Geometry: An Interactive Journey to Mastery

Lesson 19 Draw an additional radius and label points as shown in Figure S.19.1. +OAC is isosceles, so the two angles labeled a ...

Solutions See Figure S.19.4. 22 + x^2  ^2 gives x 192. Draw one line as shown in Figure S.19.5. The inscribed/central angle ...

PAQB is a kite. Recall that the diagonals of a kite are perpendicular, and one diagonal bisects the other. Label the lengths h, ...

Solutions D௘ͽ xy 2222 81. E௘ͽ xy 32 2 16. F௘ͽ ̈ ̧ ̈ ̧§·§·©¹©¹xy 231611122. G௘ͽ xy 0.67^22 9.21 0.01. D௘ͽ Cr  2 ...

D௘ͽ r 1 = 5 and r 2 = 10. E௘ͽ C 1  ͼ௘௘ͽDQGC 2  ͼ௘௘ͽ'LVWDQFH  92212 15. F௘ͽ 1RWLFHWKDWWKHGLVWDQFHEHWZH ...

 Solutions D௘ͽ &HQWHU PLGSRLQW ͼ௘௘ͽ Radius =^1122 AB    ^2212  (TXDWLRQͼ௘xí௘ͽ^2 ͼ௘yí௘ͽ^2 = ...

 D௘ͽ $UHD ELJVTXDUHíVPDOOVTXDUH ͼ௘x௘ͽ^2 íx^2. E௘ͽ )RXUUHFWDQJOHVDVVKRZQLQFigure S.21.2. F௘ͽ )RXURYHUODSS ...

 Solutions $GHFDJRQKDVVLGHV2QHDFKVLGHRIWKH¿JXUHGUDZDWULDQJOH with apex at the center of the decagon, as sh ...

 E௘ͽ 6HHFigure S.21.13. h^ 75 53andarea 253.^ F௘ͽ :HKDYHKDOIDQHTXLODWHUDOWULDQJOH ௘6HHFigure S.21.14௘ͽ Thus, the ...

 Solutions Lesson 22 Arc length ˜ 360315 2.SSr^72 This gives^7782 ˜ 2,r so r = 2. We have area 360 xxS2,^2  360 S and arc l ...

 So, for our question, we do the following. $UHD DUHDFLUFOHíDUHDRIHDFKZHGJH ^222360    VLQ  FRV  360    ...

 Solutions D௘ͽ $UHD î  E௘ͽ ʌU JLYHVr (^20) S, so area SSr^2  S (^2) S. F௘ͽ   rr 21 S Th ...

 I௘ͽ <HV)RUWULDQJOHVWKHVPDOODQGODUJH¿JXUHVHDFKFORVHO\UHVHPEOHDFLUFOH²VRFORVHO\WKDW we humans c ...

 Solutions 7KHEDVHZLWKVLGHVLVDULJKWWULDQJOHͼ௘^2 ^2 = 25^2 ௘ͽDQGWKHUHIRUHKDVDUHD^12 ˜˜    ...

 F௘ͽ 22 (^222) (^1122)  (^1122)         PP  SA SSSr r rh SSS S ˜ ˜  ˜ ˜  ˜ One hemisphere 12    ...

Solutions Lesson 24 D௘ͽ k^2  JLYHV k = 2. E௘ͽ k^2 = 12 gives k 2. F௘ͽ k^2 = 3 gives k 12. k = 200. D௘ͽ 2211 k^2  ...

G௘ͽ îͼ௘௘ͽ^3 = 27. H௘ͽ ©¹ ̈ ̧§· 121 3 | I௘ͽ $UHD î IW^2  î LQ^2. Number ...

Solutions D௘ͽ $PRXQWRIR[\JHQQHHGHGGHSHQGVRQYROXPH$PRXQWRIR[\JHQDEVRUEHGGHSHQGVRQVXUIDFHDUHD E௘ͽ k = 5, so SA ...

If the radius of the sphere is r, then the side of the cube is 2rͼ௘'UDZDSLFWXUHWRVHHWKLV௘ͽ Probability = 33 23  33  ...

 Solutions +ANM is similar to +ABC by SAS with scale factor k = 2. The area of +ANM is^1  the area of +ABC. The same is tru ...