Volume of solids

1. by Ludger, Dorian, Finn-Niklas, Marie,
Anna, Nico, Miran

3. A super chocolate should have a
volume of 250 cm³.
Choose a form of a solid and find
suitable values for the measurements.
What about the weight of this
piece of chocolate?

4. 1. a cube
2. a rectangular prism
3. a prism with 4 edges
4. a prism with 3 edges
5. a squared pyramid
6. a cylinder
7. a cone
8. a sphere
9. a truncated cone

5. 푉 = 푎3
250 = 푎3
3 250 = 6,299605 cm
푎 ≈ 6,3 cm
The cube will have a lenght of 6,3 cm.

6. a) We choose a= 15 cm and b= 10 cm
Then we can calculate the height of the chololate
bloc: V=a*b*c
250 cm³ = 15cm*10 cm*c , c= 1
2
3
cm≈1,66666 cm
The chocolate bloc will have a height of 1,7 cm.
b) We choose a =18 cm and b = 3cm
Then we can calculate the height of the chololate
bloc: V=a*b*c
250 cm³ = 18cm*3 cm*c , c≈ 4,63 cm
The chocolate bloc will have a height of 4,63 cm.
18 3
1,6
4,6
15 10
There are many more combinations. Choose two values, then you can calculate the third.

7.  
1. A parallolopid:
Formula:
e
a 
c
V  h  )

2
(

2
a c cm
cm cm
a c
cm
3
16
12 250 ³
2
2,5
  
  




Chosen:
a =6 cm
h =2,5 cm
e =12 cm
Calculated:
b = 10,6 cm
e
a
h
c
1. A trapeziod:
Formula:
V a hc
a  h  cm 
cm
7,5 250 ³
1
a  h 
cm
7,5 250 ³
2
3
5 6
²
3
33
cm  cm  cm 
cm
Chosen:
a =6 cm
c =7,5 cm
Calculated:
b =6,6 cm

8. Chosen:
h =8 cm
H =10 cm
Calculated:
g =6,25 cm
Chosen:
h =4 cm
H =18 cm
Calculated:
g =4,17 cm
Example
1:
H cm
g h
h H cm cm
(8  ) : 2  ² 
250 ³
h cm
H
g h
Formula V
6,25
250 ³
2
2
:

 




H=10 cm
h
g= 6,25 cm
Example
2 :
H cm
g h
 
h cm cm
(4  ) : 2  18 ² 
250 ³
h cm
36 250 ³
h cm
H
g h
Formula V
7
250 ³
2
2
:






g= 4,17 cm
h
H=18 cm
There are many more combinations. Choose two values, then you can calculate the third.

9. Example 1:
1
V     a 
h
² 3
h
   
5² 250| 3
1
3
h
25 
750|: 25
30

h
Example 2:
Chosen:
a =5 cm
Calculated:
h =30 cm
1
V     a 
h
a
   
² 15 250| 3
1
15 ² 750|:15
² 
50 |
7,1
3
² 3


a
a
a
Chosen:
h =15 cm
Calculated:
a  7,1 cm
h=15 cm
A=7,1 cm
h=15 cm
a=5 cm
There are many more combinations.
Choose the value of a or h, then you can
calculate the third.other one.

10. Chosen:
h =5 cm
Calculated:
r  4cm
Example
1:
Formula :
V r h
2

r h cm
 
2
r cm cm
  
5 250 ³
r cm
50 ² |:
 


r cm
² 
15,92 ² |
r 3,99
cm
250 ³
2
2

 
 
Example
1:
r= 4 cm
Formula :
V r h
2

r h cm
250 ³
 
2
r cm cm
  
13,5 250 ³
2
r cm
18,51851852 ² |:
 


r cm
² 
5,894629522 ² |
r 2,427885401
cm
2

 
h= 5 cm
 
Chosen:
h =13,5 cm
Calculated:
r  2,4cm
h= 13,5 cm
r= 2,4 cm
There are many more combinations. Choose one value r or h, then you can calculate the
second one.

11. Chosen:
h =10 cm
Calculated:
r  4,9 cm
Chosen:
r =3 cm
Calculated:
h  26,5 cm
r ²
h

cm m
²
h
hcm cm


9  
750 ³ |: (9 )
h cm
cm
r h
V
26,5
250 ³ | 3
3
9
250 ³
3
3

 


 
 

 
h=10 cm
r=4,9 cm
r=4,9 cm
r= 3cm
h=10 cm
There are many more combinations. Choose one value r or h, then you can calculate the
second one.

12. r=3,9 cm

14. Ludger: “It seems to be difficult. We have three
unknown variables, the height, the radius at the
bottom and the upper radius and we have only one
information, the volume has to be 250 cm³! Let’s
take 5 cm for the height. I’ll look for a formula to
calculate the volume of a truncated cone!“
1 2 2 V   h R  rR  r
( )
3

15. Ludger: “I assume for the height 5 cm. For R I
could try out 5 cm or 4 cm. Let’s start!”
h=5cm, R=5 cm:
   
5(25 5 ) 250
250 3
5
1
r r
25 5
r r
  
5 25 47,7468
5 22,74648
3
2
2
2
2
 

  
r r
r r


h=5cm, R= 4 cm:
   
5(16 4 ) 250
250 3
5
1
16 4
r r
r r
  
4 16 47,7468
4 31,74648
3
2
2
2
2
 

  
r r
r r



16. Ludger (grade 8): “So far, ok! But we did not
yet learn how to solve these equations.”
Dorian (grade 9):”I will show you how to solve
this quadratic equation by using a formula.”
r r
  
5 22,7468 0
5

   
2,5 22,7468
5,3849
5
5
2
5
  
2
2,88 7,88
22,7468
2
2
2
r
r
r
1/ 2
1 2
2
1/ 2
2
1/ 2
  
 
 
  
r r
r r
  
4 31,7468 0
   
2 22,7468
4
2
   
2 4 31,7468
  
2 5,9789
r
r
r r
3,9789 7,9789
2
1/ 2
1/ 2
1 2
2
1/ 2
  
r

17. 5,8cm
10cm
5cm
7,96cm
8 cm
5cm
This a cylinder! We got
the same value for the
radius of a cylinder.
Other assumptions for height and R lead to much more other truncated cones
and values.

18. The density of choclate ist 1,3 g/cm³.
So a solid of 250 cm³- full of chocolate- has a
mass of 1,3g/cm³*250cm³ =325 g.
We found the density of chololate here.