Question

find the total surface area of close right circular cylinder whose height is 10 m and radius is 7m.​

Answer 1

Given: Height and Radius of a closed circular cylinder is 10 m and 7 m respectively.

To find: Total surface area of closed circular cylinder?

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$\setlength{\unitlength}{1.2mm}\begin{picture}(5,5)\thicklines\multiput(-0.5,-1)(26,0){2}{\line(0,1){40}}\multiput(12.5,-1)(0,3.2){13}{\line(0,1){1.6}}\multiput(12.5,-1)(0,40){2}{\multiput(0,0)(2,0){7}{\line(1,0){1}}}\multiput(0,0)(0,40){2}{\qbezier(1,0)(12,3)(24,0)\qbezier(1,0)(-2,-1)(1,-2)\qbezier(24,0)(27,-1)(24,-2)\qbezier(1,-2)(12,-5)(24,-2)}\multiput(16,2)(0,32){2}{\sf{7 m}}\put(14,17.5){\sf{10 m}}\end{picture}$

⠀⠀

$\dag\;{\underline{\frak{As\;we\;know\;that,}}}$

$\star\;{\boxed{\sf{\pink{Total\:surface\:area_{\;(cylinder)} = 2 \pi r(h + r)}}}}$

$\sf Here\begin{cases} & \text{Radius,\:r = \bf{7\;m}} \\ & \text{Height,\:h = \bf{10\;m}} \end{cases}$

$\dag\;{\underline{\frak{Putting\: values\:in\;formula,}}}$

$:\implies\sf TSA_{\;(cylinder)} = 2 \times \dfrac{22}{7} \times 7(10 + 7)$

$:\implies\sf TSA_{\;(cylinder)} = 2 \times \dfrac{22}{ \cancel{7}} \times \cancel{7} \times 17$

$:\implies\sf TSA_{\;(cylinder)} = 2 \times 22 \times 17$

$:\implies\sf TSA_{\;(cylinder)} = 44 \times 17$

$:\implies{\underline{\boxed{\frak{\purple{TSA_{\;(cylinder)} = 748\;m^2 }}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Hence,\;Total\:surface\;area\;of\;closed\;cylinder\;is\; \bf{748\;m^2}.}}}$

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$\qquad\qquad\boxed{\underline{\underline{\pink{\bigstar \: \bf\:More\:to\:know\:\bigstar}}}}$

$\begin{array}{|c|c|c|}\cline{1-3}\bf Shape&\bf Volume\ formula&\bf Surface\ area\ formula\\\cline{1-3}\sf Cube&\tt l^3}&\tt 6l^2\\\cline{1-3}\sf Cuboid&\tt lbh&\tt 2(lb+bh+lh)\\\cline{1-3}\sf Cylinder&\tt {\pi}r^2h&\tt 2\pi{r}(r+h)\\\cline{1-3}\sf Hollow\ cylinder&\tt \pi{h}(R^2-r^2)&\tt 2\pi{rh}+2\pi{Rh}+2\pi(R^2-r^2)\\\cline{1-3}\sf Cone&\tt 1/3\ \pi{r^2}h&\tt \pi{r}(r+s)\\\cline{1-3}\sf Sphere&\tt 4/3\ \pi{r}^3&\tt 4\pi{r}^2\\\cline{1-3}\sf Hemisphere&\tt 2/3\ \pi{r^3}&\tt 3\pi{r}^2\\\cline{1-3}\end{array}$

Answer 2

Answer:

Given :-

• A close right circular cylinder whose height is 10 m and radius is 7 m.

To Find :-

• What is the total surface area of a close right circular cylinder.

Formula Used :-

$\boxed{\bold{\small{T.S.A\: of\: a\: Cylinder\: =\: 2{\pi}r(h + r)}}}$

Solution :-

Given :

• Height = 10 m
• Radius = 7 m

According to the question by using the formula we get,

⇒ T.S.A = $2 \times \dfrac{22}{7} \times 7(10 + 7)$

⇒ T.S.A = $2 \times \dfrac{22}{7} \times 7(17)$

⇒ T.S.A = $2 \times \dfrac{22}{7} \times 119$

⇒ T.S.A = $\dfrac{5236}{7}$

⇒ T.S.A = $\dfrac{\cancel{5236}}{\cancel{7}}$

➠ T.S.A = 748 m²

$\therefore$ The total surface area of close right circular cylinder is 748 m² .