Test bank and solution of algebra (3)

Chapter GEOMETRY 2.1 Lines and Angles 15 ∠BOD = 180° − 90° − 50° = 40° ∠ABE = 90° 16 ∠3 = 180° − 35° = 145° ∠QOR = 90° − 32° = 58° 17 ∠1 = 180° − 145° = 35° = ∠2 = ∠4 4; and, (a) (b) ∠BOC and ∠COA, ∠COA and ∠AOD, ∠AOD and ∠DOB, ∠DOB and ∠BOC x 7.75 = 5.65 6.50 7.75 ( 6.50 ) x= 5.65 x = 8.92 ft More vertical ∠EBD and ∠DBC are acute angles 18 ∠5 = ∠3 = 145° 19 ∠1° = 62° 20 ∠2 + ∠1 = 180° ⇒ ∠2 = 180° − ∠1 = 180° − 62° = 118° 21 ∠3 = 90° − 62° = 28° 22 ∠1 + ∠3 = 90° ∠3 + ∠4 = 180° ∠3 = 90° − 62° ∠3 = 28° ∠4 = 180° = 180° ∠4 = 152° 23 ∠BDE = 90° − 44° = 46° ∠BDF = 180° − 46° = 134° ∠ABE and ∠CBE are right angles 24 ∠BDE = 90° − 44° = 46° ∠ABE = 90° + 46° = 136° ∠ABC is a straight angle ∠ABD is an obtuse angle 25 ∠DEB = 44° The complement of ∠CBD = 65° is 25° 26 ∠DBE = 46° 10 The supplement of ∠CBD = 65° is 27 = 90° − 44° 180° − 65° = 115° = 46° 11 Sides BD and BC are adjacent to < DBC 28 ° 13 ∠AOB = 90 + 50 = 140 14 ∠AOC = 90° − 50° = 40° ADE = ADB + 90° = ( 90° − 44° ) + 90° 12 The angle adjacent to ∠DBC is ∠DBE ° DFE = 90° − FDE = 136° ° 29 a 3.05 3.05 = ⇒ a = 4.75 ⋅ = 4.53 m 4.75 3.20 3.20 Section 2.1 Lines and Angles 30 31 45 3.20 6.25 = 3.05 b b = 5.96 m 43 c 5.50 5.50 = = 3.05 a 4.53 4.53c = 15.4025 c = 3.40 m 44 = ⇒ AB = 3.225 2.15 AB AC = 2.15 + 3.23 = 5.38 cm x 590 = 860 550 x = 920 m + + = 180° , 45 32 ( 4.75 6.25 = 5.05 d d = 6.64 m 46 CGD = 25° 33 BHC = 34 AHC = CGE = 45° 35 BCH = 36 CHG = 1, 2, and + + = 180° , ( = GCD = 65° HGC = ° 37 GHA = 38 CGF = CHJ = 115° A = ( x + 10 ) , ° A 7 D + + = 180° , = B = ( x − 5) ° ° x=5 (b) x + 10 + x − = 180 The sum of the angles of ABCD = 360° x = 35 ° B = ( 3x − ) (a) x + 20 + 3x − = 90° x = 18° (b) x + 20 = 3x − x = 11° 41 ∠BCD = 180° − 47° = 133° 42 ? = 90° − 28° = 62° + + = 180° = 360° ° A = ( x + 20 ) , ° + ( + 3) + ( + ) + = 180° + 180° (a) x + 10 = x − 40 5) C B + + = 180 39 3= 48 HGE = 110 4, 47 The sum of the angles of ABD is 180° FGE = DEG = 70° HAB = JHA = form a straight line ) ° Chapter GEOMETRY 46 2.2 Triangles ∠5 = 45° ⇒ ∠3 = 45° ∠2 = 180° − 70° − 45° = 65° 1 A = bh = ( 61.2 )( 5.75 ) 2 A = 176 in.2 12 p = 0.862 + 0.235 + 0.684 = 1.781 in 1.781 s= = 0.8905 A = s ( s − a )( s − b )( s − c ) = 8905 (.8905 − 862 )(.8905 − 235 )(.8905 − 684 ) A = 0.586 in.2 13 A = 1 bh = ( 3.46 )( 2.55 ) = 4.41 ft 2 14 A = 1 bh = ( 234 )( 342 ) = 40, 000 mm 2 AC = AB + BC = 6.252 + 3.22 AC = 6.25 + 3.2 2 AC = 7.02 m 15 Area = 1.428 (1.428 − 0.9860 )(1.428 − 0.986 ) 24 h = 3.0 4.0 h = 18 ft (1.428 − 0.884 ) = 0.390 m ∠A = 180° − 84° − 40° = 56° ° ° A = 90 − 48 = 42 ° 16 s = ( 320 ) = 480 A = s (s − a) This is an isosceles triangle, so the base angles are equal ∠A = 180° − ( 66° + 66° ) = 48° ∠A = A = 10 A = (180° − 110° ) = 35° 1 bh = ( 7.6 )( 2.2 ) = 8.4 ft 2 1 bh = (16.0 )( 7.62 ) = 61.0 mm 2 11 Area = 471( 471 − 205 )( 471 − 415 )( 471 − 322 ) = 32,300 cm = 480 ( 480 − 320 ) = 44, 000 yd 17 p = 205 + 322 + 415 p = 942 cm 18 p = 23.5 + 86.2 + 68.4 p = 178 in 19 3(21.5) = 64.5 cm 20 Perimeter = ( 2.45 ) + 3.22 = 8.12 in 21 c = 13.82 + 22.7 = 26.6 ft 22 c = a + b = 2.482 + 1.452 c = 2.87 m 23 b = 5512 − 1752 = 522 cm Section 2.2 Triangles 47 31 An equilateral triangle c2 = a + b2 24 0.8362 = a + 0.474 32 Yes, if one of the angles of the triangle is obtuse a = 0.689 in 33 25 ∠B = 90° − 23° = 67° B D 26 c = 90.52 + 38.42 c = 98.3 cm 27 Perimeter = 98.3 + 90.5 + 38.4 = 227.2 cm 28 A = ( 90.5 )( 38.4 ) = 1740 cm 2 29 B A C A + B = 90° + B = 90° A= ⇒ redraw ΔBDC as B/2 B D A' A/2 C' C A/2 A or + = 90° C ΔADC ∼ ΔA ' DC ' ⇒ DA ' C ' = A / between bisectors = BA ' D B ΔBA ' C ', + ( BA ' D + A / ) = 90° ⎛ A B⎞ from which BA ' D = 90° − ⎜ + ⎟ ⎝2 2⎠ 90 ⎛ A+ B⎞ BA ' D = 90° − ⎜ = 45 ⎟ = 90 − ⎝ ⎠ D + B = 90° ⇒ 2= B and ΔADC as C A D ΔBDC and ΔADC are similar 34 Comparing the original triangle 30 F B C B A A= E D D since ΔAFD is isosceles Since AF = FD ( ΔAFD is isosceles ) and since B and C are midpoints, AB = CD which means ΔBAE and ΔCED are the same size and shape Therefore, BE = EC from which it follows that the inner ΔBCE is isosceles A C to the two smaller triangles shows that all three are similar 35 ∠LMK and ∠OMN are vertical angles and thus equal ⇒ ∠KLM = ∠MON The corresponding angles are equal and the triangles are similar Chapter GEOMETRY 48 (18.0 − y ) = y + 8.02 18.02 − (18.0 ) y + y = y + 8.02 36 ∠ACB = ∠ADC = 90° ; ∠A = ∠A; ∠DCA = ∠CBA, therefore ACB ∼ ADC 37 Since MKL ∼ MNO; KN = KM − MN ; 15 − = KM LM LM = KM ; = ; = ; LM = 72; LM = MN MO 12 38 y = 7.2 ft (two significant digits) 44 (1600 ) A = 2400 ( 2400 − 1600 ) AB 12 = 12 AB = 16 A = 1,100, 000 km 45 A = 39 p = + 25 + 29 = 60 A = 30 ( 30 − )( 30 − 25 )( 30 − 29 ) = 60 sidewalk 4.0 in curb ramp 80 in 1 bh = ( 8.0 )(15 ) = 60 ft 2 46 d = 7502 + 5502 = 930 m Yes, the triangle is perfect 40 = 2400 47 80 in street 2.50 ft L street 20.0 = ⇒ street = 4.0 ( 20.0 ) 4.0 ramp = ( 4.0 ( 20.0 ) ) x + 4.0 = 80.09993758, 10.0 ft 6.00 ft calculator ramp = 80 in (two significant digits) 10.02 = x + 6.002 41 ∠ = x = 8.00 180° − 50° = 65° L= (8.00 + 2.50 ) + 6.00 L = 12.09338662, calculator 42 angle between tower and wire = 90° − 52° = 38° L = 12.1 ft (three significant digits) 48 Taking the triangles in clockwise order and using Pythagorean Theorem together with side opposite 30° angle is half the hypotenuse gives side opposite 43 30° angle and third side, respectively tree 49 d = 182 + 122 + 82 = 23 ft 18.0 ft 50 break 18.0 - y 8.0 ft x = , x = 38 m 45.6 1.12 Section 2.3 Quadrilaterals 49 55 Redraw ΔBCP as 51 B 6.00 P 12.0 - PD ΔAPD is 4.5 5.4 = z 1.2 + z z = 6.0 m A x = z + 4.52 x = 7.5 m 10.0 P y = (1.2 + ) + 5.42 y = 9.0 m H 6.00 10.0 = 12.0 − PD PD ⇒ PD = 7.50 and PC = 12.0 − PD = 4.50 l = PB + PA = 4.502 + 6.002 + 7.502 + 10.02 l = 20.0 mi d H 1.25 D from which ΔBCP ∼ ΔADP, so 52 2.5 C 56 1.25 d = 1.252 + 5.02 d = 5.6 ft 1 wd + 160 = w ( d + 16 ) 2 1 wd + 160 = wd + 8w 2 8w = 160 w = 20 cm d = w − 12 = cm 53 l d 4.0 6.0 2.3 Quadrilaterals 4.0 4.0 ( 8.0 ) 4.0 d = ⇒d = 8.0 6.0 6.0 ⎛ 4.0 ( 8.0 ) ⎞ l = 8.02 + d = 8.02 + ⎜ ⎟ ⎝ 6.0 ⎠ l = 9.6 ft 2 L = s + 2w + 2l ED 312 = 54 80 50 ED = 499 ft = ( 21) + ( 21) + ( 36 ) = 198 in Chapter GEOMETRY 50 1 A1 = bh = ( 72 )( 55 ) = 2000 ft 2 A2 = bh = 72 ( 55 ) = 4000 ft 20 A = ( 392 + 672 )( 