( ) ) in 2 ( ) ) in 3

Similar documents
I.G.C.S.E. Volume & Surface Area. You can access the solutions from the end of each question

Right Circular Cylinders A right circular cylinder is like a right prism except that its bases are congruent circles instead of congruent polygons.

A. 180 B. 108 C. 360 D. 540

Section Volume, Mass, and Temperature

Grade 11 Mathematics Practice Test

Answer Keys for Calvert Math

Pretest. Explain and use formulas for lateral area, surface area, and volume of solids.

Geometry. A. Right Triangle. Legs of a right triangle : a, b. Hypotenuse : c. Altitude : h. Medians : m a, m b, m c. Angles :,

LESSON 2: TRIANGULAR PRISMS, CYLINDERS, AND SPHERES. Unit 9: Figures and Solids

resources Symbols < is less than > is greater than is less than or equal to is greater than or equal to = is equal to is not equal to

Turn Up the Volume and Let s Bend Light Beams Volume and Surface Area of a Prism

MENSURATION. Mensuration is the measurement of lines, areas, and volumes. Before, you start this pack, you need to know the following facts.

w hole + ½ partial = 10u + (½ )(10u )

Kansas City Area Teachers of Mathematics 2015 KCATM Math Competition GEOMETRY GRADE 7

221 MATH REFRESHER session 3 SAT2015_P06.indd 221 4/23/14 11:39 AM

Surface Areas of Prisms and Cylinders. Find the lateral area and surface area of each prism. Round to the nearest tenth if necessary.

CONNECTED RATE OF CHANGE PACK

MCAS Review - Measurement Session 4A

Chapter Start Thinking. 10π 31.4 cm. 1. 5π 15.7 cm 2. 5 π 7.9 cm π 13.1 cm Warm Up , 8π 2. 90,15π 3.

Appendices. Appendix A.1: Factoring Polynomials. Techniques for Factoring Trinomials Factorability Test for Trinomials:

Chapter 8 Solids. Pyramids. This is a square pyramid. Draw this figure and write names of edges. Height and Slant Height.

Then the other two numbers may be represented as x + 1 and x + 2. Now we can represent 3 less than the second as (x + 1) 3.

Ex 1: If a linear function satisfies the conditions of h( 3) = 1 and h(3) = 2, find h(x).

New Rochelle High School Geometry Summer Assignment

Sect Formulas and Applications of Geometry:

Math Round. Any figure shown may not be drawn to scale.

AREA. The Square Inch The Square Foot The Square Yard. 1 foot. 1 foot. The Square Mile The Square Meter The Square Centimeter. 1 meter.

Circle Theorems. Angles at the circumference are equal. The angle in a semi-circle is x The angle at the centre. Cyclic Quadrilateral

Lesson 6 Plane Geometry Practice Test Answer Explanations

May 05, surface area and volume of spheres ink.notebook. Page 171. Page Surface Area and Volume of Spheres.

Geometry Summer Assignment

Using the distance formula Using formulas to solve unknowns. Pythagorean Theorem. Finding Legs of Right Triangles

Lesson a: x 2 + y 2 = 9 b: 7

Practice Test Student Answer Document

4.! ABC ~ DEF,! AC = 6 ft, CB = 3 ft, AB = 7 ft, DF = 9 ft.! What is the measure of EF?

Days 3 & 4 Notes: Related Rates

Franklin Math Bowl 2007 Group Problem Solving Test 6 th Grade

Geometry: Hutschenreuter Semester II. Select the best answer for each question. Show expected work. MAKE A SUPPORTING SKETCH!

Study Guide for Benchmark #1 Window of Opportunity: March 4-11

CLASS X FORMULAE MATHS

AP Calculus BC Chapter 4 AP Exam Problems A) 4 B) 2 C) 1 D) 0 E) 2 A) 9 B) 12 C) 14 D) 21 E) 40

50 Keys To CAT Arithmetic, Algebra, Geometry and Modern Mathematics EASY EFFECTIVE PERSONALISED

Customary Units of Measurement

Unit 5 Test Review: Modeling with Geometry Honors

Use this space for computations. 1 In trapezoid RSTV below with bases RS and VT, diagonals RT and SV intersect at Q.

