square proof: Prove that shaded square is 1/5 of total area

[Clifford]

Points JKLM are the midpoints of the sides of square PQRS. Prove that the area of the shaded square is one-fifth the area of the square PQRS.



I plotted the image onto a graph and found coordinates for each point.
P(0, 2b)
Q(0, 0)
R(2b, 0)
S(2b, 2b)
J(b, 0)
K(2b, b)
L(b, 2b)
M(0, b)

By using the length formula, I got the length of one side of the PQRS is 2b. From there I got the area of the big square is 4b^2. I am not sure how to find the coordinates of the small square to get its length to find the area.

I don't know if this is even the proper way to tackle this problem. Some advice, tips or help would be great.

 

[galactus]

There are many ways to go about this. I like trig, so I will use that.

Let the sides of the big square, PQRS, be 'a'. Therefore, the area of the big square is \(\displaystyle a^{2}\).

The area, inside the big square, surrounding the darkened region can be broken up into 4 triangles.

Since J,K,L,M are midpoints, we can find the length of the side of the triangle that touches the dark region by \(\displaystyle acos(tan^{-1}(\frac{1}{2}))=\frac{2a}{\sqrt{5}}\)

The length of the side adjacent to it can be found by \(\displaystyle sin(tan^{-1}(\frac{1}{2}))=\frac{a}{\sqrt{5}}\)

The area of one of these triangles is \(\displaystyle \frac{(\frac{2a}{\sqrt{5}})(\frac{a}{\sqrt{5}})}{2}=\frac{a^{2}}{5}\)

There are 4 of them, so multiply by 4, \(\displaystyle \frac{4a^{2}}{5}\)

Since the area of the entire square is \(\displaystyle a^{2}\)

\(\displaystyle a^{2}-\frac{4a^{2}}{5}=\frac{a^{2}}{5}\)

And that is the area of the remaining dark area.

 

[Denis]

Hate trig, so:
let sides of bigger square = 1
then MS = QK = sqrt(1^2 + (1/2)^2) = sqrt(5) / 2

So we have quadrilateral MSKQ with sides 1/2 and sqrt(5) / 2;
and area MSKQ = 1/2; since height * base = area:
height = area / base = (1/2) / (sqrt(5) / 2) = 1 / sqrt(5)

Since that height = side of inner square, area inner square
= (1 / sqrt(5))^2
= 1/5

 

[soroban]

Hello, Clifford!

If you rearrange some pieces, the proof is obvious.

                  *
                 /:::*.
        *-------*-------*
       /|   *::/       /|
      /:|     /:*.    /:|
    *:::|    /::::::*/::|
        *   /:::::::/   *
        |::/*::::::/    |:::*
        |:/     *:/     |:/
        |/       /::*.  |/
        *-------*-------*
           *:::/
              *