Assessment for feedback and a grade.docx

Assessment for feedback and a grade Unit 3 Assessment - Geometry and Trigonometry Task 1: Knowledge and understanding (25 marks) Show all work for each question. 1. Convert each measure. (4 marks) a) 8 ft. 6 in. to the nearest centimetre. 12in= 1ft 8ft x 12= 96 in 8ft= 96 in 8ft. 6 in = 96 in +6 in = 102 in 1 in = 2.54cm 102 in => 102 in X 2.54 = 259.08 cm ∴ 259 cm b) 300 g to the nearest ounce. 1 oz=28.35g 300g X 1 oz 28.35 g = 10.58 oz ∴ 11 oz 2. Determine the surface area of the figure shown below. (5 marks)

SA= A bottom + 2 A side + 2 A end = 2.4x3 + 2x3x2 + 2x2.4x1.6 = 7.2+ 12+ 7.68 = 26.88 3. Solve for x, rounded to one decimal place. (2 marks) cos 59 ∘ = x 36 , cos 59 ° = 0.515 x= cos 59 ° x 36 = 0.515 x 36 = 18.54 = 18.5 4. Solve for x, rounded to one decimal place. (2 marks) Sin 25 ° = 34 x , sin 25 ° = 0.423 X sin 25 ° = 34, X= 34 sin25 ° = 80.37 = 80.4 5. If tan C = − 0 . 6687, find angle C rounded to the nearest degree. Angle C is between 0 ° and 180 ° . (3 marks) tan C = - 0.6687 C= tan − 1 (-0.6887)= - 33.7706 C is between 0 ° and 180 ° = 180 ° - 33.7706 = 146.23 ° ∴ C = 146 °

6. Solve for x in each of the following triangles. (7 marks) a) ∠ BAC = 90 ° , ∠ BCA= 23 ° , AC= 12ft tan 23 ° = AB AC = x 12 (ta n 23 ° = 0.424) → x= 12tan 23 ° = 5.09 ft b) ∠ BAC= 30 ° , ∠ ACB= 75 ° ∴∠ CBA= 180 ° - ( 30 ° + 75 ° ) = 75 °  The sum of the angle in a triangle is always 180 ° AC=x BC= 26 a sin A = b sin B = c sin C -> x sin75 ° = 26 sin30 ° sin 30 ° χ = 26 sin 75 °

χ = 26 sin75 ° sin30 ° = 26 0.966 0.5 = 50.232 ∴ x= 50.23 m c) The cosine law => a 2 = b 2 + c 2 – 2bccosA a= x, b=17.5, c=18.1 x 2 = 17.5 2 + 18.1 2 – 2(17.5x18.1)cos 68 ° x 2 = 306.25 + 327.61 – 633.5cos 68 ° x 2 = 633.86 – 633.5 x 0.3746= 633.86- 237.3091 = 396.5509 χ = √ 396.5509 =19.91 ∴ 19.91 ∈ ¿ 7. Solve for θ in the triangle. (2 marks) ∠ C = θ ∠ A= 136 ° a=27.3 c= 16.8 a sinA = c sinC => 27.3 sin136 ° = 16.8 sinθ 27.3 sin θ = 16.8 sin 136 °

Sin θ = 16.8 27.3 sin 136 ° = 16.8 27.3 x 0.694= 0.4274 θ = sin − 1 0.4274= 25.3 ∴ 25.3 ° Task 2: Applications (25 marks) 8. Brent wants to dig a circular pond surrounded by a flower bed as shown. The pond will have a diameter of 10 ft and the diameter of the pond and flower bed combined will be 14 ft. a) Brent needs to know the area of the pond in order to purchase a cover for the winter. What is the area of the pond, to the nearest square foot? (2 marks) inner diameter of the circle (d)= 10ft, Radius= 5ft outer diameter of the circle (D) = 14 ft , Radius= 7ft Area of the pond = π r 2 => π x 5 2 = 78.539 ft ∴ 79 ft b) Brent needs to know the area of the flower bed so he can plan the layout and order the flowers. What is the area of the flower bed, to the nearest square foot? (4 marks) area of flower bed= total area – area of pond = π 7 2 - π 5 2 = π ( 49 − 25 ) = ¿ 75.398 ∴ 75 ft

9. A car has a fuel consumption rating of 8.3 L/100 km. a) Calculate the fuel consumption rating in gallons per mile, to three decimal places. (3 marks) 1 liter is approximately equal to 0.264172 gallons. 8.3 L 100 km x 0.264172 gallons 1 litter = 2.191076 gallons 100 km Convert 100 km to mile => 1km is approximately equal to 0.621371 miles. ∴ 100 km = ¿ 62.1371 miles 2.191076 gallons 62.1371 miles = 0.03526 gallons per mile ∴ rounded to three decimal places => 0.035 gallons per mile b) How many gallons are used when the car is driven 1 mi? (1 mark) 1 mile x 0.035 gallons mile = 0.035 gallons So, when the car is driven 1 mile, it uses approximately 0.035 gallons of fuel. 0.035 gallons of fuel 10. Calculate the cost of carpeting a rectangular living room that measures 13 ft by 14 ft, at a cost of $4.95/ ft 2 . (3 marks) Area of the room Area = 13 x 14= 182 ft 2 => the cost of the carpet is $4.95 per ft 2 Cost = 182 x 4.95= $ 900. 90 11. A room in the shape of a rectangular prism measuring 3.5 m wide by 4.2 m long by 2.4 m high is to be painted. All surfaces except the floor are to be painted. How many square metrs of surface are to be painted, rounded to one decimal place? (4 marks) Width ( w) = 3.5m, Length (L) = 4.2 m, Height( H) =2.4m The floor’s surface area is W x L, but we’re not painting it. So, the other 5 faces have the flowing areas: 1. Two walls of W x H => 2 x (3.5m x 2.4m)= 16.8m

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