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Scale Drawing Questions
1. Three mountains Mikai, Kembo and Chaka in a village are situated in such a way that Kembo is 900m on a bearing of 1200 from Mikai. Mt. Chaka is 1200m on a bearing of 0300 from Kembo.
- Draw a sketch showing the position of the three mountains (1 mk)
- Calculate the distance of Mt. Chaka from Mt. Mikai (2 mks)
2. Shopping centres XY and Z are such that Y is 12km south of X and Z is 15kn from X. Z is on a bearing of N300W from Y. Calculate and give compass bearing of Z from X. (4mks)
3. Four telephone posts PQR and S stand on a level ground such that Q is 28m on a bearing of 0600 from P. R is 20m to the south of Q and S is 16m on a bearing of 1400 from P.
(a) Using a scale of 1cm represent 4m show the relative positions of the posts. (4mks)
(b) Find the distance and bearing of R from S. (3mks)
(c) If the height of post P is 15.6m. on a separate scale drawing, draw a diagram and determine the angle of depression of post R from the top of post P. (3mks)
(Same scale as above)
4. Alice chepchumba on her cycling practice cycled on a bearing of 120o for 5.5km, then on a bearing of 200o for 8km finally he turned northwards for 13.5km, by scale drawing determine her final position from starting point. (4 marks)
5. A surveyor recorded the measurement of field in a field book using lines AB = 260m as shown below.
B | ||
130 | R40 | |
70 | Q10 | |
50 | P20 | |
S50 | 10 | |
A |
a) Use a suitable scale to draw the map of the field. (2 marks)
b) Find the area of the field. (2 marks)
6. (a) In a Safari rally drivers are to follow route ABCGA. B is 250km from A on a bearing of 0750
from A. C is on a bearing of 1100 from A and 280km from B. the bearing of C from D is 1400 and at a distance of 300km. By scale drawing, show the position of the point A, B, C and D. (4 mks)
(b) Determine
(i) Distance of A from C (2 mks)
(ii) The bearing of B from C (1 mk)
(iii) The distance and bearing of A from D (3 mks)
7. Town X is 20km in the direction 060o from Y and Z is 30km in the direction 150o from Y. Using the scale 1cm represents 5km, find by scale drawing;
(a) the bearing of Y from Z
(b) the distance of X from Z (4mks)
8. A field was surveyed and its measurements recorded in a field book as shown below.
E 40
C 40 | F 100 80 60 40 20 A |
D 50
B 30 |
(a) Using a scale of 1cm to represent 10m, draw a map of the field. (4mks)
(b) Calculate the area of the field.
(i) in square metres. (4mks)
(ii) in hectares. (2mks)
9. A plane leaves town P to town Q on a bearing of 1300 and a distance of 350km. it then flies to town R 500km away and on a bearing 0600. Find by scale drawing the distance of R from P (3mks)
10. A surveyor recorded the following information in his field book after taking measurements in metres of a plot. The baseline is the straight line AH = 300m.
40 to F 120 to D | H 250 200 180 100 A |
100 to G
80 to C 60 to B |
(a) Using a scale of 1cm to represent 20m, draw an accurate diagram of the plot. (5mks)
b) Use your diagram to calculate the actual area of the field in hectares (5mks)
11. Three town P,Q and R are such that P is on a bearing of 120° and 20 km from Q. Town R is on bearing of 220o and 12km from P
a) Using a scale of 1 cm to 2 km, draw and locate the positions of the three towns. (3mks)
b) Measure
i) the distance between Q and R in kilometres. (2mks)
ii) the bearing of P from R. (1mk)
iii) the bearing of R from Q. (2mks)
c) Calculate the area of the figure bounded by PQR. (2mks)
12. The area of a forest on a map whose scale is 1:50,000 is 17cm2. Calculate the area of the forest in hectares. (2 mks)
13. Four towns P, Q, R and S are such that town Q is 120km due east of town P. Town R is 160km due North of town Q. Town S is on a bearing of 330o from P and on a bearing 300 o from R. use a ruler and a pair of compasses only for all your constructions.
