Free worked solutions · 2026 Edition

IB Maths AA SL Chapter 5: Triangles, sectors and solids worked solutions

Worked solutions to the practice questions in Chapter 5 of The Ultimate Guide for the IB: Mathematics: Analysis and Approaches SL (2026 Edition). Each solution shows every step with the mark it earns, the way an IB mark scheme does.

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Basics

One idea at a time, in the order the chapter teaches them.

Question 1

[9 marks]

Give your answers to this question in terms of \pi.

(a) A sphere has diameter 12 cm. Find

(i) its volume;

[2]

(ii) its surface area.

[2]

(b) A solid hemisphere has radius 9 cm. Find

(i) its volume;

[2]

(ii) its total surface area, including the flat face.

[3]

Show solutionHide solution

(a)(i) r=6, V=\dfrac{4}{3}\pi(6)^3

=288\pi\text{ cm}^3

(ii) 4\pi(6)^2

=144\pi\text{ cm}^2

(b)(i) \dfrac{1}{2}\times\dfrac{4}{3}\pi(9)^3

=486\pi\text{ cm}^3

(ii) curved surface \dfrac{1}{2}\times 4\pi(9)^2=162\pi

flat face (a circle) \pi(9)^2=81\pi

total 243\pi\text{ cm}^2

Note: using r=12 in (a) is a common error (M1A0). In (b)(ii) the flat face is often forgotten.

Final answer

(a) 288\pi cm^3, 144\pi cm^2
(b) 486\pi cm^3, 243\pi cm^2.

Question 2

[7 marks]

A right cone has vertical height equal to twice its base radius. Its volume is 18\pi\text{ cm}^3. Let the radius be r cm.

(a) Show that r^3=27.

[2]

(b) Write down the radius and the height of the cone.

[1]

(c) Find the exact slant height of the cone.

[2]

(d) Find the exact area of the curved surface of the cone.

[2]

Show solutionHide solution

(a) h=2r, so V=\dfrac{1}{3}\pi r^2(2r)=\dfrac{2}{3}\pi r^3

\dfrac{2}{3}\pi r^3=18\pi \Rightarrow r^3=\dfrac{3}{2}(18)

r^3=27

(b) r=3 cm, h=6 cm

(c) l=\sqrt{3^2+6^2}=\sqrt{45}

=3\sqrt{5} cm

(d) \pi rl=\pi(3)(3\sqrt{5})

=9\sqrt{5}\,\pi\text{ cm}^2

Note: FT from (c) in (d).

Final answer

(a) shown
(b) r=3, h=6 cm
(c) 3\sqrt5 cm
(d) 9\pi\sqrt5 cm^2.

Question 3

[6 marks]

(a) In triangle PQR, P\hat{Q}R=90^{\circ}, PQ=2\sqrt{3} cm and QR=2 cm.

(i) Find the exact value of \tan Q\hat{P}R and hence find Q\hat{P}R.

[2]

(ii) Find PR.

[1]

(b) In triangle UVW, U\hat{V}W=90^{\circ}, UW=5\sqrt{2} cm and VW=5 cm. Find V\hat{U}W and the length UV.

[3]

Show solutionHide solution

(a)(i) \tan Q\hat{P}R=\dfrac{QR}{PQ}=\dfrac{2}{2\sqrt{3}}=\dfrac{1}{\sqrt{3}}

Q\hat{P}R=30^{\circ}

(ii) PR=\sqrt{12+4}=4 cm

(b) \sin V\hat{U}W=\dfrac{VW}{UW}=\dfrac{5}{5\sqrt{2}}=\dfrac{1}{\sqrt{2}}

V\hat{U}W=45^{\circ}

UV=5 cm (isosceles right-angled triangle, or \sqrt{50-25})

Final answer

(a) 1/\sqrt3, 30^\circ, 4 cm
(b) 45^\circ, 5 cm.

Question 4

[4 marks]

In triangle PQR, Q\hat{P}R=45^{\circ}, P\hat{Q}R=60^{\circ} and QR=6\sqrt{2} cm.

(a) Find the exact length of PR.

