RS AGGARWAL CLASS 9 CHAPTER 15 VOLUMES AND SURFACE AREA OF SOLIDS EXERCISE 15C

 EXERCISE 15C

PAGE NO-588

Question 1:

Note Use $\mathrm{\pi }=\frac{22}{7}$, unless stated otherwise.
Find the curved surface area of a cone with base radius 5.25 cm and slant height 10 cm.

Answer 1:

Radius of the base, r = 5.25 cm

Slant height, l = 10 cm

∴ Curved surface area of the cone = $\mathrm{\pi }rl=\frac{22}{7}\times 5.25\times 10$ = 165 cm²

Thus, the curved surface area of the cone is 165 cm².

Question 2:

Note Use $\mathrm{\pi }=\frac{22}{7}$, unless stated otherwise.
Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.

Answer 2:

Slant height, l = 21 m

Radius of the base, r = $\frac{24}{2}$ = 12 m

∴ Total surface area of the cone = $\mathrm{\pi }r\left(r+l\right)=\frac{22}{7}\times 12\times \left(12+21\right)=\frac{22}{7}\times 12\times 33=\frac{8712}{7} {\mathrm{cm}}^2$

Thus, the total surface area of the cone is $\frac{8712}{7} {\mathrm{cm}}^2$.

Question 3:

Note Use $\mathrm{\pi }=\frac{22}{7}$, unless stated otherwise.
A joker's cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.

Answer 3:

Radius of the cone, r = 7 cm

Height of the cone, h = 24 cm

∴ Slant height of the cone, $l=\sqrt{r^2+h^2}=\sqrt{7^2+{24}^2}=\sqrt{49+576}=\sqrt{625}$ = 25 cm

Area of the sheet required to make one cap = Curved surface area of the cone = $\mathrm{\pi }rl=\frac{22}{7}\times 7\times 25$ = 550 cm²

∴ Area of the sheet required to make 10 such caps = 550 × 10 = 5500 cm²

Thus, the area of the sheet required to make 10 such jocker caps is 5500 cm².

Question 4:

Note Use $\mathrm{\pi }=\frac{22}{7}$, unless stated otherwise.
The curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find the radius of the base and total surface area of the cone.

Answer 4:

Slant height, l = 14 cm

Let the radius of the base be r cm.

Curved surface area of the cone = 308 cm²         (Given)

$$\begin{gathered} \therefore \mathrm{\pi }rl=308 \\ \Rightarrow \frac{22}{7}\times r\times 14=308 \\ \Rightarrow r=\frac{308\times 7}{22\times 14} \\ \Rightarrow r=7 \mathrm{cm} \end{gathered}$$
∴ Total surface area of the cone = $\mathrm{\pi }r\left(r+l\right)=\frac{22}{7}\times 7\times \left(7+14\right)=\frac{22}{7}\times 7\times 21$ = 462 cm²

Thus, the radius of the base is 7 cm and the total surface area of the cone is 462 cm².

Question 5:

Note Use $\mathrm{\pi }=\frac{22}{7}$, unless stated otherwise.
The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of whitewashing its curved surface at the rate of ₹ 12 per m².

Answer 5:

Slant height of the conical tomb, l = 25 m

Radius of the conical tomb, r = $\frac{14}{2}$ = 7 m

∴ Curved surface area of the conical tomb = $\mathrm{\pi }rl=\frac{22}{7}\times 7\times 25$ = 550 m²

Rate of whitewashing = ₹ 12 per m²

∴ Cost of whitewashing the conical tomb

= Curved surface area of the conical tomb × Rate of whitewashing

= 550 × 12

= ₹ 6,600

Thus, the cost of whitewashing the conical tomb is ₹ 6,600.

Question 6:

Note Use $\mathrm{\pi }=\frac{22}{7}$, unless stated otherwise.
A conical tent is 10 m high and the radius of its base is 24 m. Find the slant height of the tent. If the cost of 1 m² canvas is ₹ 70, find the cost of canvas required to make the tent.

Answer 6:

Radius of the conical tent, r = 24 m

Height of the conical tent, h = 10 m
 
∴ Slant height of the conical tent, $l=\sqrt{r^2+h^2}=\sqrt{{24}^2+{10}^2}=\sqrt{576+100}=\sqrt{676}$ = 26 m

Curved surface area of the conical tent = $\mathrm{\pi }rl=\frac{22}{7}\times 24\times 26 {\mathrm{m}}^2$

The cost of 1 m² canvas is ₹ 70.

∴ Cost of canvas required to make the tent

= Curved surface area of the conical tent × ₹ 70

$=\frac{22}{7}\times 24\times 26\times 70$

= ₹ 1,37,280

Thus, the cost of canvas required to make the tent is ₹ 1,37,280.

