IGCSE · Practice questions · Compound shapes: segments and composite regions

Split it, then add or subtract.

Six original Cambridge-style questions on compound shapes in radian measure: segment area and perimeter, tangents meeting radii at right angles, angles from inverse trigonometry, exact Paper 1 answers with \(\pi\) and surds, and overlapping circles.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Sketch, split, then calculate.

Sketch and label every radius, angle and right angle (a tangent meets a radius at \(\tfrac{\pi}{2}\)).
Split the shaded region into sectors \(\tfrac{1}{2}r^{2}\theta\), triangles \(\tfrac{1}{2}ab\sin C\) and rectangles; write the sum or difference first.
Find missing angles with \(\sin\), \(\cos\) or \(\tan\) in RADIAN mode; find missing lengths with trig, Pythagoras or the cosine rule.
Perimeter is only the outside edge: arcs plus the straight pieces that bound the region.
Accuracy: exact with \(\pi\) and surds when the angle is a fraction of \(\pi\); otherwise keep 5 s.f. and give 3 s.f.

Recall

[3 marks]

A circle has centre \(O\) and radius 7 cm. The chord \(AB\) subtends an angle of 0.9 radians at \(O\). Find the area and the perimeter of the minor segment.

Area: \(\tfrac{1}{2}(49)(0.9 - \sin 0.9) = 24.5(0.9 - 0.78333) = 2.8585\), so \(2.86\) cm². ✓
Arc: \(7(0.9) = 6.3\). Chord: \(14\sin 0.45 = 6.0895\). ✓
Perimeter: \(6.3 + 6.0895 = 12.4\) cm. ✓

Calculation

[4 marks]

Do not use a calculator in this question.

A sector \(OAB\) has radius 12 cm and angle \(AOB = \tfrac{2\pi}{3}\). Find, in exact form, (a) the length of the chord \(AB\), (b) the perimeter and (c) the area of the minor segment.

(a) \(AB = 2(12)\sin\tfrac{\pi}{3} = 24 \times \tfrac{\sqrt{3}}{2} = 12\sqrt{3}\) cm. ✓
(b) Arc: \(12 \times \tfrac{2\pi}{3} = 8\pi\), so the perimeter is \((8\pi + 12\sqrt{3})\) cm. ✓
(c) Sector: \(\tfrac{1}{2}(144)\tfrac{2\pi}{3} = 48\pi\). Triangle: \(\tfrac{1}{2}(144)\sin\tfrac{2\pi}{3} = 72 \times \tfrac{\sqrt{3}}{2} = 36\sqrt{3}\). ✓
Segment area: \((48\pi - 36\sqrt{3})\) cm². ✓

Calculation

[5 marks]

\(OAB\) is a sector of a circle, centre \(O\), radius 8 cm, with angle \(AOB = 0.7\) radians. The tangent to the circle at \(A\) meets \(OB\) extended at \(C\).

(a) Find the lengths \(AC\) and \(BC\).

(b) Find the perimeter and the area of the region bounded by \(AC\), \(CB\) and the arc \(AB\).

(a) Angle \(OAC = \tfrac{\pi}{2}\). \(AC = 8\tan 0.7 = 6.7383\), so \(AC = 6.74\) cm. ✓
\(OC = \dfrac{8}{\cos 0.7} = 10.4597\), so \(BC = 10.4597 - 8 = 2.46\) cm. ✓
(b) Arc \(AB = 8(0.7) = 5.6\). Perimeter: \(6.7383 + 2.4597 + 5.6 = 14.8\) cm. ✓
Triangle \(OAC\): \(\tfrac{1}{2}(8)(6.7383) = 26.953\). Sector: \(\tfrac{1}{2}(64)(0.7) = 22.4\). ✓
Area: \(26.953 - 22.4 = 4.55\) cm². ✓

Analysis

[5 marks]

A circle has centre \(O\) and radius 4 cm. The point \(C\) is such that the tangent from \(C\) touches the circle at \(A\) and \(AC = 7\) cm. The line \(OC\) meets the circle at \(B\).