[3]

(b) Write down the size of P\hat{R}Q.

[1]

Show solutionHide solution

(a) \dfrac{PR}{\sin 60^{\circ}}=\dfrac{6\sqrt{2}}{\sin 45^{\circ}}

PR=\dfrac{6\sqrt{2}\times\frac{\sqrt{3}}{2}}{\frac{\sqrt{2}}{2}}

PR=6\sqrt{3} cm

(b) 180^{\circ}-45^{\circ}-60^{\circ}=75^{\circ}

Final answer

(a) 6\sqrt3 cm
(b) 75^\circ.

Question 5

[4 marks]

Triangle ABC has AB=3, AC=8 and BC=7.

(a) Show that \cos B\hat{A}C=\dfrac{1}{2}.

[2]

(b) Write down the size of B\hat{A}C.

[1]

(c) Without further calculation, explain why A\hat{C}B is the smallest angle of the triangle.

[1]

Show solutionHide solution

(a) \cos B\hat{A}C=\dfrac{3^2+8^2-7^2}{2(3)(8)}

=\dfrac{9+64-49}{48}=\dfrac{24}{48}

=\dfrac{1}{2}

(b) B\hat{A}C=60^{\circ}

(c) A\hat{C}B is opposite AB=3, the shortest side

Final answer

(a) shown
(b) 60^\circ
(c) opposite the shortest side, AB=3.

Question 6

[5 marks]

Convert each angle, given in radians, to degrees.

(a) \dfrac{3\pi}{10} (b) \dfrac{7\pi}{4} (c) \dfrac{5\pi}{12} (d) \dfrac{\pi}{8} (e) \dfrac{8\pi}{3}

[5]

Show solutionHide solution

replace \pi by 180^\circ (equivalently multiply by \dfrac{180}{\pi})

(a) \dfrac{3\times180^\circ}{10}=54^\circ

(b) \dfrac{7\times180^\circ}{4}=315^\circ

(c) \dfrac{5\times180^\circ}{12}=75^\circ

(d) \dfrac{180^\circ}{8}=22.5^\circ

(e) \dfrac{8\times180^\circ}{3}=480^\circ

Note: in (e) the angle is more than one full turn; 480^\circ is the required answer, not 120^\circ.

Final answer

(a) 54^\circ
(b) 315^\circ
(c) 75^\circ
(d) 22.5^\circ
(e) 480^\circ.

Question 7

[4 marks]

A sector of a circle has radius 6 cm and arc length 15 cm.

(a) Find the angle of the sector, in radians.

[2]

(b) Find the area of the sector.

[2]

Show solutionHide solution

(a) 6\theta=15

\theta=2.5

(b) \dfrac{1}{2}(6^2)(2.5)

=45\ \text{cm}^2

Note: FT from (a) in (b).

Final answer

(a) 5/2 rad
(b) 45 cm^2.

Question 8

[5 marks]

A sector of a circle has angle 1.2 at the centre and area 30\ \text{cm}^2.

(a) Find the exact radius of the sector, giving your answer in the form a\sqrt{2} where a\in\mathbb{Z}.

[3]

(b) Hence find the exact arc length of the sector.

[2]

Show solutionHide solution

(a) \dfrac{1}{2}r^2(1.2)=30

0.6r^2=30 \Rightarrow r^2=50

r=\sqrt{50}=5\sqrt{2} cm

(b) l=5\sqrt{2}\times1.2

=6\sqrt{2} cm

Note: FT from (a).

Final answer

(a) 5\sqrt2 cm
(b) 6\sqrt2 cm.

Tutorial

Exam-style questions on each question type, with the marks the IB would give.

Question 9

[4 marks]

(a) Convert 150^\circ to radians, giving your answer in terms of \pi.

[1]

(b) Convert \dfrac{7\pi}{9} radians to degrees.

[1]

A sector of a circle has radius 6 cm and angle 60^\circ. A student writes A = \tfrac12 r^2\theta = \tfrac12(6)^2(60) = 1080 cm^2.

(c) Explain the student's error, and find the correct area of the sector in terms of \pi.