Georgia High School Graduation Test

Lesson 14.1 Skills Practice

University of Houston High School Mathematics Contest Geometry Exam Spring 2015

SAT Subject Test Practice Test II: Math Level I Time 60 minutes, 50 Questions

MATH 1371 Fall 2010 Sec 043, 045 Jered Bright (Hard) Mock Test for Midterm 2

Kansas City Area Teachers of Mathematics 2013 KCATM Math Competition GEOMETRY GRADES 7-8

The GED math test gives you a page of math formulas that

Mt. Douglas Secondary

Mesa Public Schools. Q3 practice test. Assessment Summary: Powered by SchoolCity Inc. Page 1 of 44

High School HS Geometry 1819 GSE Geometry Unit 5 Full Touchstone

17. The length of a diagonal of a square is 16 inches. What is its perimeter? a. 8 2 in. b in. c in. d in. e in.

Math 005A Prerequisite Material Answer Key

Section 0 5: Evaluating Algebraic Expressions

MATH 310 TEST Measure April Answers

MCA/GRAD Formula Review Packet

Cambridge International Examinations International General Certificate of Secondary Education. Paper 1 (Core) May/June minutes

Geometry Pre-Unit 1 Intro: Area, Perimeter, Pythagorean Theorem, Square Roots, & Quadratics. P-1 Square Roots and SRF

Mathematics 10C. UNIT ONE Measurement. Unit. Student Workbook. Lesson 1: Metric and Imperial Approximate Completion Time: 3 Days

13. a. right triangular prism

Geometry Honors Final Exam Review June 2018

Final Exam Review Packet

Kg hg dag g dg cg mg. Km hm dam m dm cm mm

Context-Based Multiple-Choice Questions

STUDENT NAME DATE PERIOD. Math Algebra I. Read each question and choose the best answer. Be sure to mark all of your answers.

CHAPTER 1 BASIC ARITHMETIC

Section MWF 12 1pm SR 117

1. Use. What are the vertices of A.,, B.,, C.,, D.,,

In Questions 1 through 5, match each term with the part that describes it on the diagram.

Geometry Final Exam Review

262 MHR Foundations for College Mathematics 11 Solutions

Geometry Final Exam REVIEW

Classwork. Opening Exercises 1 2. Note: Figures not drawn to scale. 1. Determine the volume for each figure below.

Chapter 2 Polynomial and Rational Functions

Lesson 4.1 (Part 1): Roots & Pythagorean Theorem

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. C) dc. D) dr dt = 2πdC dt

Right Triangles

COMMON UNITS OF PERIMITER ARE METRE

Chapter 8. Chapter 8 Opener. Section 8.1. Big Ideas Math Red Accelerated Worked-Out Solutions 4 7 = = 4 49 = = 39 = = 3 81 = 243

Find the perimeter of the figure named and shown. Express the perimeter in the same unit of measure that appears on the given side or sides.

Section 3.4. from Verbal Descriptions and from Data. Quick Review. Quick Review Solutions. Functions. Solve the equation or inequality. x. 1.

Chapter 1: Review of Real Numbers

How can you find decimal approximations of square roots that are not rational? ACTIVITY: Approximating Square Roots

Log1 Contest Round 2 Theta Geometry

WebAssign Lesson 2-2 Volumes (Homework)

Applications Using Factoring Polynomials

2.6 Related Rates. The Derivative as a Rate of Change. = ft related rates 175

The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Student Name:

Grade 11 Mathematics Practice Test

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education. Paper 1 (Core) May/June minutes

PRACTICE TEST 1 Math Level IC

Skills Practice Skills Practice for Lesson 3.1

UNC Charlotte Super Competition - Comprehensive test March 2, 2015

Academic Challenge 2009 Regional Mathematics Solution Set. #2 Ans. C. Let a be the side of the cube. Then its surface area equals 6a = 10, so

Section 5.1. Perimeter and Area

MATHS X STD. Try, try and try again you will succeed atlast. P.THIRU KUMARESA KANI M.A., M.Sc.,B.Ed., (Maths)

Transcription:

