Unit 3 Practice Test. Name: Class: Date: Multiple Choice Identify the choice that best completes the statement or answers the question.

Transcription

1 Name: lass: ate: I: Unit 3 Practice Test Multiple hoice Identify the choice that best completes the statement or answers the question. The radius, diameter, or circumference of a circle is given. Find the missing measures. Round to the nearest hundredth if necessary. Use the diagram to find the measure of the given angle. 1. d = 22.3 km, r =?, =? a. r = 44.6 km, = km b. r = km, = km c. r = km, = km d. r = 44.6 km, = km 2. Find the exact circumference of the circle. 3. PRS a. 95 b. 20 c. 50 d. 85 a. 7π cm b. 5π cm c. 10π cm d. 4π cm 4. PRQ a. 85 b. 20 c. 50 d. 95 1

2 Name: I: 5. Find x. ssume that segments that appear tangent are tangent. Find x. ssume that any segment that appears to be tangent is tangent. 7. a. 7 b. 6 c. 14 d. 5 Find the measure of the numbered angle. a. 65 b. 66 c. 68 d a. 115 b. 125 c. 120 d. 130 a. 35 b. 20 c. 25 d. 30 2

3 Name: I: 9. Find x. Round to the nearest tenth if necessary. 11. Find the area of a circle having a circumference of Round to the nearest tenth. Use 3.14 for π. a units 2 b units 2 c units 2 d units 2 Find the area of the shaded region. Round answers to the nearest tenth. ssume all inscribed polygons are regular. 12. a. 6 b. 5 c. 4 d. 3 Find x. Round to the nearest tenth if necessary. ssume that segments that appear to be tangent are tangent. 10. a units 2 b. 9.9 units 2 c. 9.2 units 2 d units 2 Find the area of the figure. Round to the nearest tenth if necessary. 13. a. 4 b. 5 c. 6 d. 7 a units 2 b units 2 c units 2 d units 2 3

4 Name: I: 14. Find the shape resulting from the cross-section of the cylinder. Find the volume of the cylinder. Use 3.14 for π. Round to the nearest tenth a. rectangle b. square c. circle d. ellipse Find the lateral area of each prism. Round to the nearest tenth if necessary. a in 3 b in 3 c in 3 d in a. 456 units 2 b. 225 units 2 c. 184 units 2 d. 429 units 2 a in 3 b. 12,258.6 in 3 c. 49,034.5 in 3 d in 3 4

5 Name: I: 18. Find the volume of the pyramid. Round to the nearest tenth if necessary. 20. Find the volume of a sphere that has a radius of 9.5 meters. Use 3.14 for π. Round to the nearest tenth. a m 3 b m 3 c m 3 d m 3 a. 11,520 m 3 b. 14,400 m 3 c m 3 d m Find the surface area of a sphere if the circumference of a great circle is centimeters. Use 3.14 for π. Round to the nearest tenth. a cm 2 b. 196 cm 2 c cm 2 d cm Suppose a snow cone has a paper cone that is 8 centimeters deep and has a diameter of 5 centimeters. The flavored ice comes in a spherical scoop with a diameter of 5 centimeters and rests on top of the cone. If all the ice melts into the cone, will the cone overflow? Explain. a. No. The volume of the ice is less than the volume of the cone. b. No. The volume of the ice is exactly the same as the volume of the cone. c. Yes. The volume of the ice is greater than the volume of the cone. d. There is not enough information given to solve this problem. Short nswer 22. has a diameter of about 116 millimeters. Find the circumference of the. 5

6 Name: I: Use the information to answer the questions that follow. 26. Find the values of y and z. nnie wants to grow flowers all around a circular garden with a radius of 6 feet. 27. Using the properties of tangents, find the value of x. lso mention the property of tangent applied to find x. 23. If the outermost circle is 2 to 3 feet farther from the center than the inner circle, find the minimum and maximum circumference of the inner circle to the nearest foot. 24. In, L = M, XY = 5x + 8, and ST = 15x 32. Find XL. 28. In, mrq = 150, mtq = 22, msr = 92. What is m P? 25. In, the diameter is 42 units long, and, m RT = 30. Find x. 29. Find the volume of the solid. Round to the nearest tenth. 6

7 I: Unit 3 Practice Test nswer Section MULTIPLE HOIE 1. NS: radius = diameter 2 ircumference = (2 radius π) or (diameter π) heck both your radius and circumference calculations. heck your circumference calculation. orrect! heck your radius calculation. PTS: 1 IF: asic REF: Lesson 10-1 OJ: Identify and use parts of circles. NT: NTM ME.2 TOP: Identify and use parts of circles. KEY: ircles Parts of ircles 2. NS: The circumference formula is diameter π. The diameter shown also happens to be the hypotenuse of the right triangle inscribed in the circle, so it can be found by using the Pythagorean Theorem. Use the Pythagorean Theorem. orrect! How did you find the diameter? Use the Pythagorean Theorem. PTS: 1 IF: verage REF: Lesson 10-1 OJ: Solve problems involving the circumference of a circle. NT: NTM GM.1 NTM GM.1a NTM ME.2 TOP: Solve problems involving the circumference of a circle. KEY: ircles ircumference 1