[2]

Show solutionHide solution

(a) 150 \times \dfrac{\pi}{180} = \dfrac{5\pi}{6}.

[1]

(b) \dfrac{7\pi}{9} \times \dfrac{180}{\pi} = 140^\circ.

[1]

(c) The formula A = \tfrac12 r^2\theta is only valid when \theta is in radians; the student used degrees.

[1]

\theta = \dfrac{\pi}{3}, so A = \tfrac12(36)\dfrac{\pi}{3} = 6\pi\ \text{cm}^2.

[1]

Final answer

(a) 5\pi/6
(b) 140^\circ
(c) radians needed, 6\pi cm^2.

Question 10

[5 marks]

A sector OAB of a circle has radius 10 cm and A\hat{O}B = \dfrac{3\pi}{5}.

(a) Find the arc length AB.

[1]

(b) Find the area of the sector.

[2]

(c) Find the perimeter of the sector, leaving your result exact.

[2]

Show solutionHide solution

(a) \ell = r\theta = 10\times\dfrac{3\pi}{5} = 6\pi cm.

[1]

(b) A = \tfrac12 r^2\theta = \tfrac12(100)\dfrac{3\pi}{5}

[1]

= 30\pi\ \text{cm}^2.

[1]

(c) Perimeter = 2r + \ell (the two straight edges as well as the arc)

[1]

= 20 + 6\pi cm.

[1]

Final answer

(a) 6\pi cm
(b) 30\pi cm^2
(c) 20+6\pi cm.

Question 11

[6 marks]

A sector of a circle has radius r cm and angle \theta radians. The perimeter of the sector is 20 cm and its area is 24 cm^2.

(a) Show that r^2 - 10r + 24 = 0.

[3]

(b) Hence find the two possible pairs of values of r and \theta.

[3]

Show solutionHide solution

(a) Perimeter: 2r + r\theta = 20, so \theta = \dfrac{20 - 2r}{r}.

[1]

Area: \tfrac12 r^2\theta = 24, so \tfrac12 r(20 - 2r) = 24.

[1]

10r - r^2 = 24 \Rightarrow r^2 - 10r + 24 = 0.

[1]

(b) (r - 4)(r - 6) = 0.

[1]

r = 4: \theta = \dfrac{20 - 8}{4} = 3.

[1]

r = 6: \theta = \dfrac{20 - 12}{6} = \dfrac43.

[1]

(Check: both give area 24 and \theta < 2\pi.)

Final answer

(a) shown
(b) (r,\theta)=(4,3) or (6,4/3).

Question 12

[4 marks]

The jib of a tower crane is a rigid arm of length 85 m, hinged at a point 24 m above level ground. The jib can be raised or lowered, but its end stays on the same vertical plane.

(a) When the jib makes an angle of 38^\circ above the horizontal, find the height of the end of the jib above the ground.

[2]

(b) The jib is then moved so that its end is 70 m above the ground. Find the angle the jib now makes with the horizontal.

[2]

Show solutionHide solution

(a) Height of the end above the hinge = 85\sin 38^\circ = 52.33\ldots m.

[1]

Height above the ground = 52.33 + 24 = 76.3 m.

[1]

(b) Height above the hinge = 70 - 24 = 46, so \sin\theta = \dfrac{46}{85}.

[1]

\theta = 32.8^\circ.

[1]

Final answer

(a) 76.3 m
(b) 32.8^\circ.

Question 13

[5 marks]

A walker leaves a camp C and walks 14 km on a bearing of 150^\circ to a point X. She then walks due west until she reaches a point Y that is due south of C.

(a) Find the distance that X is (i) south of C and (ii) east of C.

[2]

(b) Find the distances XY and CY.

[2]

(c) Write down the bearing of C from Y.

[1]

Show solutionHide solution

(a) The angle from due south is 180^\circ-150^\circ=30^\circ. Southward distance is 14\cos30^\circ=7\sqrt3 km A1; eastward distance is 14\sin30^\circ=7 km A1.

(b) Since Y is due south of C and XY is due west, XY=7 km and CY=7\sqrt3 km A1A1.