Chapter 1 Test Review Question Answers 1. Find the total surface area and volume of a cube in which the diagonal measures yards. x + x ) = ) x = x x A T.S. = bh) = ) ) = 1 yd V = BH = bh)h = ) ) ) = yd. A right cylindrical hole of diameter inches is drilled out of a cube with an edge of 7 inches. Find the surface area and volume of the remaining figure. A T.S. = A Prism - A Circle ) + A L.S. of the Inside Cylinder = bh)h - πr + πr)h ) = 7) 7) - π + π) 7) = - π + π 7 7 = + π ) in 7 V = V Prism - V Cylinder = bh)h - πr )H ) = 7) 7) 7) - π) 7) = - 17π ) in Baroody Page 1 of 11

Chapter 1 Test Review Question Answers. Find the slant height and surface area of a right circular cone with a volume of π cubic inches and a base circumference of 10π inches. C = 10π c = + 7 r = c = 7 c 7 V = 1 πr )H A T.S. = πr + πrl C Base = 10π π = 1 π)h 7 = H = π + ) 7)π = π + π 7) in. Find the volume and total surface area of a regular square pyramid that has an edge height of 1 yards and a side of the base of 10 yards. H = 1 - = 11 1 1 H A T.S. = 10 + 1 ) bh = 100 + 10) 1) = 0 yd V = 1 BH = 1 10) 10) 11 ) = 100 11 ) yd 10. The volume of a sphere in cubic feet is numerically equal to four times its surface area in square feet. What is the radius of the sphere? V Sphere = A T.S of Sphere )!r ) =!r ) r = 1r r = 1 ft Baroody Page of 11

Chapter 1 Test Review Question Answers. 7. Find the total surface area and volume of a regular tetrahedron if every edge measures feet. H H = ) ) - Base of Pyramid = - 1 = 1 ) A T.S. = s = ) u V = 1 s )H = ) ) = 1 1 ) = 1 1 1 = ) ft A frustum of a right cone has a slant height of feet. The radius of the bottom base is 1 feet and the radius of the top base is feet. Find the volume and total surface area of the frustum. Now, the small to the large by AA, so 1 = x x + x = 1 1 1 V Frustum = V Large Cone - V Smaller Cone = 1!R )H Large Cone - 1!r )H Small Cone = 1 1! ) 1 ) - 1 1! ) ) = 7! -! =! ft A T.S. =!RL -!rl +!R +!r =! 1) ) -! ) 1) + 1! + 1! = 1! ft Baroody Page of 11

Chapter 1 Test Review Question Answers. Find the volume and total surface area of the frustum shown Now, the small to the large by AA, so x x + = 1 x = 1 1 V Frustum = V Large Cone - V Smaller Cone = 1 πr )H Large Cone - 1 πr )H Small Cone = 1 1π) 1) - 1 π) 1) 1 = π - π = 0π u L large cone slant) = 1 + 1 = 1 l small cone slant) = 1 + = 1 A T.S. = πrl - πrl + πr + πr = π 1) 1) - π ) 1) + 1π + π = 0π 1 + 0π ft. Find the volume and the surface area. Measurements are in centimeters. 1 V = 1 )!r = 1 )!1 = 7! = 0! cm A T.S. =!r =! 1 ) = 7! cm Baroody Page of 11

Chapter 1 Test Review Question Answers 10. Find the volume and total surface area of a hemisphere with a 0 wedge cut out if the radius is inches. 0 V = 70 0 V Hemisphere) = 1 )) πr = 1 π ) = π in A TS = A LS + A B ) 1 ) πr + 1 ) πr = 1 ) + πr ) = π + π + π = π + π + π = 17π in 11. Find the radius of the base. The volume of the solid is 10π cm. V = 1 )) πr = πr 10π = πr ) 0 17 = r r = 1 cm Baroody Page of 11

Chapter 1 Test Review Question Answers 1. Find the volume. Measurements are in centimeters. 1 1 V = BH + BH 10 = ) 1) 10) + ) ) 10) = 100 cm 1. Find the radius of the base. The volume of the cone is 17! cm. V = 17! = 1 BH = 1!r )H 17! = 1!r ) r 17! = 1!r r = 17! 1! = 1 r = 1 cm Baroody Page of 11