201) = 107, 000 cm 2 21 p = 2b + 4a 1 h ( b1 + b2 ) = ( 55 )( 72 + 35 ) 2 = 2900 ft The total lawn area is about 8900 ft A3 = 22 p = a + b + b + a + ( b − a ) + ( b − a ) = 4b 23 A = b × h + a = bh + a ( w + 3.0 ) + w = 26.4 24 A = ab + a ( b − a ) = 2ab − a 2 w + 6.0 + 2w = 26.4 w = 20.4 w = 5.1 mm w + 3.0 = 8.1 mm 25 The parallelogram is a rectangle 26 The triangles are congruent Corresponding sides and angles are equal p = s = ( 65 ) = 260 m p = ( 2.46 ) = 9.84 ft 27 p = ( 0.920 ) + ( 0.742 ) = 3.324 in .0 24 p = (142 ) + (126 ) = 536 cm p = 2l + 2w = ( 3.7 ) + ( 2.7 ) = 12.8 m s s 10 p = ( 27.3) + (14.2 ) = 83.0 in s + s = 24.02 11 p = 0.362 + 0.730 + 0.440 + 0.612 = 2.144 ft 2s = 24.02 24.02 2 A = s = 288 cm s2 = 12 p = 272 + 392 + 223 + 672 = 1559 cm 13 A = s = 2.7 = 7.3 mm 14 A = 15.62 = 243 ft 28 15 A = 0.920 ( 0.742 ) = 0.683 in B A A C B 16 A = 142 (126 ) = 17,900 cm 2 A B 17 A = bh = 3.7 ( 2.5 ) = 9.3 m At top B + A = 180° 18 A = 27.3 (12.6 ) = 344 in B + A = 90° In triangle A + B + C = 180° 19 A = (1 / )( 29.8 )( 61.2 + 73.0 ) = 2000 ft 90° + C = 180° C = 90° Section 2.3 Quadrilaterals 51 29 36 l w = l - 18 sum of interior angles p = 2l + ( l − 18 ) = 180 from which = 1+ + + + + l = 54 in = 180° + 180° w = l − 18 = 54 − 18 = 360° w = 36 in 30 S = 180 ( n − ) 37 S (a) n = +2 180 3600° (b) n = +2 180° n = 22 w + 2.5 = 4w − 4.7 31 A = area of left rectangle + area of right rectangle A = ab + ac A = area of entire rectangle A = a ( b + c ) which illustrates the distributive w = 2.4 ft w = 9.6 ft 38 A = 1.80 × 3.50 = 6.30 ft 39 A = ( area of trapezoid − area of window ) property 32 A = ( a + b )( a + b ) = ( a + b ) ⎛1 ⎞ = ⎜ ( 28 + 16 ) ⋅ − 12 ( 3.5 ) ⎟ ⎝2 ⎠ = 268 ft x gal = ht of trapezoid 320 ft 268 ft 2 A = ab + ab + a + b A = a + 2ab + b which illustrates that the square of the sum is the square of the first term plus twice the product of the two terms plus the square of the second term 33 The diagonal always divides the rhombus into two congruent triangles All outer sides are always equal 34 162 + 122 = 400 = 20 p 320 = = 80 For the 4 outer edge of the walkway: 35 (a) For the courtyard: s = p = ( 80 + ) = 344 m (b) A = 86 − 80 = 996 2 A = 1000 m (2 significant digits) ⎛ 28 − 16 ⎞ = 102 − ⎜ ⎟ ⎝ ⎠ = 8.0 ft x = 0.84 gal l 40, w = 70 yd 130 yd l = 1302 − 702 p = 2l + 2w p = 1302 − 702 + ( 70 ) p = 360 yd Chapter GEOMETRY 52 41 .3 43 h ( 2.27 )(1.86 ) + s ( s − 1.46 )( s − d )( s − 1.74 ) A = 3.04 km A= cm 44 w w = 1.60 ⇒ w = 1.60h h 43.32 = h + w2 = h + (1.6h ) h = 22.9 cm w = 1.60h = 36.7 cm 50 ( 2w ) + ( w ) + 5w + 5w = 13, 200 w = 110 m l = w = 220 m 42 30.0 30.0 45 360° A diagonal divides a quadrilateral into two triangles, and the sum of the interior angles of h each triangle is 180° 15.0 60.0 ⎛ d2 ⎞ ⎛ d2 ⎞ d1 + d1 ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ A = d1d 2 46 A = h = 30.02 + 15.02 ( 30.0 + 60.0 ) ⋅ 30.02 + 15.02 A = 9060 in.2 A = 6⋅ _ 2.4 Circles ∠OAB + OBA + ∠AOB = 180° 43 ∠OAB + 90° + 72° = 180° 1.74 ∠OAB = 18° 1.86 1.46 d A = π r = π ( 2.4 ) A = 18 km 2.27 d = 2.27 + 1.86 2 ( 2.27 )(1.86 ) 1.46 + 1.74 + d For obtuse triangle, s = For right