Chapter Geometry. Lines and ngles. BE 90 because it is a vertically opposite angle to CBD which is also a right angle.. ngles POR and QOR are complementary angles, so sum to 90 POR + QOR 90 + QOR 90 QOR 90 QOR 58. 4 pairs of adjacent angles: BOC and CO share common ray OC CO and OD share common ray O OD and DOB share common ray OD DOB and BOC share common ray OB 4. x 775 mm 565 mm 655 mm (a) Using Eq. (.), we have 775 x 565 655 775( 655) x 565 x 898 mm (the same answer as Example 5) (b) More vertical, since the distance along the beam is longer for the same horizontal run, which can only be achieved if the angle increases from horizontal (see sketch). Copyright 05 Pearson Education, Inc.
ISM for Washington, Basic Technical Mathematics with Calculus, SI Version, Tenth Edition 5. EBD and DBC are acute angles (i.e., < 90 ). 6. BE and CBE are right angles (i.e., 90 ). 7. BC is a straight angle (i.e., 80 ). 8. BD is an obtuse angle (i.e., between 90 and 80 ). 9. The complement of CBD 65 is DBE CBD + DBE 90 65 + DBE 90 DBE 90 65 DBE 5 0. The supplement of CBD 65 is BD CBD + BD 80 65 + BD 80 BD 80 65 BD 5. Sides BD and BC are adjacent to DBC.. The angle adjacent to DBC is DBE since they share the common side BD, and DBE is acute because it is less than 90. OB OE + EOB but OE 90 because it is vertically opposite to DOF a given right angle, and EOB 50 because it is vertically opposite to COF a given angle of 50, so OB 90 + 50 40 4. OC is complementary to COF a given angle of 50, OC + COF 90 OC + 50 90 OC 90 50 OC 40 Copyright 05 Pearson Education, Inc.
Chapter : Geometry 5. BOD is vertically opposite to OC a found angle of 40 (see Question 4), so BOD OC BOD 40 6. is supplementary to 45, so 80 45 5 5 is supplementary to, so 80 80 5 45 7. is supplementary to 45, so 80 45 5 5 4 is vertically opposite to, so 4 4 5 8. is supplementary to 45, so 80 45 5 5 5 is supplementary to, so 5 80 5 80 5 5 45 9. 6 since they are vertically opposite 0. 6 since they are vertically opposite is a corresponding angle to 5, so 5 since and 5 are supplementary angles, 5 + 80 + 80 80 80 6 8 Copyright 05 Pearson Education, Inc.
ISM for Washington, Basic Technical Mathematics with Calculus, SI Version, Tenth Edition. 6 90 6 since they are complementary angles 6 8 is an alternate-interior angle to 6, so 6 8. 8 (see Question ) since 4 and are supplementary angles, 4 + 80 4 80 4 80 8 4 5. EDF BD 44 because they are corresponding angles BDE 90 BDF BDE + EDF BDF 90 + 44 BDF 4 4. CBE BD 44 because they are corresponding angles DBE and CBE are complementary so DBE + CBE 90 DBE 90 CBE DBE 90 44 DBE 46 and BE BD + DBE DEB 44 BE 90 + 46 BE 6 5. CBE BD 44 because they are corresponding angles DEB and CBE are alternate interior angles, so DEB CBE Copyright 05 Pearson Education, Inc. 4
Chapter : Geometry 6. CBE BD 44 because they are corresponding angles DBE and CBE are complementary so DBE + CBE 90 DBE 90 CBE DBE 90 44 DBE 46 7. EDF BD 44 because they are corresponding angles ngles DB, BDE, and EDF make a straight angle DB + BDE + EDF 80 DB 80 BDE EDF DB 80 90 44 DB 46 DFE DB because they are corresponding angles DFE 46 8. DE DB + BDE DB 46 (see Question 7) BDE 90 DE 46 + 90 DE 6 9. Using Eq. (.), a.05 4.75.0.05 a 4.75.0 a 4.5 m 0. Using Eq. (.), b.05 6.5.0.05 b 6.5.0 b 5.96 m. Using Eq. (.), c 5.05.0 4.75 (.0)(5.05) c 4.75 c.40 m Copyright 05 Pearson Education, Inc. 5
ISM for Washington, Basic Technical Mathematics with Calculus, SI Version, Tenth Edition. Using Eq. (.), d 5.05 6.5 4.75 (6.5)(5.05) d 4.75 d 6.64 m. BCE 47 since those angles are alternate interior angles. BCD and BCE are supplementary angles BCD + BCE 80 BCD 80 47 BCD 4. C Laser beam 4. O E Reflected beam B B CD Refracted beam D F COE 4 COE + EOF + DOF 80 4 + EOF + 80 COD is a straight angle, so the total angle between the reflected and refracted beams, EOF is EOF 80 46 EOF 4 This angle is a reflex angle. Copyright 05 Pearson Education, Inc. 6
Chapter : Geometry 5. Using Eq. (.), x 555 85 59 (85)(555) x 59 x 88 m 6. Using Eq. (.), B BC (.5) B B.5 cm C B + BC C.5 cm +.5 cm C 5.75 cm C 5.8 cm 7. + + 80, because,, and form a straight angle. 8. 4 since they are alternate interior angles 5 since they are alternate interior angles + + 80, because,, and form a straight angle, so 4 + + 5 80 9. The sum of the angles with vertices at B,, and Dis 80. Since those angles are unknown quantities, the sum of interior angles in a closed triangle is 80. Copyright 05 Pearson Education, Inc. 7
