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

Go back to  'Surface Areas and Volumes'

## 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.

### Solution

Reasoning:

Dimensions of the cupboard and thickness wood.

What is  known?

Rate for polishing and painting.

What is  unknown?

Total expenses for painting and polishing

Steps:

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 $$= 6275.$$

## 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}.

### Solution

Reasoning:

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.

Steps:

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 $$= 2784.26$$

## Question 3

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

### Solution

Reasoning:

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.

Steps:

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}

\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 $$= 43.75\%$$