Assessment for feedback and a grade

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 °

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

2. Two walls of L x H => 2 x ( 4.2m x 2.4m)= 20.16m 3. The ceiling of W x L => 3.5m x 4.2m= 14.7m ∴ 16.8 + 20.16 + 14.7 = 51.66 square meters  51.7 squaremeters 12. You have been hired to design a box (rectangular prism) with a volume of 24 m 3 . The height of the box must be 2 m. a) Create a table like the one below and fill in the blanks. Use whole numbers for dimensions (6 marks) Length ( m) Width( m) Height (m) Volume( m 3 ) Surface Area( m 2 ¿ 1 12 2 24 76 2 6 2 24 56 3 4 2 24 52 4 3 2 24 52 6 2 2 24 56 12 1 2 24 76 Volume= Length x Width x Height, Width= volume Length x height Surface are for rectangular prism => A= 2lw + 2lh + 2wh 1. Length=1 Width : 24 1 x 2 = 12 Surface area= 2(1x12) + 2(1x2)+2(12x2)=24+4+48=76 2. Length= 2 Width; 24 2 x 2 = 6 Surface are= 2(2x6)+2(2x2)+2(6x2)= 24+8+24=56 3. Length=3 Width: 24 3 x 2 =4 Surface area=2(3x4)+2(3x2)+2(4x2)=24+12+16=52 4. Length=4 Width; 24 4 x 2 =3 Surface area=2(4x3)+2(4x2)=2(3x2)=24+16+12=52 5. Length=6 Width; 24 6 x 2 =2

Surface area=2(6x2)+2(6x2)+2(2x2)=24+24+8=56 6. Length=12 Width; 24 12 x 2 =1 Surface area= 2(12x2)+2(12x1)+2(1x2)=48+24+4=76 b) What dimensions would you use to minimize the costs of making the box? Explain. (2 marks) To minimize the cost of making the box, you typically want to minimize the surface area, as this is directly related to the amount of material used. The cost of making the box is often proportional to the surface are. choose the dimensions with the minimum surface area. Looking at the table, we can see that when: the length is 3, the width is 4, and the height is 2 or, the length is 4, the width is 3, and the height is 2 The surface area is the smallest (52 m²). Therefore, to minimize the costs of making the box, use dimensions of: length 6 meters, width 2 meters, and height 2 meters or, length 4 meters, width 3 meters, and height 2 meters. Task 3: Communication (25 marks) Show all work for each question. 13. Define the following terms. (5 marks) a) Obtuse Angle An obtuse angle is a type of angle that is always larger than 90° but less than 180°. In other words, it lies between 90° and 180°. b) Oblique Triangle An oblique triangle is any triangle that is not a right triangle. It could be an acute triangle (all three angles of the triangle are less than right angles) or it could be an obtuse triangle (one of the three angles is greater than a right angle). c) Angle of Elevation

V= L (Length) x W(Width) x H(Height) a) 2m, 3m, 6m V=2 x 3 x 6= 36 m 3 b) 36m, 1m, 1m V=36x1x1=36 m 3 c)9m, 2m, 2m V=9x2x2=36 m 3 d) 12m,3m,1m V=12x3x1=36 m 3 e) 18m,2m, 1m V=18x2x1=36 m 3 f)4m, 3m, 3m V=4x3x3=36 m 3 16. State sinB, cosB and tanB as ratios (fractions) for the following triangle. (3 marks) Sin θ = opposite hypotenus cos θ = adjacent hypotenuse tan θ = opposite adjacent When angle C is the reference angle, ´ AC = opposite =15, ´ AB =hypotenuse=17, ´ BC =adjacent=8 So, here using formula, ∴ Sin B= 15 17 , cos B = 8 17 , tan B = 15 8 17. The promotion manager of a new band is deciding how much to charge for concert tickets. She has calculated that if tickets are $28 each, then 220 people will come to the concert. For every $1 increase in price, however, 10 fewer people come. (11 marks) Complete the table to investigate the ticket price if the goal is to maximize the revenue. Ticket Number of People Total Money from tickets( revenue)

$22 280 $ 22 x 280 = $ 6160 $23 270 $23 x 270 = $ 6210 $24 260 $24 x 260 = $ 6240 $25 250 $25 x 250 = $ 6250 $26 240 $26 x 240 = $ 6240 $27 230 $27 x230 = $ 6210 $28 220 $28 x 220 = $6160 $29 210 $29 x 210 = $6090 Conclusion: The maximum revenue of $ ____6250__ occurs when the ticket price is $ _25___. As we can see from bottom price reduced by $1 like 29, 28,27 therefore for next column price will be $26, 25,24, 23, 22. Similarly, every $1 increase in price, 10 fewer people come. Therefore, number of people from bottom will be 210, 220, 230, 240, 250, 260, 270, 280. So, the maximum revenue is $ 6250. We can see column 5 where 25 x 250 =$ 6250. Task 4: Thinking (25 marks) Show all work for each question. 18. The first floor of a house is shaped like a rectangular prism and the second floor is shaped like a triangular prism. The dimensions are shown on the following diagram. If a 1000 BTU air conditioner can cool a house with a volume of air up to 1000 m 3 , can this air conditioner cool the house sufficiently? (5 marks) Volume of total shape = volume of cuboid + volume of prism. Volume of cuboid= l x w x h ( l : length, w: width, h: height)

Lateral surface area of two cube= 4 x 1600= 6400 Top surface of cube Area of square – area of circle( π r 2 ) = (40 inch x 40 inch) – π 4 2 = 1600 – 50.265= 1549.734 ∴ surface area giving off heat = 753.982+ 6400+1549.734=8703.716 20. You are planning to start a home business. You decide to build a workroom in the shape of a rectangular prism the backyard. It will give you solitude to work and a place for customer to see your products without disturbing your family in the house. You price the materials needed and decide that the perimeter of the floor can be 64 ft in total. Maximizing the floor area inside the workshop is your only objective. What dimensions would you choose for the floor, rounded to the nearest whole numbers? You may know the answer already but provide a fully documented algebraic solution. (5 marks) Length= l , width=w , from the question given that, Perimeter P = 64ft P= 2 l +2w= 2( l +w) => 2( l +w)=64 l +w= 32 w= 32- l the area of rectangle is => A= l x w A= l (32- l ) = 32 l - l 2 Maximum area is at the point at which d dl (A) = 0 d dl (A) = d dl (32 l - l 2 )= 0 32-2 l =0 2 l =32 l =16 ft width= 32- l =16ft Length= 16ft

SO, the dimension for the floor is 16 ft x 16 ft 21. A store wants to mount a security camera to record video of the door and the cash register. The best non-moving cameras can be equipped with lenses that will cover a specific angle--an angle of up to 180º. The camera will be mounted 30 ft from the door and 20 ft from the cash register. If the door is 15 ft from the cash register, what angle must the camera cover to watch both sites? Include a diagram. (5 marks) Given, door to camera: 30 ft, camera to cash register: 20ft, door to cash register: 15 ft. using the law of cosines to find the camera cover to watch both site, cosine law => a 2 = b 2 + c 2 -2bccosA cos A= b 2 + c 2 − a 2 2 bc => cos A = 20 2 + 30 2 − 15 2 2 x 20 x 30 = 400 + 900 − 225 1200 = 1075 1200 (Camera) A ( 30 ft) c (20ft) b (15ft) a (Cash register) C (door) B