ISM for Washington, Basic Technical Mathematics with Calculus, SI Version, Tenth Edition 40. B C 7 4 5 6 7 D + + 80 since those angles form a straight angle 4 since they are alternate interior angles 4 + + 80 5 + 6 + 7 80 since those angles form a straight angle 7 alternate interior angle is shown For the interior angles of the closed geometric figure BCD, sum 4+ + + 7+ 5+ 6 sum 4+ + + 5+ 6+ 7 sum 80 + 80 sum 60 Copyright 05 Pearson Education, Inc. 8
Chapter : Geometry. Triangles. 5 45 45 since and 5 are alternate interior angles.,, and make a stright angle, so + + 80 70 + + 45 80 65. bh 6. 5.75 76 cm. C B + BC C 6.5 +.0 C 6.5 +.0 C 7.0 m 4. Using Eq. (.), h 4.0.00 4.00 (.00)(4.0) h 4.00 h 8.0 m 5. + B+ C 80 + 40 + 84 80 56 6. + B+ C 80 + 48 + 90 80 4 Copyright 05 Pearson Education, Inc. 9
ISM for Washington, Basic Technical Mathematics with Calculus, SI Version, Tenth Edition 7. This is an isosceles triangle, so the base angles are equal. B C 66 + B+ C 80 + 66 + 66 80 80 66 + 66 48 + B+ C 80 + + 0 80 70 5 8. This is an isosceles triangle, so the base angles are equal. B 9. bh 7.6. 8.4 m 0. bh 6.0 7.6 6.0 mm. By Hero s formula, p 05 + + 45 94 cm 94 s 47 cm s s a s b s c 47 cm 47 05 cm 47 cm 47 45 cm 4 47 66 49 56 cm 00 cm Copyright 05 Pearson Education, Inc. 0
Chapter : Geometry. By Hero s formula, p 86. +.5 + 68.4 78. m 78. s 89.05 m s s a s b s c 89.05 89.05 86. 89.05.5 89.05 68.4 586 m. One leg can represent the base, the other leg the height. bh.46.55 4.4 cm 4. One leg can represent the base, the other leg the height. bh 4 4 40 00 mm 5. By Hero s formula, p 0.986 + 0.986 + 0.884 s m.48 m s s a s b s c.48.48 0.986.48 0.986.48 0.884 0.90 m 6. By Hero s formula, s 48 dm s s a 48 48 44 900 dm 7. We add the lengths of the sides to get p 05 + + 45 p 94 cm Copyright 05 Pearson Education, Inc.
ISM for Washington, Basic Technical Mathematics with Calculus, SI Version, Tenth Edition 8. We add the lengths of the sides to get p.5 + 86. + 68.4 p 78. m 9. We add the lengths of the sides to get p (.5) 64.5 cm 0. We add the lengths of the sides to get p.45 +. 8. mm. c a + b c a + b c.8 +.7 c 6.6 mm. c a + b c a + b c.48 +.45 c.87 m. c a + b b c a b 55 75 b 5 cm 4. c a + b a c b a 0.86 0.474 a 0.689 km 5. ll interior angles in a triangle add to 80 + B + 90 80 B 80 90 B 67 Copyright 05 Pearson Education, Inc.
Chapter : Geometry 6. c a + b c a + b c 8.4 + 90.5 c 98. cm 7. Length c is found in Question 6, c 98.09 77 cm p 98.09 77 + 90.5 + 8.4 7. cm 8. bh 90.5 8.4 740 cm 9. B B/ ' / D C' / C DC ~ ' DC ' D' C ' / B' D between bisectors From B' C ', and all angles in a triangle must sum to 80 B + ( B' D + / ) + 90 80 B B' D 90 + + B B' D 90 But BC is a right triangle, and all angles in a triangle must sum to 80, so + B 90 90 B' D 90 B' D 45 Copyright 05 Pearson Education, Inc.
ISM for Washington, Basic Technical Mathematics with Calculus, SI Version, Tenth Edition 0. F B C D E D since FD is isosceles. Since F FD FD is isosceles and since B and C are midpoints, B CD E DE because E is a midpoint of D, so if two of the three sides are identical, the last side is the same too. so BE ECD Therefore, BE EC from which it follows that the inner BCE is isosceles. lso, since B CD FB FC BE ECD BFC BCE and all four triangles are similar triangles to the original FD So, BCE is also /4 of the area of the original FD.. n equilateral triangle.. Yes, if one of the angles of the triangle is obtuse. For example, see BC below. C B Copyright 05 Pearson Education, Inc. 4
Chapter : Geometry. D B C + B 90 + B 90 redraw BDC as B + 90 + B 90 C D B and DC as C D BDC and DC are similar. 4. B C Comparing the original triangle to the two smaller triangles (see Question ) shows that all three are similar. 5. LMK and OMN are vertically opposite angles and thus equal. Since each triangle has a right angle, the remaining angle in each triangle must be the same. KLM MON. The triangles MKL and MNO have all the same angles, so therefore the triangles are similar: MKL ~ MNO Copyright 05 Pearson Education, Inc. 5
