Solved 2 questions given below ...i want to answer must please..
Anand kumar , 7 Years ago
Grade 10
10 grade maths> Solved 2 questions given below ...i want ...
1 Answers
VISWA TEJA
Last Activity: 7 Years ago
- LET CENTRE OF SEMICIRCLE BE ‘O’,RADIUS OF SEMICIRCLE BE ‘R’. LET ‘r’ IS RADIUS OF SMALL CIRCLE THEREFORE,R=AD=BC=DC/2=AB/2. FROM TRIANGLE OBC, OB2 =OC2 +BC2 BUT OB=R+2r;OC=BC=R. SUBSTITUING IN ABOVE EQUATION (R+2r)2 =R2+R2=2R2 R+2r=(2R2)^1/2=1.414R i.e.,(ROOT 2*R) (CONSIDER # AS ROOTOVER SYMBOL SINCE WE DONT HAVE ONE) R+2r=#2 * R 2r=(#2-1)R ; r=(#2-1)/2*R AREA OF SMALL CIRCLE=3.14*r2; AREA OF SEMICIRCLE=(3.14*R2)/2 AREA RATIO IS r2 :R2/2 OR 2r2:R2 SUBSTITUTING VALUE OF r,WE GET RATIO AS 2*((#2-1)/2*R)2:R2=(#2-1)2:2. OPTION ‘C’ IS ANSWER
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