Problem source

\begin{center}
    \begin{tikzpicture}[scale = 3]
        \draw[domain = 0:180, samples=50, variable = \x]  plot ({1.5*cos(\x)},{1.5*sin(\x)});
        \draw[domain = 180:360, samples=50, variable = \x]  plot ({0.5*cos(\x)},{0.5*sin(\x)});
        \draw[domain = 0:180, samples=50, variable = \x]  plot ({0.5*cos(\x)-1},{0.5*sin(\x)});
        \draw[domain = 0:180, samples=50, variable = \x]  plot ({0.5*cos(\x)+1},{0.5*sin(\x)});
        \node at (-1.5, 0) [below left] {$A$};
        \node at (1.5, 0) [below right] {$G$};
        \node at (0, -.5) [below] {$D$};
        \node at (-0.5, 0) [left] {$C$};
        \node at (0.5, 0) [right] {$E$};
        \node at (1, 0.5) [above] {$F$};
        \node at (-1, 0.5) [above] {$B$};
        \node at (0, 1.5) [above] {$H$};
    \end{tikzpicture}
\end{center}
In the above diagram, $ABC,CDE,EFG$ and $AHG$ are semicircles and $A,C,E,G$ lie on a straight line. The radii of $ABC,EFG,AHG$ are $h$, $h$ and $r$ respectively, where $2h < r$. Show that the area enclosed by $ABCDEFGH$ is equal to that of a circle with diameter $HD$. 
Each semicircle is now replaced by a portion of a parabola, with vertex at the midpoint of the semicircle arc, passing through the endpoints (so, for example, $ABC$ is replaced by part of a parabola passing through $A$ and $C$ and with vertex at $B$). Find a formula in terms of $r$ and $h$ for the area enclosed by $ABCDEFGH$.

Solution source

$AG = r$, therefore the area is:

\begin{align*}
A &= [AHG] - 2*[ABC] + [CDE] \\
&= \frac12 \pi r^2 - \pi h^2 + \frac12 \pi (r-2h)^2 \\
&= \frac12 \pi \l r^2 - 2h^2 + r^2 -4rh+4h^2  \r \\
&= \frac12 \pi \l 2r^2 -4rh + 2h^2\r \\
&= \pi (r-h)^2
\end{align*}

This is the same area as a circle radius $r-h$

But $HD = r + (r-2d)  = 2(r-d)$, ie the circle with diameter $HD$ has radius $r-h$ as required.

Suppose $A = (-h, 0), C = (h, 0), B = (0, h)$ then our parabola is $y = \frac1{h}(h^2-x^2)$

\begin{center}
    \begin{tikzpicture}[scale = 3]
        \draw[domain = 0:180, samples=50, variable = \x]  plot ({1.5*cos(\x)},{1.5*sin(\x)});
        \draw[domain = 180:360, samples=50, variable = \x]  plot ({0.5*cos(\x)},{0.5*sin(\x)});
        \draw[domain = 0:180, samples=50, variable = \x]  plot ({0.5*cos(\x)-1},{0.5*sin(\x)});
        \draw[domain = 0:180, samples=50, variable = \x]  plot ({0.5*cos(\x)+1},{0.5*sin(\x)});

        \node at (-1.5, 0) [below left] {$A$};
        \node at (1.5, 0) [below right] {$G$};
        \node at (0, -.5) [below] {$D$};
        \node at (-0.5, 0) [left] {$C$};
        \node at (0.5, 0) [right] {$E$};
        \node at (1, 0.5) [above] {$F$};
        \node at (-1, 0.5) [above] {$B$};
        \node at (0, 1.5) [above] {$H$};


        \draw[domain = -1.5:1.5, samples=50, variable = \x, red]  plot ({\x},{1/1.5 * ( 1.5^2 - (\x)^2)});
        \draw[domain = -0.5:0.5, samples=50, variable = \x, red]  plot ({\x},{-1/0.5 * ( 0.5^2 - (\x)^2)});
        \draw[domain = -0.5:0.5, samples=50, variable = \x, red]  plot ({\x-1},{1/0.5 * ( 0.5^2 - (\x)^2)});
        \draw[domain = -0.5:0.5, samples=50, variable = \x, red]  plot ({\x+1},{1/0.5 * ( 0.5^2 - (\x)^2)});
    \end{tikzpicture}
\end{center}

The area of $ABC$ would then be $\int_{-h}^h \frac{1}{h}(h^2-x^2) \d x = \frac1{h} \left [ h^2x - \frac{x^3}{3} \right] = \frac1{h} \l 2h^3-2\frac{h^3}{3} \r = \frac{4}{3}h^2$

so we have:

\begin{align*}
A &= [AHG] - 2*[ABC] +[CDE] \\
&= \frac43 r^2-\frac83h^2+\frac43(r-2h)^2 \\
&= \frac43 \l r^2 -2h^2+r^2-4rh+4h^2) \\
&= \frac43 (r-h)^2
\end{align*}