Pearson Foundations and Pre-calculus Mathematics 10

Transcription

1 Cumulative Review Chapters and (pages 0-) Chapter. a) Since yd. = ft., to convert feet to yards, divide by ft. yd. 70 ft. yd. 70 ft. yd. ft. b) The perimeter of the pen is greater than yd., so Andrea must buy 4 yd. of fencing material. The cost is: 4($.49) $59.76 Before taxes, the material will cost $ The map scale is cm represents cm. 6.5 cm represents 6.5( cm) = cm Divide by 00 to convert cm to metres: Divide by 000 to convert m to kilometres: The distance between Edmonton and Calgary is approximately 76 km.. Sample response: I would use calipers. To measure the radius of the pipe in imperial units, I would use calipers that showed the measure in inches. To measure the radius of the pipe in SI units, I would use calipers that showed the measure in centimetres. The calipers will measure the diameter, so I divide by to determine the radius. Cumulative Review Chapters and Copyright 0 Pearson Canada Inc.

2 4. Answers will vary, depending on the conversion ratios used. a) yd cm 9 yd. = 9(9.44 cm) 8.96 cm A length of 9 yd. is approximately 8 cm. b) in..54 cm So, in (.54 cm) cm Divide by 00 to convert to metres m m A length of in. is approximately m. c) 6 km = mi. 0 6 So, 5 km 5 mi. 0 0 mi. 0 mi. A length of 5 km is approximately mi. d).54 cm = in. 60 So, 60 cm in in. Divide by to convert to feet cm = ft. = 5 ft. 5ft.in. A length of 60 cm is approximately 5 ft. in. Cumulative Review Chapters and Copyright 0 Pearson Canada Inc.

3 5. Answers will vary, depending on the conversion ratios used. Calculate an exact conversion. Convert the height of the road above the Fraser River to inches. Use the conversion:.54 cm = in. So, 54 m 54(00 cm) = cm cm in cm in cm ft cm ft. The measurement is less than 50 ft., so the Tacoma Narrows Bridge is higher above the water. The difference in heights is: 50 ft ft ft. The Tacoma Narrows Bridge is higher above the water by approximately 5 ft. Cumulative Review Chapters and Copyright 0 Pearson Canada Inc.

4 6. a) The regular tetrahedron has 4 congruent faces. Each face is a triangle with base 4.00 m and height. m. The area, A, of each face is: A (4.00 m)(. m) The surface area, SA, is: SA 4 (4.00 m)(. m) SA 4.88 m The surface area of the regular tetrahedron is approximately 4 m. b) Determine the slant height. The radius, r, of the cone is the diameter. r (7 ft.) 7 r ft., or ft. Sketch a diagram. Let s represent the slant height. Visualize cutting the cone in half through a diameter of its base. This produces an isosceles triangle with a base that is equal to the diameter of the cone and a height that is equal to the height of the cone. Use the Pythagorean Theorem in right ACD. s 5.5 s 5.5 s 7.5 s 7.5 Use the formula for the surface area of a right cone: SA rs r Substitute: r.5 and s 7.5 SA (.5) 7.5 (.5) SA The surface area of the cone is approximately 08 square feet. Cumulative Review Chapters and Copyright 0 Pearson Canada Inc. 4

5 7. Use the formula for the volume of a cone. V r h Substitute: r.5 and h5 (.5) V (5) V 9.45 The volume of the cone is approximately 9 cubic feet. 8. The formula for the volume of a cone is: V r h The radius of the cone is the diameter of the cone: yd. = 6 yd. So, substitute V = 4 and r = 6 into the formula V r h, then solve for h to determine the height of the cone. Then give the answer to the nearest whole number of yards. 4 (6) h ()(4) h (6) h The height of the cone is approximately 6 yd. 9. Determine the volume and surface area of the first pyramid. Sketch the pyramid and label its vertices. The pyramid has a square base, so the 4 triangular faces of the pyramid are congruent. Sketch and label the right triangle formed by the height, slant height, and base of the pyramid. Use the formula for the volume of a pyramid. V lwh Substitute: l 0, w0, and h8 V (0)(0)(8) V In ΔEFG, FG is the length of DC, so FG is 5 cm. Use the Pythagorean Theorem in right ΔEFG to determine the slant height, EG. EG EF FG EG 8 5 EG 89 EG 89 Area, A, of ΔBCE is: Cumulative Review Chapters and Copyright 0 Pearson Canada Inc. 5