ISM for Washington, Basic Technical Mathematics with Calculus, SI Version, Tenth Edition 6. CB DC 90 DC BC since they share the common vertex. Since all angles in any triangle sum to 80, DC 80 90 BC, BC 80 90 BC, Therefore, all the angles in CB and DC are equivalent, so CB ~ DC 7. 8. 9. KM KN MN KM 5 9 KM 6 Since MKL ~ MNO LM OM KM MN LM 6 9 (6)() LM 9 LM 8 Since DC ~ CB B C C D B 9 ()() B 9 B 6 C 50 B BC is isosceles, so CB CB But all interior angles in a triangle sum to 80 CB + CB + 50 80 CB 0 CB 65 Copyright 05 Pearson Education, Inc. 6
Chapter : Geometry 40. But all interior angles in a triangle sum to 80 angle between tower and wire 80 90 5 8 4. p ( 76.6) + 0.6 s 9.9 cm By Hero s formula, s s a s b s c 9.9 9.9 76.6 9.9 76.6 9.6 0.6 50 cm 4. p ( 600) s 400 km By Hero s formula, s s a s b s c 400 400 600,00,000 km 4. One leg can represent the base, the other leg the height. bh. 6.0 9.6 m 44. c a + b c a + b c 750 + 550 c 90 m Copyright 05 Pearson Education, Inc. 7
ISM for Washington, Basic Technical Mathematics with Calculus, SI Version, Tenth Edition 45. Wall 6.0 Ladder.8 Floor c a + b b c a b 6.0.8 b 5.7 m 46. B C.00 m D E O x On BO the idea that the side opposite the 0 angle is half the hypotenuse gives B.00 m Using Pythagorean theorem gives O B + BO BO O B BO BO m Using an identical technique on each successive triangle moving clockwise, BC m CO 4 CO.50 m CD 0.750 m DO.50 (0.750) DO.0 m DE 0.650 m x.0 (0.650) x.5 m x. m Copyright 05 Pearson Education, Inc. 8
Chapter : Geometry 47. C 8.00 m B 8.0 m Diagonal B Diagonal C 8 468 m B + C B + 8 C 468 + 64 m C 5 m C. m.0 m 48. By Eq. (.), 45.6 cm.0 cm x.00 m 45.6 cm (.00 m) x.0 cm x 8.0 m 49. By Eq.., z. 4.5 0.9 (4.5)(.) z 0.9 z 6.0 m x z + 4.5 x 56.5 m x 7.5 m y. + 6 + 5.4 x 8.0 m y 9.0 m Copyright 05 Pearson Education, Inc. 9
ISM for Washington, Basic Technical Mathematics with Calculus, SI Version, Tenth Edition 50. 0.85 m 0.85 m w 0.85 + 0.85 w.445 m w. m w 5. l.5 d By Eq. (.), d.5.75.0.75(.5) d.0 d.065 m l.75 + d l.75 +.065 l.4 m 5. ED DC B BC ED 80.0 50.0 (80.0)() ED 50.0 ED 499 m 5. Redraw BCP as B.0 0.75 6.00 P.0 - PD C Copyright 05 Pearson Education, Inc. 0
Chapter : Geometry PD is 0.0 P D since BCP DP 6.00 0.0.0 PD PD 6PD 0 0PD 6PD 0 PD 7.50 km PC.0 PD PC 4.50 km l PB + P l 4.50 + 6.00 + 7.50 + 0.0 l 7.50 +.5 l 0.0 km 54. Original area: o bh o xx ( ) New area: n bh n xx ( + 6) n xx ( + 4) 60 n o If the new area is 60 cm larger than the original, xx ( + 4) xx ( ) 60 x + x x + 6x 60 8x 60 x 0 cm is the orignal width d x d 8 cm is the original depth Copyright 05 Pearson Education, Inc.
ISM for Washington, Basic Technical Mathematics with Calculus, SI Version, Tenth Edition. Quadrilaterals.. L 4s+ w+ l L 4 540 + 540 + 90 L 5080 mm. bh ( 7 )( 55 ) 980 000 m bh 7( 55) 960 4000 m hb ( + b) ( 55 )( 7 + 5 ) 94.5 900 m 980 + 960 + 94.5 8900 m tot 4. ( w+.0) + w 6.4 w+ 6.0 + w 6.4 4w 0.4 w 5. mm w +.0 8. mm 5. p 4s 4 65 60 m 6. p 4(.46) 9.84 km 7. p + ( 0.90) ( 0.74).4 mm 8. p + ( 4) ( 6) 56 cm 9. p l+ w.7 +.7.8 m 0. p + ( 7.) ( 4.) 8.0 mm Copyright 05 Pearson Education, Inc.
Chapter : Geometry. p 6. + 7.0 + 44.0 + 6. 4.4 dm. p 7 + 9 + + 67 559 cm. s.7 7. mm 4. 5.6 4 m 5. lw 0.90 0.74 0.68 km 6. lw 4 6 7 900 cm 7. bh.7.5 9. m 8. bh 7..6 44 mm 9. hb+ b ( ) ( 9.8)( 6. 7.0) +.00 0 dm 0. hb+ b ( ) ( 0)( 9 67) + 07 000 cm. p b+ 4a. p a+ b+ b+ a+ b a + b a p a+ b+ b a p 4b. bh + a Copyright 05 Pearson Education, Inc.
ISM for Washington, Basic Technical Mathematics with Calculus, SI Version, Tenth Edition 4. ab+ b a+ ab+ b a ab a + ab a ab a [ ] [ ] 5. The parallelogram is a rectangle. 6. The triangles are congruent. Corresponding sides and angles are equal. 7. 4.0 s s s + s 4.0 8. s 576 576 s s 88 cm D B F C E G Copyright 05 Pearson Education, Inc. 4
Chapter : Geometry CD and FC are alternate interior angles, and so CD FC CB CD because of the angle bisector E CB FC CE and FC are supplementary angles, so CE 80 FC CB CE because of the angle bisector CD CB 90 FC CB + BC + CB 80 FC + BC + 90 FC 80 BC 90 9. The diagonal always divides the rhombus into two congruent triangles. ll outer sides are always equal. 0. The hypotenuse of the right triangle is c a + b c 6 + c 400 c 0 In a rhombus, all four sides are equivalent, so p 4 0 80 mm. For the courtyard p 4 s 8.0 m 4 4 For the outer edge of the walkway, each side will be x 8.0 + 6.00 87.0 m p 4x p 4( 87.0) p 48 m nalysing BC, all interior angles should sum to 80 Copyright 05 Pearson Education, Inc. 5
