Exercise E13.9 Surface Areas and Volumes NCERT Solutions Class 9

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Chapter 13 Ex.13.9 Question 1

A wooden bookshelf has external dimensions as follows: Height \(= 110\rm\, cm,\) Depth \(= 25\rm\,cm,\) Breadth \(= 85 \rm\,cm\) (see Fig. \(13.31\)). The thickness of the plank is \(5\rm\, cm\) everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is \(20\) paise per \(\begin{align}\rm\,c{m^2} \end{align}\) and the rate of painting is \(10\) paise per \(\begin{align}\rm\,c{m^2} \end{align}\). Find the total expenses required for polishing and painting the surface of the bookshelf.


Video Solution


Dimensions of the cupboard and thickness wood.

What is  known?

Rate for polishing and painting.

What is  unknown?

Total expenses for painting and polishing


length \((l)=25 \rm\,cm\)

breadth \((b)=85 \rm\,{cm}\)

height \((h)=110 \rm\,{cm}\)

Surface area to be polished 

\[\begin{align}  & =\left[ \begin{array}  & \left( h\times b \right)+2\left( h\times l \right)+2\left( b\times l \right)+ \\  \,2\left( h\times 5 \right)+4\left( 75\times 5 \right) \\ \end{array} \right] \\  & =\left[ \begin{array}  & \left( 110\times 85 \right)+2\left( 110\times 25 \right)+ \\  2\left( 85\times 25 \right)+2\left( 110\,\times 25 \right)+ \\   4\left( 75\times 5 \right) \\ \end{array} \right] \\  & =\left( \begin{array} 9350+5500+4250+ \\  1100+1500 \\ \end{array} \right)\,c{{m}^{2}} \\  & =21700\,c{{m}^{2}} \\ \end{align}\]

Expense required for polishing at the rate of \(20\) paise per \(\rm\, cm^2\)

\(\therefore\) Total expense

\[\begin{align}&=\frac{21700 \times 20}{100} \\ &=\operatorname{Rs} 4340 \end{align}\]

Surface area to be painted

\[\begin{align}  & =\left[ \begin{array}2(20\times 90)+6(75\times 20)+ \\  (75\times 90) \\ \end{array} \right] \\ & =(3600+9000+6750)c{{m}^{2}} \\  & =19350\text{c}{{\text{m}}^{2}} \\ \end{align}\]

At the rate of \(10\) paise per \(\rm sq. cm = 1935\)

Expense required for polishing and painting the surface of the bookshelf.

\(= 4340 + 1935 = 6275\)


Total expense required for polishing and painting the surface of the bookshelf \(= 6275.\)

Chapter 13 Ex.13.9 Question 2

The front compound wall of a house is decorated by wooden spheres of diameter \(21\rm\,cm,\) placed on small supports as shown in Fig 13.32. Eight such spheres are used for this purpose and are to be painted silver. Each support is a cylinder of radius \(1.5 \rm\,cm\) and height \(7\rm\, cm\) and is to be painted black. Find the cost of paint required if silver paint costs \(25\) paise per \(\begin{align}\rm\,c{m^2} \end{align}\) and black paint costs \(5\) paise per \(\begin{align}\rm\,c{m^2} \end{align}\).


Video Solution


Surface area of sphere \(=4 \pi r^{2}\)

Surface area of the cylinder \( =2 \pi r h\)

What is  known?

Diameter of the sphere.

Measurement of the cylinder.

Cost per \(\rm\, cm^2 \) for silver and black paint.

What is  unknown?

Cost of the paint.