a) Using a scale of 1cm to represent 50km, construct a scale drawing showing the positions P, Q, R and S. (6 mks)
b) Use the scale to determine
- The distance from town S to town P. (1 mk)
- The distance from town S to town R. (1 mk)
- The bearing of town S from town Q. (2 mks)
14. The actual area of an estate is 3510 hectares. The estate is represented by a rectangle measuring 2.6cm by 1.5cm on the map whose scale is l:n. Find the value of n (3 mks)
15. The following measurements were recorded in a field book of a farm in metres (xy = 400m)
C60
B 100 A 120 | y 400 340 300 240 220 140 80 x |
120 D 100 E 160 F |
a) Using a scale of 1cm representing 4000 cm, draw an accurate map of the farm.
b) If the farm is on sale at Kshs.80,000.00 per hectare, find how much it costs. (10 mks)
16. Four points A, B, C and D are situated on a horizontal plane such that B is 250 m on a bearing of 0700 from A. C is 325 m on a bearing of 1500 from B. D is due west of C and on a bearing of 2100 from B. (6 marks)
- Using a scale of 1 cm to 50 m draw an accurate drawing to show the position of A, B, C and D.
- Use your scale drawing fo find the :
- The distance between A and D (2 marks)
- The bearing of A from D (2 marks)
17. Town X is 13.5km from town Y on a bearing of 028o. A matatu leaves y at 7:35a.m
towards a bearing of 080o. The matatu is at point Z due south of X at 8:55a.m
(a) Calculate the average speed of the matatu from Y to Z
(b) If the matatu continues on the same bearing, calculate the distance it covers from Z
when it is East of X
18. Three towns X, Y and Z are such that Y is 500km on a bearing of 315o from X. Z is on
a bearing of 230o from X. given that the distance between Y and Z is 800km.
(a) using a scale of 1cm to represent 100km, draw a scale diagram to show the position
of the Towns
(b) Find the bearing of;
(i) X from Z
(ii) Z from Y
(c) Use the scale drawing to find the distance from X to Z
19. Two aeroplanes S and R leave an airport at the same time. S flies on the bearing of 240o
at 750Km/h while R flies due East at 600Km/hr..
(a) (i) Calculate the distance of each aeroplane after 30minutes
(ii) Using a scale of 1cm to represent 50km make an accurate scale drawing to show
the positions of the aeroplanes after 30minutes
(b) (i) Use the scale drawing to find the distance between the two aeroplanes after 30minutes
(ii) If each aeroplane landed after 30minutes and S received a signal to join R in 45minutes.
Find its speed
(c) Determine the bearing of :
(i) S from R
(ii) R from S
20. The table below gives a field book showing the results of a survey of a section of a piece of land
between A and E. All measurements are in metres.
D33
C21 B 42
| E 95 90 70 30 25 A |
F 36
G 25 H 40 |
(a) Draw a sketch of the land.
(b) Calculate the area of this piece of land.
21. Three towns A B and C are situated such that town A is 40km from B on a bearing of 280o.
C is 60km from B on a bearing of 130o. Another town D is only 10km from C on a bearing of 210o.
(a) Drawing accurately and using a scale of 1cm to 10km find the:-
(b) Distance from A to C and the bearing of A from C
(c) (i) Distance of B from D
(ii) Distance of A from D
(iii) Bearing of A from D
(iv) Bearing of C from D
22. A train left Naivasha for Nakuru at 1000hours. It traveled at an average speed of 45km/h
and reached Gilgil after 40minutes. It then covered the remaining 50km in 1½ hours. A second
train left Nakuru for Naivasha at 1015 hours and arrived at Gilgil at the same time as the first
train arrived at Nakuru.
a) Using a scale of 1cm to represent 10minutes in the time axis and 1cm to represent 10km
on the distance axis, draw on the same axes the graphs to show the movement of the two
trains
b) use your graph to find;
i) the distance between Naivasha and Nakuru
ii) the time at which the train met
c) calculate the average speed, in km/h of the second train
23. On a certain map, a road 20km long is represented by a line 4cm long. Calculate the area
of a rectangular plot represented by dimensions 2.4cm by 1.5cm on this map – leaving
your answer in hectares
24. A port B is on a bearings of 080o from a port A and at a distance of 95km. a submarine is
stationed at a port D, which is on a bearing of 200o from A, and a distance of 124km from B.
A ship leaves B and moves directly southwards to an island P, which is on a bearing of 140o
from A. the submarine at D on realizing that the ship was heading for the island P, decides to
head straight for the island to intercept the ship.
