140}{50}=2.8\] Required Area\[={{(2.8)}^{2}}-\pi {{(1.4)}^{2}}\] \[=7.84-\frac{22}{7}\TIMES \frac{1.4}{10}\times \frac{1.4}{10}\] \[=7.84-\frac{616}{100}=7.84-6.16=1.68\,\,{{m}^{2}}\]

"> 140}{50}=2.8\] Required Area\[={{(2.8)}^{2}}-\pi {{(1.4)}^{2}}\] \[=7.84-\frac{22}{7}\TIMES \frac{1.4}{10}\times \frac{1.4}{10}\] \[=7.84-\frac{616}{100}=7.84-6.16=1.68\,\,{{m}^{2}}\]

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Direction for question: Read the passage below and solve the questions based on it. A 20 m long piece of wire is cut into two pieces, one of which is bent into a circle and the other into the square enclosing it. The area of square outside the circle is

10th Class Mathematics in 10th Class 11 months ago

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\[4x+\pi a=20\] \[a=\frac{20}{4+\pi }\Rightarrow a=\frac{20}{4+\frac{22}{7}}\Rightarrow a=\frac{140}{50}=2.8\] Required Area\[={{(2.8)}^{2}}-\pi {{(1.4)}^{2}}\] \[=7.84-\frac{22}{7}\TIMES \frac{1.4}{10}\times \frac{1.4}{10}\] \[=7.84-\frac{616}{100}=7.84-6.16=1.68\,\,{{m}^{2}}\]

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