of 0

Our problem

A basketball player has made a basket from the center of the circle of the division line on the court. He challenges another player to make a similar shot but from the side line.

Q---Is this fair why or why not?

Q---How much further would it be to shoot from the sidelines of the middle of the court then the middle of the court?

To solve this we had to use the Pythagorean Theorem.

So basically

A(squared)+ B(squared) = C(squared)

or 25 * 25 + 37 * 37= C * C

625 + 1369 = C * C

1994 = C * C

44.65 ft. = C

This shows that the second player has to shoot from  a further distance than the first player.

Player 1: 37 ft.

Player 2: 44.65 ft.

37 < 44.65

The distance from the first player to the hoop is 37 ft. The distance from the second player to the hoop is equal to the distance from the hoop to the sideline squared, plus the distance from the second basket ball player to the front of the court, squared. After that, you have to square root it to equal the distance.

But what if they were shooting from forty feet from the hoop instead of thirty seven?

Then second equation is the same is like the first one except that it changes b from 37 to 40:

A(squared)+ B(squared) = C(squared)

or 25 * 25 + 40 * 40 = C * C

625 + 1,600 = C * C

2,225 = C * C

47.17 ft. = C

This shows that once again, the second player has to shoot from  a further distance than the first player.

Player 1: 40 ft.

Player 2: 47.17 ft.

So it seems that there is a pattern that when player two is on the edge instead of the middle, he has to shoot from further. To prove this though, a third test needs to be conducted. So . . .

The third equation is the like the second one except that it changes b from 40 to 60:

A(squared)+ B(squared) = C(squared)

or 25 * 25 + 60 * 60 = C * C

625 + 3,600 = C * C

4,225 = C * C

65 ft. = C

This proves that the second player has to shoot from  a further distance than the first player as long as he is on the edge of the court and player one is in the middle.

Player 1: 60 ft.

Player 2: 65 ft.