Question

Find area of a circular ring whose inner and outer radii are 6 cm and 14 cm respectively​

Answer 1

Solution :

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\qbezier(2.3,0)(2.121,2.121)(0,2.3)\qbezier(-2.3,0)(-2.121,2.121)(0,2.3)\qbezier(-2.3,0)(-2.121,-2.121)(0,-2.3)\qbezier(2.3,0)(2.121,-2.121)(-0,-2.3)\qbezier(1.2,0)(1.121,1.121)(0,1.2)\qbezier(1.2,0)(1.121,-1.121)(0,-1.2)\qbezier(0,-1.2)(-1.121,-1.121)(-1.2,0)\qbezier(-1.2,0)(-1.121,1.121)(0,1.2)\put(-0,0){\vector(-1,0){2.3}}\put(0,0){\vector(0,1){1.2}}\put(-1.9,0.2){ \bf R = 14 \: cm }\put(0.2,0.3){ \bf r = 6 \: cm }\end{picture}$

Area of the ring :

πR² - πr²

> π(R+r)(R-r)

> π(14+6)(14-6)

>> 20 × 8π

>> 160π cm² .

This is the required answer.

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

Answer 2

Answer: 

329.7 square m. Hence the area of circular ring is 329.7 square m.

$329.7 ^{2} \: \: m$