ISM for Washington, Basic Technical Mathematics with Calculus, SI Version, Tenth Edition. h w h - 450 p h+ h 450 4500 h+ h 900 5400 4h h 50 mm w h 450 w 900 mm. If width increases by 500 mm and length decreases by 4500 mm the dimensions will be equal (a square). w+ 500 4w 4500 6000 w w 000 mm 4w 8000 mm 4. bh.80(.50) 6.0 m Copyright 05 Pearson Education, Inc. 6
Chapter : Geometry 5. The trapezoid has lower base 900 mm and upper base 500 mm, making the lower side 4000 mm longer than the upper side. This means that a right triangle in each corner can be built with hypotenuse c of 00 mm and horizontal leg (base b) of 000 mm c b + h paint ( ) ( ) paint paint paint h c b h 00 000 h 6 890 000 area of trapezoid area of window h( b + b ) lw V V 6 890 000 900 + 500 00 400 8 4 6 mm m L paint 8 4 6 mm paint 000 mm m.4 L of paint (to two significant digits) 6. 75.0 75.0 h 75.0/ 50 75 7.5 + h 75.0 7.5 h h 64.959 cm area of 6 identical trapezoids 6 hb ( + b) (64.959 cm) 75.0 + 50 cm 4 800 cm Copyright 05 Pearson Education, Inc. 7
ISM for Washington, Basic Technical Mathematics with Calculus, SI Version, Tenth Edition 7..74 km.46 km d.86 km d.7 +.86 d.94706 km For the right triangle, bh (.7)(.86). km For obtuse triangle,.46 +.74 + d s.46 +.74 +.94706 s s.0675 km (.46)(.74) s s s d s ( 706)(.0675.74).0675.0675.46.0675.94 0.9707 km quadrilateral Sum of areas of two triangles. km + 0.9707 km.04 km.7 km 8. Cost cost of wall + cost of fence 00 50 w + 5 w + 5w+ 5w 00 0w w 0 m l w l 0 m Copyright 05 Pearson Education, Inc. 8
Chapter : Geometry 9. 60. diagonal divides a quadrilateral into two triangles, and the sum of the interior angles of each triangle is 80. 40. d d The rhombus consists of four triangles, the areas of which are equal since the sides are all consistently 4 ( bh ) ( ) 4 d d dd d and d Copyright 05 Pearson Education, Inc. 9
ISM for Washington, Basic Technical Mathematics with Calculus, SI Version, Tenth Edition.4 Circles. OB + OB + OB 80 OB + 90 + 7 80. πr π.4 8 km OB 8. πs πs p s+ s+ 4 π (.5) p (.5) + p.6 in. (.5) π s π 4 4 8.0 in 4. C BC 5 50 5. (a) D is a secant line. (b) F is a tangent line. 6. (a) EC and BC are chords. (b) ECO is an inscribed angle. 7. (a) F OE. (b) OCE is isosceles. 8. (a) EC and EC enclose a segment. (b) Radii OE and OB enclose a sector with an acute central angle. 9. c πr π 75 70 cm 0. c πr π 0.56.54 m. d r c πd π ;. 7.6 mm. c πd π 8. 6 dm. πr π 0.095 0.085 km Copyright 05 Pearson Education, Inc. 0
Chapter : Geometry 4. πr π 45.8 6590 cm 5. π ( d ) π /. / 4.6 m πd π 56 9 000 mm 4 4 6. 7. CBT 90 BC 90 65 5 8. BCT 90, any angle such as BC inscribed in a semicircle is a right angle and BCT is supplementary to BC. 9. tangent to a circle is perpendicular to the radius drawn to the point of contact. Therefore, BT 90 CBT BT BC 90 65 5 ; CB 5 0. BTC 65 ; CBT 5 since it is complementary to BC 65. ( CBT ) 5 + BTC 90 Therefore BTC 65. BC» 60 0. BC»» B + 80 + 0 60 60 0» B 60. BC / 80 40 since the measure of an inscribed angle is one-half its intercepted arc. 60 80 4. CB 5. 6. 7. π rad 0.5 0.5 0.9 rad 80 π rad 60.0 60.0.05 rad 80 π rad 5. 5..85 rad 80 Copyright 05 Pearson Education, Inc.
ISM for Washington, Basic Technical Mathematics with Calculus, SI Version, Tenth Edition 8. π rad.0.0 5.64 rad 80 π r Perimeter r + r + r 4 9. ( π ) 0.. Perimeter a+ b+ π r+ r 4 rea π r r 4 rea ar + π r 4.. ll are on the same diameter. 4. 45 0 r s B» 45 B s 45 pr 60 p s r 4 5. 6.00 6.00 of sector of quarter circle of triangle π 4 0. cm ( 6.00) ( 6.00)( 6.00) 6. CB DCE (vertical angles) BC DEC and BC CDE (alternate interior angles) The triangles are similar since corresponding angles are equal. Copyright 05 Pearson Education, Inc.
Chapter : Geometry 7. C πr π 8. ( r) 675 40 060 km π 09 r.58 mm 9. basketball hoop 0.5 π 0.445 45.7 π 40. 4. 4. volume π rl flow rate time t πr πr flow rate t t r r r r c c π d d c/ π / π 5.7 cm 5.8 π 96 cm π 9500 cm 4. ( 90 45 ) 44. Let D diameter of large conduit, then D d, where d diameter of smaller conduit F F area large conduit area 7 small conduits d 7π 4 D π 4 7d 7d 7d D 9d 7 9 ( d ) 7 The smaller conduits occupy of the larger 9 conduits. Copyright 05 Pearson Education, Inc.
