Hard Geometry Problem with Solution for the SAT

Hard Geometry Problem with Solution

for the SAT

Yesterday I gave you a Level 4 Geometry problem for the revised SAT to try. Today I will provide a solution for this problem. If you have not yet attempted the problem go back and take a look at it first so you can try it on your own. Here is the link: Hard Geometry Problem for the Revised SAT

Level 4 – Geometry

Here is the problem once again followed by a solution:

The head of a copper “hexagon head screw bolt” (one cross section of which is shown above) has the shape of a cylinder with a hole shaped like a regular hexagon. The cylindrical head is 2 cm thick with a base diameter of 3 cm. The hexagonal hole is only half the thickness of the entire head, and each side of a hexagonal cross section has a length of 1 cm. Given that the density of copper is 8.96 grams per cubic cm, and density is mass divided by volume, find the mass of the head to the nearest gram.

Solution: We first compute the volume of the head. There are two parts to the volume.

The bottom half of the head is a cylinder with height 2/2 = 1 cm and base radius 3/2. It follows that the volume is

V = πr²h = π(3/2)²(1) = 9π/4 cm³.

The top half of the head consists of the same cylinder as the bottom half, but this time we have to subtract off the volume of a hexagonal prism. The regular hexagonal face can be divided into 6 equilateral triangles, each with area

A = s²√3/4 = 1²√3/4 = √3/4.

So the volume of the hexagonal prism is

V = Bh = (6√3/4)(1) = 3√3/2 cm³

and the volume of the top half of the head is

9π/4 – 3√3/2 cm³

It follows that the total volume of the head is

9π/4 + (9π/4 – 3√3/2) = 18π/4 – 3√3/2 = (9π – 3√3)/2 cm³.

Finally,

D = M/V → 8.96 = M / [(9π – 3√3)/2] → M = 8.96 ∙ (9π – 3√3)/2 ≈ 103.39 grams.

To the nearest gram, the answer is 103.

For an even more detailed solution to this problem check out New SAT Math Problems arranged by Topic and Difficulty Level.

Feel free to add your own solutions to the comments.