Question

)The curved surface area of a right circular cylinder of height 7 cm is 44 cm square .Find the diameter of the base of the cylinder.​

Answer 1

h=7cm

surface area =44 cm

curved surface area of cylinder=pie.d.h

=22/7.d.7= 44

22d=44

d=44/22

d=2cm

Answer 2

$\huge\underline{\sf{Solution-}}$

$\begin{gathered}\rm \blue{Let \: assume \: that \: radius \: of \: cylinder \: be \: r \: cm.}\end{gathered}\\\begin{gathered}\rm \blue{So, Diameter \: of \: cylinder, d \: = 2 r \: cm}\end{gathered} \\\begin{gathered}\rm \blue{Given \: that,}\end{gathered}\begin{gathered}\rm \blue{Height \: of \: cylinder, h = 7 \: cm}\end{gathered}\\\begin{gathered}\rm \blue{Curved \: Surface \: Area \: of \: cylinder = 44cm^2}\end{gathered}\\\begin{gathered}\rm \blue{We \: know \: that}\end{gathered} \\\begin{gathered}\rm \blue{Curved \: Surface \: Area \: of \: cylinder \: of \: radius \: r \: and \: height \: h \: is \: given \: by}\end{gathered}\\\begin{gathered}\color{gray}\boxed{ \rm{ \:CSA_{(Cylinder)} = \: 2 \: \pi \: r \: h \: \: }} \end{gathered}\\\begin{gathered}\rm \blue{So, on \: substituting \: the \: values, we \: get}\end{gathered} \\\begin{gathered}\rm \blue{2 \times \dfrac{22}{7} \times r \times 7 = 44 }\end{gathered} \\\begin{gathered}\rm \blue{22 \times (2r) = 44 }\end{gathered} \\\begin{gathered}\rm \blue{ d \: =\dfrac{44}{22}}\end{gathered} \\\begin{gathered}\rm \blue{d \: = \: 2 }\end{gathered} \\ \begin{gathered}\rm \blue{Hence,}\end{gathered}\\\begin{gathered}\color{gray}\rm\implies \:\boxed{ \rm{ \:Diameter_{(Cylinder)} = 2 \: cm \: }} \end{gathered} \\\begin{gathered}\rm \red{Additional \: Information :-} \end{gathered} \\\begin{gathered}\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$