6.02d, 10.03d - Circle and sector area

6.02d, 10.03d - Circle and sector area

A sprinkler covers a circular garden, but sometimes turns through only part of a full revolution. Use the circle-area formula from 6.02d and 10.03d to find how much ground it waters, calculate sector areas, and work backwards to an unknown radius or angle.

The area of a circle

A sprinkler reaches 6 m6\,\mathrm{m} from its centre and turns through 120∘120^\circ. Does it water 12π12\pi or 24π24\pi square metres? First find the area it would water in a full turn.

Area measures how much surface a region covers. It uses square units. This is different from circumference, which measures the length of the boundary.

One way to understand the circle-area rule is to cut a circle into many equal narrow sectors and arrange them alternately pointing up and down. Their curved edges form the long edges of a shape which approaches a rectangle as the sectors become narrower.

[DIAGRAM: asset_name: Rearranging a circle into narrow sectors; asset_slug: circle-area-rearrangement; caption: Rearranging preserves area. With more narrow sectors, the strip approaches a rectangle of width pi times r and height r.; description: A circle of radius r is divided into sixteen equal sectors in alternating pale blue and sage. The same sixteen exact sectors are rearranged alternately up and down without overlap. The long edge tends to half the circumference, pi times r, and the height tends to r. The finite strip retains visibly curved edges so it is not labelled an exact rectangle.]
Rearranging preserves area. With more narrow sectors, the strip approaches a rectangle of width pi times r and height r.

Each long edge uses half the circumference, so its limiting length is 12×2πr=πr\frac12\times2\pi r=\pi r. The height approaches rr. Rearranging has not changed the area, giving πr×r=πr2\pi r\times r=\pi r^2.

Area of a circle

For a circle of radius rr,

A=πr2.A=\pi r^2.

Square the radius, not the diameter. If rr is in centimetres, AA is in square centimetres.

Supplied on both tiers' June/November 2027 formula sheets. Learn the relationship as well as how to use the supplied formula.

The square acts on rr: πr2\pi r^2 means π×r×r\pi\times r\times r, not (πr)2(\pi r)^2.

Worked example 1: a given radius

Question. A circle has radius 4 cm4\,\mathrm{cm}. Find its area in terms of π\pi.

Plan. Square the centre-to-edge distance, then multiply by pi.

A=π×42use the radius=π×16square before multiplying=16π cm2\begin{aligned} A&=\pi\times4^2 && \text{use the radius} \\ &=\pi\times16 && \text{square before multiplying} \\ &=16\pi\,\mathrm{cm}^2 \end{aligned}

Answer. 16π cm216\pi\,\mathrm{cm}^2.

Check. A square of side 8 cm8\,\mathrm{cm} just encloses the circle and has area 64 cm264\,\mathrm{cm}^2. Since 16π≈50.316\pi\approx50.3, the circle occupies less than the square, as expected.

Change only what the given 4 cm4\,\mathrm{cm} measures: it is now a diameter. The radius must be halved before squaring.

Worked example 2: a given diameter

Question. A circle has diameter 4 cm4\,\mathrm{cm}. Find its area in terms of π\pi.

Plan. Find the radius first, then use the area formula.

r=4÷2=2 cmhalve the diameterA=π×22square the radius=4π cm2\begin{aligned} r&=4\div2=2\,\mathrm{cm} && \text{halve the diameter} \\ A&=\pi\times2^2 && \text{square the radius} \\ &=4\pi\,\mathrm{cm}^2 \end{aligned}

Answer. 4π cm24\pi\,\mathrm{cm}^2.

Check. Halving the radius quarters the area: 14×16π=4π\frac14\times16\pi=4\pi. Both length directions have halved, so the area factor is 12×12\frac12\times\frac12.

That square relationship explains why doubling a sprinkler's reach would water four times as much ground in a full turn.

Watch out

For diameter 4 cm4\,\mathrm{cm}, the line A=π×42A=\pi\times4^2 is wrong: it uses the diameter as a radius. Write r=2r=2 first. Also, 2πr2\pi r gives a length, not an area.

Find the radius before calculating the next area.

Sector area as a fraction

A half-circle has half the full circle's area, and a quarter-circle has a quarter. Equal central angles cut equal areas from a fixed circle.

Just as for arc length, an angle of θ∘\theta^\circ selects θ360\frac{\theta}{360} of the whole. This time multiply that fraction by the circle's area, not its circumference.

[DIAGRAM: asset_name: Minor and major sectors of one circle; asset_slug: circle-sector-areas; caption: The angle fraction selects the same fraction of the full area. The 60 and 300 degree sectors together make the circle.; description: Two congruent circles of radius six metres. A sixty-degree minor sector is shaded in one, and the complementary three-hundred-degree major sector is shaded in the other. Their areas are respectively six pi and thirty pi square metres. The two bounding radii and each selected central angle are clearly marked.]
The angle fraction selects the same fraction of the full area. The 60 and 300 degree sectors together make the circle.

Sector area in degrees

For radius rr and sector angle θ∘\theta^\circ at the centre,

Asector=θ360×πr2.A_{\text{sector}}=\frac{\theta}{360}\times\pi r^2.

Use the angle of the required region: a major sector has 180<θ<360180<\theta<360. A semicircle uses θ=180\theta=180.

Not separately supplied on the 2027 formula sheets: derive it as a fraction of the supplied circle-area formula.

Both tiers receive the full-circle formula for the target 2027 exams. The fraction connecting a full circle to a sector is still yours to choose and apply.

Worked example 3: a minor sector

Question. A sprinkler reaches 6 m6\,\mathrm{m} and turns through 60∘60^\circ. Find the area watered, in terms of π\pi.

Plan. Find one sixth of the full-circle area.

Asector=60360×π×62angle fraction of circle area=16×36π=6π m2\begin{aligned} A_{\text{sector}}&=\frac{60}{360}\times\pi\times6^2 && \text{angle fraction of circle area} \\ &=\frac16\times36\pi \\ &=6\pi\,\mathrm{m}^2 \end{aligned}

Answer. 6π m26\pi\,\mathrm{m}^2.

Check. Six such sectors fit around the centre: 6×6π=36π m26\times6\pi=36\pi\,\mathrm{m}^2, exactly the full-circle area.

Change only the turn to the reflex angle between the same two directions. The major sector covers the rest of the circle.

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