(a) Using a scale of 1cm to represent 10km draw a diagram to show the positions of A,B,D, and P
(b) Hence;
Determine
(i) the distance from A to D
(ii) the bearing of the submarine from the ship when the ship was setting off from B
(iii) the bearing of the island P from D
(iv) the distance the submarine had to cover to reach the island P
25. Use a scale of 1cm represents 50km in these questions. Five towns A, B, C, D and E are
situated such that A is 200 km from B on a bearing of 050° from E. C is 300 km from B on
a bearing of 150° from B. D is 350km on a bearing of 240° from C. E is 200km from D and the bearing of D from E is 100°
a) Draw the diagram representing the positions of the towns
b) From the diagram, determine;
i) The distance in km of A from E
ii) The bearing of D from B
26. Four towns P, Q, R & S are such that P is 280 km North of R, S is190 km from R on a
bearing of 310o and Q is 240 km from P on a bearing of 105o.
a) Using scale of 1 cm rep. 50 km, locate the four towns.
b) Find; (i) distance SQ.
(ii) Bearing of S from Q.
(iii) The shortest distance between P and side QR.
27. Four ships are at sea such that a streamliner S is 150km on a bearing of 025° from a cargo
ship C. A trawler T is 300km on a bearing of 145° from the cargo ship and a yacht Y is due
West of C and on a bearing of 300° from T.
a) Using a scale of 1cm= 50km, draw on accurate scale drawing showing the positions of S, C, T
and Y
b) By measurement from your scale drawing determine:
i) The distance and bearing of Y from S
ii) The distance ST
iii) The distance YT
28. A tea farm in Kakamega forest was surveyed and the results were recorded in the surveyors
note book as shown below. The measurements are in meters
250 | Y | |
C80
A60 | 240 170 70 50 | D70
B60 |
X | 0 |
Using a scale of 1: 25, draw the map of the plot and hence calculate the area of the plot in Hectares
29. The information below shows the entries in a surveyor’s field book after a survey of a farm.
XY = 280m is the baseline. All measurements are in metres
280 | Y | |
B 105
A 100 | 230 190 160 90 40 | 110E
45E
95G |
X | O |
(a) Use a scale of 1cm represents 20m to draw the map of the farm
(b) Estimate the area of the farm in hectares
(c) If the point Y lies due north of X, find correct to 1 decimal place, the :
(i) Bearing of E from X
(ii) Distance of E from X
30. The measurements of a flower garden were recorded in a surveyor’s field book as shown.
| 250 | Y |
C80 | 240 170 70 | D 70
B 60 |
X | 0 |
Draw a sketch of the field and find its area. (Measurements are in m)
31. A map has a scale 1:40,000:
(a) Calculate the distance between two points on the ground if the corresponding distance
shown on the map is 3.25cm
(b) Calculate the area in the map of woodland which occupies 36ha on the ground
32. Three scouts John, Peter and Samwel stand on three adjacent peaks of equal altitude
on mountain range. The distance between John and Peter is 800metres and the bearing
of Peter from John is 020o. The distance between John and Samwel is 1500metres, and the
bearing of Samwel from John is 320o.
(a) Calculate the bearing of John from Peter
(b) Calculate:- (i)
the distance
(ii) the bearing of Samwel from Peter
33. The figure below represents a surveyor’s sketch of a plot of land. Calculate the area of the plot in
square metres given that XY = 50m, XK = 20m, XM = 25m, XL = 35m, KA = 40m, MD = 38m
and LB = YC = 60m.
34. Two boats P and Q are located 30km apart; P being due North of Q. An observer at P
spots a ship whose bearing he finds as S 56oE from Q, the bearing of the same ship is 038o. Calculate the distance of the ship from Q to 2 decimal places
35. A map is drawn to scale of 1:100,000. What area in km², is represented by a rectangle
measuring 4.5cm by 5.4 cm
36. Two places A and B are 900km apart on the earth’s surface. If A is due North of B and
given that the latitude of A is 5oN. Find the latitude of B. (Take radius of the earth to be 6370km)
37. A car starts from rest and build up a speed of 40m/s in 1min 40seconds. It then travels
at this steady speed for 5minutes. Brakes are then applied and the car is brought to rest
in 2minutes.