6 A (0) 89 A 5 89 Since the triangular faces of the pyramid are congruent, the lateral area, A L, of the pyramid is: A L 4(5) 89 AL 0 89 Area, B, of the base of the pyramid is: B (0)(0) B 00 Surface area, SA, of the pyramid is: SA SA The volume of the first pyramid is approximately 67 cm and its surface area is approximately 89 cm. Sketch the pyramid and label its vertices. The pyramid has a square base, so the 4 triangular faces of the pyramid are congruent. Sketch and label the right triangle formed by the height, slant height, and base of the pyramid. Determine the volume and surface area of the second pyramid. Use the formula for the volume of a pyramid. V lwh Substitute: l 8, w8, and h V (8)(8)() V 77. In ΔQRS, RS is the length of PN, so RS is 4 cm. Use the Pythagorean Theorem in right ΔQRS to determine the slant height, QS. QS QR RS QS 4 QS 85 QS 85 Area, A, of ΔMNQ is: A (8) 85 A 4 85 Cumulative Review Chapters and Copyright 0 Pearson Canada Inc. 6

7 Since the triangular faces of the pyramid are congruent, the lateral area, A L, of the pyramid is: A L 4(4) 85 AL 6 85 Area, B, of the base of the pyramid is: B (8)(8) B 64 Surface area, SA, of the pyramid is: SA SA 8.65 The volume of the second pyramid is approximately 77 cm and its surface area is approximately 8 cm. The second pyramid has the greater volume, but the lesser surface area. So, the pyramid with the greater volume does not have the greater surface area. 0. a) Determine the surface area of the hemisphere. SA of hemisphere = SA of one-half a sphere + area of a circle SA (4 r ) r SA r r SA r Substitute: r 0 SA (0) SA The surface area of the hemisphere is approximately 770 square inches. Determine the surface area of the sphere. Use the formula for the surface area of a sphere. SA 4r Substitute: r 7 SA 4 (7) SA 6.68 The surface area of the sphere is approximately 6 square inches. Difference in surface areas: The hemisphere has the greater surface area, by approximately 8 square inches. b) Determine the volume of the hemisphere. V of hemisphere = V of one-half a sphere Cumulative Review Chapters and Copyright 0 Pearson Canada Inc. 7

8 4 V r V r Substitute: r 0 (0) V V The volume of the hemisphere is approximately cubic inches. Determine the volume of the sphere. Use the formula for the volume of a sphere. 4 V r Substitute: r 7 4 (7) V V The volume of the sphere is approximately cubic inches. Difference in volumes: The sphere has the greater volume, by approximately 84 cubic inches.. Surface area of composite object is: Lateral area of pyramid + surface area of right square pyramid overlap The lateral area of the pyramid is the sum of the areas of its triangular faces. (5.0)(5.8) (6.0)(5.6) 6.6 The surface area of the right rectangular prism is the sum of the areas of its rectangular faces. (5.0)(.0) (5.0)(8.0) (8.0)(.0) 58.0 The overlap is the base of the right rectangular pyramid. (6.0)(5.0) 0.0 Surface area of the composite object is: SA = SA = 90.6 The surface area of the composite object is approximately 9 m. The volume of the composite object is: Volume of right rectangular pyramid + volume of right rectangular prism Cumulative Review Chapters and Copyright 0 Pearson Canada Inc. 8

9 Sketch the pyramid and label its vertices. FG is the length of DC, or.0 m. Use the Pythagorean Theorem in right EFG to determine the height of the pyramid, EF. EF FG EG EF EF EF 4.64 EF 4.64 Use the formula for the volume of a right rectangular pyramid. V lwh Substitute: l 6.0, w5.0, h 4.64 V (6.0)(5.0)( 4.64) V Use the formula for the volume of a right rectangular prism. V lwh Substitute: l 8.0, w5.0, h.0 V (8.0)(5.0)(.0) V 0 Volume of composite object: = The volume of the composite object is approximately 70 m. Cumulative Review Chapters and Copyright 0 Pearson Canada Inc. 9

10 . Sketch and label the pyramid. FG is the length of DC, or 4 in. Use the Pythagorean Theorem in right EFG to determine the slant height, EG. EG EF FG EG 40 4 EG 76 EG 76 The lateral area of the pyramid is the sum of the areas of its 4 triangular faces. 4 (48)( 76) The lateral area of the pyramid is approximately 4478 square inches.. Let r represent the radius of the base of the hemisphere. Use the formula for the circumference of a circle. C r Substitute: C r 0.5 r r SA of a hemisphere = SA of one-half a sphere + area of a circle SA (4 r ) r SA r r SA r r Substitute: SA (4.854 ) SA.0808 Volume of a hemisphere = volume of one-half a sphere Cumulative Review Chapters and Copyright 0 Pearson Canada Inc. 0

11 4 V r V r Substitute: r (4.854 ) V V 9.56 The surface area of the hemisphere is approximately. mm and its volume is approximately 9.6 mm. Chapter 4. a) Use the tangent ratio. tan A adjacent Atan b) Use the tangent ratio. tan A adjacent Atan Cumulative Review Chapters and Copyright 0 Pearson Canada Inc.

