, find the outside perimeter of the figure.
Ten identical circles are arranged as shown. If the total area of the shaded region is
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A r e a b e t w e e n t h r e e t o u c h i n g u n i t c i r c l e s = 3 − 2 1 π . I n t h e c i r c l e s s h o w n t h e r e a r e 9 s u c h a r e a s . S o a r e a o f e a c h c i r c l e i s a l s o = 9 9 3 − 2 9 π = 3 − 2 1 π . . ⟹ E a c h o f t h e c i r c l e s h o w n i s a n u n i t c i r c l e c i r c u m f e r e n c e 2 π . M i d d l e t w o c i r c l e s o f e a c h o f t h e t h r e e s i d e s c o n t r i b u t e h a l f c i r c l e t h a t i s π . S o t h e s e s i x c i r c l e s c o n t r i b u t e 6 π . T h r e e c o r n e r c i r c l e s h a v e 3 1 π s e g m e n t n o t o n p e r i m e t e r . S o e a c h h a s 3 5 π c o n t r i b u t i o n . T o t a l 3 ∗ 3 5 π = 5 π . S o o u t s i d e p e r i m e t e r o f t h e f i g u r e = ( 6 + 5 ) π = 1 1 π .