(a) Draw a velocity-time graph to show the journey
(b) Use your graph to find;
(i) the initial acceleration
(ii) the deceleration when the car is brought to rest
(iii) the distance traveled
38. The diagram below represents two vertical watch-towers AB and CD on a level ground.
P and Q are two points on a straight road BD. The height of the tower AB is 20m and
road BD is 200m
(a) A car moves from B towards D. At point P, the angle of depression of the car from
point A is 11.3o. Calculate the distance BP to 4 significant figures
(b) If the car takes 5 seconds to move from P to Q at an average speed of 36km/.hr. Calculate
the angle of depression of Q from A to 2 decimal places
(c) Given that QC = 50.9m, calculate;
(i) the height of CD in metres to 2 decimal places
(ii) the angle of elevation of A from C to the nearest degree
39. Town B is 180 km on a bearing of 0500 from town A. Another town C is on a bearing of 1100
from town A and on a bearing of 1500 from town B. A fourth town D is 240 km on a bearing of
3200 from A. Without using a scale drawing, calculate to the nearest kilometer.
(a) The distance AC
- The distance CD
Scale drawing Answers
1 | (i)
(ii) MC |
B1
M1
A1 3 |
sketch not on scale | ||
2. | Sin Z = 0.4 = 126.420 Compass bearing N53.580W |
M1
A1 M1 A1 | |||
04 | |||||
3. | Distance of R from S 3.8cm 0.1 3.8 x 4 = 15.2m Bearing of R from S 0680 10
Angle of depression = 330 | B1
B1
B1
B1
B1 B1
B1
B1
B1
B1 | 600 bearing from P and 7cm drawn South of Q and 5cm drawn from P 1400 bearing from P and 4cm drawn Completed diagram.
Award of 3 digits only
Posts P drawn
Position P and R shown and triangle completed
Angle of depression given | ||
10 | |||||
4 | 1cm represent 1km
N D 208o 4KM
120o
5.5km
B 200o
13.5km 8KM
C
Bearing 030o 4km from starting point |
1M
1M
1M
A1 |
Bearing of starting point A
Use of scale correctly and plotting of points
Use of bearing correctly | ||
5 |
P d 4cm B Q C A 1CM A 1cm 4cm 2cm 6cm 3cm B F e 5cm
A = ½ x 5 x 2 = 5cm2 B = ½ X 2(2+ 1) = 3cm2 C = ½ x 6 (1+4) = 15cm2 D = ½ x 3 x 4 = 6cm E= ½ x 5 x 15 = 75/2 = 37.5
f = ½ x 1 x 5 = 2.5 Total = 69cm2
Area = 69 x 1000000 10000
= 690m2
| B1
B1
M1
A1 | Correct scales
Correct drawing | ||
7. | (a) 324o (b) (7.2 x 5)km = 36km |
B1
B1
B1
B1 |
Z accurately located wrt Y
X accurately located wrt Y
Bearing of X from Z
Distance of X from Z | ||
04 | |||||
8. | (a) (b) Area1 = ½ x 2 x 3 = 3cm2 Area2 = ½ x 4(5+3) = 16cm2 Area3 = ½ x 5 x 4 = 10cm2 Area4 = ½ x2x4 = 4cm2 Area5 = 4 x 4 = 16cm2 Area6 = ½ x 4 x 4 = 8cm2 Total area = (3+16+10+4+16+8)cm2 = 57cm2 Actual area = (57×100)m2 = 5700m2 (c) 10,000m2 = 1ha 5700m2 = ? 1 x 5700 10,000 = 0.57ha |
S1
B1
B2
B1
B1
M1
A1
M1
A1 |
Scale
Base line
Offsets (all – offsets) A want B1 for at least 2
(3 areas)
(3 areas)
Addition of all six areas | ||
10 | |||||
9. | B1
B1
B1 | For North line at 600 may be simplified lacation of R
for 700km 10km | |||
03 | |||||
