Area of real shapes and the trapezoidal rule

Areas and perimeters of sectors and annuluses, composite shapes made by adding and subtracting, and the trapezoidal rule for estimating the area of irregular land.

You already know the areas of rectangles, triangles, trapeziums and circles from Years 9 and 10. Real things are rarely that tidy. A sprinkler waters a slice of a circle, a sports field has rounded ends, and a farm boundary follows a creek. This section is about breaking real shapes into pieces you can handle, and estimating the ones you can't.

Sectors and arcs

A sector is a slice of a circle between two radii, like a slice of pizza. The angle at the centre, $\theta$, tells you what fraction of the whole circle you have: $\dfrac{\theta}{360}$. So the sector's area and its curved edge (the arc) are that fraction of the full circle's area and circumference.

Sector

$$A = \frac{\theta}{360} \times \pi r^2 \qquad l = \frac{\theta}{360} \times 2\pi r$$

Example 1●

A garden sprinkler sprays water to a distance of $6$ m and turns through $120°$. Find the area it waters and the perimeter of the wet region, each correct to two decimal places.

$A = \dfrac{120}{360} \times \pi \times 6^2 = 12\pi \approx 37.70$ m²

120° is one third of a full turn.

$l = \dfrac{120}{360} \times 2\pi \times 6 = 4\pi \approx 12.57$ m

The arc is one third of the full circumference.

Perimeter $\approx 12.57 + 6 + 6 = 24.57$ m

The perimeter is the arc plus the two straight radii.

Composite shapes

A composite shape is made from simpler shapes. Either add the pieces, or start with a big shape and subtract the holes. Sketch it and label each piece before calculating.

Two semicircles with diameter $50$ m make one full circle of radius $25$ m.

Example 2●●

A school oval is a rectangle $80$ m by $50$ m with a semicircle on each $50$ m end, as shown. Find its area and perimeter, each to the nearest whole number.

$A = 80 \times 50 + \pi \times 25^2 = 4000 + 1963.50 \approx 5963$ m²

The two semicircles together make a full circle with radius 50 ÷ 2 = 25 m.

$P = 2 \times 80 + \pi \times 50 \approx 317$ m

The perimeter is the two straight sides plus the full circle's circumference. The dashed lines are inside the shape, so they don't count.

Example 3●

A circular pond of radius $4$ m has a paved path $1$ m wide around it. Find the area of the path, correct to two decimal places.

$A = \pi \times 5^2 - \pi \times 4^2 = \pi(25 - 16)$

The path is the big circle minus the pond: a ring called an annulus. The outer radius is 4 + 1 = 5 m.

$A = 9\pi \approx 28.27$ m²

Simplify, then use the calculator once at the end.

Annulus

$$A = \pi(R^2 - r^2)$$

The trapezoidal rule

Land along a creek or a lake has no formula. Surveyors draw a straight line across it, called a traverse, and measure the width at equally spaced points. Each strip between two widths is roughly a trapezium, so its area is the average width times the strip width.

Trapezoidal rule (one application)

$$A \approx \frac{h}{2}\left(d_f + d_l\right)$$

Here $h$ is the distance between the two measurements, $d_f$ is the first width and $d_l$ is the last. With more measurements, apply the rule to each strip and add.

Example 4●●

A block of land runs $60$ m along a traverse. Its widths at $0$ m, $30$ m and $60$ m are $18$ m, $25$ m and $14$ m. Estimate its area using two applications of the trapezoidal rule.

$A_1 \approx \dfrac{30}{2}(18 + 25) = 645$ m²

Two strips, each 30 m wide. First strip: widths 18 and 25.

$A_2 \approx \dfrac{30}{2}(25 + 14) = 585$ m²

Second strip: widths 25 and 14.

$A \approx 645 + 585 = 1230$ m²

Add the strips.

One application: $\dfrac{60}{2}(18 + 14) = 960$ m², an underestimate.

Compare: one application ignores the 25 m width in the middle and gives a much smaller answer.

InteractiveSurvey a block between two creeks

Try this

Start with $1$ application. Why is the estimate so far from the true area?
Increase to $2$, then $4$, then $6$ applications. Write down the error each time. What happens to it?
With $2$ applications, check the estimate yourself from the yellow widths using $\dfrac{h}{2}(d_f + d_l)$ for each strip.
Press New block a few times. Can you find a block where one application already does well? What do its creek banks look like?

Your turn

A circular garden bed has a diameter of $4$ m. It is surrounded by a path $1$ m wide. Find the area of the path, correct to two decimal places.

Show the working

Inner radius $= 2$ m, outer radius $= 3$ m. Path area $= \pi(3^2 - 2^2) = 5\pi \approx 15.71$ m².

Flashcards

Area of a sector

$\dfrac{\theta}{360} \times \pi r^2$

Arc length

$\dfrac{\theta}{360} \times 2\pi r$

Trapezoidal rule

$A \approx \dfrac{h}{2}(d_f + d_l)$ for each strip

Practice

Find the area of a circle with radius $7$ cm, correct to two decimal places.
Find the area of a sector with radius $10$ m and angle $90°$, correct to two decimal places.
Find the arc length of a sector with radius $12$ cm and angle $60°$, correct to two decimal places.
Find the area of an annulus with outer radius $10$ cm and inner radius $6$ cm, correct to two decimal places.
Use one application of the trapezoidal rule with $h = 20$ m, $d_f = 15$ m and $d_l = 21$ m.
Widths of $10$ m, $16$ m and $12$ m are measured $25$ m apart. Estimate the area with two applications of the trapezoidal rule.
The area of a semicircle with radius $r$ is:
A quarter circle of radius $10$ cm is cut from a $10$ cm by $10$ cm square, with its centre at one corner. Find the area left, correct to two decimal places.
Exam style: A paddock lies along a $200$ m traverse. Its widths at $0$ m, $100$ m and $200$ m are $80$ m, $110$ m and $60$ m. Use two applications of the trapezoidal rule to estimate its area in hectares.
Exam style: A school oval is a rectangle $80$ m by $50$ m with a semicircle on each $50$ m end. Find its area to the nearest square metre.