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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 090 o T. 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 030 o T. 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 240 o T. 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 090 o T and a yacht at a bearing of 150o T. 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 060 o T while the goanna walks at a bearing of 150o T. 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 137 o 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 210o T at 5 knots (1 knot = 1 nautical mile per hour) for 4 hours, what is its new bearing from the lighthouse? Ans wers: Q.1. (i) 180o T, (ii) 270o T, (iii) 225o T, (iv) 315o T, (v) 030o T, (vi) 110o T, (vii) 280o T, (viii) 253o T, (ix) 197o T, (x) 056o T 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) 053o T (iii) 233o T Q.5. (i) 17.32 nautical miles (ii) 180o T (iii) 000o T Q.6. (i) 566m (ii) 015o T (iii) 195o T Q.7. 317o T Q.8. (i) 671 km (ii) 995 km Q.9. 234o T 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 30 o . 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 depressio n 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 60 o . 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) 13 km = 3606metres (ii) 273o 41’T (iii) 093o 41’T 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. Ans wers: Q.1. (i) AB = 12 cm, BAC = 36o 52’, BCA = 53o 8’, Area = 54 cm2 (ii) AC = 9.22 cm, BAC = 49o 24’, BCA = 40o 36’, Area = 21 cm2 (iii) BC = 4.62 cm, AB = 9.24 cm, BCA = 60o , Area = 18.48 cm2 Q.2. (i) ABC = 41o 49’, ACB = 108o 11’, AB = 11.4 cm, Area = 22.8 cm2 (ii) BAC = 48o 45’, ABC = 61o 15’, AC = 9.33 cm, Area = 35.07 cm2 (iii) BAC = 42o 05’, ABC = 87o 55’, AC = 10.44 cm, Area = 27.99 cm2 Q.3. (i) BAC = 38o 56’, ABC = 70o 32’, ACB = 70o 32’, Area = 25.45 cm2 (ii) AC = 5.79 cm, ACB = 88o 57’, BAC = 51o 03’, Area = 20.25 cm2 (iii) AB = 13.76 cm, BAC = 33o 50’, ABC = 16o 10’, Area = 19.15 cm2 SIMILAR FIGURES 1. 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? 2. 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? 3. 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? 4. Explain, in words, how you could determine the height of a telegraph pole. 5. 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 Sun’s image on Ryan’s 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. 6. 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 = Length _ of _ telegraph _ pole _ shadow _ x _ length _ of _ ruler (1metre ) Length _ of _ ruler _ shadow 5. Dia. = 1.506 million km = 1.506 x 106 km Im age _ Area 48 2 64 2 6. Original _ Area 6 1 Ratio = 64:1