triangle, A = and A = s ( s − 1.46 )( s − d )( s − 1.74 ) A of quadrilateral = Sum of areas of two triangles, 2π s πs = 2s + π ( 3.25 ) p = ( 3.25 ) + p = 11.6 in p = 2s + π ( 3.25 ) = 4 A = 8.30 in A= π s2 Section 2.4 Circles 53 ∠ABT = 90° AC = ⋅ ∠ABC = ( 25° ) ∠CBT = ∠ABT − ∠ABC = 90° − 65° = 25° ; ∠CAB = 25° = 50° 20 ∠BTC = 65° ; ∠CBT = 35° since it is (a) AD is a secant line (b) AF is a tangent line complementary to ∠ABC = 65° (a) EC and BC are chords (b) ∠ECO is an inscribed angle 21 ARC BC = ( 60° ) = 120° BC = ( 60° ) = 120° 22 (a) AF ⊥ OE (b) OCE is isosceles AB + 80° + 120° = 360° AB = 160° (a) EC and EC enclose a segment (b) radii OE and OB enclose a sector with an acute central angle 23 ∠ABC = (1/ ) ( 80° ) = 40° since the measure of an c = 2π r = 2π ( 275 ) = 1730 ft 24 ∠ACB = 10 c = 2π r = 2π ( 0.563) = 3.54 m 11 d = 2r ; c = π d = π ( 23.1) = 72.6 mm inscribed angle is one-half its intercepted arc (160° ) = 80° ⎛ π ⎞ = 0.393 rad 25 022.5° ⎜ ° ⎟ ⎝ 180 ⎠ π rad 12 c = π d = π ( 8.2 ) = 26 in 26 60.0° = 60.0° ⋅ 13 A = π r = π ( 0.09522 ) = 0.0285 yd 27 125.2° = 125.2π rad/180° = 2.185 rad 14 A = π r = π ( 45.8 ) = 6590 cm 28 3230° = 3230° ⋅ 15 A = π ( d / ) = π ( 2.33 / ) = 4.26 m 2 1 16 A = π d = π (1256 ) = 1, 239, 000 ft 4 17 ∠CBT = 90° − ∠ABC = 90° − 65° = 25° ° 18 ∠BCT = 90 , any angle such as ∠BCA inscribed in a semicircle is a right angle and ∠BCT is supplementary to ∠BCA 19 A tangent to a circle is perpendicular to the radius drawn to the point of contact Therefore, 29 P = 180° = 1.05 rad π rad 180° = 56.4 rad πr ( 2π r ) + 2r = + 2r 30 Perimeter = a + b + ⋅ 2π r + r 1 31 Area = π r − r 32 Area = 1 ( ar ) + π r 2 33 All are on the same diameter Chapter GEOMETRY 54 34 A A 39 s 45 r B r AB = 45° s 2π r = s= 45° 360° π O ⋅r B r Asegment = area of quarter circle − area of triangle 1 Asegment = π r − ⋅ r ⋅ r ⋅ 2 r (π − ) Asegment = 35 6.00 40 A 6.00 2r h r- h A of sector = A of quarter circle − A of triangle 1 A = ⋅ π ( 6.00 ) − ( 6.00 )( 6.00 ) A = 10.3 in.2 B 36 ∠ACB = ∠DCE (vertical angles) ∠BAC = ∠DEC and ∠ABC = ∠CDE (alternate interior angles) Therefore, the triangles are similar since corresponding angles are equal AB = r + r AB = 2r AC = 2r AB = in right triangle OAC 2 37 c = 2π r ⇒ π = π= ⎛ 2r ⎞ r = ⎜⎜ ⎟⎟ + ( r − h ) from which ⎝ ⎠ c d ; d = 2r ⇒ r = from which 2r c ⋅ d2 c ⋅ π is the ratio of the circumference to d the diameter h= (2 − ) r π= 41 h = 11.5 km d 38 d 5 = = ⋅ = c 4 16 16 16 c = d ⇒π = = 3.2 5 which is incorrect since π = 3.14159 ⋅ ⋅⋅ r = 6378 - 11.5 km r = 6378 km Section 2.4 Circles ( 6378 + 11.5 ) 55 = d + 63782 using A = d = 383 km A= 42 πd2 , π (12.0 + ( 0.60 ) ) − π (12.0 ) A = 23.8 m 0.346 km d 45 C = 2π r = 2π ( 3960 ) = 24,900 mi 6378 km 46 11( 2π r ) = 109 6378 km r = 1.58 mm ( 6378 + 0.346 ) = d + 6378 2 47 d = 66.4 km 43 d = 682 + 582 d = 89 > 85 48 68 km Abasketball Ahoop ⎛ 12.0 ⎞ ⎟ ⎠ = ⎝ = ⎛ 18.0 ⎞ π⎜ ⎟ ⎝ ⎠ π⎜ volume π r12 L = t time 2 2π r1 πr ⋅ flow rate = = t t r22 = ⋅ r12 flow rate = r2 = 2r1 58 km d 49 c = 112; c = π d ; d = c / π = 112 / π = 35.7 in Signal cannot be received 44 Using A = A= πd2 , π (12.0 + ( 0.60 ) ) A = 23.8 m − π (12.0 ) 2 51 Let D = diameter of large conduit, then D = 3d where d = diameter of smaller conduit F= 0.60 m π D2 = ⋅ π ⋅d2 4 7d 7d 7d F= = = 2 D ( 3d ) 9d F= d = 12.0 m π ( 902 − 452 ) A = 9500 cm 50 A = The smaller conduits occupy conduits of the larger Chapter GEOMETRY 56 52 A of room = A of rectangle + A of circle A = 24 ( 35 ) + π ( 9.0 ) = 1030.85174 A = 1000 ft , two significant digits 53 Length = ( ) ( 2π )( 5.5) + ( )( 5.5) = 73.8 in 54 d = ⋅ 2π 12.52 − 2.252 d = 309 km r plane 12.5 2.25 ITC 2.5 Measurement of Irregular Areas The use of smaller intervals improves the approximation since the total omitted area or the total extra area is smaller Using data from the south end gives 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 upper curve The calculated area would be too high since each trapezoid would include more area than that under the curve AP Atrap = 55 Horizontally and opposite to original direction 56 Let A 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 20° angle Call the intersection of this line and the extension of the upper dashed line B, then 360° 15° = ⇒ ∠ACB = 7.5° 24 teeth tooth 20° ∠ABC = 180° − = 170° ∠ x + ∠ABC + ∠ACB = 180° ∠ x + 170° + 7.5° = 180° ∠ x = 2.5° x = 5° 2.0 ⎡0.0 + ( 6.4 ) + ( 7.4 ) + ( 7.0 ) + ( 6.1) ⎦⎤ ⎣ ⎡⎣ +2 ( 5.2 ) + ( 5.0 ) + ( 5.1) + 0.0 ⎤⎦ Atrap = 84.4 = 84 m to two significant digits Asimp h ( y0 + y1 + y2 + y3 + ⋅⋅⋅ + yn − + yn −1 + yn ) 20 (0 + ( 6.4 ) + ( 7.4 ) + ( 7.0 ) + ( 6.1) + ( 5.2 ) = + ( 5.0 ) + ( 5.1) + 0) = 88 m = Asimp = 1.00 (0 + ( 0.52 ) + ( 0.75 ) + (1.05 ) + (1.15) + (1.00 ) + 0.62) = 4.9 ft Atrap = (0 + ( 0.52 ) + ( 0.75 ) + (1.05 ) + (1.15 ) + (1.00 ) + 0.62) = 4.8 ft Atrap = 0.5 ⎡0.6 + ( 2.2 ) + ( 4.7 ) + ( 3.1) + ( 3.6 ) ⎤⎦ ⎣ ⎡⎣ +2 (1.6 ) + ( 2.2 ) + (1.5 ) + 0.8⎤⎦ Atrap = 9.8 m Section 2.5 Measurement of Irregular Areas 10 Asimp = 0.5 (0.6 + ( 2.2 ) + ( 4.7 ) + ( 3.1) + ( 3.6 ) + (1.6 ) + ( 2.2 ) + (1.5 ) + 0.8) = 9.3 mi 11 A = 10 (38 + ( 24 ) + ( 25 ) + (17 ) + ( 34 ) + ( 29 ) + ( 36 ) + ( 34 ) + 30) ⎛ 23 km ⎞ A = 2330 mm ⎜ 2 ⎟ ⎝ 10 mm ⎠ A = 12, 000 km 2 12 A = ⋅ 4.0 ⎡ ( 55.0 ) + ( ( 54.8 ) ) + ( ( 54.0 ) ) ⎣ + ( ( 53.6 ) ) + ( ( 51.2 ) ) + ( ( 49.0 ) ) + ( ( 31.1) ) + ( ( 21.7 ) ) + ( 0.0 ) ⎤⎦ A = 7500 m 45 [170 + ( 360 ) + ( 420 ) + ( 410 ) + ( 390 ) + ( 350 ) + ( 330 ) + ( 290 ) + 230] Atrap = 120, 000 ft 14 Asimp = 45 (230 + ( 290 ) + ( 330 ) + ( 340 ) + ( 390 ) + ( 410 ) + ( 420 ) + ( 360 ) + 170) 50 (5 + (12 ) + (17 ) + ( 21) + ( 22 ) + ( 25 ) + ( 26 ) + (16 ) + (10 ) + ( ) + 0) = 8100 ft 16 Atrap = = 0.500 ⎡0.0 + (1.732 ) + ( 2.000 ) + (1.732 ) ⎣ + 0.0] = 2.73 in.2 This value is less than 3.14 in.2 because all of the trapezoids are inscribed 18 Atrap = 2.0 (3.5 + ( 6.0 ) + ( 7.6 ) + (10.8 ) + (16.2 ) + (18.2 ) + (19.0 ) + (17.8 ) + (12.5 ) + 8.2) = 229 in.2 Acircle = π r = π ( d / ) = π ( 2.5 / ) = 4.9in.2 Atotal = 229 − ( 4.9 ) = 219 in.2 0.250 (0.000 + (1.323) + (1.732 ) + (1.936 ) + ( 2.000 ) + (1.936 ) = 3.00 in.2 The trapezoids are small so they can get closer to the boundary 19 Asimp = 0.500 (0.000 + (1.732 ) + ( 2.000 ) + (1.732 ) + 0.000) = 2.98 in.2 The ends of the areas are