ISM for Washington, Basic Technical Mathematics with Calculus, SI Version, Tenth Edition 45. of room of rectangle + of circle 4 800( 000) + π ( 0) 4 7 9.7 0 mm Length 5.5 + 4 5.5 7.8 cm 4 46. ( π ) 47. Horizontally and opposite to original direction 48. Let be the left end point at which the dashed lines intersect and C be the center of the gear. Draw a line from C bisecting the 0 angle. Call the intersection of this line and the extension of the upper dashed line B, then 60 5 CB 7.5 4 teeth tooth 0 BC 80 70 x + BC + CB 80 x + 70 + 7.5 80 x.5 x 5 49. s θr 450 s (.8) km s 60 km Copyright 05 Pearson Education, Inc. 4
Chapter : Geometry.5 Measurement of Irregular reas. The use of smaller intervals improves the approximation since the total omitted area or the total extra area is smaller. lso, since the number of intervals would be 0 (an even number) Simpson's Rule could be employed to achieve a more accurate estimate.. Using data from the south end as stated gives only five intervals. Therefore, the trapezoidal rule must be used since Simpson's rule cannot be used for an odd number of intervals.. Simpson's rule should be more accurate in that it accounts better for the arcs between points on the curve, and since the number of intervals (6) is even, Simpson's Rule can be used. 4. The calculated area would be too high since each trapezoid would include more area than under the curve. The shape of the curve is such that a straight line approximation for the curve will always overestimate the area below the curve (the curve dips below the straight line approximation). 5. h trap [ y0 + y+ y +... + yn + yn ].0 trap 0.0 ( 6.4 ) ( 7.4 ) ( 7.0 ) ( 6. ) ( 5. ) ( 5.0 ) ( 5. ) 0.0 + + + + + + + + 84.4 84 m to two significant digits trap 6. h simp ( y0 + 4y + y + 4y + + yn + 4yn + yn) simp 0 + 4 ( 6.4 ) + ( 7.4 ) + 4 ( 7.0 ) + ( 6. ) + 4 ( 5. ) + ( 5.0 ) + 4 ( 5. ) + 0 simp simp 87.8667 m 88 m (to two significant digits) 7. h simp ( y0 + 4y + y + 4y + + yn + 4yn + yn) 0.0 simp 0 4 ( 0.6 ) ( 0. ) 4 ( 0. ) ( 0.5 ) 4 ( 0.0 ) 0.0 + + + + + + 0.448 m 0.45 m (rounded to significant digits) Copyright 05 Pearson Education, Inc. 5
ISM for Washington, Basic Technical Mathematics with Calculus, SI Version, Tenth Edition 8. h trap [ y0 + y + y +... + yn + yn] 0. trap 0 ( 0.6 ) ( 0. ) ( 0. ) ( 0.5 ) ( 0.0 ) 0.0 + + + + + + 0.48 m 0.44 m (rounded to significant digits) trap 9. h trap [ y0 + y + y +... + yn + yn] 0.5 trap 0.6 (. ) ( 4.7 ) (. ) (.6 ) (.6 ) (. ) (.5 ) 0.8 + + + + + + + + 9.8 km trap 0. h simp ( y0 + 4y + y + 4y + + yn + 4yn + yn) 0.5 simp 0.6 4 (. ) ( 4.7 ) 4 (. ) (.6 ) 4 (.6 ) (. ) 4 (.5 ) 0.8 + + + + + + + + 9. km 9. km (rounded to significant digits) simp. h trap [ y0 + y + y +... + yn + yn] 6.0 trap 7 + ( 5 ) + ( 7 ) + ( ) + ( ) + ( 0 ) + ( 9 ) + ( ) + ( 8 ) + trap trap simp 6 km ( 540 mm ) mm 9 000 km. h simp ( y0 + 4y + y + 4y + + yn + 4yn + yn).5 simp 0 4 ( 5.0 ) ( 7. ) 4 ( 8. ) ( 8.6 ) 4 ( 8. ) ( 7. ) 4 ( 5.0 ) 0.0 + + + + + + + + 76. m 76 m. h trap [ y0 + y + y +... + yn + yn].0 trap 0 5. 4. 9.9.0.4.6.5 + + + + + + + + 7.9 + 6.5 +.5 + 9. + 0 trap ] 75.4 km 80 km Copyright 05 Pearson Education, Inc. 6
Chapter : Geometry 4. h simp ( y0 + 4y + y + 4y + + yn + 4yn + yn) simp 0 + 4 5. + 4. + 4 9.9 +.0 + 4.4 +.6 + 4.5 + 7.9 + 4 6.5 +.5 + 4 9. + 0 simp simp [ ] 79.07 km 80 km 5. h simp ( y0 + 4y + y + 4y + + yn + 4yn + yn) 50 simp 5 4 ( ) ( 7 ) 4 ( ) ( ) 4 ( 5 ) ( 6 ) 4 ( 6 ) ( 0 ) 4 ( 8 ) 0 + + + + + + + + + + 8050 m 8.0 0 m 6. h trap [ y0 + y + y +... + yn + yn].0 trap.5 ( 6.0 ) ( 7.6 ) ( 0.8 ) ( 6. ) ( 8. ) ( 9.0 ) ( 7.8 ) (.5 ) 8. + + + + + + + + + 8.7 cm trap circles circles total tot trap π d 4 π (.50 cm) 9.87477 cm 8.7 cm 9.87477 cm al 8.88 cm 0 cm 7. h trap [ y0 + y + y +... + yn + yn] 0.500 trap 0.0 (.7 ) (.000 ) (.7 ) 0.0 + + + +.7 cm This value is less than.4 cm because all of the trapezoids are inscribed. 8. h trap [ y0 + y + y +... + yn + yn] 0.50 0.000 + (.) + (.7) + (.96) + (.000) trap + (.96) + (.7) + (.) + 0.000 trap.00 cm The trapezoids are smaller so they can get closer to the boundary, and less area is missed from the calculation. Copyright 05 Pearson Education, Inc. 7
ISM for Washington, Basic Technical Mathematics with Calculus, SI Version, Tenth Edition 9. h simp ( y0 + 4y + y + 4y + + yn + 4yn + yn) 0.500 simp 0.000 4 (.7 ) (.000 ) 4 (.7 ) 0.000 + + + +.98 cm simp The ends of the areas are curved so they can get closer to the boundary, including more area in the calculation. 0. h simp ( y0 + 4y + y + 4y + + yn + 4yn + yn) 0.50 simp 0.000 + 4 (. ) + (.7 ) + 4 (.96 ) + (.000 ) + 4(.96) + (.7) + 4(.) + 0.000 simp.08 cm The areas are smaller so they can get closer to the boundary. Copyright 05 Pearson Education, Inc. 8