8 I: 3. NS: PRS and QRT are vertical angles and are therefore congruent. So, set the two expressions equal and solve for x. orrect! That's the answer for x, you need the measure of PRS. id you use vertical angles? How are vertical angles related? PTS: 1 IF: verage REF: Lesson 10-2 OJ: Recognize major arcs, minor arcs, semicircles, and central angles, and their measures. NT: NTM ME.2 TOP: Recognize major arcs, minor arcs, semicircles, and central angles, and their measures. KEY: Major rcs Minor rcs Semicircles entral ngles 4. NS: PRQ and QRT are a linear pair and are therefore supplementary. First find m QRT. PRS and QRT are vertical angles and are therefore congruent. So, set the two expressions equal and solve for x. orrect! That's the answer for x, you need the measure of PRQ. heck over your work. How many degrees are in a linear pair? PTS: 1 IF: verage REF: Lesson 10-2 OJ: Recognize major arcs, minor arcs, semicircles, and central angles, and their measures. NT: NTM ME.2 TOP: Recognize major arcs, minor arcs, semicircles, and central angles, and their measures. KEY: Major rcs Minor rcs Semicircles entral ngles 5. NS: The triangle shown is a right triangle since the tangent segment, FE, intersects a radius, E, which always results in a right angle. So to solve for x, use the Pythagorean Theorem. Note that me = x since they are both radii of the same circle. Use the Pythagorean Theorem and me = x. orrect! Is the triangle a right? Use the Pythagorean Theorem and me = x. PTS: 1 IF: verage REF: Lesson 10-5 OJ: Use properties of tangents. NT: NTM GM.1 NTM GM.1a TOP: Use properties of tangents. KEY: Tangents 2

9 I: 6. NS: When two secants intersect in the interior of a circle, then the measure of an angle formed by this intersection is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. In this diagram, the measures of the intercepted arcs for 6 are not given, but they have a sum of 250 since the arcs shown have a sum of 110 ( = 250). dd the intercepted arcs and divide by 2. orrect! id you divide correctly? How do you find the measures of the other two arcs? PTS: 1 IF: verage REF: Lesson 10-6 OJ: Find measures of angles formed by lines intersecting on or inside a circle. NT: NTM GM.1 NTM GM.1b NTM ME.2 ST: M2P1.c TOP: Find measures of angles formed by lines intersecting on or inside a circle. KEY: Measure of ngles ircles 7. NS: When two secants intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. orrect! What is the measure of the other intercepted arc? heck your subtraction. id you find the positive difference of the measures of the intercepted arcs? PTS: 1 IF: verage REF: Lesson 10-6 OJ: Find measures of angles formed by lines intersecting outside the circle. NT: NTM GM.1 NTM GM.1b NTM ME.2 ST: M2P1.c TOP: Find measures of angles formed by lines intersecting outside the circle. KEY: Measure of ngles ircles 8. NS: When a secant and tangent intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. heck your subtraction. id you subtract carefully? orrect. heck your subtraction. PTS: 1 IF: verage REF: Lesson 10-6 OJ: Find measures of angles formed by lines intersecting outside the circle. NT: NTM GM.1 NTM GM.1b NTM ME.2 ST: M2P1.c TOP: Find measures of angles formed by lines intersecting outside the circle. KEY: Measure of ngles ircles 3

10 I: 9. NS: The products of the segments for each intersecting chord are equal. Multiply the segments and set them equal to each other. id you factor correctly? Use multiplication, not addition. orrect! PTS: 1 IF: verage REF: Lesson 10-7 OJ: Find measures of segments that intersect in the interior of a circle. NT: NTM GM.1 NTM GM.1b NTM ME.2 ST: M2P1.c TOP: Find measures of segments that intersect in the interior of a circle. KEY: ircles Interior of ircles 10. NS: When two secant segments intersect in the exterior of a circle, set an equality between the product of each external segment and the entire segment. heck your multiplication. heck the segments in your multiplication. orrect! heck your multiplication. PTS: 1 IF: verage REF: Lesson 10-7 OJ: Find measures of segments that intersect in the exterior of a circle. NT: NTM L.2 NTM L.2c NTM RE.2 ST: M2P1.c TOP: Find measures of segments that intersect in the exterior of a circle. KEY: ircles Exterior of ircles 11. NS: The area formula for a circle is π radius 2. To find the radius, use the formula for the circumference, which is circumference = 2 π radius or radius = circumference. 2 π How do you find the area of a circle? What is the formula for the area of a circle? orrect! The area of a circle is pi times the square of the radius. PTS: 1 IF: verage REF: Lesson 11-3 OJ: Find areas of circles. NT: NTM ME.2 NTM ME.2b ST: M2P5.b TOP: Find areas of circles. KEY: rea ircles rea of ircles 4

