IB Math Studies Year 1 Name:__________________________Topic 5: Geometry and Trigonometry – Formative Quiz 7

1. The quadrilateral ABCD shown below represents a sandbox. AB and BC have the same length. AD is 9 m

long and CD is 4.2 m long. Angles CDA and CBA are 95° and 130° respectively.

diagram not to scale (a) Find the length of AC.

(3)

(b) (i) Write down the size of angle A.CB

(ii) Calculate the length of AB.

(4) (b) Show that the area of the sandbox is 31.1 m2 correct to 3 s.f.

(4)

The sandbox is a prism. Its edges are 40 cm high. The sand occupies one third of the volume of the sandbox.

(c) Calculate the volume of sand in the sandbox.

(3)

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(Total 14 marks)2. Three right pyramids Andal, Batsu and Cartos were discovered in the dense jungle of Marhartmasol. Each

pyramid has a square base with centres A, B and C respectively.

A

B

C

C a rto s

B a tsu

A n d a l

Diagram not to scale

A surveying team was lowered from a helicopter to the top of Andal to take measurements of the area. Andal is 40 metres high. The angle of elevation from the top of Andal to the top of Batsu is 3°. The horizontal distance from A, the centre of the base of Andal, to B, the centre of the base of Batsu is 600 metres.

(a) Use the diagram below to find the height of Batsu.(3)

Diagram not to scale

3 º

A n d a l B a tsu

A B6 0 0 m

4 0 m

(b) Cartos is found to be 92 metres high and the angle of elevation from the top of Andal to the top of Cartos is 4°.

(i) Draw a diagram similar to the diagram in part (a) to show the relationship between Andal and Cartos.

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(ii) What is the horizontal distance from A to C?

(4) (c) The diagram below represents measurements relative to the centres of the bases of the pyramids. The

surveyors determined the angle at A to be 110°, and the distance AB to be 600 m.Diagram not to scale

A

B

C

6 0 0 m

11 0 º

(i) What is the distance between B and C? Give your answer to the nearest metre.

(ii) What is the size of angle ACB?

(iii) What is the area of the land inside triangle ABC?

(8)

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(Total 15 marks)3. A child’s toy is made by combining a hemisphere of radius 3 cm and a right circular cone of slant height l as

shown on the diagram below.

diagram not to scale

(a) Show that the volume of the hemisphere is 18 cm3.

(2)

The volume of the cone is two-thirds that of the hemisphere.

(b) Show that the vertical height of the cone is 4 cm.

(4) (c) Calculate the slant height of the cone.

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(2)(d) Calculate the angle between the slanting side of the cone and the flat surface of the hemisphere.

(3) (e) The toy is made of wood of density 0.6 g per cm3. Calculate the weight of the toy.

(3) (f) Calculate the total surface area of the toy.

(5)(Total 19 marks)

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4. A path goes around a forest so that it forms the three sides of a triangle. The lengths of two sides are 550 m and 290 m. These two sides meet at an angle of 115°.A diagram is shown below.

diagram not to scale (a) Calculate the length of the third side of the triangle. Give your answer correct to the nearest 10 m.

(4) (b) Calculate the area enclosed by the path that goes around the forest.

(3)

Inside the forest a second path forms the three sides of another triangle named ABC.

Angle CAB is 53°, AC is 180 m and BC is 230 m.

diagram not to scale

(c) Calculate the size of angle B.CA

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(4)(Total 11 marks)

IB Math Studies Year 1 Name:__________________________Topic 5: Geometry and Trigonometry – Formative Quiz 8

1. A farmer has a triangular field, ABC, as shown in the diagram.AB = 35 m, BC = 80 m and BÂC =105°, and D is the midpoint of BC.

diagram not to scale (a) Find the size of BĈA.

(3) (b) Calculate the length of AD.

(5)

The farmer wants to build a fence around ABD.

(c) Calculate the total length of the fence.

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(2) (d) The farmer pays 802.50 USD for the fence. Find the cost per metre.

(2) (e) Calculate the area of the triangle ABD.

(3) (f) A layer of earth 3 cm thick is removed from ABD. Find the volume removed in cubic metres.

(3)(Total 18 marks)

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2. An office tower is in the shape of a cuboid with a square base. The roof of the tower is in the shape of a square based right pyramid.The diagram shows the tower and its roof with dimensions indicated. The diagram is not drawn to scale.

O1 0 m

H

E

G

F

D

A B

C

4 0 m

6 m(a) Calculate, correct to three significant figures,

(i) the size of the angle between OF and FG;

(3) (ii) the shortest distance from O to FG;

(2) (iii) the total surface area of the four triangular sections of the roof;

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(3) (iv) the size of the angle between the slant height of the roof and the plane EFGH;

(2) (v) the height of the tower from the base to O.

