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Plane mensuration · Lesson 3 of 3

Segments, chords and composite shapes

A segment is a sector minus a triangle: its area, the chord, the perimeter of a segment, and shapes built from rectangles, triangles and parts of circles.

16 minYou should already know: Angles, triangles & polygons
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A chord cuts a circle into two segments. The smaller one, between the chord and the minor arc, is the minor segment. Segments don’t have a formula of their own: they’re built from a sector and a triangle, which you already know from the last lesson and trigonometry.

Try it

Segments: sector minus triangleChange the angle and radius
OAB90°r = 14
154sector, 90/360 × πr²98triangle, ½r² sin 90°56segment area = sector − triangle41.8perimeter = arc + chord = 22 + 19.8
The segment is the part between the chord and the arc. Its area is the sector (154) take away triangle OAB (98), whose area is ½ × r × r × sin θ. Its perimeter is the arc plus the chord, 22 + 19.8; the chord is 2r sin(θ/2) = 19.8.

The segment is the gold part: the whole sector take away triangle OABOAB. Press Trace the perimeter: a segment’s edge is the arc and the chord, with no radii.

The three facts

θchordsegment
Segmentsector −- triangle

For a chord that makes an angle θ\theta at the centre of a circle of radius rr:

triangle OAB=12r2sin⁡θ\text{triangle } OAB = \frac12 r^2 \sin\theta segment area=θ360πr2−12r2sin⁡θ\text{segment area} = \frac{\theta}{360}\pi r^2 - \frac12 r^2 \sin\theta chord AB=2rsin⁡θ2\text{chord } AB = 2r\sin\frac{\theta}{2}

The chord formula comes from the perpendicular from OO to ABAB: it halves both the chord and the angle, making two right-angled triangles with hypotenuse rr.

Oθ/2r½ chord
Half the chord12AB=rsin⁡θ2\frac12 AB = r\sin\frac{\theta}{2}
perimeter of a segment=arc+chord\text{perimeter of a segment} = \text{arc} + \text{chord}

Worked example · WAEC 2022

WAEC 2022 · Paper 2 · Q3

A chord subtends an angle of 72∘72^\circ at the centre of a circle of radius 24.5 m24.5\text{ m}. Calculate the perimeter of the minor segment. [Take π=227]\left[\text{Take }\pi = \frac{22}{7}\right]

Perimeter (m, 1 d.p.)

  1. What goes round the edge?

    The arc and the chord.

    Think first. Which two lengths make up the perimeter of a segment?

  2. The arc

    72360×2×227×24.5=30.8 m\frac{72}{360} \times 2 \times \frac{22}{7} \times 24.5 = 30.8\text{ m}

  3. The chord

    Half the chord is 24.5sin⁡36∘≈24.5×0.5878≈14.4024.5\sin 36^\circ \approx 24.5 \times 0.5878 \approx 14.40 m, so the chord is about 28.8028.80 m.

    Think first. The perpendicular from the centre halves the 72∘72^\circ. What is half the chord?

  4. The perimeter

    30.8+28.80≈59.630.8 + 28.80 \approx 59.6 m.

Composite shapes

Many questions are shapes made of pieces: a running track (a rectangle with a semicircle at each end), a gate with a semicircular arch, a board with a segment cut out. The method:

  1. Split the shape into pieces you know: rectangles, triangles, sectors, semicircles.
  2. For an area, add the pieces (or take away a piece that has been cut out).
  3. For a perimeter, go round the outside edge only. Lines where two pieces join aren’t part of it.

Worked example · WAEC 2018

WAEC 2018 · Paper 2 · Q3 (a)

The diagram shows an athletics track with two parallel sides and two semicircular ends. Each of the parallel sides is 60 metres long and the diameter of each semicircular end is 120 metres. Calculate the distance covered by an athlete who runs round the track two times. [Take π=227]\left[\text{Take }\pi = \frac{22}{7}\right]

60 m120 m

  1. What is one lap?

    Two straights of 60 m, and two semicircular ends of diameter 120 m. The two semicircles together make one whole circle of diameter 120 m.

    Think first. Which parts of the track does the athlete run along?

  2. One lap

    2×60+227×120=120+377.14=497.14 m2 \times 60 + \frac{22}{7} \times 120 = 120 + 377.14 = 497.14\text{ m}

  3. Two laps

    2×497.14≈994.32 \times 497.14 \approx 994.3 m.

Your turn

WAEC 2014 · Paper 2 · Q4 (a)

  1. (a)

    In the diagram, OO is the centre of the circle of radius r cmr\text{ cm} and ∠XOY=90∘\angle XOY = 90^\circ. If the area of the shaded segment XKYXKY is 504 cm2504\text{ cm}^2, calculate the value of rr. [Take π=227]\left[\text{Take }\pi = \frac{22}{7}\right]

    rrOXYK
Worked solution (try it first)

(a)

  1. The shaded segment is the quarter-circle sector XOYXOY minus the right-angled triangle XOYXOY.
  2. Sector =90360×227×r2= \frac{90}{360} \times \frac{22}{7} \times r^2
    =11r214= \frac{11r^2}{14}.
  3. Triangle =12r2= \frac12 r^2.
  4. Segment =11r214−7r214= \frac{11r^2}{14} - \frac{7r^2}{14}
    =4r214= \frac{4r^2}{14}
    =2r27= \frac{2r^2}{7}.
  5. So 2r27=504\frac{2r^2}{7} = 504, r2=1764r^2 = 1764 and r=42r = 42 cm.

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