Chapter : Geometry.6 Solid Geometric Figures. V lwh ( )( ) V l w h V 4lwh V 4V The volume increases by a factor of 4.. s r + h h s r h 7.5.9 h.8 cm. V π rh.9 cm V π 0.4 cm V 77 cm 4. 4 V πrh+ πr 40 V π + ( ) (.0) π (.0) V 666.90 m V 700 m 5. V s V 7.5 cm V 66 cm 6. V π rh V π.5 cm 48.4 cm V 897. cm 4 V 8.40 0 cm Copyright 05 Pearson Education, Inc. 9
ISM for Washington, Basic Technical Mathematics with Calculus, SI Version, Tenth Edition 7. πr + πrh π π 689 + 689 99 444 m.99 0 m 8. 4π r 6 4π 0.067 mm 0.056 mm 9. 4 V π r 4 V π V.8 m ( 0.877 m) 0. V π rh V π V 70 mm ( 5. mm) ( 5.66 mm). S π rs S π 78.0 cm 8.8 cm S 0 54.7 cm S 0 500 cm. S ps S 45 m 7 m S 46 900 m. V Bh V 76 cm 0 cm V 50 9 cm 5 V.5 0 cm Copyright 05 Pearson Education, Inc. 40
Chapter : Geometry 4. V Bh V ( 9.0 cm) (. cm) V 949. cm V 940 cm 5. S ph (.09 m)(.05 m) S S.58 m 6. S π rh d S π h S π ( 50 mm)( 47 mm) S 7 5 mm S 70 000 mm S.7 0 mm 7. V 4 5 π π d V π 0.8 cm V r V 0.4969 cm V 0.5 cm Copyright 05 Pearson Education, Inc. 4
ISM for Washington, Basic Technical Mathematics with Calculus, SI Version, Tenth Edition 8. To analyze the right triangle formed by the center of the pyramid base, the top of the pyramid, and any lateral facelength s, notice that the bottom of that triangle has width of half the square base side length..4 b. s h + b h s b h 4.. h 8.796 m V Bh V.4 m 8.796 m V 459.998 m V 460 m 9. s h + r s h + r s 0.74 +.9 πr πrs s.40055 cm + (.9 cm) π (.9 cm)(.40055 cm) π + 7. cm Copyright 05 Pearson Education, Inc. 4
Chapter : Geometry 0. There are four triangles in this shape, all having the same area. Using Hero's formula for each triangle: s ( a+ b+ c) s (.67 dm) s 5.505 dm ss ( a)( s b)( s c) 5.505(.85) 5.505(.85) 5.805 dm The total surface area consists of four of these triangles, 4 5.805 dm. dm Or, we could determine the lateral side length h (triangle heights) from the Pythagorean Theorem a h + a.67 h.67 h.78 dm There are four triangles of the same area, so total surface area is: 4 bh (.67 dm)(.78 dm). dm. 4 V π r 4 d V π 4 d V π 8 V π d 6. + flat curved πr + 4 πr πr + πr π r Copyright 05 Pearson Education, Inc. 4
ISM for Washington, Basic Technical Mathematics with Calculus, SI Version, Tenth Edition. Let r radius of cone, V Let h height of the cone ( r) h cylinder Vcone π rh V cylinder rh Vcone π rh V cylinder V 4. cone conebase π 6 π 4 π π π 4 4πr πr + πrs πr πrs r r + rs r s 5. final original final original final original ( r) 4π 4π r 6π r 4π r 4 6. w weight density volume w γv w N m 000 m 9800 (.00 cm) (.00 km ) m 00 cm km 8.94 0 N w 7. + + lw + wh + lh.0 9.50 + 9.50 8.75 +.0 8.75 604 cm base ends sides Copyright 05 Pearson Education, Inc. 44
Chapter : Geometry 8. The volume of pool can be represented by a trapezoidal right prism V width trapezoid V hb ( + b) w V (4.0) (.60 +.00 ) (5.0) V 648 m 9. V π rh V π d π V 4 V 44 969 m V.4 0 m h ( 0.76 m) ( 540 000 m) 5 0. There are three rectangles and two triangles in this shape. The triangles have hypotenuse c a + b c + 4 c 5.00 cm + rec tan gles triangles (8.50)(5.00) + (8.50)(.00) + (8.50)(4.00) + (4.00)(.00) 4 cm. V BH V 0 50 V 645 000 m 6 V.6 0 m Copyright 05 Pearson Education, Inc. 45
ISM for Washington, Basic Technical Mathematics with Calculus, SI Version, Tenth Edition. Use the Pythagorean Thoerem s h + r s h + r s 8.90 + 4.60 s 0.085 cm S π rs S π ( 4.60 cm)( 0.085 cm) S 45 cm. 4 V π r 4 d V π 4 50. V π V 66 65 m V 66 600 m 4. 4 + 4 V π + V.7 m V πr πrh ( 0.6) π ( 0.6) (.98) 5. The lateral side length can be determined from the Pythagorean Theorem s 8.0 + h s 8.0 + 40.0 s 40.79 mm x + ps + 560 mm 6 (4 6)(40.79) Copyright 05 Pearson Education, Inc. 46
Chapter : Geometry 6. 0.06 0.96 Let n number of revolutions of the lateral surface area S n S 76 n π rh 76 76 n π h d 76 n π dh 76 m n π (0.60 m)(0.96 m) n 4 revolutions 7. c π r 75.7 π r 75.7 r π 4 V π r 4 75.7 V π π V 70 cm 8. S π rh d S π h S π dh ssuming the label overlaps on both ends of the can by 0.50 cm S π (8.50 cm).5 + 0.5 + 0.5 S 4 cm Copyright 05 Pearson Education, Inc. 47
ISM for Washington, Basic Technical Mathematics with Calculus, SI Version, Tenth Edition 9. V V + V cylinder V πr h + πr h V π + V.0995 cm cylinder cylinder cone cone ( 0.65 / ) (.75) π (.5 / ) ( 0.65) V.0 cm 40. cone π r cone cone ( x) fluid 9 ( x) ( x) 9 V 9 y x y cone h 4 To achieve half the volume of the cone V π r ( x) ( x) 4 ( x) ( x) 486 0.565 864 864 x h fluid fluid π π 4 864 x x. 48 cm Copyright 05 Pearson Education, Inc. 48
Chapter : Geometry Review Exercises. CGH and given angle 48 are corresponding angles, so CGH 48 CGE and CGH are supplementary angles so CGE + CGH 80 CGE 80 48 CGE. CGE from Question CGE and EGF are complementary angles so CGE + EGF 90 EGF 90 EGF 58 DGH. CGE from Question CGE and DGH are vertically opposite angles DGH CGE 4. CGE from Question EGI CGE + 90 EGI + 90 EGI 5. c a + b c 9 + 40 c 68 c 4 6. c a + b c 4 + 48 c 500 c 50 Copyright 05 Pearson Education, Inc. 49