(c) C is due north of Y: bearing 000^\circ A1.

Final answer

(a) 7\sqrt3 km south, 7 km east
(b) XY=7 km, CY=7\sqrt3 km
(c) 000^\circ.

Question 14

[7 marks]

In triangle ABC, AB = AC = 5 and BC = 6. Let B\hat{A}C = 2\theta, where 0 < \theta < \dfrac{\pi}{2}.

(a) Use the cosine rule to find \cos 2\theta.

[2]

(b) Hence find the exact value of \sin\theta.

[3]

(c) Find the exact area of triangle ABC.

[2]

Show solutionHide solution

(a) BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos 2\theta: 36 = 25 + 25 - 50\cos 2\theta.

[1]

\cos 2\theta = \dfrac{14}{50} = \dfrac{7}{25}.

[1]

(b) \cos 2\theta = 1 - 2\sin^2\theta, so \dfrac{7}{25} = 1 - 2\sin^2\theta.

[1]

\sin^2\theta = \dfrac12\left(1 - \dfrac{7}{25}\right) = \dfrac{9}{25}.

[1]

\theta is acute so \sin\theta > 0: \sin\theta = \dfrac35.

[1]

(c) \sin 2\theta = 2\sin\theta\cos\theta with \cos\theta = \dfrac45: \sin 2\theta = 2\cdot\dfrac35\cdot\dfrac45 = \dfrac{24}{25}.

[1]

Area = \tfrac12(5)(5)\sin 2\theta = \dfrac{25}{2}\cdot\dfrac{24}{25} = 12.

[1]

(Check: the height from A to BC is \sqrt{25 - 9} = 4 and \tfrac12(6)(4) = 12.)

Final answer

(a) 7/25
(b) 3/5
(c) 12.

Question 15

[7 marks]

In triangle ABC, AB = 4, AC = 6 and \cos A = -\dfrac14.

(a) Find the length of [BC].

[3]

(b) Find the exact value of \sin A.

[2]

(c) Hence find the exact area of triangle ABC.

[2]

Show solutionHide solution

(a) BC^2 = 4^2 + 6^2 - 2(4)(6)\left(-\dfrac14\right)

[1]

= 16 + 36 + 12 = 64.

[1]

BC = 8.

[1]

(b) \sin^2 A = 1 - \cos^2 A = 1 - \dfrac{1}{16} = \dfrac{15}{16}.

[1]

\sin A = \dfrac{\sqrt{15}}{4} (positive, since A is an angle of a triangle).

[1]

(c) Area = \tfrac12(4)(6)\sin A = 12 \times \dfrac{\sqrt{15}}{4}

[1]

= 3\sqrt{15}.

[1]

Final answer

(a) 8
(b) \sqrt{15}/4
(c) 3\sqrt{15}.

Question 16

[4 marks]

A solid cone has base radius 6 cm and curved surface area 60\pi cm^2.

(a) Find the slant height of the cone.

[1]

(b) Find the perpendicular height of the cone.

[1]

(c) Find the volume of the cone, in terms of \pi.

[1]

(d) Write down the exact value of the sine of the angle between a slant edge of the cone and its base.

[1]

Show solutionHide solution

(a) \pi r l = 60\pi \Rightarrow 6l = 60, so l = 10 cm.

[1]

(b) h = \sqrt{10^2 - 6^2} = 8 cm.

[1]

(c) V = \tfrac13\pi(6^2)(8) = 96\pi\ \text{cm}^3.

[1]

(d) The slant edge, the height and the base radius form a right-angled triangle; the angle at the base has \sin = \dfrac{h}{l} = \dfrac{8}{10} = \dfrac45.

[1]

Final answer

(a) 10 cm
(b) 8 cm
(c) 96\pi cm^3
(d) 4/5.

Question 17

[6 marks]

An arc of radius r joins A to B; its centre is O. The chord AB cuts off a minor segment, and A\hat{O}B=\theta radians, where 0<\theta<\pi.

(a) Show that the perimeter of the minor segment is r\left(\theta+2\sin\dfrac{\theta}{2}\right).

[2]

Now let r=6 and \theta=\dfrac{2\pi}{3}.