12 5. Sketch and label a diagram to represent the information in the problem. In right ABC, AC is B and BC is adjacent to B. tan B adjacent AC tan B BC b tan 5 Solve for b. 0 0tan 5b b The height of the tree is approximately 6 yd. 6. Sketch and label a diagram to represent the information in the problem. Let c represent the vertical distance travelled. In right CDE, DE is C and CE is adjacent to C. tan C adjacent DE tan C CE c tan 6.4 Solve for c tan 6.4c c The vertical distance Jay Cochrane travelled was approximately 0 ft. Cumulative Review Chapters and Copyright 0 Pearson Canada Inc.

13 7. a) In right WXY, WX is Y and WY is the hypotenuse. sin Y hypotenuse WX sin Y WY 5 sin Y 7 Y Y is approximately 6.9. b) In right CDE, CE is adjacent to E and DE is the hypotenuse. adjacent cos E hypotenuse CE cos E DE 7 cos E 9 E E is approximately Sketch and label a diagram to represent the information in the problem. The angle between the ladder and the wall is Q. In right PQR, PR is Q and PQ is the hypotenuse. sin Q hypotenuse PR sin Q PQ 4.5 sin Q Q.04 The angle between the ladder and the wall is approximately. Cumulative Review Chapters and Copyright 0 Pearson Canada Inc.

14 9. Sketch and label a diagram to represent the information in the problem. In right ACD, CD is A and AC is the hypotenuse. sin A hypotenuse CD sin A AC CD sin sin 8CD CD In right ACD, AD is adjacent to A and AC is the hypotenuse. adjacent cos A hypotenuse AD cos A AC AD cos cos8AD AD The basketball court is approximately 50 ft. by 94 ft. Cumulative Review Chapters and Copyright 0 Pearson Canada Inc. 4

15 0. a) Determine the measure of S first. S90T Determine the length of RS. sin T hypotenuse RS sin T ST RS sin sin 66RS RS Determine the length of RT. adjacent cost hypotenuse RT cost ST RT cos cos66RT RT S is 4, RS is approximately 4.4 m, and RT is approximately 6.4 m. Cumulative Review Chapters and Copyright 0 Pearson Canada Inc. 5

16 b) Determine the measure of M first. S90T Determine the length of MN. tan P adjacent MN tan P NP MN tan tan 44MN MN Determine the length of MP. Use the Pythagorean Theorem. MP NP MN MP 7.4 (7.460 ) MP M is 46, MN is approximately 7. cm, and MP is approximately 0. cm.. Sketch and label a diagram to represent the information in the problem. The required angle is B. In right ABC, AC is B and BC is adjacent to B. tan B adjacent AC tan B BC 5 tan B 9 B The path the helicopter will take to fly directly to its base is at an angle of approximately 59 to the path it took flying south. Cumulative Review Chapters and Copyright 0 Pearson Canada Inc. 6

17 . Use the Pythagorean Theorem in right PQR. PR PQ RQ RQ PR PQ x x SQ SR RQ 4 cm 0 cm 4 cm Use the Pythagorean Theorem in right PQS. PS PQ SQ y y Determine the measure of R in right PQR. sin R hypotenuse PQ sin R PR sin R 9 R Determine the measure of P in right PQR. P90R Determine the measure of S in right PQS. tans adjacent PQ tans SQ tans 4 S.704 Cumulative Review Chapters and Copyright 0 Pearson Canada Inc. 7

18 Determine the measure of P in right PQS. P90S QPR RPS QPS RPS QPS QPR RPS RPS PRS PRQ QRS PRS QRS PRQ PRS x is 0.0 cm, y is approximately 40.0 cm, QPR is approximately 4.6, RPS is approximately 4.7, QPS is approximately 58., PRQ is approximately 46.4, PRS is approximately.6, and S is approximately.7. Cumulative Review Chapters and Copyright 0 Pearson Canada Inc. 8