(2)

A parrot’s nest is perched at a point, P, on the edge, BF, of the tower. A person at the point A, outside the building, measures the angle of elevation to point P to be 79°.

(b) Find, correct to three significant figures, the height of the nest from the base of the tower.

(2)(Total 14 marks)

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3. In the diagram below A , B and C represent three villages and the line segments AB, BC and CA represent the roads joining them. The lengths of AC and CB are 10 km and 8 km respectively and the size of the angle between them is 150°.

diagram not to scale (a) Find the length of the road AB.

(3) (b) Find the size of the angle CAB.

(3)

Village D is halfway between A and B. A new road perpendicular to AB and passing through D is built. Let T be the point where this road cuts AC. This information is shown in the diagram below.

diagram not to scale (c) Write down the distance from A to D.

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(1) (d) Show that the distance from D to T is 2.06 km correct to three significant figures.

(2)

A bus starts and ends its journey at A taking the route AD to DT to TA.

(e) Find the total distance for this journey.

(3)

The average speed of the bus while it is moving on the road is 70 km h–1.The bus stops for 5 minutes at each of D and T.

(f) Estimate the time taken by the bus to complete its journey. Give your answer correct to the nearest minute.

(4)(Total 16 marks)

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4. The figure below shows a rectangular prism with some side lengths and diagonal lengths marked. AC = 10 cm, CH = 10 cm, EH = 8cm, AE 8 cm.

A

B C

D

E

FG

H

1 0 cm

8 cm(n o t to sc a le )

1 0 cm

8 cm

(a) Calculate the length of AH.

(2)

(b) Find the size of angle H.CA

(3)(c) Show that the total surface area of the rectangular prism is 320 cm2.

(3)(d) A triangular prism is enclosed within the planes ABCD, CGHD and ABGH. Calculate the volume of

this prism.

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(3)(Total 11 marks)

IB Math Studies Year 1 Topic 5: Geometry and Trigonometry – Formative Quiz 7 – MS

1. Unit penalty (UP) is applicable in (a), (b)(ii) and (d)(a) AC2 = 92 + 4.22 – 2 × 9 × 4.2 × cos 95° (M1)(A1)

UP AC = 10.3 m (A1)(G2)Note: (M1) for correct substituted formula and (A1) for correct substitution If radians used answer is 6.59. Award at most (M1)(A1)(A0)

(b) (i) ACB = 25° (A1)

(ii)

130sin...258.10

sin25AB

(M1)(A1)AB = 5.66 m (A1)(ft)(G2)

Note: (M1) for correct substituted formula and (A1) for correct substitution. (A1) for correct answer.

Follow through with angle ACB ˆ and their AC.Allow AB = 5.68 if AC = 10.3 used.If radians used answer is 0.938 (unreasonable answer). Award at most (M1)(A1)(A0)(ft)

ORUsing that ABC is isosceles

cos 25° = AB

...258.1021

(or equivalent) (A1)(M1)(ft)UP AB = 5.66 m (A1)(ft)(G2)

Note: (A1) for 21

of their AB seen, (M1) for correct trigonometric ratio and correct substitution, (A1) for correct answer.

If 21

AB seen and correct answer is given award (A1)(G1).Allow AB = 5.68 if AC = 10.3 used.If radians used answer is 3.32. Award (A1)(M1)(A1)(ft).If sin65 and radians used answer is 3.99. Award (A1)(M1)(A1)(ft)

(c) Area = 21

× 9 × 4.2 × sin 95° + 21

× (5.6592...)2 × sin 130° (M1)(M1)(ft)(M1)

UP = 31.095... = 31.1 m2 (correct to 3 s.f.) (A1)(AG)Note: (M1)(M1) each for correct substitution in the formula of the area of each triangle, (M1) for adding both areas. (A1) for unrounded answer.Follow through with their length of AB but last mark is lost if they do not reach the correct answer.

(d) Volume of sand = 31

(31.09... × 0.4) (M1)(M1)

= 4.15 m3 (A1)(G2)Note: (M1) for correct formula of volume of prism and for correct

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substitution, (M1) for multiplying by 31

and last (A1) for correct answer only.

[14]2. (a)

3 º6 0 0 m

x

(M1)

tan 3° = 600x

x = 600 tan 3°x = 31.4447x = 31.4 m (A1)Therefore, height = 40 m + 31.4 m

= 71.4 m (A1) 3 (b) (i)

4 º

A Cx

4 0

92

(A1)Note: For (A1) the candidate must have the 40, the 92 and the 4° in the appropriate places.