11 I: 12. NS: The shaded area here represents two-fifths of the area of the circle after subtracting the area of the regular pentagon. Re-check all of your calculations. Re-check all of your calculations. orrect! It s two-fifths of the leftover area. PTS: 1 IF: verage REF: Lesson 11-3 OJ: Solve problems involving segments of circles. NT: NTM PS.1 NTM PS.2 NTM PS.3 TOP: Solve problems involving segments of circles. KEY: ircles Segments of ircles 13. NS: This figure consists of a rectangle and two semicircles which combine to make one circle with radius 6.5. Use the radius, not the diameter in the area formula for the circle. orrect! The area of the rectangle is There are two semicircles that make one full circle. PTS: 1 IF: verage REF: Lesson 11-4 OJ: Find areas of composite figures. NT: NTM ME.2 NTM ME.2b ST: M2P5.b TOP: Find areas of composite figures. KEY: rea omposite Figures rea of omposite Figures 14. NS: cross-section is the intersection of a three-dimensional body with a plane. orrect! heck your answer. It will be the vertical cross-section of the cylinder shown. Refer to the hint and try again. PTS: 1 IF: asic REF: Lesson 12-1 OJ: Investigate cross sections of three-dimensional figures. NT: NTM GM.4 NTM GM.4a NTM GM.4b ST: M2P4.a TOP: Investigate cross sections of three-dimensional figures. KEY: ross Sections Three-imensional Figures 5

12 I: 15. NS: The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base. Lateral area is perimeter of the base times the height. The height of this prism is 4. Lateral area is perimeter of the base times the height. What is the height of this prism? orrect! Lateral area is perimeter of the base times the height. The height of this prism is 4. PTS: 1 IF: verage REF: Lesson 12-2 OJ: Find lateral areas of prisms. NT: NTM ME.2 NTM ME.2b ST: M2P5.b TOP: Find lateral areas of prisms. KEY: Lateral rea Prisms Lateral rea of Prisms 16. NS: The volume of a cylinder is found by the formula π radius 2 height. In this figure, the height and diameter are given. To find the radius, divide the diameter by 2. What is the formula for the volume of a cylinder? How do you find the volume of a cylinder? What is the radius of the base? orrect! PTS: 1 IF: verage REF: Lesson 12-4 OJ: Find volumes of cylinders. NT: NTM ME.2 NTM ME.2b ST: M2P5.b TOP: Find volumes of cylinders. KEY: Volume ylinders Volume of ylinders 17. NS: The volume of a cylinder is found by the formula π radius 2 height. In this figure, the height and radius are given. id you square the radius? orrect! id you use the correct formula? id you include pi in the formula? PTS: 1 IF: verage REF: Lesson 12-4 OJ: Find volumes of cylinders. NT: NTM ME.2 NTM ME.2b ST: M2P5.b TOP: Find volumes of cylinders. KEY: Volume ylinders Volume of ylinders 6

13 I: 18. NS: The volume formula for a pyramid is 1 h, where is the area of the base and h is the height of the pyramid. In 3 the figure, the length and width of the base are given. Since the base is a rectangle, the area is found by multiplying the length and width. To find the height of the pyramid, it is necessary to use the Pythagorean Theorem. s shown in the figure, there is a right triangle formed with legs that are the height and half the width. The hypotenuse is the slant height of the pyramid s face. So the height is equal to slant 2 (width 2) 2. You need to divide your answer by three. The formula you need is V = 1 h. 3 You need to find the height of the pyramid. orrect! PTS: 1 IF: verage REF: Lesson 12-5 OJ: Find volumes of pyramids. NT: NTM ME.2 NTM ME.2b ST: M2P5.b TOP: Find volumes of pyramids. KEY: Volume Pyramids Volume of Pyramids 19. NS: The surface area of a sphere is found by the formula 4πr 2, where r is the radius of the sphere. This problem gives the circumference, so begin by computing the radius which can be found by the formula c. For example, if the 2π circumference is given as 12.56, then the radius is = 2 and the surface area would be = If the solid is a hemisphere, then the formula for the surface area is 3πr 2, since a hemisphere has half the surface Ê 1 area of the sphere 2 4πr2 = 2πr 2 ˆ Ë Á plus the area of the great circle on its base Ê 2πr 2 ˆ Ë Á. oes the formula indicate cubing the radius? id you leave pi out of the formula? id you use the correct formula? orrect! PTS: 1 IF: verage REF: Lesson 12-6 OJ: Find surface areas of spheres. NT: NTM PS.1 NTM PS.2 NTM PS.3 ST: M2P5.b TOP: Find surface area of spheres. KEY: Surface rea Spheres Surface rea of Spheres 7