curved so they can get closer to the boundary 20 Asimp = = 120, 000 ft 15 Asimp = 17 Atrap + (1.732 ) + (1.323) + 0.000) + ( ( 45.8 ) ) + ( ( 42.0 ) ) + ( ( 37.2 ) ) 13 Atrap = 57 0.250 (0.000 + (1.323) + (1.732 ) + (1.936 ) + ( 2.000 ) + (1.936 ) + (1.732 ) + (1.323) + 0.000) = 3.08 in.2 The areas are smaller so they can get closer to the boundary Chapter GEOMETRY 58 2.6 Solid Geometric Figures 15 S = 16 S = 2π rh = 2π ( d / ) h = 2π ( 250 / )( 347 ) V1 = lwh1 , V2 = ( 2l )( w )( 2h ) = 4lwh = 4V1 = 273, 000 ft The volume is four times as much 1 ps = ( × 1.092 )(1.025 ) = 3.358 m 2 s = r + h2 17 V = 17.52 = 11.92 + h h = 12.8 cm 22.4 = 11.2; h = s − b = 14.2 − 11.2 2 = 8.73 m 1 V = Bh = ( 22.4 ) ( 8.73) = 1460 m3 3 18 b = 1 ⎛ 11.9 ⎞ V = π r h = π ⎜ ⎟ ( (10.4 ) ) 3 ⎝ ⎠ V = 771 cm3 2 ⎛ 122 ⎞ V = π ( 40.0 ) ⎜ ⎟ + π ( 40.0 ) ⎝ ⎠ V = 441, 000 ft 19 s = h + r = 0.2742 + 3.392 = 3.40 cm A = π r + π rs = π ( 3.39 ) + π ( 3.39 )( 3.40 ) = 72.3 cm V = e3 = 7.153 = 366 ft V = π r h = π ( 23.5 ) ( 48.4 ) = 84, 000 cm 2 20 There are four triangles in this shape ⎛ 3.67 ⎞ s = 3.67 − ⎜ ⎟ = 3.18, A = ps ⎝ ⎠ = ( × 3.67 )( 3.18 ) = 23.3 in.2 A = 2π r + 2π rh = 2π ( 689 ) + 2π ( 689 )( 233) = 3,990, 000 mm A = 4π r = 4π ( 0.067 ) = 0.056 in.2 4 V = π r = π ( 0.8773 ) = 2.83 yd 3 1 10 V = π r h = π ( 25.1) ( 5.66 ) = 3730 m3 3 4 ⎛d ⎞ d3 21 V = π r = π ⎜ ⎟ = π 3 ⎝2⎠ V = πd3 22 A = Aflat + Acurved = π r + ⋅ 4π r 2 = 3π r 11 S = π rs = π ( 78.0 )(83.8 ) = 20,500 cm 1 12 S = ps = ( 345 )( 272 ) = 46,900 ft 2 1 13 V = Bh = ( 0.762 ) (1.30 ) = 0.25 in.3 3 14 V = Bh = ( 29.0 ) (11.2 ) = 9420 cm3 ⎛ ⎞ ⎛ 0.83 ⎞ ⎜ πr ⎟ = π ⎜ ⎟ = 0.15 yd 2⎝3 ⎠ ⎝ ⎠ 23 Vcylinder Vcone π ( 2r ) h = 2 = 3πr h Section 2.6 Solid Geometric Figures 59 (π r + π rs ) 4r = r + rs 24 π r = 34 s = h + r = 3.50 + 1.80 = 3.94 in S = π rs = π (1.80 )( 3.94 ) = 22.3 in.2 3r = rs r = s 4π ( 2r ) final surface area = = original surface area 4π r 2 25 lb 5280 ft ft ⋅1 mi ⋅ ⋅ ⋅1 in 12 in ft mi ton =1.45 × 108 lb ⋅ 2000 lb = 72,500 ton 26 62.4 4 35 V = π r = π ( d / ) 3 = π (165 / ) = 2.35 × 106 ft 36 V = π r + π r h = π ( 2.00 ) + π ( 2.00 ) ( 6.5 ) = 115 ft 27 A = 2lh + 2lw + wh = (12.0 )( 8.75 ) + (12.0 )( 9.50 ) + ( 9.50 )( 8.75 ) 1 ps = 162 + ( )(16 ) 82 + 40 2 = 1560 mm 37 A = l + A = 604 in.2 28 V1 = 50.0 × 78.0 × 3.50 = 13, 650 ft 38 V2 = × 78.0 × 5.00 × 50.0 = 9750 ft V = V1 + V2 = 13, 650 + 9750 = 23, 400 ft 0.06 0.96 29 V = π r h = π ( d / ) h = π ( 4.0 / ) ( 3,960, 000 ) 2 = 5.0 × 107 ft or 0.00034 mi3 30 V = h ( a + ab + b ) = ( 0.750 ) ( 2.502 + 2.50 ( 3.25 ) + 3.252 ) = 6.23 m3 ⎛ 0.06 ⎞ N ⋅π ⎜ ⎟ ( 0.96 ) = 76 ⎝ ⎠ N = 280 revolutions 39 c = 2π r = 29.8 ⇒ r = 29.8 2π 31 V = 1.80 3.932 − 1.80 (1.50 ) = 9.43 ft 4 ⎛ 29.8 ⎞ V = π r3 = π ⎜ ⎟ 3 ⎝ 2π ⎠ 32 There are three rectangles and two triangles in this shape V = 447 in.3 ⎛1⎞ A = ⎜ ⎟ ( 3.00 )( 4.00 ) + 3.00 ( 8.50 ) + 4.00 ( 8.50 ) ⎝2⎠ + 8.50 3.002 + 4.002 = 114 cm 33 V = 1 BH = ( 2502 ) (160 ) = 3,300, 000 yd 3 3 40 Area = (π ⋅ + 0.25 )( 4.25 ) = 41 in.2 Chapter GEOMETRY 60 41 V = cylinder + cone (top of rivet) = π r 2h + π r 2h 2 = π ( 0.625 / ) ( 2.75 ) + π (1.25 / ) ( 0.625 ) = 1.10 