(b) Find the exact area of the minor segment.

[2]

(c) Use the result of part (a) to find the exact perimeter of the minor segment.

[2]

Show solutionHide solution

(a) Draw the perpendicular from O to AB. It bisects AB and the angle AOB, so AB = 2r\sin\dfrac{\theta}{2}.

[1]

Perimeter = arc + chord = r\theta + 2r\sin\dfrac{\theta}{2} = r\left(\theta+2\sin\dfrac{\theta}{2}\right).

[1]

(b) Area of segment =\tfrac12 r^2(\theta-\sin\theta)=\tfrac12(36)\left(\dfrac{2\pi}{3}-\dfrac{\sqrt3}{2}\right)

[1]

=12\pi-9\sqrt3.

[1]

(c) Perimeter =6\left(\dfrac{2\pi}{3}+2\sin\dfrac{\pi}{3}\right)=6\left(\dfrac{2\pi}{3}+\sqrt3\right)

[1]

=4\pi+6\sqrt3.

[1]

Final answer

(a) shown
(b) 12\pi-9\sqrt3 cm^2
(c) 4\pi+6\sqrt3 cm.

Question 18

[6 marks]

In triangle ABC, AB = 10 cm, AC = 8 cm and its area equals 20 cm^2. Let B\hat{A}C=A.

(a) Find the two possible values of A.

[3]

(b) Show that the two possible values of BC^2 are 164 - 80\sqrt3 and 164 + 80\sqrt3.

[3]

Show solutionHide solution

(a) Area = \tfrac12(AB)(AC)\sin A, so \tfrac12(10)(8)\sin A = 20.

[1]

40\sin A = 20, so \sin A = \dfrac12.

[1]

A = 30^\circ or A = 150^\circ (both are valid angles of a triangle).

[1]

(b) Cosine rule: BC^2 = 10^2 + 8^2 - 2(10)(8)\cos A = 164 - 160\cos A.

[1]

A = 30^\circ: BC^2 = 164 - 160\cdot\dfrac{\sqrt3}{2} = 164 - 80\sqrt3.

[1]

A = 150^\circ: \cos 150^\circ = -\dfrac{\sqrt3}{2}, so BC^2 = 164 + 80\sqrt3.

[1]

Final answer

(a) 30^\circ or 150^\circ
(b) shown.

Mastery

Full exam questions that mix question types, at Paper 1 and Paper 2 standard.

Question 19

[6 marks]

A circle has centre O and radius 9 cm. A chord [AB] subtends an angle \theta radians at O, where 0 < \theta < \pi. The area of the minor segment cut off by the chord is exactly one quarter of the area of the circle.

(a) Show that the area of the segment is 40.5(\theta - \sin\theta).

[2]

(b) Find the value of \theta.

[2]

(c) Find the perpendicular distance from O to the chord [AB].

[2]

Show solutionHide solution

(a) Sector = \tfrac12(81)\theta and triangle OAB = \tfrac12(81)\sin\theta.

[1]

Segment = \tfrac12(81)\theta - \tfrac12(81)\sin\theta = 40.5(\theta - \sin\theta).

[1]

(b) The circle has area 81\pi, so 40.5(\theta - \sin\theta) = \dfrac{81\pi}{4}, i.e.; \theta - \sin\theta = \dfrac{\pi}{2}.

On the GDC (radian mode) graph y = x - \sin x and y = \dfrac{\pi}{2} and find the intersection.

[1]

\theta = 2.31 (3 s.f.; unrounded 2.3099\ldots).

[1]

(c) The perpendicular from O to the chord bisects \angle AOB and forms a right-angled triangle with hypotenuse 9: distance = 9\cos\dfrac{\theta}{2}.

[1]

= 9\cos(1.155\ldots) = 3.64 cm.

[1]

Final answer

(a) shown
(b) 2.31 rad
(c) 3.64 cm.

Question 20

[6 marks]

A solid metal cone has base radius 9 cm and perpendicular height 12 cm. It is melted down and recast, without loss of metal, as a solid sphere.