14 I: 20. NS: The volume of a sphere is found by the formula 4 3 π radius 3. In this problem, the radius or diameter is given. To find the radius from the diameter, simply divide by 2. If the solid is a hemisphere, then it has one-half the volume of a sphere. So, divide the volume of the sphere by 2 to get the volume of a hemisphere. orrect! You need to multiply your answer by 4 3. You need to cube the radius, not square it. The volume formula has 4 3 in it, not 1 3. PTS: 1 IF: verage REF: Lesson 12-6 OJ: Find volumes of spheres. NT: NTM ME.2 NTM ME.2b ST: M2P5.b TOP: Find volumes of spheres. KEY: Volume Spheres Volume of Spheres 21. NS: The volume of the cone is = 52.3 cm 3. The volume of the ice (a sphere) is = 65.4 cm 3. So the volume of the ice is greater than the volume of the cone, causing it to overflow the cone. heck your volume calculations. heck your volume calculations. orrect! This problem can be solved with the information given. PTS: 1 IF: verage REF: Lesson 12-6 OJ: Solve problems involving volumes of spheres. NT: NTM GM.1 NTM GM.1b ST: M2P5.b TOP: Solve problems involving volumes of spheres. KEY: Volume Spheres Volume of Spheres SHORT NSWER 22. NS: about mm ircumference = (2 radius π) or (diameter π) PTS: 1 IF: asic REF: Lesson 10-1 OJ: Solve multi-step problems. NT: NTM GM.1 NTM GM.1a NTM ME.2 TOP: Solve multi-step problems. KEY: Solve Multi-Step Problems 8

15 I: 23. NS: 18.8 ft; 25.1 ft Find the range of values for the radius of the inner circle. The minimum radius will give the minimum circumference and the maximum radius will give the maximum circumference. PTS: 1 IF: dvanced REF: Lesson 10-1 OJ: Solve multi-step problems. NT: NTM GM.1 NTM GM.1a NTM ME.2 TOP: Solve multi-step problems. KEY: Solve Multi-Step Problems 24. NS: 14 In a circle, two chords are congruent if they are equidistant from the center. In a circle, if a diameter is perpendicular to a chord then it bisects the chord. PTS: 1 IF: verage REF: Lesson 10-3 OJ: Solve multi-step problems. NT: NTM GM.1 NTM GM.1b NTM ME.2 ST: M2P4.a TOP: Solve multi-step problems. KEY: Solve Multi-Step Problems 25. NS: 1.5 sin RT = T R PTS: 1 IF: verage REF: Lesson 10-3 OJ: Solve multi-step problems. NT: NTM GM.1 NTM GM.1b NTM ME.2 ST: M2P4.a TOP: Solve multi-step problems. KEY: Solve Multi-Step Problems 26. NS: 70 ; 105 If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. PTS: 1 IF: verage REF: Lesson 10-4 OJ: Solve multi-step problems. NT: NTM GM.1 NTM GM.1b NTM ME.2 NTM GM.1a ST: M2P1.c TOP: Solve multi-step problems. KEY: Solve Multi-Step Problems 27. NS: 16; Two segments from the same exterior point are tangent to the circle. If two segments from the same exterior point are tangent to a circle, then they are congruent. PTS: 1 IF: asic REF: Lesson 10-5 OJ: Solve multi-step problems. NT: NTM GM.1 NTM GM.1a NTM GM.1b NTM ME.2 ST: M2P1.c TOP: Solve multi-step problems. KEY: Solve Multi-Step Problems 9

16 I: 28. NS: 35 When two secants intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. m P = 1 Ê ˆ msr mtq 2 Ë Á PTS: 1 IF: asic REF: Lesson 10-6 OJ: Solve multi-step problems. NT: NTM GM.1 NTM GM.1b NTM ME.2 ST: M2P1.c TOP: Solve multi-step problems. KEY: Solve Multi-Step Problems 29. NS: in 3 V = π radius 2 height 2 3 π radius 3 PTS: 1 IF: dvanced REF: Lesson 12-6 OJ: Solve multi-step problems. NT: NTM ME.2 NTM ME.2b NTM PS.1 NTM PS.2 NTM PS.3 NTM GM.1 NTM GM.1b ST: M2P5.b TOP: Solve multi-step problems. KEY: Solve Multi-Step Problems 10