Chapter 9.3 — Exercise 9.2 — Circle Areas

Formulas for Rings, Sectors, and Their Relatives

Several shapes in this exercise are built directly from the circle formula rather than needing new ones of their own:

Every problem below reduces to combining two or more rows from this table — usually by adding, subtracting, or halving areas built from the same πr² core.

Punching Circles Out of a Rectangle

A 36 cm × 25 cm acrylic sheet has 56 circular buttons cut from it, each 3.5 cm in diameter (radius 1.75 cm). The remaining sheet's area is simply the rectangle's area minus the total area removed:

Using 22/7 for π and choosing 1.75 = 7/4 keeps every intermediate step a clean fraction, which is exactly why textbook problems favour diameters like 3.5, 7, 14, or 21 — they divide evenly against the 7 in 22/7.

A Circle Fitted Inside a Square

A circle inscribed in a 28 cm square touches all four sides, which fixes its diameter at exactly 28 cm (the square's side) and its radius at 14 cm:

The inscribed circle's radius is always half the square's side — no other measurement is needed once "inscribed in a square" is read correctly.

Shaded Regions Built from Semicircles

Several problems ask for a shaded area left over after semicircles are added to or removed from a simpler shape. In the first, four semicircles of diameter d are built on the four sides of a square of side d, and the constraint r + d + r = 42 cm (from the figure) pins down every measurement:

In a second version, two semicircles of diameter 10.5 m sit inside a circle of diameter 21 m, and the shaded region is what's left of the big circle after the two semicircles are removed:

A third combines two big semicircles (radius 42 cm) with four small ones (radius 21 cm), where the shaded region alternates between adding and subtracting semicircle areas around the figure:

The four small semicircles don't vanish from the working — they cancel out in pairs, two contributing to the shaded region and two being removed from it in equal measure, leaving the total exactly equal to the two big semicircles added together. The general pattern across all three problems is the same: identify every semicircle in the figure, work out whether the shaded region needs each one added or subtracted based on the diagram, and let the radii — not fresh formulas — do the remaining work.

Quarter Circles Doing the Same Job

A figure built from four half circles and two quarter circles, all sharing centre distances OA = OB = OC = OD = 14 cm, has a shaded region found the same subtract-and-add way:

A related figure has four equal circles centred at the corners A, B, C, D of a 7 cm square, touching each other externally in pairs. The shaded region here is what's left of the square once the four quarter-circle corners are removed:

Circles Drawn From a Triangle's Corners

An equilateral triangle of area 49√3 sq.cm has a circle of radius 7 cm (half the triangle's side) drawn from each corner. Since every angle of an equilateral triangle is 60°, each corner circle contributes a 60° sector sitting inside the triangle:

Only the portion of each circle that actually falls inside the triangle counts here — that's exactly what a 60° sector captures, rather than the circle's full area.

Four Circles Around a Square's Corners

Two related problems ask for the gap left between circles arranged around a square's corners. When four equal circles of radius a touch one another, the square joining their centres has side 2a:

When the same idea is scaled up to a 24 cm square with circles of radius 12 cm at each corner:

The two problems are really the same construction at two different scales — one left in terms of the variable a, the other worked out fully in centimetres.

A Quarter Circle Removed From a Trapezium

A trapezium ABCD (AB ∥ CD, ∠BCD = 90°) with AB = BC = 3.5 cm and DE = 2 cm has a quarter circle removed from one corner. The parallel side CD isn't given directly — it's built from CD = DE + EC = 2 + 3.5 = 5.5 cm — and the quarter circle's radius equals BC = 3.5 cm, the trapezium's right-angled side:

A Grazing Horse Traces a Sector

A horse tethered at one corner of a 70 m × 52 m rectangular field by a 21 m rope can only graze within a quarter circle — the field's corner forces a 90° angle, and the rope length fixes the radius at 21 m:

The field's overall 70 m × 52 m size never enters the calculation at all — since the 21 m rope is shorter than either side of the field, the horse's reach never comes near the field's actual boundary, so the field's own dimensions turn out to be irrelevant information.

Where the Ring and Sector Formulas Go Next

Every problem in this exercise combines the same short list of circle-based formulas from the earlier table with the composite-figure thinking practised in Exercise 9.1 — the only genuinely new idea here is that a "piece" of a figure can now be a sector, semicircle, or quadrant instead of only a triangle or trapezium. These same ring and sector formulas return in Circles in Class 9, where they're extended to arcs, chords, and tangent constructions built on the same πr² foundation.