(a) Show that angle \(AOC = 1.05\) radians, correct to 3 significant figures.

(b) Find the area and the perimeter of the region bounded by \(AC\), \(CB\) and the minor arc \(AB\).

(a) Angle \(OAC = \tfrac{\pi}{2}\), so \(\tan\theta = \dfrac{7}{4}\) and \(\theta = 1.05165 = 1.05\) radians. ✓
(b) Triangle: \(\tfrac{1}{2}(4)(7) = 14\). Sector: \(\tfrac{1}{2}(16)(1.05165) = 8.4132\). ✓
Area: \(14 - 8.4132 = 5.59\) cm². ✓
\(OC = \sqrt{16 + 49} = \sqrt{65} = 8.0623\), so \(CB = 4.0623\). Arc \(AB = 4(1.05165) = 4.2066\). ✓
Perimeter: \(7 + 4.0623 + 4.2066 = 15.3\) cm. ✓

Analysis

[6 marks]

Do not use a calculator in this question.

\(ABCD\) is a rectangle with \(AB = 12\) cm and \(AD = 6\) cm. An arc of a circle, centre \(A\), radius 12 cm, joins \(B\) to a point \(E\) on \(DC\).

(a) Show that angle \(DAE = \tfrac{\pi}{3}\) and find the exact length of \(DE\).

(b) Find, in exact form, the area and the perimeter of the region bounded by \(BC\), \(CE\) and the arc \(BE\).

(a) Triangle \(ADE\) is right-angled at \(D\) with \(AE = 12\): \(\cos(DAE) = \dfrac{6}{12} = \dfrac{1}{2}\), so angle \(DAE = \tfrac{\pi}{3}\). ✓
\(DE = \sqrt{144 - 36} = \sqrt{108} = 6\sqrt{3}\) cm. ✓
(b) Angle \(EAB = \tfrac{\pi}{2} - \tfrac{\pi}{3} = \tfrac{\pi}{6}\). Sector \(EAB\): \(\tfrac{1}{2}(144)\tfrac{\pi}{6} = 12\pi\). Triangle \(ADE\): \(\tfrac{1}{2}(6)(6\sqrt{3}) = 18\sqrt{3}\). ✓
Area: rectangle minus triangle minus sector \(= (72 - 18\sqrt{3} - 12\pi)\) cm². ✓
\(CE = 12 - 6\sqrt{3}\), \(BC = 6\), arc \(BE = 12 \times \tfrac{\pi}{6} = 2\pi\). ✓
Perimeter: \((18 - 6\sqrt{3} + 2\pi)\) cm. ✓

Analysis

[6 marks]

Two circles, each of radius 5 cm, have centres \(P\) and \(Q\) with \(PQ = 8\) cm. The circles meet at \(X\) and \(Y\).

(a) Show that angle \(XPY = 1.287\) radians, correct to 4 significant figures.

(b) Find the area and the perimeter of the region common to both circles.

(a) By symmetry \(XY\) bisects \(PQ\) at right angles at \(M\), with \(PM = 4\). In triangle \(PMX\), \(\cos(XPM) = \dfrac{4}{5}\), so angle \(XPM = 0.64350\). ✓
Angle \(XPY = 2(0.64350) = 1.2870\) radians. ✓
(b) The region is two equal segments with \(r = 5\), \(\theta = 1.2870\). \(\sin 1.2870 = 0.96\) (exactly \(2 \times 0.6 \times 0.8\)). ✓
One segment: \(\tfrac{1}{2}(25)(1.2870 - 0.96) = 4.0875\). Area: \(2 \times 4.0875 = 8.18\) cm². ✓
Perimeter: two arcs, \(2 \times 5(1.2870) = 12.9\) cm. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.