Let d1 and d2 be the diameters of two circles respectively then,d1+d2=14cm ⇒d1=14-d2 (1)Also Circumference of circle one=π×d1 cmand Circumference of circle two=π×d2 cmthen,πd1-πd2=8⇒d1-d2=8π⇒14-d2-d2=8π⇒14-2d2=8π⇒7-d2=4π⇒d2=7-4π⇒d2=7-4×722⇒d2=7-2×711⇒d2=77-1411⇒d2=6311So, d1=14-6311⇒d1=154-6311⇒d1=9111So, Circumference of circle one=π×9111=227×9111=26cmCircumference of circle two=π×6311=227×6311=18cm

Sum number 13

Let d_{1} and d_{2} be the diameters of two circles respectively then, d_{1} + d_{2} = 14 cm \Rightarrow d_{1} = 14 - d_{2} . . . . . (1) Also Circumference of circle one = π \times d_{1} cm and Circumference of circle two = π \times d_{2} cm then, {πd}_{1} - {πd}_{2} = 8 \Rightarrow d_{1} - d_{2} = \frac{8}{π} \Rightarrow 14 - d_{2} - d_{2} = \frac{8}{π} \Rightarrow 14 - 2 d_{2} = \frac{8}{π} \Rightarrow 7 - d_{2} = \frac{4}{π} \Rightarrow d_{2} = 7 - \frac{4}{π} \Rightarrow d_{2} = 7 - \frac{4 \times 7}{22} \Rightarrow d_{2} = 7 - \frac{2 \times 7}{11} \Rightarrow d_{2} = \frac{77 - 14}{11} \Rightarrow d_{2} = \frac{63}{11} So, d_{1} = 14 - \frac{63}{11} \Rightarrow d_{1} = \frac{154 - 63}{11} \Rightarrow d_{1} = \frac{91}{11} So, Circumference of circle one = π \times \frac{91}{11} = \frac{22}{7} \times \frac{91}{11} = 26 cm Circumference of circle two = π \times \frac{63}{11} = \frac{22}{7} \times \frac{63}{11} = 18 cm

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