Areas without formulas · Plane geometry I

Competition figures are rarely shapes with names. A crescent between two circles, the shaded remainder of a triangle, the part of a square no circle reaches - none of these is in a formula table, and none needs to be. Two moves compute them all: cut the figure into pieces that have names, or embed it in something that does and subtract. Circles join in through a single rule: everything a central angle cuts off is the same fraction of the whole.

Decomposition and subtraction

Principle

An area problem is bookkeeping. If the figure splits along the lines already drawn into triangles, rectangles and sectors, add up the pieces. If it is what remains when simple pieces are removed from a simple whole, compute the whole and subtract. Decide the direction before computing anything: the shaded region itself is often the one thing you should not attack directly.

Example

In a rectangle \(ABCD\) with \(|AB| = 4\) and \(|BC| = 3\), the point \(P\) lies on the side \(BC\) with \(|BP| = 1\). The triangle \(APD\) is awkward head-on, but its complement is not: the right triangles \(ABP\) (legs \(4\) and \(1\)) and \(PCD\) (legs \(2\) and \(4\)) fill the rest of the rectangle, so \(p_{APD} = 12 - 2 - 4 = 6\). The direct route confirms it: base \(|AD| = 3\), height \(4\).

Sliding a vertex

Principle

In \(p = \tfrac{1}{2}\,\text{base} \times \text{height}\) the opposite vertex contributes exactly one number: its distance to the line of the base. Fix a base and let the vertex move. Along a parallel to the base the area does not change at all; along any straight segment the distance to the base line is a linear function of the position, so the extreme areas sit at the segment's endpoints.

Example

A median splits a triangle into two parts of equal area: seen from the shared apex, the two halves of the base are equal and the height is one and the same. In the other direction, two triangles \(ABP\) and \(ABQ\) on the common base \(AB\) have equal areas exactly when \(P\) and \(Q\) are equidistant from the line \(AB\) - for instance whenever \(PQ \parallel AB\).

Fractions of a circle

Theorem

A central angle \(\varphi\) in a circle of radius \(r\) cuts off a sector and an arc, and each is the fraction \(\frac{\varphi}{360^\circ}\) of the whole:

p_{\text{sector}} = \frac{\varphi}{360^\circ}\,\pi r^2 , \qquad \ell_{\text{arc}} = \frac{\varphi}{360^\circ}\, 2\pi r .

The disc and the circumference are shared out in proportion to the angle at the centre; we use this without proof. A quarter-circle, a sixth, a half - all are read straight off the angle.

Definition

A circular segment is the region between a chord and its arc. It has no formula of its own: it is the sector minus the isosceles triangle enclosed by the two radii and the chord - subtraction again, this time inside the circle.

The right angle at the centre claims a quarter of everything: of the full turn, of the circumference, of the disc. The darker region is the segment - the quarter sector minus the triangle \(OAB\).

Warning

The fraction is always taken of the full circle, because the central angle is measured at the centre and the full turn is \(360^\circ\). A \(60^\circ\) sector drawn inside a semicircle is still one sixth of the whole disc, not one sixth of the half - the semicircle is scenery, never the denominator.

Example

A chord equal to the radius. The two radii and the chord then form an equilateral triangle, so the central angle is \(60^\circ\) and the segment has area

\frac{\pi r^2}{6} - \frac{r^2\sqrt{3}}{4}

- a sixth of the disc minus an equilateral triangle of side \(r\). Both terms stay exact; whether such an expression is large or small is settled by algebra, not decimals.

Warm-up

Example

A square of side \(a\) is inscribed in a circle. Compute the area of one of the four circular segments between a side of the square and the circle.

Find the radius first

The diagonal of the square is a diameter of the circle, so \(2R = a\sqrt{2}\), giving \(R = \frac{a\sqrt{2}}{2}\) and the tidier \(R^2 = \frac{a^2}{2}\).

Which fraction of the disc does one side cut off?

The four vertices divide the circle into four equal arcs, so the central angle over one side is \(90^\circ\): the sector is a quarter of the disc, \(\frac{1}{4}\pi R^2 = \frac{\pi a^2}{8}\).

Sector minus triangle

The triangle enclosed by the two radii and the side is right isosceles with legs \(R\), area \(\frac{R^2}{2} = \frac{a^2}{4}\). So the segment is \(\frac{\pi a^2}{8} - \frac{a^2}{4} = \frac{a^2}{8}(\pi - 2)\). Check the books: four segments plus the square give \(\frac{a^2}{2}(\pi - 2) + a^2 = \frac{\pi a^2}{2} = \pi R^2\), the whole disc.