# Mensuration problem: IGCSE MathematicsPlease answer me question number 7 on page 211 from following web address...

Mensuration problem: IGCSE Mathematics

Please answer me question number 7 on page 211 from following web address

http://kennso.wikispaces.com/file/view/T5_Notes.pdf

Please it is urgent

Many thanks

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### 2 Answers

The problem no. 7

If the ball fits tightly inside the box then the ball is tangent to the walls of the box and the radius of the ball is half of the length of the cubic box.

If the radius of the ball is of 10 cm then the length of each side of the box is of 20 cm.

a) The volume of the ball is `V = 4pi*10^3/3 =gt V = 4000pi/3 cm^3`

b) The volume of the box: `V_1` = Area of the square base * height

`V_1 = 20*20*20 = 8000 cm^3`

c) To find the volume that remains not occupied you should subtract the volume of the ball from the volume of the box.

`V_1 - V = 8000 - 4000pi/3 = 3813.(3) cm^3`

To find the percentage volume of the box not occupied by the ball you should use the computation method called the rule of three.

If the volume of the box represents the complete percentage of 100% then the volume that is not occupied represents the x percentage that you need to find.

`x = (V_1 - V)*100/V_1 =gt x = 3813.(3)*100/8000 = 3813.(3)/80`

x = 47.(6)%

**The volume of the box is`V_1 = 8000 cm^3` , the volume of the ball is `V = 4000pi/3 cm^3` and the percentage volume of the box that is not `occupied by the ball x = 47.(6)%` .**

Part c is not very clear to me, if you can elaborate more, it will be good