I realize I'll get a reply in the morning so...
A squash ball must have a diameter either 0.5 mm either above or below 40 mm.
a) What is the minimum and maximum surface area of a squash ball? Express answer to nearest square millimetre.
Minimum: 4902 Maximum: 5153 mm
b)If a cube box is to be sized so that the ball fits exactly inside box, what is the minimum and maximum dimensions of box?
min diam = 39.5
max diam = 40.5
area = pi d62, so plug in the values for d.
Use the two values for d to get the dimensions of the box. The box's sides must equal the ball's diameter, right?
Thanks
To find the minimum and maximum surface area of a squash ball, you need to calculate the surface area of a sphere with a diameter of 40.5 mm and a diameter of 39.5 mm.
The formula for the surface area of a sphere is A = 4πr^2, where r is the radius of the sphere.
a) To find the minimum surface area, use the diameter of 39.5 mm:
First, find the radius by dividing the diameter by 2: r = 39.5 mm / 2 = 19.75 mm
Then, calculate the surface area:
A = 4π(19.75)^2 = 4901.95 mm² (rounded to the nearest square millimeter)
Similarly, for the maximum surface area, use the diameter of 40.5 mm:
r = 40.5 mm / 2 = 20.25 mm
A = 4π(20.25)^2 = 5153.63 mm² (rounded to the nearest square millimeter)
Therefore, the minimum surface area of the squash ball is approximately 4902 mm², and the maximum surface area is approximately 5153 mm².
b) To find the minimum and maximum dimensions of the box that can fit the squash ball exactly, you need to consider the diameter of the ball.
For the minimum dimension, use the diameter of 40.5 mm:
The minimum dimension of the box would be 40.5 mm by 40.5 mm by 40.5 mm.
For the maximum dimension, use the diameter of 39.5 mm:
The maximum dimension of the box would be 39.5 mm by 39.5 mm by 39.5 mm.
Therefore, the minimum dimensions of the box would be 40.5 mm by 40.5 mm by 40.5 mm, and the maximum dimensions would be 39.5 mm by 39.5 mm by 39.5 mm.