Algebra Problems, Math Problems

Problem of the Day: 2/15/13

If xy=3, yz=1, and xz=\frac{1}{27}, what is the value of y?

Solution to yesterday’s problem:

There are two parts to this problem: the area of the hexagon and the area of the circles. First, let’s find the area of the hexagon.

We can divide any regular hexagon into six equilateral triangles. The formula for the area of an equilateral triangle is A = \frac{\sqrt{3}}{4}s^2, where s is the side length of the triangle. (Try deriving this formula by drawing the height of the triangle – you’ll see two 30-60-90 triangles.) Substituting 3 into this formula, we get




Multiplying this by six for the entire hexagon,




Now that we’ve found the area of the hexagon, let’s move on to the area of the circles. Notice that three circles fit into the height of two of the triangles. In one of these triangles, the height is \frac{s\sqrt{3}}{2} (find this by splitting the triangle into two 30-60-90 triangles). So the height of the hexagon is 3\sqrt{3}, substituting for s and multiplying by 2. Since three circles fit into this length, we can say that the diameter of each circle is \sqrt{3}, and the radius \frac{\sqrt{3}}{2}. Using the circle area formula,

A=\pi(\frac{\sqrt{3}}{2})^2  latex A=\pi(\frac{3}{4})$


Multiplying this by seven for the seven chocolates, we get



Now we have our two parts. To find the area left over on the box, simply subtract these two with your calculator.


Finally, we find that there are about 6.89 square inches open on the bottom of the box.


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s