# Tag Archives: law of cosines

## Two Squares, Two Triangles, and some Circles

Here’s another fun twist on another fun problem from the Five Triangles ‘blog.  A month ago, this was posted.

What I find cool about so many of the Five Triangles problems is that most permit multiple solutions.  I also like that several Five Triangles problems initially appear to not have enough information.  This one is no different until you consider the implications of the squares.

I’ve identified three unique ways to approach this problem.  I’d love to hear if any of you see any others.  Here are my solutions in the order I saw them.  The third is the shortest, but all offer unique insights.

Method 1: Law of Cosines

This solution goes far beyond the intended middle school focus of the problem, but it is what I saw first.  Sometimes, knowing more gives you additional insights.

Because DEF is a line and EF is a diagonal of a square, I know $m\angle CEF=45^{\circ}$, and therefore $m\angle CED=135^{\circ}$.  $\Delta CEF$ is a 45-45-90 triangle with hypotenuse 6, so its leg, CE has measure $\frac{6}{\sqrt{2}}=3\sqrt{2}$.  Knowing two sides and an angle in $\Delta DEC$ means I could apply the Law of Cosines.

$DC^2 = 4^2 + (3\sqrt{2})^2 - 2\cdot (3\sqrt{2}) \cdot \cos(135^{\circ})=58$

Because I’m looking for the area of ABCD,  and that is equivalent to $DC^2$, I don’t need to solve for the length of DC to know the area I seek is 58.

Method 2: Use Technology

I doubt many would want to solve using this approach, but if you don’t see (or know) trigonometry, you could build a solution from scratch if you are fluent with dynamic geometry software (GeoGebra, TI-Nspire, GSP).  My comfort with this made finding the solution via construction pretty straight-forward.

1. Construct segment EF with fixed length 6.
2. Build square CEGF with diagonal EF.  (This can be done several ways.  I was in a transformations mood, so I rotated EF $90^{\circ}$ to get the other endpoints.)
3. Draw line EF  and then circle with radius 4 through point E.
4. Mark point D as the intersection of circle and line EF outside CEGF .
5. Draw a segment through points and C.  (The square of the length of CD is the answer, but I decided to go one more step.)
6. Construct square ABCD with sides congruent to CD.  (Again, there are several ways to do this.  I left my construction marks visible in my construction below.)
7. Compute the area of ABCD.

Here is my final GeoGebra construction.

Method 3: The Pythagorean Theorem

Sometimes, changing a problem can make it much easier to solve.

As soon as I saw the problem, I forwarded it to some colleagues at my school.  Tatiana wrote back with a quick solution.  In the original image, draw diagonal, CG, of square CEGF. Because the diagonals of a square perpendicularly bisect each other, that creates right $\Delta DHC$ with legs 3 and 7.  That means the square of the hypotenuse of $\Delta DHC$ (and therefore the area of the square) can be found via the Pythagorean Theorem.

$DC^2 = 7^2+3^2 = 58$

Method 4: Coordinate Geometry

OK, I said three solutions, and perhaps this approach is completely redundant given the Pythagorean Theorem in the last approach, but you could also find a solution using coordinate geometry.

Because the diagonals of a square are perpendicular, you could construct ECFG with its center at the origin.  I placed point C at (0,3) and point E at (3,0).  That means point D is at (7,0), making the solution to the problem the square of the length of the segment from (0,3) to (7,0).  Obviously, that can be done with the Pythagorean Theorem, but in the image below, I computed number i in the upper left corner of this GeoGebra window as the square of the length of that segment.

Fun.

## Generalized Pythagoras through Vectors

Here’s a proof of the Pythagorean Theorem by way of vectors.  Of course, if your students already know vectors, they’re already way past the Pythagorean Theorem, but I thought Richard Pennington‘s statement of this on LinkedIn gave a pretty and stunningly brief (after all the definitions) proof of one of mathematics’ greatest equations.

Let O be the origin, and let $\overrightarrow A$ and $\overrightarrow B$ be two position vectors starting at O. The vector from $\overrightarrow A$ to $\overrightarrow B$ is simply $\overrightarrow {B-A}$, which I will call $\overrightarrow C$. Using properties of dot products,

$\overrightarrow C\cdot\overrightarrow C = \overrightarrow {B-A}\cdot\overrightarrow {B-A} = \overrightarrow B\cdot\overrightarrow B-2\overrightarrow A\cdot\overrightarrow B+\overrightarrow A\cdot\overrightarrow A$

The dot product of a vector with itself is the square of its magnitude, so

$|\overrightarrow C|^2=|\overrightarrow B|^2-2|\overrightarrow A||\overrightarrow B|cos\theta+|\overrightarrow A|^2$

where $\theta$ is the angle between $\overrightarrow A$ and $\overrightarrow B$.

This is the Law of Cosines–in my classes, I call it the generalized Pythagorean Theorem for all triangles.  If $\theta=\frac{\pi}{2}$, then $\overrightarrow A$ and $\overrightarrow B$ are the legs of a right triangle with hypotenuse $\overrightarrow C$ which makes $cos(\theta)=cos(\frac{\pi}{2})=0$ and

$|\overrightarrow C|^2=|\overrightarrow A|^2+|\overrightarrow B|^2$

Pretty.