Geometry: An Interactive Journey to Mastery

(Greg DeLong) #1


Solutions


Lesson 22


  1. Arc length ˜ 360315 2.SSr^72 This gives^7782 ˜ 2,r so r = 2.

  2. We have area 360 xxS2,^2  360 S and arc length ˜ 360 xx 22 S  360 S.
    They are the same numerical value.

  3. D௘ͽ $UHD DUHDVHFWRUíDUHDWULDQJOH u ^11 SS 62 6 6 9 18.
    E௘ͽ $UHDRIVKDGHGUHJLRQ u  ^313131 area circle 1 1 SS 12.
    ௘6HHFigure S.22.1௘ͽ
    F௘ͽ %\7KDOHV¶VWKHRUHPWKHWULDQJOHKDVDULJKWDQJOH,WKDVDƒDQJOHDQG
    therefore, half an equilateral triangle.
    Its sides are 5, 10, and 53, and its area is^122 ˜u 553 25 3.
    The shaded region is half the circle minus this triangle and, therefore,
    has area^1252222 S5.^2  25 3 S 25 3
    G௘ͽ )LUVWQRWLFHWKDWIRUDFLUFOHRIUDGLXVr, the area of the shaded wedge
    shown in Figure S.22.2 is given as follows.
    $UHDVKDGHG DUHDVHFWRUíDUHDWULDQJOH
    2
    2
    2
    22


360 12 2


360


360 sin 2 cos 2
360 sin 2 cos 2.

x rbh
x rbh
xxxrr r
xxxrr

S


S


S


S


˜˜





˜§· §· ̈ ̧ ̈ ̧©¹ ©¹


§· §· ̈ ̧ ̈ ̧©¹ ©¹


90°


11


Figure S.22.1


b hb

Figure S.22.2
Free download pdf