# 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.

### 9 responses to “Two Squares, Two Triangles, and some Circles”

1. Colin Beveridge

How about extending FC into the large square, and drawing the perpendicular to FC (extended) through D? Call their intersection X.

Then, DXF is a right-angled isosceles triangle, with hypotenuse 10cm; |DX| and |FX| are both 5√2. |CX| is 2√2 (|FX| – |FC| = 5√2 – 3√2). Now, DXC is a right-angled triangle with CD as a hypotenuse; |CD|^2 = |CX|^2 + |DX|^2 = 8 + 50 = 58, the area of ABCD.

2. Mike Lawler

My first solution also used the Pythagorean theorem, but was a little different than #3 (but not as elegant as #3 either).

Extend line segment CF into the larger square and drop a perpendicular from point D. Call the intersection point X.

XDF is a 45-45-90 right triangle with a hypotenuse of 10 and legs of length 5*\sqrt(2).

XDC is also a right triangle with legs of length 5 \sqrt(2) and 2 \sqrt(2), and thus the length of DC is the square root of 58 by the Pythagorean theorem.

3. chrisharrow

Thanks, Mike & Colin for the nearly simultaneous, independent, but identical solutions. I hadn’t thought about constructing $CD^2$ from the “inside”. Lovely.

4. Colin Beveridge

Another application of the law of cosines: in triangle DFC, we know two sides ( |DF| = 10 and |FC| = 3√2) and an angle (DFC = 45º). That gives: |DC|^2 = 10^2 + (3√2 )^2 – (2)(10)(3√2)cos(45º) = 100 + 18 – 60 = 58.

5. Five Triangles

What if this problem were given to a year 6 student who hasn’t yet learned constructions, coordinate geometry, square roots, the Pythagorean theorem, or trigonometry?

• chrisharrow

I wonder if it would be possible for a year 6 student to discover Pick’s Theorem and then use that to determine area(ABCD)=58?
I’ve been trying to accomplish this without overtly resorting to constructions.

Does anyone out there have an answer to Five Triangles’ year 6 challenge?

6. David Radcliffe

I used complex numbers. Label C=0, F= $3\sqrt{2}$, E = 3i $\sqrt{2}$, and D=z.
Then $\displaystyle \frac{DE}{DF} = \frac{4}{10} = \frac{z-3i\sqrt{2}}{z-3\sqrt{2}}$. Solving this equation yields $z=\sqrt{2}(5i-2)$, so the area of square ABCD is $|z|^2 = 58$.

• chrisharrow

This rocks, David. Thanks!

• Colin Beveridge

After playing with Geogebra a bit, I like option 3 the best – especially if you rotate the smaller square so the origin is at its centre and the marked diagonal is vertical (i.e., its vertices are at C(-3,0), E(0,3), G(3,0), and H(0,-3)). D is then at (0,7) and the right-angled triangle on the left leaps out at you.

The reason I like it? If you let |DH| = y and |CE| = x, the general solution to puzzles set up like this also jumps out: the area of the larger square is (y-x/2)^2 + (x/2)^2.