Question 7:

A bus stop is barricated from the remaining part of the road by using 50 hollow cones made of recycled cardboard. Each one has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is ₹ 25 per m², what will be the cost of painting all these cones? (Use π = 3.14 and $\sqrt{1.04}$ = 1.02).

Answer 7:

Radius of each cone, r = $\frac{40}{2}$ = 20 cm = 0.2 m         (1 m = 100 cm)

Height of each cone, h = 1 m
 
∴ Slant height of each cone, $l=\sqrt{r^2+h^2}=\sqrt{{\left(0.2\right)}^2+1^2}=\sqrt{0.04+1}=\sqrt{1.04}$ = 1.02 m

Curved surface area of each cone = $\mathrm{\pi }rl=3.14\times 0.2\times 1.02$ = 0.64056 m²

∴ Curved surface area of 50 cones = 0.64056 × 50 = 32.028 m²

Cost of painting = ₹ 25 per m²

∴ Total cost of painting all the cones

= Curved surface area of 50 cones × ₹ 25

= 32.028 × 25

= ₹ 800.70

Thus, the cost of painting all the cones is ₹ 800.70.

Question 8:

Note Use $\mathrm{\pi }=\frac{22}{7}$, unless stated otherwise.
Find the volume, curved surface area and the total surface area of a cone having base radius 35 cm and height is 12 cm.

Answer 8:

Radius of the cone, r = 35 cm

Height of the cone, h = 12 cm
 
∴ Slant height of the cone, $l=\sqrt{r^2+h^2}=\sqrt{{35}^2+{12}^2}=\sqrt{1225+144}=\sqrt{1369}$ = 37 cm

(i) Volume of the cone = $\frac{1}{3}\mathrm{\pi }{r}^{2}h=\frac{1}{3}\times \frac{22}{7}\times {\left(35\right)}^2\times 12$ = 15400 cm³

(ii) Curved surface area of the cone = $\mathrm{\pi }rl=\frac{22}{7}\times 35\times 37$ = 4070 cm²

(iii) Total surface area of the cone = $\mathrm{\pi }r\left(r+l\right)=\frac{22}{7}\times 35\times \left(35+37\right)=\frac{22}{7}\times 35\times 72$ = 7920 cm²

Question 9:

Find the volume, curved surface area and the total surface area of a cone whose height and slant height are 6 cm and 10 cm respectively.

Answer 9:

Height of the cone, h = 6 cm
Slant height of the cone, l = 10 cm
Radius, r = $\sqrt{l^2-h^2}=\sqrt{100-36} =\sqrt{64} =8 \mathrm{cm}$

Volume of the cone = $\mathrm{\pi }{r}^{\mathit{2}}h\mathit{ }$
                              $$\begin{gathered} =\frac{1}{3}\times 3.14\times 8^2\times 6 \\ =401.92 {\mathrm{cm}}^3 \end{gathered}$$

Curved surface area of the cone = $\mathrm{\pi r}l$
                                                $$\begin{gathered} =3.14\times 8\times 10 \\ =251.2 {\mathrm{cm}}^2 \end{gathered}$$

∴ Total surface area = $\mathrm{\pi }rl+\mathrm{\pi }{r}^2$
                               $$\begin{gathered} =251.2 + 3.14\times 8^2 \\ =452.16 {\mathrm{cm}}^2 \end{gathered}$$

Question 10:

Note Use $\mathrm{\pi }=\frac{22}{7}$, unless stated otherwise.
A conical pit of diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
HINT 1 m³ = 1 kilolitre.

Answer 10:

Radius of the conical pit, r = $\frac{3.5}{2}$ m 

Depth of the conical pit, h = 12 m
 
∴ Capacity of the conical pit = $\frac{1}{3}\mathrm{\pi }{r}^{2}h=\frac{1}{3}\times \frac{22}{7}\times {\left(\frac{3.5}{2}\right)}^2\times 12$ = 38.5 m³ = 38.5 kL         (1 m³ = 1 kilolitre)

Thus, the capacity of the conical pit is 38.5 kL.

Question 11:

A heap of wheat is in the form of a cone of diameter 9 m and height 3.5 m. Find its volume. How much canvas cloth is required to just cover the heap? (Use π = 3.14).

