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1989 Paper 1 Q1
D: 1500.0 B: 1500.0
coordinate geometry area semicircle parabola quadratic curve integration geometric proof circle

TikZ diagram
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: \(AG = r\), 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 \(y = \frac1{h}(h^2-x^2)\)

TikZ diagram
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*}

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