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 = 962. (i) C = 4(20 + 15 + 10) –16= 4(45) – 16 = 164 (ii) C = 4(50 + 30 + 20) – 16= 4(100) – 16 = 384All 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 – 83.In the given skeleton, A = 4(96) – 8 = 3764. (i) A = 4(164) – 8 = 648 (ii) A = 4(384) – 8 = 15285. No. of cubes = C = 4(L + B + H) – 16Total surface areaA = 4C – 8= 4(4(L + B + H) – 16))A = 16(L + B + H) – 72