Answer 11:

Radius of the heap, r = $\frac{9}{2}$ m = 4.5 m

Height of the heap, h = 3.5 m
 
​∴ Volume of the heap of wheat = $\frac{1}{3}\mathrm{\pi }{r}^{2}h=\frac{1}{3}\times 3.14\times {\left(4.5\right)}^2\times 3.5$ = 74.1825 m³

Now, 

Slant height of the heap, $l=\sqrt{r^2+h^2}=\sqrt{{\left(4.5\right)}^2+{\left(3.5\right)}^2}=\sqrt{20.25+12.25}=\sqrt{32.5}$ ≈ 5.7 m

∴ Area of the canvas cloth required to just cover the heap of wheat

= Curved surface area of the heap of wheat

$$\begin{gathered} =\mathrm{\pi }rl \\ \approx 3.14\times 4.5\times 5.7 \\ \approx 80.541 {\mathrm{m}}^2 \end{gathered}$$
Thus, the area of canvas cloth required to just cover the heap is approximately 80.541 m² .

Question 12:

Note Use $\mathrm{\pi }=\frac{22}{7}$, unless stated otherwise.
A man uses a piece of canvas having an area of 551 m², to make a conical tent of base radius 7 m. Assuming that all the stitching margins and wastage incurred while cutting, amount to approximately 1 m², find the volume of the tent that can be made with it.

Answer 12:

 

Area of the canvas = 551 m²
​
Area of the canvas used in stitching margins and wastage incurred while cutting = 1 m² 

∴ Area of the canvas used in making the tent = 551 − 1 = 550 m²

Radius of the tent, r = 7 m

Let the slant height and height of the tent be l m and h m, respectively.

Area of the canvas used in making the tent = 550 m²

$$\begin{gathered} \therefore \mathrm{\pi }rl=550 \\ \Rightarrow \frac{22}{7}\times 7\times l=550 \\ \Rightarrow l=\frac{550}{22}=25 \mathrm{m} \end{gathered}$$

Now,

Height of the tent, $h=\sqrt{l^2-r^2}=\sqrt{{25}^2-7^2}=\sqrt{625-49}=\sqrt{576}$ = 24 m

∴ Volume of the tent = $\frac{1}{3}\mathrm{\pi }{r}^{2}h=\frac{1}{3}\times \frac{22}{7}\times {\left(7\right)}^2\times 24$ = 1232 m³

Thus, the volume of the tent that can be made with the given canvas is 1232 m³.

PAGE NO-589

Question 13:

How many metres of cloth, 2.5 m wide, will be required to make a conical tent whose base radius is 7 m and height 24 metres?

Answer 13:

Radius of the conical tent, r = 7 m
Height of the conical tent, h = 24 m
 $$\begin{gathered} \mathrm{Now}, l=\sqrt{r^2+h^2} \\ =\sqrt{49 +576} \\ =\sqrt{625} \\ =25 \mathrm{m} \end{gathered}$$

 $$\begin{gathered} \mathrm{Curved} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cone} =\mathrm{\pi }rl \\ =\frac{22}{7}\times 7\times 25 \\ = 550 {\mathrm{m}}^2 \\ \mathrm{Here}, \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cloth} =\mathrm{curved} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cone} = 550 {\mathrm{m}}^2 \end{gathered}$$

Width of the cloth = 2.5 m
∴ Length of the cloth = $\frac{\mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cloth}}{\mathrm{width} \mathrm{of} \mathrm{the} \mathrm{cloth}}=\frac{550}{2.5}=220 \mathrm{m}$

Question 14:

Two cones have their heights in the ratio 1 : 3 and the radii of their bases in the ratio 3 : 1. Show that their volumes are in the ratio 3 : 1.

Answer 14:

Let the heights of the first and second cones be h  and 3h, respectively .
Also, let the radius of the first and second cones be 3r and r,  respectively.

∴ Ratio of volumes of the cones = $\frac{\mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{first} \mathrm{cone}}{\mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{second} \mathrm{cone}}$
                            
                                               $$\begin{gathered} =\frac{{\displaystyle \frac{1}{3}}\mathrm{\pi }\times {\left(3\mathrm{r}\right)}^2\times \mathrm{h}}{{\displaystyle \frac{1}{3}}\mathrm{\pi }\times {\mathrm{r}}^2\times 3\mathrm{h}} \\ =\frac{9r^{2}h}{3r^{2}h}=3 : 1 \end{gathered}$$

Question 15:

A cylinder and a cone have equal radii of their bases and equal heights. If their curved surface areas are in the ratio 8 : 5., show that the radius and height of each has the ratio 3 : 4.

