Thank you Andrew Clark for this solution.

(a) It is known that the area of the largest equilateral triangular section of a cube is 140cm 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√3/2) = a2√3/4 . As the area of the triangle is 140 cm2 , a2 = 560/√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, 2b2 = 560/√3 so that b=12.7 to three significant figures.

(b) The distances between the centres of two adjacent faces of another cube is 8cms. 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 2x2 = 82 , so that
x =
√
 
32
 

, and the length of the side is 8√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 cm2 . Thus the area left is 144−32 = 112 cm2 . There are six such faces giving an area of 672 cm2 .

In addition, there are eight equilateral triangular faces of side length 4√2 . The total area of these is 8 ×8√3 = 64√3 cm2 . Thus the total surface area is 672 + 64√3 = 783 cm2 to the nearest square centimetre.

The volume of a tetrahedron is (1/3) ×(area of base) ×height . Consider one of the tetrahedra, and take one of the right-angled triangles as its base; then its volume is (1/3) ×8 ×4 = 32/3 . Thus the volume of the remaining solid is 123 − (8×32)/3 = 1643 cm3 to the nearest cubic centimetre.