ISM for Washington, Basic Technical Mathematics with Calculus, SI Version, Tenth Edition 7. c a + b c 400 + 580 c 496 400 c 704.5565985 c 700 8. c a + b a c b a 65 56 a 089 a 9. c a + b c 6.0 +.80 c 54. c 7.57099 c 7.6 0. c a + b c 6 + 5. c 6506.0 c 8.475775 c 8. c a + b a c b a 6. 9. a 444.7 a.08889 a. Copyright 05 Pearson Education, Inc. 50
Chapter : Geometry. c a + b b c a b 0.885 0.78 b 0.770 b 0.44689 b 0.44. p s p 8.5 mm p 5.5 mm 4. p 4s p 4 5. cm p 60.8 cm 5. bh.5 m.88 m.06 m 6. s ( a+ b+ c) s 75 + 8 + 9 s 6 cm s s a s b s c 6 6 75 6 8 6 9 6(4)(78)(97) 67 004 496 885.67404 cm 890 cm 7. c π r c π d c π ( 98.4 mm) c 09.77 mm c 09 mm Copyright 05 Pearson Education, Inc. 5
ISM for Washington, Basic Technical Mathematics with Calculus, SI Version, Tenth Edition 8. p l+ w p.98 dm +.86 dm p 9.68 dm 9. hb ( + b) 4. cm 67. cm + 6.7 cm 5.69 cm 0 cm 0. π r d π π d 4 π (.8 m) 4 844.96760 m 845 m. V Bh V bl h V 6.0 cm 4.0 cm 4.0 cm V 688 cm V 690 cm. V π rh V π ( 6.0 cm) (.40 cm) V 977.60979 cm V 9770 cm Copyright 05 Pearson Education, Inc. 5
Chapter : Geometry. V Bh V 850 m 5 m V 6046.6667 m V.60 0 m 4. 4 V π r 5 4. mm V π V 565.65404 mm V 5650 mm 5. 6s 6 0.50 m.64 m.6 m 6. πr + πrh d d π + π h π d + π dh π (.0 cm) + π (.0 cm)(5.80 cm) 4.7458 cm 4. cm 7. s r + h s.8 +.5 s 5.564 s.6467 mm S π rs S π.8 mm (.6467 mm) S 66.5788974 mm S 66.6 mm Copyright 05 Pearson Education, Inc. 5
ISM for Washington, Basic Technical Mathematics with Calculus, SI Version, Tenth Edition 8. 4π r 760 km 4π 5 506 576 km 5.5 0 km 8 9. 50 BT 5 0. TB + BT + TB 80 TB 90 since an angle inscribed in a semicircle is 90 BT 5 from Question 9 ll angles in BT must sum to 80 TB 80 90 5 TB 65. BTC is a complementary angle to BT BT 5 from Question 9 BTC + BT 90 BTC 90 5 BTC 65. BT 90 since any angle inscribed in a semi-circle is 90. BE and DC are corresponding angles since BE ~ DC BE DC BE 5 4. D C + CD D (4 + 4) + 6 D 00 D 0 Copyright 05 Pearson Education, Inc. 54
Chapter : Geometry 5. since BE ~ DC BE B CD D BE 4 6 0 6(4) BE 0 BE.4 6. since BE ~ DC E B C D E 4 8 0 4(8) E 0 E. 7. p base of triangle + hypotenuse of triangle + semicircle perimeter p b+ b + ( a) + π ( a) p b+ b + 4a + π a 8. p perimeter of semicircle + 4 square lengths p ( π s) + 4 s p π s+ 4s 9. area of triangle + area of semicircle b( a) + π ( a) ab + π a 40. area of semicircle + area square ( π s ) + s Copyright 05 Pearson Education, Inc. 55
ISM for Washington, Basic Technical Mathematics with Calculus, SI Version, Tenth Edition 4. square is a rectangle with four equal sides. rectangle is a parallelogram with perpendicular intersecting sides so a square is a parallelogram. rhombus is a parallelogram with four equal sides and since a square is a parallelogram, a square is a rhombus. 4. If two triangles share two angles that are the same, then the third angle must also be the same in both triangles. The triangles are similar to each other because they all have the same angles, and the sides must be proportional. 4. π r If the radius of the circle is multplied by n, then the area of the new circle is: π π ( nr) ( nr) ( π ) n r The area of the circle is multiplied by n, when the radius is multplied by n. ny plane geometric figure scaled by n in each dimension will increase its area by n. 44. V s If the length of a cube's side is multplied by n, then the volume of the new cube is: V ( ns) V ns V n s The volume of the cube is multiplied by, when the length of the side is multplied by. n This will be true of any geometric figure scaled by n in all dimensions. n 45. B a E d C b c D BEC ED, since they are vertically opposite angles BC DB, both are inscribed in B CBD CD, both are inscribed in CD which shows ED BEC a b d c Copyright 05 Pearson Education, Inc. 56
Chapter : Geometry 46. B C D We are given BD 6. The two angles BC and DC of the quadrilateral at the point where the tangents touch the circle are each 90. The four angles of the quadrilateral will add up to 60. BC + DC + BD + BCD 60 90 + 90 + 6 + BCD 60 BCD 44 47. The three angles of the triangle will add up to 80. If the tip of the isosceles triangle is 8, find the other two equal angles. (base angle) + 8 80 (base angle) 4 base angle 7 48. The two volumes are equal V sphere V sheet 4 πrsphere πrsheet t 4 dsphere dsheet t dsphere dsheet t 6 4 d t d sphere sheet (.50 cm) t (4.0 cm) t 0.047959 cm t 0.05 cm The flattened sphere is 0.05 cm thick. Copyright 05 Pearson Education, Inc. 57
