if A shortstop is standing in the base path between second base and third base when she fields the ball she is 35 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 10th there are 3 bases and she is right next tot eh second one and the base 2 and 3 are 35 feet and 1st and second are
Since the field is square and the shortstop is standing between second base and third base, we can assume that each base is equidistant from each other.
The distance between second base and third base is given as 35 feet. Therefore, the distance between second base and first base would also be 35 feet.
To find the distance the shortstop needs to throw the ball to first base, we can use the Pythagorean theorem. The distance between second base and first base is the hypotenuse of a right triangle with sides of 35 feet.
By applying the Pythagorean theorem: a^2 + b^2 = c^2, where a and b are the lengths of the sides and c is the length of the hypotenuse, we can find the distance the shortstop needs to throw the ball.
a = b = 35 feet (distance between second base and first base)
c^2 = a^2 + b^2
c^2 = 35^2 + 35^2
c^2 = 2450 + 2450
c^2 = 4900
c = √4900
c ≈ 70 feet
Therefore, the shortstop needs to throw the ball approximately 70 feet to reach first base.
Explain how it is 96.1
Apologies for the confusion. Let's re-evaluate the problem based on the new information that the distance between the bases is 35 feet.
If the shortstop is standing between second base and third base, the distance from the shortstop to second base is 35 feet. Since the shortstop needs to throw the ball to first base, we need to find the distance from second base to first base.
Using the Pythagorean theorem, we can determine this distance. Let's label the sides of the right triangle formed:
1. The distance from the shortstop to second base: 35 feet.
2. The distance from second base to first base: unknown (let's call it x).
3. The distance from the shortstop to first base: unknown (let's call it c).
According to the Pythagorean theorem, a^2 + b^2 = c^2, where a and b are the lengths of the legs of the triangle and c is the length of the hypotenuse.
In our case, a = 35 feet, b = x, and c is the distance from the shortstop to first base.
Plugging in the values into the formula:
35^2 + x^2 = c^2
980 + x^2 = c^2
To find the value of c, we need to know the value of x. However, without further information about the relationship between the bases, we cannot determine the exact value of x or c. Therefore, we cannot provide an accurate answer and round it to the nearest 10th as requested.
To find the distance the shortstop needs to throw the ball to first base, we can use the Pythagorean theorem.
Given that the distance between second base and third base is 35 feet, and the shortstop is standing between those two bases, we can consider this distance as the base of a right triangle. Let's call this distance "x".
The other two sides of the triangle are the distances from the shortstop to second base and from second base to first base. We already know that the distance from the shortstop to second base is 35 feet.
To find the distance from second base to first base, we need more information. But since you mentioned that the field is square, we can assume that all the bases are equidistant. In other words, the distance from second base to first base is the same as the distance from second base to third base, which is 35 feet.
Now, we have a right triangle with the base (x) and the two sides (35 feet). We can use the Pythagorean theorem, which states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
Thus, we can set up the equation:
x^2 = 35^2 + 35^2
Simplifying, we have:
x^2 = 1225 + 1225
x^2 = 2450
To find x, we take the square root of both sides:
x = √2450
x ≈ 49.5
Therefore, the shortstop needs to throw the ball approximately 49.5 feet to first base. Rounded to the nearest 10th, the answer is 49.5 feet.