Math, asked by shagun160904, 10 months ago

guys please someone solve question no 22​

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Answers

Answered by Anonymous
3

❏ 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.

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