Cubes in a Skeleton Cuboid
An Investigation
Here is a picture of a skeleton cuboid that has been
made by sticking together a lot of cubes, each of edge 1cm.
You can easily check that the base of the skeleton cuboid
shown above measures 12cm by 6cm and that the
height of the cuboid is 10cm.
Questions:1) How many 1cm cubes have been used to make this skeleton
cuboid?
2) How many 1cm cubes would be needed to make a skeletons
with the following outside measurements?
(i) 20cm × 15cm × 10cm
(ii) 50cm × 30cm × 20cm
3) What is the total surface
area of the skeleton
shown ?
4) What are the total surface areas of the other
sizes of skeletons given in question 2?
5) Can you find a formula for the number of
cubes required if the skeleton
measures Lcm by Bcm by Hcm
6) Can you find a formula for the total surface
area in terms of L, B and H? How many cubes
in a skeleton cuboid?
Solution
Let the dimensions of the skeleton cuboid be
Lcm, Bcm and Hcm.
The skeleton can be divided up into a number of
parts :8 corners
4 square rods of length (L – 2)
4 square rods of length (B – 2)
4 square rods of length (H – 2)
Let C = number of cubes in the model.
Then C = 8 + 4(L – 2) + 4(B – 2) + 4(H – 2)
C = 4(L + B + H) – 16 or C = 4(L + B + H – 4)
continued on next slide
1. C = 4(12 + 6 + 10) – 16= 4(24) – 16 = 96
2. (i) C = 4(20 + 15 + 10) –16= 4(45) – 16 = 164
(ii) C = 4(50 + 30 + 20) – 16= 4(100) – 16 = 384
All the cubes in a skeleton have four exposed
faces, except for the eight corners which have
three exposed faces.
Let A = total surface area. A = 4C – 8
3.In the given skeleton, A = 4(96) – 8 = 376
4. (i) A = 4(164) – 8 = 648
(ii) A = 4(384) – 8 = 1528
5. No. of cubes = C = 4(L + B + H) – 16
Total surface area
A = 4C – 8= 4(4(L + B + H) – 16))
A = 16(L + B + H) – 72