Let the length of each edge of the cube be \(a\) cm \(\therefore\) The total surface area of the cube = \(6a^2\) square cm If the edge length is increased by 20%, the new edge length = \(a + a \times \cfrac{20}{100}\) cm \(= a + \cfrac{a}{5} = \cfrac{6a}{5}\) cm \(\therefore\) New total surface area of the cube = \(6\left(\cfrac{6a}{5}\right)^2\) square cm = \(\cfrac{216a^2}{25}\) square cm \(\therefore\) Percentage increase in surface area \(= \cfrac{\cfrac{216a^2}{25} - 6a^2}{6a^2} \times 100\%\) \(= \cfrac{(216a^2 - 150a^2) \times 100}{25 \times 6a^2}\%\) \(= \cfrac{4 \times 66a^2}{6a^2}\%\) \(= 44\%\)

Q.Let the length of each edge of the cube be \(a\) cm \(\therefore\) The total surface area of the cube = \(6a^2\) square cm If the edge length is increased by 20%, the new edge length = \(a + a \times \cfrac{20}{100}\) cm \(= a + \cfrac{a}{5} = \cfrac{6a}{5}\) cm \(\therefore\) New total surface area of the cube = \(6\left(\cfrac{6a}{5}\right)^2\) square cm = \(\cfrac{216a^2}{25}\) square cm \(\therefore\) Percentage increase in surface area \(= \cfrac{\cfrac{216a^2}{25} - 6a^2}{6a^2} \times 100\%\) \(= \cfrac{(216a^2 - 150a^2) \times 100}{25 \times 6a^2}\%\) \(= \cfrac{4 \times 66a^2}{6a^2}\%\) \(= 44\%\)

\(\therefore \cfrac{7+x-2+10+x+3}{4}=9\) Or, \(\cfrac{2x+18}{4}=9\) Or, \(2x+18=36\) Or, \(2x=36-18=18\) Or, \(x=9\)