(ii)

4 º5 2

x (A1)

tan 4° = x52

(M1)

x = 4tan52

x = 743.6346453 = 744 m (A1) 4 (c) (i)

A

B C

6 0 0 m 7 4 4 m11 0 º

BC2 = 6002 + 7442 – 2 × 600 × 744 cos 110° (M1)BC2 = 1218891.584 (or 1218198.119) (A1)BC = 1104.034231 (or 1103.720...)BC = 1104 (to the nearest metre) (A1)

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(ii) 1104110sin

600sin

c

(M1)

sin c = 1104110sin600

(M1)c = 30.710635°c = 30.7° (3 s.f.) (A1)

(iii) area = 21

× 600 × 744 sin 110° (M1)= 209739.393= 210000 m2 (3 s.f.) (A1) 8

[15]

3. (a) V = 3π

34

21 r

For using 3π

34 r

(with or without 21

) (M1)

= 32

× × 33 For using 21

(their sphere formula) (M1)= 18 cm3 (AG) 2

(b) V = 32

× 18 For using 32

× their answer to (a) (M1)= 12 (A1)

12 = 31

× 32 × h For equating the volumes (M1)

27.28097.113

9π36π h

(A1)h = 4 cm (AG) 4

(c) l2 = 42 + 32 For using Pythagoras theorem (M1)

l = 5 (A1) 2 (d) For identifying the correct angle (M1)

tan = 34

or sin = 54

or cos = 53

(M1) = 53.1° (0.927 radians) (A1) 3

(e) For summing volume of cone and hemisphere. (M1)

Volume = 12 + 18= 30 cm3 (94.2 cm3)For multiplying the volume by 0.6 (M1)Weight = 0.6 × 30= 56.5 g (A1) 3

(f) Surface area of cone = rl= × 3 × 5 = 15 (M1)(A1)

Surface area of a hemisphere = 21

× 4r2

16

= 21

× 4 × × 32

= 18 (M1)(A1)Total surface area = 15 + 18= 103.67= 104 cm2 (A1) 5

[19]

4. UP applies in this question (a) l2 = 2902 + 5502 – 2 × 290 × 550 × cos 115° (M1)(A1)

Note: Award (M1) for substituted cosine rule formula, (A1) for correct substitution.

l = 722 (A1)(G2)UP = 720 m (A1)

Note: If 720 m seen without working award (G3).The final (A1) is awarded for the correct rounding of their answer.

(b) Area = 21

× 290 × 550 × sin 115 (M1)(A1)Note: Award (M1) for substituted correct formula (A1) for correct substitution.

UP = 72 300 m2 (A1)(G2)

(c) 53sin230

Bsin180

(M1)(A1)

Note: Award (M1) for substituted sine rule formula, (A1) for correct substitution.

B = 38.7° (A1)(G2)BCA = 180 – (53° + 38.7°)

= 88.3° (A1)(ft)[11]

IB Math Studies Year 1 Topic 5: Geometry and Trigonometry – Formative Quiz 8 – MS

1. Note: Unit penalty (UP) applies in parts (b)(c) and (e)

(a) 80105sin

35BCAsin

(M1)(A1)

Note: Award (M1) for correct substituted formula, (A1) for correct substitutions.

ACB = 25.0° (A1)(G2) (b) Length BD = 40 m (A1)

Angle ABC = 180° – 105° – 25° = 50° (A1)(ft)Note: (ft) from their answer to (a).

AD2 = 352 + 402 – (2 × 35 × 40 × cos 50°) (M1)(A1)(ft)

Note: Award (M1) for correct substituted formula,(A1)(ft) for correct substitutions.

UP AD = 32.0m (A1)(ft)(G3)17

Notes: If 80 is used for BD award at most (A0)(A1)(ft)(M1)(A1)(ft)(A1)(ft) for an answer of 63.4 m.If the angle ABC is incorrectly calculated in this part award at most (A1)(A0)(M1)(A1)(ft)(A1)(ft).If angle BCA is used award at most (A1)(A0)(M1)(A0)(A0).

(c) length of fence = 35 + 40 + 32 (M1)

UP = 107m (A1)(ft)(G2)Note: (M1) for adding 35 + 40 + their (b).

(d) cost per metre = 10750.802

(M1)Note: Award (M1) for dividing 802.50 by their (c).

cost per metre = 7.50 USD (7.5 USD) (USD not required) (A1)(ft)(G2)

(e) Area of ABD = 21

× 35 × 40 × sin 50° (M1) = 536.2311102 (A1)(ft)

UP = 536m2 (A1)(ft)(G2)Note: Award (M1) for correct substituted formula,(A1)(ft) for correct substitution, (ft) from their valueof BD and their angle ABC in (b).

(f) Volume = 0.03 × 536 (A1)(M1)

= 16.08 = 16.1 (A1)(ft)

(G2)Note: Award (A1) for 0.03, (M1) for correct formula.(ft) from their (e).If 3 is used award at most (A0)(M1)(A0).

