Answer:
Radius of outer circle (R) = 4 cm \[\therefore \] Area of the outer circle \[\pi {{R}^{2}}\] \[=\text{3}.\text{14(4}{{\text{)}}^{\text{2}}}\text{c}{{\text{m}}^{\text{2}}}\] \[=\text{3}.\text{14}\times \text{16 c}{{\text{m}}^{\text{2}}}\] \[=\text{5}0.\text{24 c}{{\text{m}}^{\text{2}}}.\] Radius of inner circle (r) \[~=\text{3 cm}\] \[\therefore \] Area of the inner circle \[=\pi {{r}^{2}}\] \[\text{= 3}\text{.14 }\!\!\times\!\!\text{ (3}{{\text{)}}^{\text{2}}}\text{ c}{{\text{m}}^{\text{2}}}\] \[\text{= 28}\text{.26 c}{{\text{m}}^{\text{2}}}\] \[\therefore \] Area of the remaining sheet = Area of the outer circle ? Area of the inner circle \[=\text{5}0.\text{24 c}{{\text{m}}^{\text{2}}}-\text{ 28}.\text{26 c}{{\text{m}}^{\text{2}}}\] \[=\text{21}.\text{92 c}{{\text{m}}^{\text{2}}}.\]
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