in.3 Chapter Review Exercises ∠CGE = 180° − 148° = 32° ∠EGF = 180° − 148° − 90° = 58° ∠DGH = 180° − 148° = 32° 42 p = 18 = 2r + π r 18 r= π +2 ∠EGI = 180 − 148° + 90° = 122° V= ⎛ 18 ⎞ ⋅π ⎜ ⎟ ( 0.075 ) ⎝π +2⎠ V = 1.4 m c = 92 + 402 = 41 c = a + b = 14 + 482 ⇒ c = 50 ⎛4 ⎞ 43 V2 = π r23 = 0.92V1 = 0.92 ⎜ π r13 ⎟ 3 ⎝ ⎠ r = 0.97r1 radius decreased by 3% c = a + b = 4002 + 5802 ⇒ c = 700 c = a + b ⇒ 6500 = a + 56002 ⇒ a = 3300 a = 0.7362 − 0.3802 = 0.630 44 10 a = 1282 − 25.12 = 126 11 c = a + b ⇒ 36.12 = a + 29.32 ⇒ a = 21.1 12 c = a + b ⇒ 0.8852 = 0.7822 + b ⇒ b = 0.414 y = 12 12 − x y = (12 − x ) 2 12 π ⋅ ⋅ ⎛3 ⎞ 3⋅ = ⋅ π ⋅ ⎜ (12 − x ) ⎟ ⋅ (12 − x ) ⎝4 ⎠ x = 12 − 864 13 P = 3s = ( 8.5 ) = 25.5 mm 14 p = 45 = (15.2 ) = 60.8 in 15 A = 1 bh = ( 0.125 )( 0.188 ) = 0.0118 ft 2 16 s = 1 ( a + b + c ) = (175 + 138 + 119 ) = 216 2 x = 2.50 cm A = s ( s − a )( s − b )( s − c ) = 216 ( 216 − 175 )( 216 − 138 )( 216 − 119 ) A = 8190 ft 17 C = π d = π ( 98.4 ) = 309 mm 18 p = 2l + 2w = ( 2980 ) + (1860 ) = 9680 yd Chapter Review Exercises 19 A = 61 1 h ( b1 + b2 ) = ( 34.2 )( 67.2 + 126.7 ) = 3320 in.2 2 ⎛ 32.8 ⎞ 20 A = π r = π ⎜ ⎟ = 845 m ⎝ ⎠ 21 V = Bh = ( 26.0 )( 34.0 )(14.0 ) = 6190 cm3 22 V = π r h = π ( 36.0 ) ( 2.40 ) = 9770 in.3 23 V = 1 Bh = ( 3850 )(125 ) = 160, 000 ft 3 34 AD = 62 + ( + ) = 10 35 BE = ⇒ BE = 2.4 10 36 AE = ⇒ AE = 3.2 10 37 P = b + b + ( 2a ) + π ( 2a ) = b + b + 4a + π a 38 p = ( 2π s ) + 4s = π s + 4s 39 A = 1 b ( 2a ) + ⋅ π ( a ) = ab + π a 2 2 40 A = (π s ) + s 2 4 ⎛ 2.21 ⎞ 24 V = π r = π ⎜ ⎟ = 5.65 mm 3 ⎝ ⎠ 25 A = 6e = ( 0.520 ) = 1.62 m ⎛ ⎛ 12.0 ⎞2 ⎛ 12.0 ⎞ ⎞ 26 A = 2π r + 2π rh = 2π ⎜ ⎜ 58.0 ) ⎟ + ⎜ ⎝ ⎟⎠ ⎜⎝ ⎟⎠ ( ⎟ ⎝ ⎠ A = 2410 ft 27 s = r + h = 1.822 + 11.52 ⇒ s = 1.822 + 11.52 S = π rs = π (1.82 ) 1.82 + 11.52 S = 66.6 in.2 ⎛ 12, 760 ⎞ 28 A = 4π r = 4π ⎜ ⎟ = 5.115 × 10 km ⎝ ⎠ 29 ∠BTA = 50° = 25° A rhombus is a parallelogram with four equal sides and since a square is a parallelogram, a square is a rhombus 42 If two angles are equal then so is the third and the triangles are similar 43 A = π r , r ⇒ nr ⇒ A = π ( nr ) = n (π r ) The area is multiplied by n 30 ∠TBA = 90° , ∠BTA = 25° ⇒ ∠TAB = 90° − 25° = 65° 31 ∠BTC = 90° − ∠BTA = 90° − 25° = 65° 32 ∠ABT = 90° (any angle inscribed in a semi-circle is 90° ) 33 ∠ABE = 90° − 37° = 53° 41 A square is a rectangle with four equal sides and a rectangle is a parallelogram with perpendicular intersecting sides so a square is a parallelogram 44 V = e3 ; e ⇒ ne ⇒ V = ( ne ) = n3 e3 The volume is multiplied by n3 Chapter GEOMETRY 62 2 ⎛ 18.0 ⎞ ⎛ 18.0 ⎞ 52 A = ⎜ ⎟ + 2π ⎜ ⎟ = 52.1 cm ⎝ ⎠ ⎝ ⎠ 45 B C a b c E 53 d D AB 14 = 13 18 AB = 10 m 54 A BEC = AED, vertical BCA = ADB, both are inscribed in AB A CBE = CAD, both are inscribed in CD 140 ft a b which shows ΔAED ∼ ΔBEC ⇒ = d c B = 144° C B A ° 47 2(base angle) + 38 = 180 ° base angle = 71° 48 The two volumes are equal ⎛ 1.50 ⎞ ⎛ 14.0 ⎞ π⎜ ⎟ =π ⎜ ⎟ ⋅t ⎝ ⎠ ⎝ ⎠ t = 0.0115 in 49 L = 0.482 + 7.82 = 7.8 m 50 c = 2100 + 95002 = 9700 ft ⎛ 2.4 ⎞ 51 p = ⎜ ⎟ = 10 cm ⎝ 2⎠ B h 84 ft h 120 = ⇒ h = 192 140 + 84 140 area of A = (140 )(120 ) = 8400 ft 2 area of B = (120 + 192 )( 84 ) = 13, 000 