Chapter 14 Answers - BISD Moodle

Chapter , Lesson 34. (π – )x 2 . [Because ∆ABC is a 30°-60° righttriangle, AB = 2x and AC = . BecauseAB = 2x, OB = x; so the area of the circle isπx 2 . Because the legs of ∆ABC are x andSlicing a Circle.39., its area is x( ) = x 2 . Theshaded area is πx 2 – x 2 = (π – )x 2 .]•35. (5π – 8)x 2 . [The legs of each small righttriangle are 2x and x, and soOB = = = .The radius of circle O is ; so its area isπ( ) 2 = 5πx 2 . The area of the rectangle is(4x)(2x) = 8x 2 . The shaded area is5πx 2 – 8x 2 = (5π – 8)x 2 .]•36. ( – π)x 2 . [The small triangle is a30°-60° right triangle with shorter leg x, andso its longer leg is and AB = 2x .The area of equilateral ∆ABC isAB 2 = (2x ) 2 = x 2 . The radiusof the circle is x and so its area is πx 2 . Theshaded area is x 2 – πx 2 = ( – π)x 2 .]Cable Disaster.37. Scale drawing:38. The strength of the cable depends, not on itsdiameter, but on the area of its cross section.The area of a cross section of a 1-inch cableis π( ) 2 = π 2 in 2 . The total area of thecross sections of two-inch cables is2π( ) 2 = π 2 in 2 . The 1-inch cable shouldhave been replaced by four-inch cables.40. The height of the fifth strip is a leg of a righttriangle with hypotenuse 10 and base 5.h 2 + 5 2 = 10 2 ; soh = = = .The width of the strip is 1; so its area is1⋅ = .41. , , , , = 8, ,= 6, and . ( = ,•42. 72.6.•43. 290. [4(72.6) ≈ 290.]= , = , etc.)44. It would be less than the actual area of thecircle because the rectangular strips do notfill the quarter circle.45. Narrower strips should fill more of thecircle.46. Keep dividing the circle into narrower andnarrower strips.Tangent Circles.•47. a + b. [Its diameter is 2a + 2b = 2(a + b); soits radius is a + b.]•48. πa 2 + πab or πa(a + b). [The blue area is halfthe area of the largest circle plus half thearea of the left circle minus half the area ofthe right circle. π(a + b) 2 + πa 2 – πb 2 =π(a 2 + 2ab + b 2 ) + πa 2 – πb 2 =πa 2 + πab = πa(a + b).]