, the shortstop needs to throw the ball approximately 89.9 feet from second base to first base.