## Monday, June 18, 2012

### Problem of the Week Day 1: Week of 6/18/12 - 6/22/12

Today, we will officially begin 2012’s first problem of the week. Before I start, please do not comment with your answers (please feel free to comment with any tips, or cool stuff with the types of problems I’m giving) because I want everyone to have the chance to figure it out themselves. If you have any questions or want to know if your answers are correct, please email me directly at Ethan@EthanMath.com.
You may think that since this is a five day long problem, it must be extremely hard; even the easy one. That is not the case. The easy problem is definitely doable for anyone who has taken fifth-sixth grade math and understands the directions. The hard problem is on the difficult side, but still definitely possible. With some extra effort, you will be able to get through it.
Easy Problem: I like to start the week with some triangle calculations, because the things you can do with them really are fascinating. With triangles, you can measure the height of a building without measuring the building, find the weight of a nearby planet, and many other incredible calculations. For the easy problem, we will keep it down to the basics of triangles.

Look at the triangle above. It has side lengths 2, √5, and √5. You will need to do two things with the triangle: determine the height h, and then find the area a.
For the height, you will use something called the Pythagorean Theorem. If a right triangle’s shortest side is a, it’s middle side is b, and it’s longest side (which is called the hypotenuse) is c, then a^2 + b^2 = c^2.
Say you have a triangle with sides 6, 5, and 5. To figure out the height, it must be a right triangle. However, we can make it a right triangle by spitting it down the middle. This gives us sides:
a = 6/2 = 3
b = height
c = 5
3^2 + b^2 = 5^2
9 + b^2 = 25
b^2 = 16
b = 4
To figure out the area, you do:
(base x height)/2
In this case, it would be:
(6 x 4)/2
24/2
12
So, the area of this is 12 square units. Do the same thing with the other triangle.
Once you’ve finished, make sure you jot down a and h for tomorrow’s problem.
Hard Problem:

Look at the triangle above. In this problem, two of the sides are missing. You will also have to figure out the missing angle. To do this, you can use something called cosines. Cosines say that if you divide the two sides next to the angle (the shorter one is called the adjacent and the longer is called the hypotenuse), that you will get the cosine of that angle.
1. 