In the diagram above, shape ABCD is a square and M is the midpoint on side AB.
What fraction of the square is shaded?
In attempting to solve this, I first drew lines at known midpoints: The shaded area below is $\frac{1}{64}$, so two of them make $\frac{1}{32}$: The shaded area below is $\frac{1}{32}$: There are two, but I need to subtract two of this shape: And that's where I'm stuck.
1 Answer
$\begingroup$If you make it a unit square with $C$ the origin and $D$ being $(1,0)$ you can write the equations for lines $CM$ and $AD$. The intersection point is $(\frac 13,\frac 23)$ so the kite is composed of two isosceles triangles with base $(\frac 13,\frac 23)$ to $(\frac 23,\frac 23)$. The one that has peak $M$ has altitude $\frac 13$. The upside down one has altitude $\frac 16$. The total area is then $\frac 12\cdot \frac 13(\frac 13+\frac 16)=\frac 1{12}$
$\endgroup$3
