Cubes in a Skeleton Cuboid

1. Cubes in a Skeleton Cuboid An Investigation

2. 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.

3. 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. 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?

5. 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

6. 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