(a) Find the exact volume of the cone, in terms of \pi.

[2]

(b) Find the radius of the sphere.

[2]

(c) Find the percentage decrease in the total surface area when the cone is recast as the sphere.

[2]

Show solutionHide solution

(a) V = \tfrac13\pi(9^2)(12) = 324\pi\ \text{cm}^3.

[2]

(b) \tfrac43\pi R^3 = 324\pi \Rightarrow R^3 = 243.

[1]

R = 6.24 cm.

[1]

(c) Cone: slant height = \sqrt{9^2 + 12^2} = 15, total surface area = \pi(9)(15) + \pi(9^2) = 216\pi = 678.6. Sphere: 4\pi R^2 = 489.3.

[1]

Decrease = \dfrac{678.6 - 489.3}{678.6} \times 100 = 27.9\%.

[1]

Final answer

(a) 324\pi cm^3
(b) 6.24 cm
(c) 27.9\%.

Question 21

[9 marks]

In triangle ABC let a = BC, b = CA and c = AB.

(a) Use the cosine rule to show that \cos B = \dfrac{a^2 + c^2 - b^2}{2ac}, and write down the corresponding expression for \cos C.

[2]

(b) Hence show that a = b\cos C + c\cos B.

[3]

In a particular triangle ABC, AB = 15, AC = 13, \cos B = \dfrac35 and \cos C = \dfrac{5}{13}.

(c) Use part (b) to find the length of [BC].

[2]

(d) Find the exact area of triangle ABC.

[2]

Show solutionHide solution

(a) The cosine rule for the side opposite B: b^2 = a^2 + c^2 - 2ac\cos B.

[1]

Rearranging: \cos B = \dfrac{a^2 + c^2 - b^2}{2ac}, and similarly \cos C = \dfrac{a^2 + b^2 - c^2}{2ab}.

[1]

(b) b\cos C + c\cos B = b\cdot\dfrac{a^2 + b^2 - c^2}{2ab} + c\cdot\dfrac{a^2 + c^2 - b^2}{2ac}.

[1]

= \dfrac{a^2 + b^2 - c^2}{2a} + \dfrac{a^2 + c^2 - b^2}{2a}.

[1]

= \dfrac{2a^2}{2a} = a.

[1]

(c) BC = b\cos C + c\cos B = 13\left(\dfrac{5}{13}\right) + 15\left(\dfrac35\right)

[1]

= 5 + 9 = 14.

[1]

(d) \sin B = \sqrt{1 - \dfrac{9}{25}} = \dfrac45 (positive, since B is an angle of a triangle).

[1]

Area = \tfrac12(BC)(AB)\sin B = \tfrac12(14)(15)\dfrac45 = 84.

[1]

Final answer

(a), (b) shown
(c) 14
(d) 84.

Question 22

[9 marks]

In triangle ABC, AB = 7, BC = x and B\hat{A}C = 30^\circ, where x > 0.

(a) Show that \sin A\hat{C}B = \dfrac{7}{2x}.

[1]

(b) Find all values of x that allow two distinct triangles. Justify your answer.

[4]

Now let x = 5.

(c) Find the two possible values of the area of triangle ABC.

[4]

Show solutionHide solution

(a) Sine rule: \dfrac{\sin\angle ACB}{7} = \dfrac{\sin 30^\circ}{x} = \dfrac{1}{2x}, so \sin\angle ACB = \dfrac{7}{2x}.

[1]

(b) A triangle exists only if \dfrac{7}{2x} < 1 for two values of C: x > 3.5.

[2]

Two triangles need both the acute angle C and the obtuse angle 180^\circ - C to be possible, which requires 30^\circ + (180^\circ - C) < 180^\circ, i.e.; C > 30^\circ.

[1]

For acute C, C > 30^\circ means \sin C > \dfrac12, i.e.; \dfrac{7}{2x} > \dfrac12, so x < 7. Hence 3.5 < x < 7.

[1]

(c) \sin C = \dfrac{7}{10} = 0.7, so C = 44.4^\circ or C = 135.6^\circ.

[1]

Then B = 180^\circ - 30^\circ - C = 105.6^\circ or 14.4^\circ.

