Finding the Area of Squares and Triangles

[nivek516]

Does anyone have any ideas on how to find the area of the entire region given in this image? The areas of each square are labeled. I am trying to find the area of the the blue squares and white triangles. I think I have found the area of the middle triangle \(\displaystyle (\frac{\sqrt{324}}{2})\). How do I find the outer triangle areas?

 

[NadroJV13]

Two questions for you to consider. One: is this all the information you are given; Is there anything about triangles being right triangles or anything given about the heights or angles? Two: Are any of the lines shown as congruent?

If neither of these two questions can be answered from the diagram, I appologize but I cannot help. There is a formula in existence called heron's Calculation. But I am not familar with the formula.

 

[nivek516]

All of the Blue boxes are squares, so I know the sides are the sq. root of that number. All I need to know now is how to figure out what the outer lengths are of the triangle (the blue lines). I was told that graphing is a good way to do this, but I would like to calculate it algebraically.

 

[Denis]

Does anyone have any ideas on how to find the area of the entire region given in this image? The areas of each square are labeled. I am trying to find the area of the the blue squares and white triangles. I think I have found the area of the middle triangle \(\displaystyle (\frac{\sqrt{324}}{2})\).

That's correct: sqrt(324) / 2 = 18 / 2 = 9

Label your diagram (starting at top, going clockwise) ABCDEF.
Label middle triangle XYZ, such that you have other 3 triangles labelled ABX, CDY and EFZ.

Use Law of Cosines to calculate the middle triangle's angles.
With triangle ABX:
angleAXB = 90 - angle YXZ
From that, calculate AB using Law of Sines.

Similarly, calculate CD and EF

You can now calculate the areas of those 3 triangles.

 

[soroban]

Hello, nivek516!

I worked out the areas of the four triangles ... and got a BIG surprise!

Suppose we consider that inner triangle . . .

                  * B
        __     *  *
       /18  *     *
         *        *
    A *           *  __
        *         * /26
          *       *
        __  *     *
       /20    *   *
                * *
                  * C

I used the Law of Cosines to find the angles.

. . \(\displaystyle \cos A \:=\:\frac{18 + 20 - 26}{2\sqrt{18}\sqrt{20}} \:=\:0.316227766 \quad\Rightarrow\quad A \:=\:71.56505188^o\)


\(\displaystyle \text{Then I used this formula to find the area: }\;\Delta ABC \:=\:\tfrac{1}{2}bc\sin A\)

. . \(\displaystyle \text{and I got: }\;A \;=\;\tfrac{1}{2}(\sqrt{18})(\sqrt{20})\sin 71.56505188^o \;=\;\boxed{9}\) . (which is your answer)



\(\displaystyle \text{The triangle at the far left has sides }\sqrt{18}\text{ and }\sqrt{20}\)

\(\displaystyle \text{The included angle is: }\:\theta \:=\:360^o - 90^o - 90^o - 71.56505118^o \;=\;108.4349488^o\)

\(\displaystyle \text{And its area is: }\;A \;=\;\tfrac{1}{2}(\sqrt{18})(\sqrt{20})\sin108.4349488^o \;=\;\boxed{9}\)


Again, the area is exactly 9 . . . interesting!


I used the same technique to find the area of the triangle at the upper-right.
. . And it too has an area of 9.

At this point, I finally saw what was happening and why.
The areas of the three outer triangles are always equal to that of the central triangle.
[I won't bother explaining the reasons right now.]

So the areas are: 18, 20, 26, 9, 9, 9, 9 . . . a total of 100 square units.