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Date November Example question Marks available 1 Reference code EXN.2.SL.TZ0.5
Level Standard Level Paper Paper 2 Time zone Time zone 0
Command term State Question number 5 Adapted from N/A

Question

The living accommodation on a university campus is in the shape of a rectangle with sides of length 200m200m and 300m300m.

There are three offices for the management of the accommodation set at the points AA, BB and CC. These offices are responsible for all the students in the areas closest to the office. These areas are shown on the Voronoi diagram below. On this coordinate system the positions of AA, BB and CC are (100, 160)(100, 160), (100, 40)(100, 40) and (250, 100)(250, 100) respectively.

The equation of the perpendicular bisector of [AC][AC] is 5x-2y=6155x2y=615.

The manager of office CC believes that he has more than one third of the area of the campus to manage.

Find the area of campus managed by office CC.

[3]
b.

Hence or otherwise find the areas managed by offices AA and BB.

[3]
c.

State a further assumption that must be made in order to use area covered as a measure of whether or not the manager of office CC is responsible for more students than the managers of offices AA and BB.

 

[1]
d.

A new office is to be built within the triangle formed by AA, BB and CC, at a point as far as possible from the other three offices.

Find the distance of this office from each of the other offices.

[2]
e.

Markscheme

Divides area into two appropriate shapes

For example,

Area of triangle (12×200×40=)4000m2(12×200×40=)4000m2      (A1)

Area of rectangle (200×97)=19400m2(200×97)=19400m2      (A1)

23400(m2)23400(m2)      A1

 

Note: The area can be found using different divisions. Award A1 for any two correct areas found and A1 for the final answer.

 

[3 marks]

b.

EITHER

200×300-234002=18300m2200×300234002=18300m2      (M1)A1

 

OR

12×100×(203+163)=18300m212×100×(203+163)=18300m2      (M1)A1

 

THEN

Area managed by both offices AA and BB is 18300m218300m2      A1

 

[3 marks]

c.

Density of accommodation/students is uniform      R1

 

[1 mark]

d.

250-163=87(m)250163=87(m)        (M1)A1

 

Note: M1 is for an attempt to find the distance from the intersection point to one of the offices.

 

[2 marks]

e.

Examiners report

[N/A]
b.
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c.
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d.
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e.

Syllabus sections

Topic 3—Geometry and trigonometry » SL 3.6—Voronoi diagrams
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Topic 3—Geometry and trigonometry

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