Geometry: An Interactive Journey to Mastery



Solutions

6. 7KHEDVHZLWKVLGHVLVDULJKWWULDQJOHͼ௘^2 ^2 = 25^2 ௘ͽDQGWKHUHIRUHKDVDUHD^12 ˜˜   
         
      
     

LA

SA

V

u  u  u



˜

7. D௘ͽ %DVHHGJH VSA ˜ ˜^1132   ^22 ˜˜˜     
   ௘6HHFigure S.23.5௘ͽ
   E௘ͽ 6ODQWKHLJKW V = 11,760; LA 
   ௘6HHFigure S.23.6௘ͽ
   F௘ͽ +HLJKW  6222 32 ;VSA^13 16 32 ; 16 8 32.
   G௘ͽ /HWe be the base edge. Then, 10 13 ee^2 6 gives 5.
   Slant height: s   636.^2 ©¹ ̈ ̧§·e^25 
   SA u ˜  ˜ e^2 ^12 e s     


8. Area base u˜˜ 61 3^1122332.
   WDQ   VR WDQ  ^ |^1 h h^
   Thus,
   V|˜ ˜ |^13233 2.05 1.8.


9. D௘ͽ RISSr^22 u    NP  2
   E௘ͽ $UHDRIÀRRU ʌU^2. Area of hemisphere = 21 SSrr^22 GRXEOHWKHÀRRU6RZHQHHGJDOORQV

8 10

6

Figure S.23.5

20 29

21

Figure S.23.6