Chapter 1 Test Review Question Answers 1. Find the volume and total surface area of the solid shown below. V = 00 0 V Cylinder) =!r )H in =! ) 1) = 0! in 1 in A T.S. = A Base ) + A Lateral Surface of Cylinder) + A Rectangles ) 0 = )!r +!rh) + bh) =! ) +! ) 1) + ) 1) = 1! + 0! + 7 = 7! + 7) in 1. Identify the surface of rotation generated in the diagram below and compute its total surface area and volume. The surface of rotation is a cylinder with a cylindrical hole through its middle 1 A T.S. = A T.S. of Cylinder - A Holes + A L.S. of Inside Cylinder = πr ) + πrh - πr ) + πrh = 1π) + π ) ) - π) + π ) ) = π + 0π - 1π + 0π = π u V = V Large Cylinder - V Small Cylinder = πr )H - πr )H = π ) 1 - ) = π) 7) = π u Baroody Page 7 of 11

Chapter 1 Test Review Question Answers 1. A right cylindrical log was cut parallel to the axis. Find the volume and the total surface area of the piece shown. 0 10 V = BH = A Segment ) H) = = ) - s [ )] H 0 0!r ) - 10 [ )] 0 1!) 10 ) ) 0 0! = [ - ] 0) 0 0 = 00! - 70 ) u A T.S. = A Segment ) + A Rectangle + 1 A L.S. of the Cylinder) ) + 10 = 0! - ) 0) + 1! ) 10) 0) = 100! - 0 + 00 + 100! 00! = - 0 + 00 ) u 17. A cylinder with diameter mm is cut twice at a angle as shown. Find the volume of this cylindrical solid. If you're interested I won't test you on this!), you can also find the surface area if you know that the ends of the "cannoli" are ellipses. V = V Cylinder - V Cutout Half Cylinder ) V = πr H 1-1 πr H ) = π 1) - π ) = 1π - π = 0π mm A TS = A LS + A B = 1 πrh Ends [ ) + πrh Center Cylinder ] + π r 1 r ), where r 1 is the semi-major axis and r is the semi-minor axis of the parabola forming the ends of the "cannoli." = π ) + π ) + π ) ) = π + π + 1π = 0π + 1π ) mm Baroody Page of 11

Chapter 1 Test Review Question Answers 1. Find the volume and surface area of a hexagonal prism with height 1 cm, and the length of one side of the regular hexagonal base being 10 cm. 1 V = BH [ )] H = s ) = [ ) ] 1) = 100 ) cm 10 [ )] + bh A T.S. = s ) = 00 + 10) 1) = 00 + 70) cm 1. A cube has a volume of 1 cm. A corner is cut off as shown in the diagram. Find the volume of this triangular pryamid and determine what fractional part of the cube it represents. V Cube = BH 1 = s )s = s H s = H = ) - ) = 1 - = 1 = V Pyramid = 1 BH = 1 ) ) = 1 = u ) V Pyramid V Cube = 1 = 1 Baroody Page of 11

Chapter 1 Test Review Question Answers 0. Each edge of the cube shown has a length of. The midpoint of each face of the cube is joined to form an eight-sided regular octahedron. Find the volume of the octahedron. Each edge of the regular octadedron measures as each is the hypotenuse of a --0 triangle with legs of length formed by connecting two face midpoints with the midpoint of an edge of the cube). Now, V = 1 BH ) = 1 ) H) = 1H...if only we knew the height! To find the height of half of the octahedron, we'll use the triangle shown. You should be able to figure out the two lengths shown below and then use the Pythagorean Theorem to find H or you can just recognize that it's since it's half the height of the cube!). H H = ) - ) = - 1 = = Show Stuff So, V = 1H = 1 ) = u 1. A "segment" of a grapefruit has diameter 1 cm and the flat faces form a 0 angle. Assuming the grapefruit was spherical, what are the volume and surface area of the segment? 1 V = 0 0 ) πr = 1 1 A TS = 1 1 πr ) + 1 πr ) = 7 1π π ) = π ) = 1π cm ) + π = π cm Baroody Page 10 of 11

Chapter 1 Test Review Question Answers. Find the volume and total surface area of the solid generated by rotating the figure below around the dashed line. All dimensions are in centimeters. 10 V = 1 πr H + πr H + 1 = 1 π ) πr ) ) ) + π ) 10 ) + 1 ) ) π = 1π + 10π + π = π + 10π) cm A TS = πrl + πrh + πr = π ) + π )10 + π ) = 1π + 0π + π = 7π + π) cm Baroody Page 11 of 11