Answer 15:

Suppose that the respective radii and height of the cone and the cylinder are r and h.
Then ratio of curved surface areas = 8 : 5
Let the curved surfaces areas be 8x and 5x.

   $$\begin{gathered} \mathrm{i}.\mathrm{e}., 2\mathrm{\pi }rh = 8\mathrm{x} \mathrm{and} \mathrm{\pi }rl = 5\mathrm{x} \Rightarrow \mathrm{\pi }r\sqrt{h^{\mathit{2}}\mathit{+}{r}^2}=5x \\ \mathrm{Hence} 4{\mathrm{\pi }}^{\mathit{2}}{r}^{\mathit{2}}{h}^2=64x^2 \mathrm{and} {\mathrm{\pi }}^2{r}^2(h^{\mathit{2}}\mathit{+}{r}^{\mathit{2}}\mathit{)}=25x^2 \\ \therefore \mathrm{Ratio} \mathrm{of} \mathrm{curved} \mathrm{surface} \mathrm{areas}=\frac{4{\mathrm{\pi }}^2{r}^{\mathit{2}}{h}^2}{{\mathrm{\pi }}^{\mathit{2}}{r}^2(h^{\mathit{2}}\mathit{+}{r}^{\mathit{2}}) }=\frac{64}{25} \\ \Rightarrow \frac{4{\mathrm{\pi }}^2{r}^{\mathit{2}}{h}^2}{{\mathrm{\pi }}^{\mathit{2}}{r}^2(h^{\mathit{2}}\mathit{+}{r}^{\mathit{2}}) }=\frac{64}{25} \\ \Rightarrow \frac{4h^2}{(h^2+r^2)}=\frac{64}{25} \\ \Rightarrow 25h^2 =16h^2 +16r^2 \\ \Rightarrow 9h^2=16r^2 \\ \Rightarrow \frac{r^2}{h^2} =\frac{9}{16} \\ \Rightarrow \frac{r}{h}=\frac{3}{4} \end{gathered}$$

∴ The ratio of the radius and height of the cone is 3 : 4.

Question 16:

A right circular cone is 3.6 cm high and the radius of its base is 1.6 cm. It is melted and recast into a right circular cone having base radius 1.2 cm. Find its height.

Answer 16:

Let the cone which is being melted be denoted by cone 1 and let the cone into which cone 1 is being melted be denoted by cone 2.

Height of cone 1 = 3.6 cm
Radius of the base of cone 1 = 1.6 cm
Radius of the base of cone 2 = 1.2 cm
Let h cm be the height of cone 2.
The volumes of both the cones should be equal.

 $$\begin{gathered} \mathrm{i}.\mathrm{e}., \frac{1}{3}\mathrm{\pi }\times 1.6^2\times 3.6 =\frac{1}{3}\mathrm{\pi }\times 1.2^2\times \mathrm{h} \\ \Rightarrow \mathrm{h} = \frac{1.6\times 1.6\times 3.6}{1.2\times 1.2}=6.4 \mathrm{cm} \end{gathered}$$

∴ Height of cone 2 = 6.4 cm

Question 17:

A circus tent is cylindrical to a height of 3 metres and conical above it. If its diameter is 105 m and the slant height of the conical portion is 53 m, calculate the length of the canvas 5 m wide to make the required tent.

Answer 17:

Radius of the tent, r = 52.5 m
Height of the cylindrical portion of the tent, H = 3 m
Slant height of the conical portion of the tent, l = 53 m
The tent is a combination of a cylindrical and a conical portion.
i.e., area of the canvas = curved surface area of the cone + curved surface area of the cylinder
                                   = $\mathrm{\pi }rl +2\mathrm{\pi }rH$
                                  $$\begin{gathered} =\frac{22}{7}\times \frac{105}{2}\left(53 +2\times 3 \right) \\ =\frac{22}{7}\times \frac{105}{2}\times \left(53+6\right) \\ =9735 {\mathrm{m}}^2 \end{gathered}$$

But area of the canvas = length $\times$ breadth

∴ Length of the canvas = $\frac{\mathrm{area}}{\mathrm{breadth}}=\frac{9735}{5}=1947 \mathrm{m}$
                       

Question 18:

An iron pillar consists of a cylindrical portion 2.8 m high and 20 cm in diameter and a cone 42 cm high is surmounting it. Find the weight of the pillar, given that 1 cm³ of iron weights 7.5 g.

