basketball courts and three point lines?

Question

ok i don't understand what to do

Not only have the rules of basketball changed over time, the dimensions of the court have changed as well. It wasn’t until 1968 that a 3-point line became a regular part of basketball. Today, the high schools and colleges set their 3-point line at 19 feet 9 inches away from the point directly below the center of the basket. The NBA on the other hand, uses two lines that are parallel to, and three feet from, the side lines and an arc that is 23 feet 9 inches from the point directly below the center of the basket that intersects each of the other two lines. If a standard basketball court is 50 feet wide and the point below the center of the hoop falls on the baseline, how many more square feet are inside the NBA 3-point line than the high school/college three-point line? Express your answer as a decimal to the nearest tenth.

Answer 1

First, lets work on the 3-pt shaded area (inside) of a high school court.

this is actually equal to a half circle

Area of circle = pie x radius^2

= 3.1416 x (19feet and 9 inches)^

= 3.1416 x (19.75 feet)^2

= 3.1416 x 390.0625 sq. ft

= 1225.4175 sq. ft.

since its a half circle then area inside 3 pt line = 1225.4175 / 2 =612.709 sq.ft.



now lets work on the NBA court which is definitely not a half circle.

Theres a right triangle formed by the

a. radius = 23 ft and 9 inches = 23.75 ft = hypotenuse

b. the line on the baseline = 50/2 - 3 = 25 -3 = 22 feet (base of triangle), and

c. the line parallel to the side line but length is unknown ---> height



lets solve for the length of the line parallel to the side line ( so we can solve for the area of this triangle)

since right triangle...... hyp^2 = base^2 + height^2

23.75^2 = 22^2 + height^2

564.0625 - 484 = height^2

80.0625 = ht^2

8.9478 = height



Now lets solve for area of triangle = 1/2(base)(height)

= 1/2(22)(8.9478) = 98.4254 sq.ft



since there'll be these imaginary traingles on both sides, we multiply the area of triangle by 2 = 98.4254 x 2 = 196.8508 sq.ft.



the only problem left is getting the area in the middle bound by the 2 triangles. to solve this we nid the angle (A) formed by the triangle at the point under the hoop.



cos A = adjacent side / hypotenuse

= 22 / 23. 75

= 22.1324 degrees

since there are 2 of these angles = 2 x 22.1324 degrees

= 44.2648 degrees



subtract it to the angle of a straight line => 180 - 44.2648 = 135.7352 degrees --------> means that the area inside this angle is --------->

135.7352 / 360 = 0.377 of the area of the circle that could have been formed given the radius of 23.75 feet



now lets solve for the area of that would be circle = pie x radius^2

A = 3.1416 x 23.75^2

= 3.1416 x 564.0625 sq.ft.

= 1772.05875 sq.ft.

actual area = 1772.05875 x 0.377 = 668.066

Add to this the combined area of the 2 triangles =

= 668.066 + 196.8508 = 864.9168 sq.ft.



Now we know the areas of both the NBA court and a college court, we can find the difference.

difference = 864.9168 - 612.709 = 252.2078 sq.ft.



Answer: 252.2078 sq.ft.

Answer 2

Basketball Court 3 Point Line

Answer 3

Get most of the way there by treating the distance given as the radius and using (1/2)(pi)r^2 to find the area of the semicircle that lies within the 3-point arc.



College: Area = (1/2)(pi)(19.75)^2 = 612.71 ft^2

NBA: Area = (1/2)(pi)(23.75)^2 = 886.03 ft^2



This area is right for college, but not quit for the NBA, since the 3-point arc is truncated near the sidelines. Then you need to subtract out the small regions near the sideline of the NBA court. To do that, use trig functions to find the angle of a right triangle with its base along the baseline from the center of the basket to the three point line near the sideline, and its hypotenuse at the point of intersection of the arc with the 3-point sideline. The base distance is 22 feet (3 feet less than half the court width). The hypotenuse distance is 23 ft 9 in, or 23.75 feet. The angle then is arccos(22/23.75) = 22.13 degrees. The remaining leg of the triangle is 23.75 x sin(22.13) = 8.95 ft.



Now, use the angle to find the area within the pie slice that would exist if the lines near the sideline didn't truncate the 3-point arc:



Area = (22.13/360)(pi)(23.75)^2 = 108.93 ft^2



Then, find the area of the actual triangle (1/2 x base x height) = (0.5)(8.95)(22)= 98.45 ft^2.



The small areas on the side of the NBA court that are lost because the 3-point arc is truncated can be found from the difference between these areas. There are two of them, so multiply by 2:



Area = 2(108.93 - 98.45) = 20.96 ft^2



To find the actual NBA area within the three point line, subtract this from the NBA arc area:



Actual NBA area = 886.03 - 20.96 = 865.07 ft^2



The difference between the college and NBA areas within the 3-point lines is then 865.07 - 612.71 = 252.36 ft^2, or 252.4 ft^2 to the nearest tenth.

Answer 4

I live in Ohio, but every park I've ever been to has had a three point line as well as a free throw line.

Answer 5

basketball courts point lines