10. | (a) Sin = 8/12 DOC = 41.81 x 2 = 83.620 (b) Area of APCO = (16×20) – ( ½ x 122x sin 83.62) = 320 – 71.15 = 248.45 (c) 83.62 x 22 x 122 360 7 = 105.09cm2 (d) 248.45 – 105.09 = 353.54 | M1 M1 A1 M1 M1 A1 M1
A1 M1 A1 | |||
10 | |||||
11. |
|
B1 B1 B1
M1A1 B1 B2
M1
A1 |
Locating Q Locating P Locating R | ||
10 | |||||
12. |
1:50,000 L.S.F. 1:25,000,000 A.S.F. 17cm2: 425000000 cm2 42500 m2
42500 10,000
= 4.25 ha |
M1
A1 2 |
13. Positions
Q B1
R B1
S B1
Const 300 B1
- B1
Scale B1
b.) i. 7.8 x 50 = 390 km. B1
ii. 7.10 x 50 = 355 km B1
iii. 3200 B2
10
14. | Area = 35100000m2 = 351000000000cm2 Area = 2.6×1.5 = 3.9cm2 Scale = 3.9:351000000000 = 90000000000 n = 9×1010 |
M1
A1 B1 |
finding area
area (actual) | ||
03 |
| ||||
15. |
Areas. Total area = 62000m2 = 62000 = 6.2ha 10000 1ha = 80,000 6.2ha = 80000 x 6.2 1 = ksh 496,000.00 |
B3
M1
M1
M1
M1
B1
M
A1 |
3 for at least 6. 2 for at least 4, 1 for at least 2 | ||
10 | |||||
16 |
|
B1
B1
B1
B1
B1
M1A1 M1A1 B1
|
Locating A
Locating B
Locating C
Locating D
North at D | ||
| 10 |
17. a) YZ = 13.5
Sin 28o sin 100o
Duration of travel = 8:55a.m – 7.35a.m
= 4/3
Speed = 6.436
4/3
= 4.827km/hr
(b) 13.5 = 6.436 + ZQ
Sin 10o Sin 118o
6.436 + ZQ = 13.5 x sin118o= 68.659
ZQ= 68.659-6.436
= 62.223
18. 1cm rep 100km
b) i) 049 1
ii) 190 1
c) 6.7 0.1
670 10
19. a) (i) Distance covered by s
= (750 x ½ )km = 375 km
Distance covered by R
= (600 x ½ ) km = 300 km
(b) (i) Distance between the two aeroplanes
= 12.5 x 50 = 625 + 5 km
(ii) Speed = 625 x 60 km/hr
45
= 833 1/3 km /h
(c) (i) Bearing of S from R = 225o
(ii) The bearing of R from S = 72o
20.
Area A: ½ x 25 (33 + 21) = 675
Area B: ½ x 40 (21 x 42) = 1260
Area C: ½ x 30 x 42 = 630
Area D: ½ x 25 x 40 = 500
Area E: ½ x 5 (40 + 25) = 162.5
Area F: ½ X 60 (25 + 36) = 1830
Area G: ½ x 5 x 36 = 90 √
= 5,147.5m2
21. A to C = 96 ± 1 km
Bearing = 300o
- 62 1km
- 97 1 km
- 304o
- 304o
030o
22. Graph
b) i) 80 km
ii) 11.06a.m
c) Average speed of the 2nd train
Time taken = 80 111/12 = 80 x 12
23
= 41.74km/h
23. L.S.F = 4 = 1
2000000 500000
A.S.F = 1
2 = 1
5 x 105 2.5 x 1011
Area of rectangle = (2.4 x 1.5) cm2
= 3.6cm2
Actual area = 3.6 x 2.5 x 1011 ha
100 x 10000
= 9 x 105
= 900,000ha
24. a) Δ ABD ly constructed
Δ ABP
b) i) AD = 4.5 + 0.1cm
Distance A + D
= 4.5 X 10 = 45km
ii) Bearing of (i) from B
= 241 + 1
iii) Bearing P from D
= 123 = 2
iv) Dp = 12.9 + 0.2 am
Distance D + P = 12.9 X 10
= 129 km
25. a)
b) i) 6.8 + 0.1cm
Distance Ae = 340 + 5 km
ii) 180 + 18 = 198 + 2
26. a)
b) (i) SP = 7.8 x 50 = 390 km + 5 km
(ii) S & Q = 255o
+ 1o
(iii) 4 x 50 = 200 km + 5 km
27. (a) Scale = 50km
Drawing accurately
Lines drawn //
(b)By measurement:
(i) Distance SY = 6.9 x 50 = 345 5km
Bearing Y For S = 360o – 114 = 2461o
(ii) distance ST = 7.9 x 50 = 39.5 5km
- distance YT = 9.8 x 50 = 490 5km
XY = 250m
Area of A = ½ x 50 x 60 = 1500m2
B = ½ x 70 x 60 = 2100m2
C = ½ (60 +80) x 120=11050m2
D = ½ x 80 x 80 = 3200m2
F = ½ x 10 x 70 = 350m2
Total area = 26600m2
Ha = 26600 = 2.66ha
10,000
29.