Answer 18:

Height of the cylindrical portion of the iron pillar, h = 2.8 m = 280 cm
Radius of the cylindrical portion of the iron pillar, r = 20 cm
Height of the cone which is surmounted on the cylindrical portion, H = 42 cm
Now, volume of the pillar = volume of the cylindrical portion + volume of the conical portion
                                       = $\mathrm{\pi }{r}^{\mathit{2}}h +\frac{1}{3}\mathrm{\pi }{r}^{\mathit{2}}H$
                                      $$\begin{gathered} =\frac{22}{7}\times {10}^2\left(280+\frac{1}{3}\times 42\right) \\ =\frac{22}{7}\times \left(280+14\right) \\ =\frac{22}{7}\times 100\times 294 \\ =92400 {\mathrm{cm}}^3 \end{gathered}$$

∴ Weight of the pillar = volume of the pillar $\times$ weight per cubic cm
                                 $$\begin{gathered} =92400\times \frac{7.5}{1000} \\ =693 \mathrm{kg} \end{gathered}$$

Question 19:

From a solid right circular cylinder with height 10 cm and radius of the base 6 cm, a right circular cone of the same height and base is removed. Find the volume of the remaining solid.

Answer 19:

Height of the cylinder = 10 cm
Radius of the cylinder = 6 cm
The respective heights and radii of the cone and the cylinder are the same.
∴ Volume of the remaining solid = volume of the cylinder − volume of the cone
                                                 $$\begin{gathered} =\mathrm{\pi }{r}^{\mathit{2}}h\mathit{ }-\frac{1}{3}\mathrm{\pi }{r}^{\mathit{2}}h\mathit{ } \\ =\frac{2}{3}\mathrm{\pi }{r}^{\mathit{2}}h \\ =\frac{2}{3}\times 3.14\times 6^2\times 10 \\ =753.6 {\mathrm{cm}}^3 \end{gathered}$$

Question 20:

Water flows at the rate of 10 metres per minute through a cylindrical pipe 5 mm in diameter. How long would it take to fill a conical vessel whose diameter art the surface 40 cm and depth 24 cm?

Answer 20:

Radius of the cylindrical pipe = 2.5 mm = 0.25 cm
Water flowing per minute = 10 m = 1000 cm
Volume of water flowing per minute through the cylindrical pipe = $\mathrm{\pi }(0.25{)}^{2}1000 {\mathrm{cm}}^3$ = 196.4 cm³
Radius of the the conical vessel = 40 cm
Depth of the vessel = 24 cm
Volume of the vessel = $\frac{1}{3}\mathrm{\pi }(20{)}^{2}24=10057.1 {\mathrm{cm}}^3$
Let the time taken to fill the conical vessel be x min.
Volume of water flowing per minute through the cylindrical pipe $\times$ x = volume of the conical vessel
$\Rightarrow x=\frac{10057.1}{196.4}=51 \min 12 \sec$
∴ The cylindrical pipe would take 51 min 12 sec to fill the conical vessel.

Question 21:

Note Use $\mathrm{\pi }=\frac{22}{7}$, unless stated otherwise.
A cloth having an area of 165 m² is shaped into the form of a conical tent of radius 5 m. (i) How many students can sit in the tent if a student, on an average, iccupies $\frac{5}{7} {\mathrm{m}}^2$ on the ground? (ii) Find the volume of the cone.

Answer 21:

Radius of the conical tent, r = 5 m

Area of the base of the conical tent = $\mathrm{\pi }{r}^2=\frac{22}{7}\times 5^2=\frac{550}{7} {\mathrm{m}}^2$

Average area occupied by a student on the ground = $\frac{5}{7}$ m²

∴ Number of students who can sit in the tent

$$\begin{gathered} =\frac{\mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{base} \mathrm{of} \mathrm{the} \mathrm{conical} \mathrm{tent}}{\mathrm{Average} \mathrm{area} \mathrm{occupied} \mathrm{by} \mathrm{a} \mathrm{student} \mathrm{on} \mathrm{the} \mathrm{ground}} \\ =\frac{{\displaystyle \frac{550}{7}}}{{\displaystyle \frac{5}{7}}} \\ =110 \end{gathered}$$
Thus, the number of students who can sit in the tent is 110.

Let the slant height of the tent be l m.

Curved surface area of the tent = 165 m²

$$\begin{gathered} \therefore \mathrm{\pi }rl=165 \\ \Rightarrow \frac{22}{7}\times 5\times l=165 \\ \Rightarrow l=\frac{165\times 7}{22\times 5} \\ \Rightarrow l=10.5 \mathrm{m} \end{gathered}$$
Let the height of the tent be h m.

$h=\sqrt{l^2-r^2}=\sqrt{{\left(10.5\right)}^2-5^2}=\sqrt{110.25-25}=\sqrt{85.25}$ ≈ 9.23 m

∴Volume of the tent = $\frac{1}{3}\mathrm{\pi }{r}^{2}h=\frac{1}{3}\times \frac{22}{7}\times {\left(5\right)}^2\times 9.23$ ≈ 241.74 m³

Thus, the volume of the conical tent is approximately 241.74 m³.