ISM for Washington, Basic Technical Mathematics with Calculus, SI Version, Tenth Edition 49..0 m c c a + b (.0 m) ( 7.80 m) c + c 6.8 m c 7.8976786 m c 7.89 m 7.80 m 50. 640 m c c a + b ( 640 m) ( 00 m) c + c 0 649 600 m c 6.7489 m c 00 m 5. n equilateral triangle has equal sides, so all edges of the triangle and square are cm p 6 cm cm 5. rea of square + rea of 4 semi-circles π r s + 4 s + π r s s + π π s s + π (4.50 m) (4.50 m) + 5.058656 m 5. m 00 m Copyright 05 Pearson Education, Inc. 58
Chapter : Geometry 5. F 4 m Since line segments BC, D, and EF are parallel, the segments B and CD are proportional to F and DE B F CD DE B 4 m 8 m 54 m 8 m( 4 m) B 54 m B 9.55555556 m B 0 m 54. B C D E 8 m 54 m 5 m B h 45 m 84.0 m Since the triangles are similar, their sides are proportional. h 5 45 + 84 45 5(9) h 45 h 97.479 m Lot is a triangle ( 45 m )( 5 m ) 9060 m Lot B is a trapezoid B ( 5 m 97.479 m )( 84.0 m ) 500 m + Copyright 05 Pearson Education, Inc. 59
ISM for Washington, Basic Technical Mathematics with Calculus, SI Version, Tenth Edition 55. Since the triangles are proportional BF MB E M BF 4.5 m.6 m. m 4.5 m.6 m BF. m BF 6.0 m 56. The triangles are proportional so, DE D BC B DE 6.0 cm.0 cm 4.0 cm 6.0 cm(.0 cm) DE 4.0 cm DE.0 cm 57. The longest distance between points on the photograph is c a + b ( 0.0 cm) ( 5.0 cm) c + c 05 cm c.0569 cm Find the distance in km represented by the longest measure on the map 8 450 m km x (.0569 cm) 00 cm 000 m x 5.90688 km x 5.9 km Copyright 05 Pearson Education, Inc. 60
Chapter : Geometry 58. π r M π r d d L S L S diameter of large piston in cm diameter of small piston in cm dl π M ds π M d.0 M.5 M.8987605 M.90 59. L ds 700 km 45 km The diameter of the satellite's orbit is Earth diameter plus two times its distance from the surface of Earth. c π D c π + ( 700 km ( 45 km) ) c π ( 700 km +690 km) c 4 00 km 60. m x x + 5 Copyright 05 Pearson Education, Inc. 6
ISM for Washington, Basic Technical Mathematics with Calculus, SI Version, Tenth Edition lw xx ( + 5) x + 5x The diagonal is given, so a + b c ( x+ 5) + x x + 0x+ 5 + x 76 x + 0x 696 The left side is twice the area! ( x + 5 x) 696 696 696 8470 m 6. rea of the drywall is the area of the rectangle subtract the two circular cutouts. lw ( π r ) π d lw 4 π d lw π (50 mm) ( 00 mm)( 400 mm) 687 577.45 mm 6.7 0 mm 6..8 0 6 8.50 0.7 0 4 d Copyright 05 Pearson Education, Inc. 6
Chapter : Geometry d The triangles are similar so, d + 50 000 000 700 80 000 d d + 50 000 000 650 690 000 690 000 d 650 + 50 000 000 ( d ) 690 000 d 650 d + 95 500 000 000 68 650 d 95 500 000 000 95 500 000 000 d 68 650 d 9 56.78 km d 6.9 0 km 6. h [ y0 + 4y+ y + 4 y +... + yn + 4yn + yn] 50 0 + 4( 50) + ( 480) + 4( 0 + 90 + 60) + ( 50) + 4( 50) + ( 70) + 4( 560) + 40 50 ( 740) 06 666 m 6. 0 m 64. h V [ y0 + y+ y +... + yn + yn] 50 V 560 780 + + 4650 + 670 + 5600 + 680 + 60 + 0 50 V (55 90) V 6 9 750 m 6 V 6.9 0 m 65. V π rh π d V h 4 π (4. m) V. m 4 V 9.47787 m V 9 m ] Copyright 05 Pearson Education, Inc. 6
ISM for Washington, Basic Technical Mathematics with Calculus, SI Version, Tenth Edition 66. h.50 m.50 m.50 m rea of cross-section is the area of six equilateral triangles with sides of.50 m each Using Hero's formula, s ( a+ b+ c) s (.5 +.5 +.5) s.75 m ss ( a)( s b)( s c).75(.75.5).706 m V area of cross section height V 6.706 m (6.75 m) V 09.60640 m V.0 0 m 67. 000.0 m c a + b a c b ( 500.0 m) ( 500.00 m) a a 00.0 m a 0.000 m 68. c distance apart in km c a + b (.4 km) (.7 km) c + c 9.45 km c 4.40544 km c 4.4 km a 000.00 m Copyright 05 Pearson Education, Inc. 64
Chapter : Geometry 69. V V + V cylinder dome 4 V πrh+ πr Note the height if the cylinder is the total height less the radius of the hemisphere. V π + V 0.8754 m ( 0.80 m) (.05 m 0.80 m) π ( 0.80 m) Convert m to L, V 0.8754 m V 87.54 L V 87 L 70. 000 L m s.5 - x x x x Given x.50 m Find the lateral height s of the pyramid's triangles s a + b s s s x (.5 x) + ( 0.75 m) (.5 m) +.5 m s.45777974 m tent surface area surface area of pyramid + surface area of cube tent surface area 4 triangles + 4 squares 4 xs + 4x (.50 m)(.45778 m) + 4.50 m.887 m. m Copyright 05 Pearson Education, Inc. 65
ISM for Washington, Basic Technical Mathematics with Calculus, SI Version, Tenth Edition 7. w 6 h 9 6h w 9 07 w + h 6h 449 + h 9 56 449 h + h 8 7 449 h 8 8(449) h 7 h 75.86795 cm h 5.45795645 cm h 5.5 cm 6h w 9 6 5.45795645 cm w 9 w 9.585847 cm w 9. cm 7. 60 590 r r r 60 590 96 00 km π r π (96 00 km ) 0 55.75 km 0 000 km Copyright 05 Pearson Education, Inc. 66
Chapter : Geometry 7. 8 m 8 m 454 m 454 m 8 m 8 m 8 m The area is the sum of the areas of three triangles, one with sides 454, 454, and 8 and two with sides 8, 8, and 454. The semi-perimeters are given by 8+ 8+ 454 s 508 454 + 454 + 8 s 594.5 508( 508 8)( 508 8)( 508 454) + 594.5( 594.5 454)( 594.5 454)( 594.5 8) 6 000 m Copyright 05 Pearson Education, Inc. 67