Question

guys please someone solve question no 22​

Answer 1

❏ Question:-

The decorative solid trophy is made of two solids of shining glass as shown in figure given below . Find the cost of the glass for making the trophy , if one cubic centimetre of glass costs 4.

$\setlength{\unitlength}{0.7 cm}\begin{picture}(12,4)\thicklines\put(7.5,9.675){\circle{1.3}}\put(7.5,12.675){\circle{1.3}}\put(7.5,9.675){\line(0,1){3}}\put(6.89,9.675){\line(0,1){3}}\put(8.1,9.675){\line(0,1){3}}\put(7.5,12.675){\line(1,0){0.6}}\put(8.2 , 11.2){ 28\:cm }\put(5.5,5.5){ A }\put(11.1,5.8){ B }\put(11.08,8.9){ C }\put(5.35,8.5){ D }\put(3.45,10.15){ E }\put(3.4,7.15){ F }\put(9.14,10.235){ H }\put(9.14,7.3){ G }\put(7.75,6.2){ 28\:cm }\put(6,6){\line(1,0){5}}\put(6,9){\line(1,0){5}}\put(11,9){\line(0,-1){3}}\put(6,6){\line(0,1){3}}\put(4,7.3){\line(1,0){5}}\put(4,10.3){\line(1,0){5}}\put(9,10.3){\line(0,-1){3}}\put(4,7.3){\line(0,1){3}}\put(6,6){\line(-3,2){2}}\put(6,9){\line(-3,2){2}}\put(11,9){\line(-3,2){2}}\put(11,6){\line(-3,2){2}}\end{picture}$

❏ Solution:-

✏ Given:-

• side of the cube = 28 cm.

• height of the cylinder = 28 cm.

• diameter of the cylinder = 28 cm.

✏ To Find:-

• Volume of the whole trophy
• cost of making the whole body(trophy).

✏ Explanation:-

➩ for the cube,

$\sf\longrightarrow Volume_{cube}=28^{3}\:cm^3$

$\sf\longrightarrow Volume_{cube}=21,952\:cm^3$

➩ for the cube,

$\sf\longrightarrow Volume_{\red{cylinder}}=\frac{22}{\cancel7}\times 14{}^{2}\times\cancel{28}\:cm^3$

$\sf\longrightarrow Volume_{\red{cylinder}}=(22\times 14{}^{2}\times4)\:cm^3$

$\sf\longrightarrow Volume_{\red{cylinder}}=17,248\:cm^3$

∴ Volume of the Trophy is

$\sf\implies (21,952+17,248)\:cm^3$

$\sf\implies 39,200\:cm^3$

∴ Cost of making the trophy at 4 Rs/cm³.

$\sf\implies Cost=\frac{39,200}{4}\:Rs$

$\sf\implies Cost=\frac{39,200}{4}\:Rs$

$\sf\implies \boxed{\red{\large{Cost=9,800\:Rs}}}$

❏ Formula Used :-

$\setlength{\unitlength}{0.7 cm}\begin{picture}(12,4)\thicklines\put(5.5,5.5){ A }\put(11.1,5.8){ B }\put(11.08,8.9){ C }\put(5.35,8.5){ D }\put(3.45,10.15){ E }\put(3.4,7.15){ F }\put(9.14,10.235){ H }\put(9.14,7.3){ G }\put(7.75,6.2){ a\:cm }\put(6,6){\line(1,0){5}}\put(6,9){\line(1,0){5}}\put(11,9){\line(0,-1){3}}\put(6,6){\line(0,1){3}}\put(4,7.3){\line(1,0){5}}\put(4,10.3){\line(1,0){5}}\put(9,10.3){\line(0,-1){3}}\put(4,7.3){\line(0,1){3}}\put(6,6){\line(-3,2){2}}\put(6,9){\line(-3,2){2}}\put(11,9){\line(-3,2){2}}\put(11,6){\line(-3,2){2}}\end{picture}$

For a Cube of Side a ,

$\sf\longrightarrow\boxed{ Diagonal=\sqrt{3}\times side}$

$\sf\longrightarrow\boxed{T.S.A.=6\times side{}^{2}}$

$\sf\longrightarrow\boxed{ Volume=Side^{3}}$

Where, •T.S.A.=Total Surface area.

For a right circular cylinder of base radius r and

height h,

$\setlength{\unitlength}{1 cm}</p><p>\begin{picture}(12,4)\thicklines\put(2.7,2.7){ . }\put(6,3){\circle{2}}\put(6,6){\circle{2}}\put(6,3){\line(0,1){3}}\put(5.3,3){\line(0,1){3}}\put(6.7,3){\line(0,1){3}}\put(6,3){\line(1,0){0.7}}\put(6,6){\line(1,0){0.7}}\put(6.2,6.13){ r }\put(6.8,4.5){ h }\end{picture}$

$\sf\longrightarrow\boxed{ L.S.A=2\pi r h}$

$\sf\longrightarrow\boxed{ T.S.A.=2\pi r (r+h)}$

$\sf\longrightarrow \boxed{Volume=\pi r{}^{2}h}$

Where, •L.S.A.=Curved Surface area.

•T.S.A.=Total Surface Area.