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Stage: 4 Challenge Level: Challenge Level:1

Thank you Andrew Clark for this solution.

(a) It is known that the area of the largest equilateral triangular section of a cube is $140$cm$^2$ . What is the side length of the cube?

Cube with equilateral triangle section.

Let the sides of the equilateral triangle have length $ a $; then the area of the triangle is $ (a/2)(a\sqrt{3}/2) = a^2\sqrt{3}/4 $. As the area of the triangle is $ 140 {\rm cm}^2 $, $ a^2 = 560/\sqrt{3} $. The length of the diagonal of a face of the cube is $ a $; let $ b $ be the length of a side of the cube. Then, by Pythagoras' Theorem, $ 2b^2 = 560/\sqrt{3} $ so that $ b=12.7 $ to three significant figures.

(b) The distances between the centres of two adjacent faces of another cube is $8$cms. What is the side length of this cube?

Cube .

Let the side length of the cube be $ 2x $. From the centre points A and B of both faces to the line at which their planes meet is $ x $, and AB ($ = 8 $) is the hypotenuse of a right angled triangle whose other two sides have length $ x $ . Therefore $ 2x^2 = 8^2 $ , so that $ x= \sqrt{32} $ , and the length of the side is $ 8\sqrt{2} $ .

(c) Another cube has an edge length of $12$cms. At each vertex a tetrahedron with three mutually perpendicular edges of length $4$cms is sliced away. What is the surface area and volume of the remaining solid?

Cube with corners chopped off.

From each face of the cube we remove four triangles whose total area is $ 32 {\rm cm}^2 $ . Thus the area left is $ 144-32 = 112 {\rm cm}^2 $ . There are six such faces giving an area of $ 672 {\rm cm}^2 $ . \par In addition, there are eight equilateral triangular faces of side length $ 4\sqrt{2} $ . The total area of these is $ 8 \times 8\sqrt{3} = 64\sqrt{3} {\rm cm}^2 $ . Thus the total surface area is $ 672 + 64\sqrt{3} = 783 {\rm cm}^2 $ to the nearest square centimetre.

The volume of a tetrahedron is $ (1/3) \times \hbox{(area of base)} \times \hbox{height} $ . Consider one of the tetrahedra, and take one of the right-angled triangles as its base; then its volume is $ (1/3) \times 8 \times 4 = 32/3 $ . Thus the volume of the remaining solid is $ 12^3 - (8\times 32)/3 = 1643 {\rm cm}^3 $ to the nearest cubic centimetre.