[18]

2. Note: Throughout this question watch out for and accept other alternative approaches e.g. in part (a)(i), the cosine formula may not necessarily be used.

1 0 1 0

O

6F G

(a) (i) = GFO = arccos

)6)(10)(2(10610 222

(M2)Notes: Award (M1) for any correct method (formulae).Award (M1) for substituting correctly in the formula used.

= 72.5° (3 s.f.) (A1) 3Note: Award (A1) for correct answer only.

(ii) h = slant height or shortest distance from O to FG = 3 tan (M1)

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= 9.53939... = 9.54 m (3 s.f.) (A1) 2Note: Follow through with candidate’s

(iii) Area of OFG = 21

(10)(6)(sin ) (M2)

therefore total surface area of roof = 4 × 21

(10)(6)(sin ) = 114.4727... = 114 m2 (3 s.f.) (A1) 3

Notes: Award (M1) for using any correct method (formulae).Award (M1) for substituting correctly in the formulae used.Follow through with candidate’s . Accept 115 m2.

(iv) Angle between slant height (line) and plane EFGH = arccos

h3

(M1) = 71.7° (3 s.f.) (A1) 2

Note: Follow through with candidate’s h. (v) H = Height of tower from base to O

= 40 + 22 3h (M1)

= 49.055385... = 49.1 m (3 s.f.) (A1) 2Note: Follow through with candidate’s h

(b) Height (BP) = )7990sin(79sin6

(M1)= 30.9 m (3 s.f.) (A1) 2

[14]

3. Unit penalty (UP) applies in parts (a), (c) and (e).(a) AB2 = 102 + 82 – 2 × 10 × 8 × cos 150° (M1)(A1)

UP AB = 17.4 km (A1)(G2)Note: Award (M1) for substitution into correct formula, (A1) for correct substitution, (A1) for correct answer.

(b)

150sin4.17

BACsin8

(M1)(A1)CÂB = 13.3° (A1)(ft)(G2)

Notes: Award (M1) for substitution into correct formula, (A1) for correct substitution, (A1) for correct answer. Follow through from their answer to part (a).

UP (c) AD = 8.70 km (8.7 km) (A1)(ft)

Note: Follow through from their answer to part (a). (d) DT = tan(13.29...°) × 8.697... = 2.0550... (M1)(A1)

= 2.06 (AG)Notes: Award (M1) for correct substitution in the correct formula, award (A1) for the unrounded answer seen. If 2.06 not seen award at most (M1)(A0).

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UP (e)22 06.270.8 + 8.70 + 2.06 (A1)(M1)

= 19.7 km (A1)(ft)(G2)Note: Award (A1) for AT, (M1) for adding the three sides of the triangle ADT, (A1)(ft) for answer. Follow through from their answer to part (c).

(f) 707.19

× 60 + 10 (M1)(M1)= 26.9 (A1)(ft)

Note: Award (M1) for time on road in minutes, (M1) for adding 10, (A1)(ft) for unrounded answer. Follow through from their answer to (e).

= 27 (nearest minute) (A1)(ft)(G3)

Note: Award (A1)(ft) for their unrounded answer given to the nearest minute.

[16]

4. (a) AH = 12888 22 (or 11.3 or 28 ) (M1)(A1)(G2)

For use of Pythagoras theorem then for correct answer.

(b) 10102)128(1010HCAcos

222

(M1)(A1)(ft)

OR

128 = 102 + 102 – 2 × 10 × 10 × cos HCAFor use of cosine rule (M1)with correct values substituted (A1)(ft)

HCA = 68.9 (6854) (A1)(ft)(G2)Note: Allow 68.8 (68 48) following through with AH =11.3 from part (a).

ORTriangle ACH is isosceles. Let M be the midpoint of AH.

1024MCAsin

, hence MCA = 34.45For use of sine rule in appropriate triangle (M1)with correct values substituted (A1)(ft)

HCA = 2 34.45 = 68.9 (6854) (A1)(ft)(G2)

(c) Height AB (or equivalent) = 6810 22 (A1)Award (A1) for 6 seen.

Area = 2 × (8 × 8) + 4 × (6 × 8) (M1)(A1)= 320 cm2 (AG)

Notes:Award (M1) for sum of 2 faces and 4 faces. Final (A1) must result in an answer of 320.If candidate starts with 320 and works backwards to find height of 6, award (M1)(A1). Only award final (A1) if the candidate comments or proves that

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the height of 6 is consistent with the dimensions of the prism.

(d) Volume = 2688

(M1)(M1)Notes:Award (M1) for correct numerator and (M1) for division by 2 Follow through value for height from (c).

=192 cm3 (A1)(ft)(G3)[11]

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