NCERT Solutions for Class 9 Maths Chapter 13: Surface Areas and Volumes - Exercise 13.9

Exercise 13.9

1. A wooden bookshelf has external dimensions as follows: Height = \[110  \text{cm}\] , Depth = $25  \text{cm}$, Breadth = \[85  \text{cm}\] (see the given figure). The thickness of the plank is $5  \text{cm}$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 $\text{c}{{\text{m}}^{2}}$and the rate of painting is $10$ paise per $\text{c}{{\text{m}}^{2}}$, find the total expenses required for polishing and painting the surface of the bookshelf.

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Ans:

External length \[\left( l \right)\]of book self\[=85  \text{cm}\].

External breadth $\left( b \right)$of book self\[=\]$25  \text{cm}$.

External height \[\left( h \right)\]of book self\[=\]\[110  \text{cm}\].

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External surface area of shelf while leaving out the front face of the shelf = \[lh\text{ }+\text{ }2\left( lb\text{ }+\text{ }bh \right)\]

\[=\text{ }\left[ 85\text{ }\times \text{ }110\text{ }+\text{ }2\text{ }\left( 85\text{ }\times \text{ }25\text{ }+\text{ }25\text{ }\times \text{ }110 \right) \right]\text{ c}{{\text{m}}^{2}}\]

\[=\text{ }\left( 9350\text{ }+\text{ }9750 \right)\text{ c}{{\text{m}}^{2}}\]

\[=\text{ }19100\text{ c}{{\text{m}}^{2}}\]

Area of front face \[=\text{ }\left[ 85\text{ }\times \text{ }110\text{ }-\text{ }75\text{ }\times \text{ }100\text{ }+\text{ }2\text{ }\left( 75\text{ }\times \text{ }5 \right) \right]\text{ c}{{\text{m}}^{2}}\]

\[=\text{ }1850\text{ }+\text{ }750\text{ c}{{\text{m}}^{2}}\]

\[=\text{ }2600\text{ c}{{\text{m}}^{2}}\]

Area to be polished \[=\text{ }\left( 19100\text{ }+\text{ }2600 \right)\text{ }c{{m}^{2}}\]

\[=\text{ }21700\text{ c}{{\text{m}}^{2}}\]

Cost of polishing \[1\text{ c}{{\text{m}}^{2}}\]area \[=\text{ Rs}.\text{ }0.20\]

Cost of polishing \[21700\text{ c}{{\text{m}}^{2}}\] area Rs. \[\left( 21700\text{ }\times \text{ }0.20 \right)\text{ }=\text{ Rs}.\text{ }4340\]

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It can be observed that length \[\left( l \right)\] , breadth$\left( b \right)$, and height \[\left( h \right)\]of each row of the bookshelf is\[75 \text{cm}\],$20  \text{cm}$ and $30  \text{cm}$ respectively.

Area to be painted in $1$ row \[=\text{ }2\left( l\text{ }+\text{ }h \right)b\text{ }+\text{ }lh\]

\[=\text{ }\left[ 2\text{ }\left( 75\text{ }+\text{ }30 \right)\text{ }\times \text{ }20\text{ }+\text{ }75\text{ }\times \text{ }30 \right]\text{ c}{{\text{m}}^{2}}\]

\[=\text{ }\left( 4200\text{ }+\text{ }2250 \right)\text{ c}{{\text{m}}^{2}}\]

\[=\text{ }6450\text{ c}{{\text{m}}^{2}}\]

Area to be painted in $3$ rows \[=\text{ }\left( 3\text{ }\times \text{ }6450 \right)\text{ c}{{\text{m}}^{2}}\text{ }=\text{ }19350\text{ c}{{\text{m}}^{2}}\]

Cost of the painting \[1\text{ c}{{\text{m}}^{2}}\]area \[=\text{ Rs}.\text{ }0.10\] 

Cost of the  painting \[19350\text{ c}{{\text{m}}^{2}}\] area \[=\text{ Rs}.\text{ }\left( 19350\text{ }\times \text{ }0.1 \right)\]

\[=\text{ }Rs.\text{ }1935\]

Total expense required for polishing and painting \[\text{ Rs}.\text{ }\left( 4340\text{ }+\text{ }1935 \right)\]

\[=\text{ }Rs.\text{ }6275\]

 Therefore, the Total expense for  polishing and painting the surface of the bookshelf cost is  \[\text{Rs}\text{. }6275\] .