[1]

Area = \tfrac12(AB)(BC)\sin B = \tfrac12(7)(5)\sin B = 17.5\sin B.

[1]

Areas: 16.9\ \text{cm}^2 (for B = 105.6^\circ) and 4.36\ \text{cm}^2 (for B = 14.4^\circ).

[1]

Final answer

(a) shown
(b) 3.5<x<7
(c) 16.9 or 4.36.

Question 23

[14 marks]

A circular inspection track has centre O and radius 14 m. Point A is due north of O. Point B is reached by moving clockwise round the track through \theta radians, with 0<\theta<\frac\pi2. An inspector walks along the minor arc from A to B and returns along the straight chord. This route is 38 m long.

(a) Show that 14\theta+28\sin(\theta/2)=38.

[3]

(b) Find \theta.

[2]

(c) Find the area enclosed by the route.

[3]

(d) Find the bearing of B from A.

[3]

(e) A radio antenna stands 9 m vertically above O. Find its angle of elevation from the midpoint of the chord.

[3]

Show solutionHide solution

arc =14\theta; chord =28\sin(\theta/2); add to obtain the stated equation. M1A1A1 The route has one curved and one straight part.

Radian-mode solve on (0,\pi/2) gives \theta=1.41. M1A1 Use the stated interval; store the full angle.

A=98(\theta-\sin\theta)=41.8 m^2 M1A1A1 The enclosed region is the minor segment.

O\hat AB=(\pi-\theta)/2; bearing =180^\circ-(90^\circ-\theta\,90/\pi)=131^\circ. M1A1A1 At A, AO points south; the chord lies east of it.

OM=14\cos(\theta/2); \tan\alpha=9/OM; \alpha=40.2^\circ. M1M1A1 Project onto the horizontal ground before finding elevation.

Final answer

(a) shown
(b) 1.41 rad
(c) 41.8 m^2
(d) 131^\circ
(e) 40.2^\circ.

Question 24

[16 marks]

A calibration vessel is made from a cylinder and a hemispherical roof, both of radius r m. It has a flat circular floor, a cylindrical height h m and an enclosed volume of 180 m^3. Ignore the thickness of the metal. For a cylinder, curved area is 2\pi rh.

(a) Write an equation connecting r and h, and hence express h in terms of r.

[3]

(b) Show that the external surface area is S=360/r+5\pi r^2/3.

[3]

(c) Explain why the design requires 0<r<(270/\pi)^{1/3}.

[2]

(d) Use a GDC graph on this interval to find the radius and surface area at the minimum.

[3]

(e) Find the cylindrical height at this minimum.

[2]

(f) Find the fraction of the volume contained in the roof at this minimum.

[3]

Show solutionHide solution

\pi r^2h+\frac23\pi r^3=180; h=\frac{180}{\pi r^2}-\frac{2r}{3}. M1A1A1 Conserve the enclosed volume.

S=2\pi rh+2\pi r^2+\pi r^2=\frac{360}{r}+\frac{5\pi r^2}{3}. M1M1A1 Include the floor but exclude the joining disc.

r>0, h>0\Rightarrow r^3<270/\pi. M1R1 A positive cylindrical height gives the upper bound.

GDC minimum: r=3.25 m and S=166 m^2. M1A1A1 The minimum lies inside the design interval.

h=\frac{180}{\pi r^2}-\frac{2r}{3}=3.25 m. M1A1 Substitute the stored radius.

roof volume =\frac23\pi r^3=72 m^3; fraction =72/180=\frac25. M1A1A1 Divide a volume by the total volume.

Final answer

(a) h=180/(\pi r^2)-2r/3
(b) shown
(c) h>0
(d) r=3.25 m, S=166 m^2
(e) h=3.25 m
(f) 2/5.

Going further

Beyond the syllabus: the worked examples and investigations of the chapter's Going further pages.

G1 Area from three sides

G1 Worked example

[10 marks]

A triangle has sides 7, 8 and x, and area A.

(a) Use Heron's formula to show that 16A^2=(x^2-1)(225-x^2).