(b) Total area = area (1) + (2) + (3) + (4) +(5) + (6) + (7)
Area (1)= ½ x 90 x 100 = 4500m2
(2) = (100 + 105)10 = 10250m2
2
(3) = ½ x 90 x 105 = 4725m2
(4) = ½ x 50 x 110 = 2750m2
(5) = ½ x (110 + 45)70 = 5425m2
(6) = (45 + 95) 120 8400m2
2
(7) = ½ x 40 x 95 = 1900m2
Total area = 37,950m2
In hectares = (37950) ha = 3.795ha
10,000
(c) (i) bearing of E from x is 0.25 1o
` (ii) Distance Ex = (12.8 0.1 x 20m) = 256 2m
30. Area A = ½ x 170 x 80 = 6800
B = ½ x 80 x 80 = 3200
C = ½ x 10 x 70 = 350
D = ½ x 170 x 130 = 11050
E = ½ x 70 x 60 = 2100
Total = 23,500 m2
31. (a) L.s.f = 1
40,000
1 = 3.25
40,000 x
x = 130,000cm
(b) A.s.f = 1 2
40,000
1 2 = x
40,000 36,000,000
x = 0.0225cm2
32.
(a) bearing = 180 + 20 = 200o
(b) a2 = 1500 +
a2 = b2 + c2 – 2bc cos A
a2 = 15002 + 8002 – 2 x 1500 x 800cos 60
= 2250000 + 640000 – 1200000
= 1690000
a = 1300m
(c) 1300 = 1500
Sin 60 sin c
1300 sin c = 1500 sin 60
Sin c = 1500 sin 60
1300
= 0.9993
c = 87.79o
c = 87.80
33.
. A of ∆ XYD = ½ x 50 x 38 = 950m2
A of XBCY = ½ (50 + 15) 60
= ½ x 65 x 60
= 1950m2
Total A = (950 + 1950)m2
= 2900m2
34. B1 for 86o
30 = Q5
Sin 86o Sin 56o
QS = 30sin 56o
Sin 86o
= 24.93km
35. 1cm for 100000cm
1cm2 = (100000cm)2
Area = 5.4 x 4.5 x 100000 cm2
= 5.4 x 4.5 x 100000 x 100000Km2
100000 x 100000
= 24.3km2
36. x 22 x 6370 x 2 = 900
360 7
= 900 x 360 x 7
22 x 6370 x 2
= 8.1o
Latitude of B = 8.1o – 5o N
= 3.5o S
37. i) acc = 40 – 20
100 – 50
= 20/50 = 0.4m/s
ii) 20 – 40 = -20
460 – 400 60 = 0.3333 m/s2
iii) Area = ½ (520 + 300) x 40 x 1/1000 = 16.4 km
38. a) Tan 11.3 = 200
x
x = 200
Tan 11.3 = 100.1m
b) (36 x 1000) m/s
60 x 60
D = (10 x 5) 50m Tan = 7.590
< of depression = 7.590
c) i) √ 50.92 – 49.92 = 10.04cm
ii) Tan = 10.04
200
= 2.874°
= 3°
39. a) Make a sketch to show positive of A, B, C and D
Use sine rule in ∆ABC
X
180
x = 180 sin 80o
Sin 80o Sin 40o sin 40o
= 275.8
Hence AC = 276 km
(b) Use the cosine rule in ∆ AD when DAC = 150o
y2 = 2402 + 2762 – 2 x 240 x 276 cos 150o
= 576000 + 76180 – 132 480 (-cos 30o)
= 133776 + 114731 = 248507
y = 248507
= 498.5
Hence CD = 499 km
(c) Using sine rule in ∆ABC we have
BC = 180
Sin 60o sin 40o
BC = 180 sin 60
Sin 40
= 242.5
= 243 km