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Standard Level · Geometry and Trigonometry

Three-dimensional geometry

Distance, surface area and volume

DifficultyAll difficulties
Exam modePaper 1 & 2

Question 1

MildNo calculator
[Maximum mark: 7]

Let A(2,−3,5)\mathrm A(2,-3,5) and B(−1,1,5)\mathrm B(-1,1,5). Find

(a) the distance between A and B.[2]

(b) the distance between O and B.[1]

(c) the coordinates of the midpoint M of the line segment [AB].[2]

(d) the coordinates of point C given that B is the midpoint of [AC].[2]

Worked solution

Question 1 mark scheme

(a)

dAB=32+42+02=5d_{AB}=\sqrt{3^2+4^2+0^2}=5

(b)

dOB=12+12+52=27=33d_{OB}=\sqrt{1^2+1^2+5^2}=\sqrt{27}=3\sqrt3

(c)

M(1/2,−1,5)\mathrm M(1/2,-1,5)

(d)

C(−4,5,5)\mathrm C(-4,5,5)

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Question 2

MildNo calculator
[Maximum mark: 16]

Complete the table

Solid Volume Surface area
cuboid
cuboid with source dimensions
cylinder
cylinder with source dimensions
cone
cone with source dimensions
sphere
sphere with source dimensions

for each shape [1+3]

Worked solution

Question 2 mark scheme

Solid Volume Surface area
Cuboid with side lengths 5, 3 and 4 V=5×3×4=60V=5\times3\times4=60 S=2×15+2×12+2×20=94S=2\times15+2\times12+2\times20=94
Cylinder of height 5 and diameter 4 r=2r=2
h=5h=5

V=π 22 5=20πV=\pi\,2^2\,5=20\pi		 S=2π (2)(5)+2×(π 22)=28πS=2\pi\,(2)(5)+2\times(\pi\,2^2)=28\pi
Cone of height 4 and diameter 6 r=3r=3
h=4h=4

V=13π 32 4=12πV=\dfrac13\pi\,3^2\,4=12\pi		 slant height: l=5l=5

S=π (3)(5)+π 32=24πS=\pi\,(3)(5)+\pi\,3^2=24\pi
Sphere of radius 3 V=43π 33=36πV=\dfrac43\pi\,3^3=36\pi S=4π 32=36πS=4\pi\,3^2=36\pi

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Question 3

MildNo calculator
[Maximum mark: 7]

For a right pyramid of square base of side 8 and vertical height 3 find

(a) the volume[2]

(b) the surface area[5]

Square-based right pyramid with height 3, base side 8, and points A, B, C, D, E, M and N

Worked solution

Question 3 mark scheme

(a)

V=13 82 3=64V=\dfrac13\,8^2\,3=64

(b)

AM2=42+32⇒AM=5AM^2=4^2+3^2\Rightarrow AM=5

S=82+4×(12×8×5)=64+80=144S=8^2+4\times\left(\dfrac12\times8\times5\right)=64+80=144

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Question 4

MildNo calculator
[Maximum mark: 4]

The diagram shows a vertical pole PQ, which is supported by two wires fixed to the horizontal ground at A and B.

Pole PQ and wires to A and B, with marked 36°, 30° and 70° angles

BQ=40 m\mathrm{BQ}=40\text{ m}

PB^Q=36∘\mathrm P\widehat{\mathrm B}\mathrm Q=36^\circ

BA^Q=70∘\mathrm B\widehat{\mathrm A}\mathrm Q=70^\circ

AB^Q=30∘\mathrm A\widehat{\mathrm B}\mathrm Q=30^\circ

Find

(a) the height of the pole, PQ;[2]

(b) the distance between A and B.[2]

Worked solution

Question 4 mark scheme

(a)

PQ40=tan⁡36∘⇒PQ≈29.1\dfrac{\mathrm{PQ}}{40}=\tan36^\circ\Rightarrow\mathrm{PQ}\approx29.1 m (3 sf)

(b)

AQ^B=80∘\mathrm A\widehat{\mathrm Q}\mathrm B=80^\circ

ABsin⁡80∘=40sin⁡70∘⇒AB=41 9.\dfrac{\mathrm{AB}}{\sin80^\circ}=\dfrac{40}{\sin70^\circ}\Rightarrow\mathrm{AB}=41\ 9. m (3 sf)

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Question 5

MediumCalculator
[Maximum mark: 14]

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.

