The area of a square with right triangle inside it

[SilverKing]

Hi everyone,

I need to find the area of the square in the following figure:

I aimed to find the length of BC, but first I had to find the unknowns of the right triangle CDE, which are EC=5m, <DCE=36.86ْ , <DEC=53.13ْ .

Then I thought that I can use some trig relation for angles to get <BDC and BCD respectively (since the angles of the square are all 90ْ).

And that's where I am stuck. Any help?

 

[HallsofIvy]

I don't think you need the angles, but, as you say, EC= 5 m. Now look at the three right triangles outside the given right triangle. If we take AE= x and ED= y then AD= BC= x+ y. If we take AF= p and FB= q (The "D" on the top of the rectangle is clearly wrong- I am relabeling it "F") then AB= CD= p+ q.

The reason for doing that is we now have $\displaystyle x^2+ p^2=9$, $\displaystyle (x+y)^2+ q^2= 16$, and $\displaystyle (p+ q)^2+ y^3= 25$. That's only three equations in four unknowns so we can't solve for actual values but it should be enough to give $\displaystyle (x+ y)(p+ q)$, the area of the rectangle.

 

[Deleted member 4993]

Halls, that can be "labelled" with 3 unknowns instead of 4:
x = BC = DC
p = AE; so DE = x-p
q = BF; so AF = x-q

p^2 + (x-q)^2 = 9
x^2 + q^2 = 16
x^2 + (x-p)^2 = 25
3 equations, 3 unknowns... but non-linear .... ouch!

Oui?

.

 

[SilverKing]

Thanks everybody for your quick responds. And as HallsofIvy did, I change the upper D to F, sorry about that.

So, basically there is no "actual" answer to this problem, since the equations are non-linear?

 

[Deleted member 4993]

Thanks everybody for your quick responds. And as HallsofIvy did, I change the upper D to F, sorry about that.

So, basically there is no "actual" answers to this problem, since the equations are non-linear?

There is "actual" answer to the problem you have posted.

You cannot solve for 4 unknowns explicitly. However, you can derive a relationship (like HoI showed) to calculate area of the square.

 

[wjm11]

Well, something kept telling me there is an easy way...there is!
2 of the inside triangles are similar.
I've labelled this way:

D         e          E  a-e   C

a-f


F
                              
                              a






A             a               B

Let angleEBF = u, angleABF = v
Then angleCBE = 90-u-v and angleBEC = u+v

This makes angleDEF = 90-u-v
Since angleCBE = angleDEF, then triangleBCE is similar to triangleEDF
Using similarity:
BE / BC = FE / DE
4/a = 3/e : e = 3a/4 (which means a-e = a/4).

Using triangle BCE:
BC^2 + EC^2 = BE^2
a^2 + (a/4)^2 = 4^2
a^2 + a^2/16 = 16
17a^2 = 16^2
a^2 = 16^2 / 17
a = 16 / SQRT(17) ..... 3 cheers!!

Nice work, Denis. I figured there must be an easier way, too. You can just use trig. If you start with the 3-4-5 triangle resting on the x-axis, then rotate it through some angle theta, it will be in the correct position to fit into a square, i.e., the x value of one vertex becomes equal to the y value of the other vertex. The result of that is

5sin(theta + alpha) = 4cos(theta)

Where alpha is the small angle in the 3-4-5 triangle (about 36.87 deg).

Solving, we get theta is approx. 14.0362 degrees.

From this the side length of the square is found to be a = 4cos(theta) = 3.88057 -- in close agreement with your result.

and the area is about 15.0588

 

[SilverKing]

What a MATH! I got it now. Thank you everyone for your help. I really appreciate it.