ft 2 B + ( 90° ) + 36° = 180° 46 120 ft ' s 55 FB 1.60 = ⇒ FB = 6.0 m 4.5 1.20 56 DE 33 = ⇒ DE = 22 in 16 24 57 The longest distance in inches between points on the photograph is, 8.00 + 10.0 = 12.8 in from which x 18, 450 = 12.8 ⎛ ft ⎞ ⎛ mi ⎞ x = (12.8 )(18, 450 ) in ⎜ ⎟⎜ ⎟ ⎝ 12 in ⎠ ⎝ 5280 ft ⎠ x = 3.73 mi ⎛ 3.10 ⎞ ⎟ ⎠ 58 MA = ⎝ = 1.90 ⎛ 2.25 ⎞ π⎜ ⎟ ⎝ ⎠ π⎜ 59 c = π D = π ( 7920 + ( 210 ) ) = 26, 200 mi Chapter Review Exercises 60 c = 2π r = 651 ⇒ r = 63 651 2π ( 500.10 ) 67 − 500 = 10 ft ⎛ 651 ⎞ A = π r2 = π ⎜ ⎟ = 33, 700 m ⎝ 2π ⎠ 68 x + 4.0 ft x 1.02 61 A = ( 4.0 )( 8.0 ) − 2π ⋅ = 30 ft 15.6 ft ( x + 4.0 ) 62 1.38 × 10 = x + 15.62 x = 28.4 1.27 × 104 x + = 32.4 ft, length of guy wire 1.50 × 108 69 d 1500 m 600 m d 1.27×104 = d + 1.50 × 108 1.38×106 d = 1.39 × 106 1700 m 250 63 A = ⎡ 220 + ( 530 ) + ( 480 ) ⎤⎦ ⎣ ⎡⎣ +4 ( 320 + 190 + 260 ) + ( 510 ) + ( 350 ) ⎤⎦ d = 17002 − 15002 + 6002 ⎡⎣ +2 ( 730 ) + ( 560 ) + 240 ⎤⎦ A = 1, 000, 000 m 64 V = 250 ⎡560 + (1780 ) + ( 4650 ) + ( 6730 ) ⎣ + ( 5600 ) + ( 6280 ) + ( 2260 ) + 230] V = 6,920, 000 ft 66 Area of cross section = area of six equilateral triangles with sides of 2.50 each triangle has 2.50 ( 3) = 3.75 V = area of cross section × 6.75 = 3.75 ( 3.75 − 2.50 ) × 6.75 = 110 m3 70 w + 44 ft w p = 2l + 2w ⎛ 4.3 ⎞ 65 V = π r h = π ⎜ ⎟ (13) = 190 m ⎝ ⎠ semi-perimeter = d = 1000 m 288 = ( w + 44 ) + w w = 50 ft l = w + 44 = 94 ft 71 V = π r h + ⋅ π r 3 ⎡ ⎛ 2.50 ⎞ ⎛ 2.50 ⎞ ⎛ 2.50 ⎞ ⎤ = ⎢π ⎜ ⎟ ⎜ 4.75 − ⎟ + ⋅ ⋅π ⎜ ⎟ ⎥ ⎠ ⎝ ⎠ ⎥⎦ ⎢⎣ ⎝ ⎠ ⎝ ⎛ 7.48 gal ⎞ ⎜ ⎟ ⎝ ft ⎠ = 159 gal Chapter GEOMETRY 64 72 75 s r r h 3.25 - 2.50 r r Vcyl = π r h = Vhemisphere = r= tent surface area = surface area of pyramid + surface area of cube ( )( 2.50 ) 3h 76 1620 = ps + 4e 2 = ⋅ πr 1590 r ( 3.25 − 2.50 ) 2 ⎛ 2.50 ⎞ +⎜ ⎟ + ( 2.50 ) ⎝ ⎠ = 32.3 m ? ⎛ ⎞ A = π r ⎜ 1620 − 15902 ⎟ = 303, 000 km ⎝ ⎠ 16h w 16 = ⇒w= h 73 ⎛ 16h ⎞ 1522 = w2 + h = ⎜ ⎟ + h ⇒ h = 74.5 cm ⎝ ⎠ 16h w= = 132 cm 74 77 Label the vertices of the pentagon ABCDE The area is the sum of the areas of three triangles, one with sides 921, 1490, and 1490 and two with sides 921, 921, and 1490 The semi-perimeters are given by 921 + 921 + 1490 = 1666 and 921 + 1490 + 1490 s2 = = 1950.5 s1 = h +2 k +3 A = 1666 (1666 − 921)(1666 − 921)(1666 − 1490 ) + 1950.5 (1950.5 − 1490 )(1950.5 − 1490 ) 3k -1 + ( 5k + ) = ( 4k + 3) + ( 3k − 1) 2 k =3 4k + = 15, 3k − = A= (8)(15 ) = 60 ft 2 Note: 1) 3k − > ⇒ k > < k < 3) For k = the triangle solution is an isosceles, but not right triangle 2) There is a solution for (1950.5 − 921) = 1, 460, 000 ft ... cm 74 77 Label the vertices of the pentagon ABCDE The area is the sum of the areas of three triangles, one with sides 921, 1490, and 1490 and two with sides 921, 921, and 1490 The semi-perimeters... r Asegment = area of quarter circle − area of triangle 1 Asegment = π r − ⋅ r ⋅ r ⋅ 2 r (π − ) Asegment = 35 6.00 40 A 6.00 2r h r- h A of sector = A of quarter circle − A of triangle 1 A = ⋅... 1.74 + d For obtuse triangle, s = For right triangle, A = and A = s ( s − 1.46 )( s − d )( s − 1.74 ) A of quadrilateral = Sum of areas of two triangles, 2π s πs = 2s + π ( 3.25 ) p = ( 3.25 )