[3]

(b) State the condition on x for such a triangle to exist.

[1]

(c) Find the value of x for which A is greatest, and this greatest area. Show that the triangle is then right-angled.

[3]

(d) Find the two values of x for which A=14\sqrt3. For each, find the angle between the sides of length 7 and 8.

[3]

Show solutionHide solution

(a) s=\dfrac{15+x}{2}, so s-7=\dfrac{x+1}{2}, s-8=\dfrac{x-1}{2}, s-x=\dfrac{15-x}{2}

A^2=\dfrac{(15+x)(x+1)(x-1)(15-x)}{16}

pair the brackets as differences of squares: 16A^2=(x^2-1)(225-x^2)

A1AG

(b) 1<x<15

Note: both brackets must be positive; this is the triangle inequality.

(c) with u=x^2, 16A^2=-u^2+226u-225, a quadratic with its vertex at u=113

x=\sqrt{113}, 16A^2=112\times112, so A=28

x^2=113=7^2+8^2, so by the cosine rule the angle between the sides 7 and 8 has cosine 0: it is a right angle

Note: the check A=\tfrac12(7)(8)\sin C\le28, with equality only at C=90^\circ, gives the same answer.

(d) 16\times588=9408=(u-1)(225-u), so u^2-226u+9633=0

u=113\pm56: x=\sqrt{57} or x=13

\cos C=\dfrac{113-u}{112}=\dfrac12 or -\dfrac12: C=60^\circ (when x=\sqrt{57}) and C=120^\circ (when x=13)

Final answer

(b) 1<x<15
(c) x=\sqrt{113}, A=28
(d) x=\sqrt{57}, 60^\circ
x=13, 120^\circ

G1 Investigation

[8 marks]

A triangle is called Heronian if its three sides and its area are all integers.

(a) Use Heron's formula to find the areas of the triangles with sides 13,14,15; 5,5,6; and 5,5,8.

[2]

(b) A triangle has sides a, b, c and s=\frac12(a+b+c). Let p=s-a, q=s-b and r=s-c. Show that p, q and r are positive, that p+q+r=s, and that the area is \sqrt{s\,pqr}.

[2]

(c) Find every Heronian triangle with perimeter 32.

[4]

Show solutionHide solution

(a) s=21: \sqrt{21\times8\times7\times6}=\sqrt{7056}=84; ; s=8: \sqrt{8\times3\times3\times2}=12; ; s=9: \sqrt{9\times4\times4\times1}=12

A1A1

Note: one mark for 84; one mark for the two areas of 12.

(b) p=\dfrac{b+c-a}{2}>0 by the triangle inequality, and the same for q and r; p+q+r=3s-(a+b+c)=3s-2s=s, and Heron's formula reads \sqrt{s\,pqr}

A1A1

(c) s=16, so p,q,r are positive integers with p+q+r=16, and A=\sqrt{16pqr}=4\sqrt{pqr}

A is an integer only if pqr is a perfect square

list p\le q\le r with p+q+r=16 (21 triples) and test pqr: only 1\times3\times12=36 and 4\times6\times6=144 are squares

sides 16-p, 16-q, 16-r: the triangle 15,13,4 (area 24) and the triangle 12,10,10 (area 48)

Final answer

(a) 84, 12, 12
(c) 4,13,15 (area 24) and 10,10,12 (area 48).

TI-84 Plus CE keystrokes

The GDC panels in the book give keystrokes for the TI-Nspire CX II and the Casio fx-CG50. These are the TI-84 Plus CE keystrokes for the same calculator jobs. The full list of TI-84 keystrokes is in Chapter 12.

Job	TI-84 Plus CE keystrokes
Solve an equation	`MATH, Numeric Solver (MATH, 0:Solver on older OS versions): enter the equation, then SOLVE. Or graph both sides and use 2nd TRACE (CALC), 5:intersect`

From the book

The Ultimate Guide for the IB: Mathematics: Analysis and Approaches SL

Chapter 5 teaches triangles, sectors and solids from the basics to full exam questions: how the IB asks it, every question type with a marked example, and these practice questions in three steps.

About the book