Measurement
Worksheets
VOLUMES & AREAS
Calculate the
volume and surface area of the following figures. They are not drawn to scale.
1.
2.
3.
4
5. 
6.
7.
8. 
9.
10.
11.
12.
Answers:
1. V = 33.5 cm3,
A = 50.3 cm2; 2. V =
64 cm3, A = 96 cm2;
3. V = 226.2 cm3,
A = 207.3 cm2 4. V =
268.1 cm3, A = 201.1 cm2
5. V = 216 cm3,
A = 216 cm2 6. V =
502.7 cm3, A = 351.9 cm2
7. V = 24 m3,
A = 68 m2 8.
V = 8640 m3, A = 2641.3 m2
9. V = 589.0 m3,
A = 464.2 m2 10. V =
18000 m3, A = 4342.2 m2
11. V = 1250 m3,
A = 848.6 m2 12. V =
1767.1 m3, A = 947.9 m2
TRIGONOMETRY: BEARINGS
Bearings are indicated as an angle measured clockwise from north and are
given in standard 3-figure notation. The letter T is usually written after the
angle to indicate the direction from True north.
Hence the direction east is written as 090oT.
Q.1. Write down the bearings of
the following directions:
(i)
South (ii) west (iii) south-west (iv) north-west
(v) 30o east
of north (vi)
20o south of east (vii) 10o
north of west
(viii) 17o
south of west (ix) 17o
west of south (x) 34o north of
east
Q.2. A boat is 10 nautical miles from a
lighthouse at a bearing 030oT. How far is the boat (i) east of the lighthouse
(ii) north of the lighthouse?
Q.3. A boat is 20 nautical miles from a
lighthouse at a bearing 240oT. How far is the boat (i) west of the lighthouse
(ii) south of the lighthouse?
Q.4. A car and a utility leave the same town. The
car is driven 150 km north while the utility is driven 200 km west. When they have
reached their destination,
(i) how
far is the car from the utility?
(ii) what is the bearing of
the car from the utility? (Answer to nearest degree)
(iii) what is the bearing
of the utility from the car? (Answer to nearest degree)
Q.5. A lighthouse keeper sees a launch at a
bearing of 090oT and a yacht at a bearing of 150oT. If
the launch is 15 nautical miles due north of the yacht,
(i) how
far is the yacht from the lighthouse?
(ii) what
is the bearing of the yacht from the launch?
(iii) what
is the bearing of the launch from the yacht?
Q.6. A bandicoot and a goanna are released into
the wild from the same place. The bandicoot walks at a bearing of 060oT
while the goanna walks at a bearing of 150oT. When they have both
walked the same distance from their release point they are 800m apart.
(i) how far has each of them walked? (to
nearest metre)
(ii) what
is the bearing of the bandicoot from the goanna? (to
nearest degree)
(iii) what
is the bearing of the goanna from the bandicoot? (to
nearest degree)
Q.7. Tom the cat is chasing Jerry the mouse. If
Jerry is at a bearing of 137o from Tom, What is the bearing of Tom
from Jerry?
Q.8. A plane flies at a bearing of 214o
at a speed of 400 km/h for 3 hours. How far is the plane (i)
west of its starting point (ii) south of its starting point (to nearest km)
Q.9. A boat is 10 nautical miles due north of a
lighthouse. If it sails at a bearing of 210oT at 5 knots (1 knot = 1
nautical mile per hour) for 4 hours, what is its new bearing from the
lighthouse?
Answers:
Q.1. (i)
180oT, (ii) 270oT,
(iii) 225oT, (iv) 315oT, (v) 030oT, (vi) 110oT,
(vii) 280oT, (viii) 253oT, (ix) 197oT,
(x) 056oT
Q.2. (i)
5 nautical miles (ii) 8.66 nautical
miles
Q.3. (i)
17.32 nautical miles (ii) 10 nautical miles
Q.4. (i)
250 km (ii) 053oT (iii) 233oT
Q.5. (i)
17.32 nautical miles (ii) 180oT (iii) 000oT
Q.6. (i)
566m (ii) 015oT (iii) 195oT
Q.7. 317oT
Q.8. (i)
671 km (ii) 995 km
Q.9. 234oT
TRIGONOMETRY: ANGLES of ELEVATION & DEPRESSION; BEARINGS
Q.1. Kulsoom was
standing on the top of a coastal cliff and looking at Laura who was in a boat
below. The angle of depression of Laura from Kulsoom
was 30o.
The cliff was measured as being 100 metres high.
(i)
What
was the angle of elevation of Kulsoom from Laura?
(ii)
Draw
a diagram of the above situation and clearly label the angles of elevation and
depression.
(iii)
How
far was Laura from the base of the cliff?
