Edwin has a cube with edge length \(8\text{ cm}\). He cuts off a corner by doing the following steps:
He finds the midpoint of each edge.
He then makes a cut through three of these points on adjacent edges.
He then removes the other seven corners by making similar cuts.
Edwin thinks that the new shape will have a smaller total surface area than the original cube. Show that Edwin is right by finding how much less the new surface area is.
We will consider one corner, determine the surface area decrease there, and then multiply the result by \(8\) to account for the eight corners. At each corner, since the cut is made through the midpoints of three adjacent edges of the cube, the surface areas of three identical isosceles right-angled triangles, each with \(4\) cm base and \(4\) cm height, are removed and replaced by the surface area of a single equilateral triangle.
Since the isosceles right-angled triangles each have \(4\text{ cm}\) base and \(4\text{ cm}\) height, each has area equal to \(\frac{1}{2}(4)(4)=8 \text{ cm}^2\).
Each side length of the equilateral triangle is formed by the hypotenuse of one of the isosceles right-angled triangles. Using the Pythagorean Theorem, we calculate the length of the hypotenuse of each right-angled triangle to be \(\sqrt{4^2+4^2}=\sqrt{ 32}=4\sqrt{2}\text{ cm}\). Thus, the remaining equilateral triangle has sides of length \(4\sqrt{2}\text{ cm}.\) Let \(h\) be the height of the equilateral triangle. Since the triangle is equilateral, the height bisects the base.
By the Pythagorean Theorem, \(h^2=(4\sqrt{2})^2 - (2\sqrt{2})^2 =32-8 =24\). Since \(h>0\), we have \(h=\sqrt{24}=2\sqrt{6}\text{ cm}.\)
Therefore, the area of the remaining equilateral triangle is \(\frac{1}{2}(4\sqrt{2})(2\sqrt{6})=4\sqrt{12}=8\sqrt{3}\text{ cm}^2\).
At each corner, the surface area is increased by the area of the equilateral triangle and decreased by the areas of the three right-angled triangles. Therefore, removing a corner changes the surface area by \(8\sqrt{3}-3(8)=(8\sqrt{3}-24)\text{ cm}^2\). Since \(8\sqrt{3} < 24\), this result is negative and the surface area is decreased in each corner. Therefore, removing a corner decreases the surface area by \((24-8\sqrt{3})\text{ cm}^2\).
Since there are eight corners, the total decrease in surface area is \[8 \times (24-8\sqrt{3})=192-64\sqrt{3} \doteq 81.1 \text{ cm}^2\]