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

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Ans:

Radius (r) of wooden sphere$=\left(\frac{21}{2}\right) \mathrm{cm}=10.5 \mathrm{~cm}$.

Surface area of wooden sphere$=4 \pi \mathrm{r}^{2}$ $=\left[4 \times \frac{22}{7} \times(10.5)^{2}\right] \mathrm{cm}^{2}=1386 \mathrm{~cm}^{2}$.

Radius $\left(\mathrm{r}_{1}\right)$ of the circular end of cylindrical support$=1.5 \mathrm{~cm}$.

Height (h) of cylindrical support $=7 \mathrm{~cm}$ Curved Surface Area of cylindrical support$=2 \pi r h$. $=\left[2 \times \frac{22}{7} \times(1.5) \times 7\right] \mathrm{cm}^{2}=66 \mathrm{~cm}^{2}$.

Area of the circular end of cylindrical support $=\pi r^{2}=\left[\frac{22}{7} \times(1.5)^{2}\right] \mathrm{cm}^{2}$ $=7.07 \mathrm{~cm}^{2}$.

Area to be painted silver $=[8 \times(1386-7.07)] \mathrm{cm}^{2}$ $=(8 \times 1378.93) \mathrm{cm}^{2}=11031.44 \mathrm{~cm}^{2}$.

Cost for painting with silver color  Rs. \[(11031.44\times 0.25)=\]\[\text{Rs}\]. \[2757.8\].

Area to be painted black$=(8 \times 66) \mathrm{cm}^{2}=528 \mathrm{~cm}^{2}$.

Cost for painting with black color \[=\text{ Rs}.\text{ }\left( 528\text{ }\times \text{ }0.05 \right)\text{ }=\text{ Rs}.\text{ }26.40\].

Total cost in painting\[=\text{ Rs}.\text{ }\left( 2757.86\text{ }+\text{ }26.40 \right)\].

\[=\text{ Rs}.\text{ }2784.26\].

Therefore, it will cost \[\text{Rs}.\text{ }2784.26\] in painting in such a way.

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

Let the diameter of the sphere be$d$.

Radius $\left(\mathrm{r}_{1}\right)$ of sphere $=\frac{d}{2}$ new radius $\left(\mathrm{r}_{2}\right)$ of sphere $=\frac{d}{2}\left(1-\frac{25}{100}\right)=\frac{3}{8} d$

CURVED SURFACE AREA $\left(\mathrm{S}_{1}\right)$ of sphere $=4 \pi \mathrm{r}_{1}^{2}$

$=4 \pi\left(\frac{d}{2}\right)^{2}=\pi d^{2}$

CURVED SURFACE AREA $\left(\mathrm{S}_{2}\right)$ of sphere when radius is decreased $=4 \pi \mathrm{r}_{2}^{2}$

$=4 \pi\left(\frac{3 d}{8}\right)^{2}=\frac{9}{16} \pi d^{2}$

Decrease in surface area of sphere $=\mathrm{S}_{1}-\mathrm{S}_{2}$

 $=\pi d^{2}-\frac{9}{16} \pi d^{2}$

$=\frac{7}{16} \pi d^{2}$

Percentage decrease in surface area of sphere $=\frac{S_{1}-S_{2}}{S_{1}} \times 100$

$=\frac{7\pi {{d}^{2}}}{16\pi {{d}^{2}}}\times 100$

$=\frac{7 \pi d^{2}}{16 \pi d^{2}} \times 100=\frac{700}{16}=43.75 \%$

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.9

Opting for the NCERT solutions for Ex 13.9 Class 9 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 13.9 Class 9 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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