For wooden sphere:

\[\begin{align} \text{Diameter} &= 2r = 21 \; \rm{mm}\\ &= r = \frac{{21}}{2}\,\,cm \end{align}\]

Surface area for wooden sphere \(= 4\)

\[\begin{align} &= 4 \times \frac{{22}}{7} \times {(\frac{{21}}{2})^2}\\ &= 1386\,\,c{m^2} \end{align}\]

Since the support is in cylinder of radius \(1.5\rm\,cm.\)

So the area of wooden sphere to be painted 

\[\begin{align} &= 1386 - \frac{{22}}{7}{(1.5)^2}\\&\text{[ Area of the sphere is ]}\\&= 1378.93\,\,\rm\,{m^2} \end{align}\]

Total number of sphere are \(8.\)

Total volume of the sphere to be Silver painted \(\begin{align} \end{align}\)

\[\begin{align} &= 8 \times 1378.93\\&= 11031.44\,\,\rm\,c{m^2} \end{align}\]

Cost of painting at the rate of \(25\) paise per \(\begin{align}\rm\,{m^2} \end{align}\).

\[\begin{align} &= \frac{{11031.44 \times 25}}{{100}}\\ &= 2757.86 \end{align}\]

For a cylindrical support.

Radius \((r) =1.5 \rm\,m\)

Height \((h) =7\rm\, m\)

Surface of the cylindrical support \( =2 \pi r h\)

\[\begin{align} = \frac{2}{3} \times \frac{{22}}{7} \times 1.5 \times 7 = 66\,\,\rm\,{m^2} \end{align}\]

Surface area of \(8\) cylindrical support

\[\begin{align} = 8 \times 66 = 528\,\,\rm\,c{m^2} \end{align}\]

Cost of black painting per \(\rm\,cm^2\) \(= 5\) paise

Total cost of black painting 

\[\begin{align}&= \frac{{528 \times 5}}{{150}}\\ &= 26.40\end{align}\]

Cost of paint

\[\begin{align}&= 2757.856 + 26.40 \\ &= 2784.26 \end{align}\]


Cost of paint \(= 2784.26\)

Chapter 13 Ex.13.9 Question 3

The diameter of a sphere is decreased by \(25\%\). By what percent does its curved surface area decrease?


Video Solution


Curved surface area of a sphere is \( 4^2 \). So, if diameter is decreased by \(25\%\), then radius also decreases by \(25\%\) so, the percentage change in curved surface area will be the ratio of difference between old area and new area by old area multiplied by \(100.\)

What is  known?

Decreased percentage of diameter.

What is  unknown?

Decreased percentage of curved surface area.


Let the radius of the sphere be\(\begin{align}\,\frac{r}{2} \end{align}\) \(cm.\)

Then it’s diameter\(\begin{align} = 2(\frac{r}{2}) \end{align}\)\(= r \rm\,cm\)

Curved surface area of the original sphere:

\[\begin{align} = 4\pi {\left( {\frac{r}{2}} \right)^2} = \pi {r^2}c{m^2} \end{align}\]

New diameter of the sphere:

\(\begin{align}r&={  r \!-\! r \!\times\! \frac{{25}}{{100}}\!\left[ \begin{array}{l}{\text{diameter of a sphere}}\\{\text{ is decreased by 25}}\% \end{array} \right]}\\&={  \frac{{3r}}{4}cm}\end{align}\)

Radius of the new sphere:

\[\begin{align} = \frac{1}{2}(\frac{{3r}}{4})= \frac{3}{8r} cm\end{align}\]

New curved surface area of the sphere.

\[\begin{align} =4 \pi\left(\frac{3 r}{8}\right)^{2}=\frac{9 \pi r^{2}}{16} \mathrm{cm}^{2}\end{align}\]

Decrease in the original curved Surface area

\[\begin{align} &= \pi {r^2} - \frac{{9\pi {r^2}}}{{16}}\\ &= \frac{{16\pi {r^2} - 9\pi {r^2}}}{{16}}\\ &= \frac{{7\pi {r^2}}}{{16}} \end{align}\]

Percentage of decrease in the original curved surface area

\[\begin{align} &= \frac{{7\pi {r^2}}}{{16}} \times 100 \\ &= 43.75 \% \end{align}\]


Hence the original curved surface area decrease by \(= 43.75\%\)

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