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

\frac{\sqrt{3}}{4}(3)^2

\frac{\sqrt{3}}{4}(9)

\frac{9\sqrt{3}}{4}

Multiplying this by six for the entire hexagon,

6(\frac{9\sqrt{3}}{4})

3(\frac{9\sqrt{3}}{2})

\frac{27\sqrt{3}}{2}

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})$

A=\frac{3}{4}\pi

Multiplying this by seven for the seven chocolates, we get

7(\frac{3}{4}\pi)

\frac{21}{4}\pi

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

\frac{27\sqrt{3}}{2}-\frac{21}{4}\pi\approx6.8893

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

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