Square-based cuboid tower with pyramid roof, base side 6 m, wall height 40 m and roof edge 10 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;[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∘79^\circ.

(b) Find, correct to three significant figures, the height of the nest from the base of the tower.[2]

Worked solution

Question 5 mark scheme

Isosceles roof triangle OFG with sides 10, 10 and 6 and angle θ at F

(a)

(i) cos⁡θ=3/10\cos\theta=3/10 or cosine rule for OF^G⇒θ=72.5∘\mathrm O\widehat{\mathrm F}\mathrm G\Rightarrow\theta=72.5^\circ (3 s.f.)

(ii) tan⁡θ=h/3⇒h=3tan⁡θ=9.53939…=9.54\tan\theta=h/3\Rightarrow h=3\tan\theta=9.53939\ldots=9.54 m (3 s.f.)

(iii) Area of △OFG=12(10)(6)(sin⁡θ)=30sin⁡θ)\triangle\mathrm{OFG}=\dfrac12(10)(6)(\sin\theta)=30\sin\theta)

total surface area of roof =4×30sin⁡θ)=114.4727…=114 m2=4\times30\sin\theta)=114.4727\ldots=114\text{ m}^2 (3 s.f.)

(iv) cos⁡φ=3/h⇒φ=71.7∘\cos\varphi=3/h\Rightarrow\varphi=71.7^\circ (3 s.f.)

(v) H=H= Height of tower from base to O =40+h2−32=49.055385…=49.1=40+\sqrt{h^2-3^2}=49.055385\ldots=49.1 m

(b)

Height (BP) =6sin⁡79∘sin⁡(90∘−79∘)=30.9=\dfrac{6\sin79^\circ}{\sin(90^\circ-79^\circ)}=30.9 m

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Question 6

MediumCalculator
[Maximum mark: 15]

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.

Three pyramids Andal, Batsu and Cartos with base centres A, B and C joined

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∘3^\circ. 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

Elevation diagram from Andal of height 40 m to Batsu, 600 m horizontally away, with angle 3°

(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∘4^\circ.

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

(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∘110^\circ, and the distance AB to be 600 m.

Triangle ABC with AB 600 m and angle A 110°

(i) What is the distance BC? 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]

Worked solution

Question 6 mark scheme

(a)

Right triangle with base 600 m, vertical side x and angle 3°

tan⁡3∘=x600⇒x=600tan⁡3∘=31.4447=31.4\tan3^\circ=\dfrac{x}{600}\Rightarrow x=600\tan3^\circ=31.4447=31.4 m

Therefore, height =40 m+31.4 m=71.4 m=40\text{ m}+31.4\text{ m}=71.4\text{ m}

(b)

(i)

Elevation diagram between A and C with heights 40 and 92, base x and angle 4°

(ii)

Right triangle with height 52, base x and angle 4°

tan⁡4∘=52x⇒x=52tan⁡4∘=743.6346453=744\tan4^\circ=\dfrac{52}x\Rightarrow x=\dfrac{52}{\tan4^\circ}=743.6346453=744 m

(c)

(i)

Triangle ABC with AB 600 m, AC 744 m and angle A 110°

BC2=6002+7442−2×600×744cos⁡110∘⇒BC=1104\mathrm{BC}^2=600^2+744^2-2\times600\times744\cos110^\circ\Rightarrow\mathrm{BC}=1104

(ii) sin⁡c600=sin⁡110∘1104⇒sin⁡c=600×sin⁡110∘1104⇒c=30.710635∘=30.7∘\dfrac{\sin c}{600}=\dfrac{\sin110^\circ}{1104}\Rightarrow\sin c=\dfrac{600\times\sin110^\circ}{1104}\Rightarrow c=30.710635^\circ=30.7^\circ

(iii) area =12×600×744sin⁡110∘=209739.393=210000 m2=\dfrac12\times600\times744\sin110^\circ=209739.393=210000\text{ m}^2 (3 s.f.)

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