Q.2. Kulsoom was on
the 100 metres high cliff and observed a buoy, 40 metres from the base of the
cliff. What was the angle of depression of the buoy?
Q.3. Laura then observed Kulsoom
at the top of a different cliff. Laura noted that the angle of elevation was 40o.
She rowed her boat 50 metres towards the cliff and noted that the angle of
elevation was 60o. What was the height of the cliff?
Q.4. Laura wanted to determine the height of a
tree. She measured 50 metres from the base of the tree and used a sight tube at
that point to determine the angle of elevation as 20o. How tall was
the tree?
Q.5. A ball is rolling in a direction 30o
south of east. What is the ball bearing?
Q.6. Laura and Kulsoom
were in opposing teams for an orienteering exercise. Both started from the same
point at the same time. Laura jogged for 20 minutes at 6.0 km/h at a bearing of
150o. Kulsoom walked for 45 minutes at 4.0
km/h at a bearing of 240o. Both then sat down to rest.
(i)
How
far apart were Laura and Kulsoom when they were
resting?
(ii)
What
was the bearing of Kulsoom from Laura?
(iii)
What
was the bearing of Laura from Kulsoom?
ANSWERS
Q.1. (i) 30o
(ii) 
(iii) 173.2 metres
Q.2. 68o 12
Q.3. 81.38 metres
Q.4.
18.2 metres
Q.5. 120To
Q.6. (i) 
km
= 3606metres
(ii) 273o41T
(iii) 093o41T
TRIGONOMETRY: SINE &
COSINE RULE
Q.1. (i) Find the length of the sides and the values of the
angles in the following triangles where such measurements are not already
shown.
(ii) Find the area of each triangle.
Q.2. (i) Find the
length of the sides and the values of the angles in the following triangles
where such measurements are not already shown.
(ii) Find the area of each triangle.
Q.3. (i) Find the
length of the sides and the values of the angles in the following triangles
where such measurements are not already shown.
(ii) Find the area of each triangle.
Answers:
Q.1. (i) AB = 12
cm, 
BAC
= 36o52, BCA
= 53o8, Area = 54 cm2
(ii) AC = 9.22 cm, BAC
= 49o24, BCA
= 40o36, Area = 21 cm2
(iii) BC = 4.62 cm, AB = 9.24 cm, BCA
= 60o, Area = 18.48 cm2
Q.2. (i) ABC
= 41o49, ACB
= 108o11, AB = 11.4 cm, Area = 22.8 cm2
(ii) BAC
= 48o45, ABC
= 61o15, AC = 9.33 cm, Area = 35.07 cm2
(iii) BAC
= 42o05, ABC
= 87o55, AC = 10.44 cm, Area = 27.99 cm2
Q.3. (i) BAC
= 38o56, ABC
= 70o32, ACB
= 70o 32, Area = 25.45 cm2
(ii) AC = 5.79 cm, ACB
= 88o57, BAC
= 51o03, Area = 20.25 cm2
(iii) AB = 13.76 cm, BAC
= 33o50, ABC
= 16o10, Area = 19.15 cm2
SIMILAR
FIGURES
- A
tree casts a shadow of 6.4 metres when a 3.0 metre shadow stick casts a shadow of 1.6 metres. How tall is the tree?
- Later
in the day, the stick in Q.1. cast a shadow of
2.1 metres. What would be the length of the
shadow cast by the tree at this time?
- A
student who is 160 cm tall casts a shadow of 40 cm at the same time as the
school building casts a shadow of 4.6 metres.
How tall is the school building?
- Explain,
in words, how you could determine the height of a telegraph pole.
- Josh
and Ryan wanted to find the diameter of the Sun. They knew that the Sun is
150 million km from Earth. Josh poked a pinhole in a piece of cardboard
while Ryan drew two lines 0.5 cm apart on a page of his exercise book.
Ryan held a metre rule above his exercise book
with the zero mark just next to the lines he drew on his book. Josh moved
the cardboard with a pinhole up and down next to the ruler and focused the
Suns image on Ryans book. When the image was focused on the lines and
was exactly 0.5 cm in diameter Josh measured the distance between the
cardboard and the book. It was 49.8 cm. Draw a diagram showing the two
similar triangles and calculate the diameter of the Sun.
- A
transparency with a square of side 6.0 cm was placed on an overhead
projector. A square of side 48 cm was produced on the screen. What was the
ratio of the area of the image to the area of the original square?
Answers:
1. 12m 2. 8.4m 3.
18.4m
4. Measure the length of the shadow cast by the
telegraph pole. Hold a 1 metre ruler vertically on
level ground. Use a second ruler to measure the length of the shadow cast by
the 1 metre ruler.
Height of telegraph pole =
5. 
Dia. = 1.506 million km = 1.506 x 106km
6. 
Ratio = 64:1