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CBSE Class 9 Answered

please solve this question
Asked by ss8121426 | 28 May, 2020, 09:53: AM
Expert Answer
Question: In the given figure, three circles each of radius 2 cm touch each other externally. These circles are circumscribed by a circle of radius x. Find the value of x and area of the shaded region.
Solution:
Triangle ABC is an equilateral triangle with each side = 4 cm
Let O be the centre of the larger circle and X be the point on it such that O-B-X.
Take D as the mid-point of BC and join OB and BD
Triangle OBD is a right triangle right angled at D and angle OBD = 45ocos 30 degree equals fraction numerator B D over denominator O B end fraction
fraction numerator square root of 3 over denominator 2 end fraction equals fraction numerator 2 over denominator O B end fraction
O B equals fraction numerator 4 over denominator square root of 3 end fraction c m
O X equals O B plus B X equals open parentheses fraction numerator 4 over denominator square root of 3 end fraction plus 2 close parentheses c m
therefore space R a d i u s space o f space t h e space l a r g e r space c i r c l e equals open parentheses fraction numerator 4 over denominator square root of 3 end fraction plus 2 close parentheses c m
A r e a space o f space t h e space s h a d e space r e g i o n equals A r e a space o f space increment A B C space minus space 3 cross times open parentheses A r e a space o f space a space s e c t o r space o f space a n g l e space 60 degree space i n space a space c i r c l e space o f space r a d i u s space 2 space c m close parentheses
equals fraction numerator square root of 3 over denominator 4 end fraction cross times 4 squared minus 3 cross times 60 over 360 cross times straight pi cross times 2 squared
equals open parentheses 4 square root of 3 minus 2 straight pi close parentheses cm squared
Answered by Renu Varma | 28 May, 2020, 11:04: AM
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