Home / IBDP Maths AI: Topic SL 3.6: sites, vertices, edges, cells: IB style Questions HL Paper 1

IBDP Maths AI: Topic SL 3.6: sites, vertices, edges, cells: IB style Questions HL Paper 1

Question 2 [Maximum mark :6]

The Voronoi diagram below shows three identical cellular phone towers, T1, T2 and T3. A fourth cellular phone tower, T4 is located in the shaded region. The dashed lines in the diagram below represent the edges in the Voronoi diagram.

Horizontal scale: 1 unit represents 1 km. Vertical scale: 1 unit represents 1 km.

Tim stands inside the shaded region.

(a) Explain why Tim will receive the strongest signal from tower T4. [1]

Tower T2 has coordinates (−9, 5) and the edge connecting vertices A and B has equation y = 3.

(b) Write down the coordinates of tower T4. [2]

Tower T1 has coordinates (−13, 3)

(c)Find the gradient of the edge of the  Voronoi diagram between towers T1 and T2. [3]

▶️Answer/Explanation

(a) every point in the shaded region is closer to tower T4 Note: Specific reference must be made to the closeness of tower T4. (b) ( 9, 1) Note: Award A1 for each correct coordinate. Accept x = −9 and y =1. Award at most A0A1 if parentheses are missing. (c) correct use of gradient formula  e.g. (m =) \(\frac{5-3}{-9-13}(= \frac{1}{2})\)  taking negative reciprocal of their m (at any point)  edge gradient = −2

Question

The diagram shows a pyramid \({\text{VABCD}}\) which has a square base of length \(10{\text{ cm}}\) and edges of length \(13{\text{ cm}}\). \({\text{M}}\) is the midpoint of the side \({\text{BC}}\).

Calculate the length of \({\text{VM}}\).[2]

a.

Calculate the vertical height of the pyramid.[2]

b.
▶️Answer/Explanation

Markscheme

Unit penalty (UP) applies in this question.

\({\text{VM}}^{2} = {13^2} – {5^2}\)     (M1)

UP     \( = 12{\text{ cm}}\)     (A1)     (C2)[2 marks]

a.

Unit penalty (UP) applies in this question.

\({h^2} = {12^2} – {5^2}\) (or equivalent)     (M1)

UP     \( = 10.9{\text{ cm}}\)     (A1)(ft)     (C2)[2 marks]

b.

Question

The right pyramid shown in the diagram has a square base with sides of length 40 cm. The height of the pyramid is also 40 cm.

a.Find the length of OB.[4]

b.Find the size of angle OBP.[2]

▶️Answer/Explanation

Markscheme

Note: Unit penalty (UP) applies in this part

\({\rm{PB}} = \frac{1}{2}\sqrt {{{40}^2} + {{40}^2}}  = \sqrt {800}  = 28.28(28.3)\)     (M1)(A1)

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

(UP)     \({\rm{OB}} = \sqrt {{{40}^2} + {{28.28}^2}}  = 49.0{\text{ cm }}\left( {\sqrt {2400} {\text{ cm}}} \right)\)     (M1)(A1)(ft)     (C4)

Note: Award (M1) for correct substitution, can (ft) from any answer to PB.[4 marks]

a.

\({\sin ^{ – 1}}\left( {\frac{{40}}{{49}}} \right)\)

OR

\({\cos ^{ – 1}}\left( {\frac{{28.28}}{{49}}} \right)\)

OR

\({\tan ^{ – 1}}\left( {\frac{{40}}{{28.28}}} \right)\)     (M1)

= 54.7 (54.8)     (A1)(ft)     (C2)

Note: Award (M1) for any correct trig. ratio.

In radians = 0.616, award (M1)(A0).

Note: Common error: (a) \(OB = \sqrt {40^2 + 20^2} = 44.7 {\text{ cm}}\). Award (M0)(A0)(M1), (A1)(ft), and (b) angle OBP = 63.4° (63.5°) (M1)(A1)(ft).[2 marks]

b.

Question

A shipping container is a cuboid with dimensions \({\text{16 m}}\), \({\text{1}}\frac{{\text{3}}}{{\text{4}}}{\text{ m}}\) and \({\text{2}}\frac{{\text{2}}}{{\text{3}}}{\text{ m}}\).

a.Calculate the exact volume of the container. Give your answer as a fraction.[3]
b.Jim estimates the dimensions of the container as 15 m, 2 m and 3 m and uses these to estimate the volume of the container.
Calculate the percentage error in Jim’s estimated volume of the container.[3]
 
▶️Answer/Explanation

Markscheme

\(V = 16 \times 1\frac{3}{4} \times 2\frac{2}{3}\)     (M1)

Note: Award (M1) for correct substitution in volume formula. Accept decimal substitution of \(2.66\) or better.

\( = 74.6666{\text{ }} \ldots \)     (A1)
\( = {\text{74}}\frac{{\text{2}}}{{\text{3}}}{\text{ }}{{\text{m}}^{\text{3}}}{\text{ }}\left( {\frac{{{\text{224}}}}{{\text{3}}}{\text{ }}{{\text{m}}^{\text{3}}}} \right)\)     (A1)    (C3)

Note: Correct answer only.[3 marks]

a.

\({\text{%  error}} = \frac{{\left( {90 – 74\frac{2}{3}} \right) \times 100}}{{74\frac{2}{3}}}\)     (A1)(M1)

Note: Award (A1) for \(90\) seen, or inferred in numerator, (M1) for correct substitution into percentage error formula.

\( = 20.5\)     (A1)(ft)     (C3) 

Note: Accept \( – 20.5\).[3 marks]

b.

Question

A rectangular cuboid has the following dimensions.

Length     0.80 metres     (AD)

Width      0.50 metres     (DG)

Height     1.80 metres     (DC)

a.Calculate the length of AG.[2]

b.Calculate the length of AF.[2]

c.Find the size of the angle between AF and AG.[2]

▶️Answer/Explanation

Markscheme

\({\text{AG}} = \sqrt {{{0.8}^2} + {{0.5}^2}} \)     (M1)

AG = 0.943 m     (A1)     (C2)[2 marks]

a.

\({\text{AF}} = \sqrt {{\text{A}}{{\text{G}}^2} + {{1.80}^2}} \)     (M1)

= 2.03 m     (A1)(ft)     (C2)

Note: Follow through from their answer to part (a).[2 marks]

b.

\(\cos {\rm{G\hat AF}} = \frac{{0.943(39 \ldots )}}{{2.03(22 \ldots )}}\)     (M1)

\(\operatorname{G\hat AF} = 62.3^\circ \)     (A1)(ft)     (C2)

Notes: Award (M1) for substitution into correct trig ratio.

Accept alternative ratios which give 62.4° or 62.5°.

Follow through from their answers to parts (a) and (b).[2 marks]

c.
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