Class 9 maths chapter 13 exercise 13.4 || class 9 surface area and volume exercise 13.4 || class 9 exercise 13.4 || ncert maths class 9 chapter 13 solutions || surface area and volume class 9 exercise13.4
Explore step-by-step solutions for Class 9 Maths Chapter 13, Exercise 13.4, which deepens students’ understanding of Surface Areas and Volumes by focusing on the surface area and volume of a right circular cone. This exercise takes learners beyond the simpler shapes introduced earlier and challenges them to work with slant height, curved surface area, total surface area and volume of conical structures. Through a variety of practical word problems and real-life scenarios, students learn how to apply standard formulas in a meaningful context. This strengthens their spatial reasoning, reinforces the connection between geometry and measurement and equips them with essential skills for advanced topics in mensuration and real-world problem-solving.
class 9 maths chapter 13 exercise 13.4 || class 9 surface area and volume exercise 13.4 || class 9 exercise 13.4 || ncert maths class 9 chapter 13 solutions || surface area and volume class 9 exercise 13.4
Exercise 13.4
1. Find the surface area of a sphere of radius:
(i) 10.5 cm (ii) 5.6 cm (iii) 14 cm
(i) 10.5 cm (ii) 5.6 cm (iii) 14 cm
Answer
(i) Radius (r) of sphere \( =10.5 {~cm} \)Surface area of sphere \( =4 \pi r^{2} \)
\(=4 \times \frac{22}{7} \times 10.5 \times 10.5\)
\(=88 \times 1.5 \times 10.5\)
\(=1386 {~cm}^{2}\)
(ii) Radius (r) of sphere \( =5.6 {~cm} \)
Surface area of sphere \( =4 \pi r^{2} \)
\(=4 \times \frac{22}{7} \times 5.6 \times 5.6\)
\(=88 \times 0.8 \times 5.6\)
\(=394.24 {~cm}^{2}\)
(iii) Radius (r) of sphere \( =14 {~cm} \)
Surface area of sphere \( =4 \pi r^{2} \)
\(=4 \times \frac{22}{7} \times 14 \times 14\)
\(=4 \times 44 \times 14\)
\(=2464 {~cm}^{2}\)
2. Find the surface area of a sphere of diameter:
(i) 14 cm (ii) 21 cm (iii) 3.5 m
(i) 14 cm (ii) 21 cm (iii) 3.5 m
Answer
(i) Radius (r) of sphere \( =\frac{\text { Diameter }}{2} \)\(=\frac{14}{2}\)
\(=7 {~cm}\)
Surface area of sphere \(=4 \pi r^{2}\)
\(=4 \times \frac{22}{7} \times 7 \times 7\)
\(=88 \times 7\)
\(=616 {~cm}^{2}\)
(ii) Radius (r) of sphere \( =\frac{\text { Diameter }}{2} \)
\(=\frac{21}{2}\)
\(=10.5 {~cm}\)
Surface area of sphere \( =4 \pi {r}^{2} \)
\(=4 \times \frac{22}{7} \times 10.5 \times 10.5\)
\(=1386 {~cm}^{2}\)
(iii) Radius (r) of sphere \( =\frac{\text { Diameter }}{2} \)
\(=\frac{3.5}{2}\)
\(=1.75 {~m}\)
Surface area of sphere \( =4 \pi {r}^{2} \)
\(=4 \times \frac{22}{7} \times 1.75 \times 1.75\)
\(=38.5 {~cm}^{2}\)
3. Find the total surface area of a hemisphere of radius 10 cm.
(Use \( \pi=3.14 \))
Answer
Radius (r) of hemisphere \( =10 {~cm} \)Total surface area of hemisphere \( = \) CSA of hemisphere \(+\) Area of circular end of hemisphere
\(=2 \pi {r}^{2}+\pi {r}^{2}\)
\(=3 \pi {r}^{2}\)
\(=[3 \times 3.14 \times 10 \times 10]\)
\(=942 {~cm}^{2}\)
class 9 maths chapter 13 exercise 13.4 || class 9 surface area and volume exercise 13.4 || class 9 exercise 13.4 || ncert maths class 9 chapter 13 solutions || surface area and volume class 9 exercise 13.4
4. The radius of a spherical balloon increases from 7 cm to 14 cm
as air is being pumped into it. Find the ratio of surface areas of
the balloon in the two cases
Answer
Radius (\( {r}_{1} \)) of spherical balloon \( =7 {~cm} \)Radius (\( r_{2} \)) of spherical balloon, when air is pumped into it \( =14 {~cm} \)
\(\text {Ratio}=\frac{\text { Initial surface area }}{\text { Surface area after pumping ballon }}\)
\(=\frac{4 \pi r_{1} \times r_{1}}{4 \pi r_{2} \times r_{2}}\)
\(=\left(\frac{r_{1}}{r_{2}}\right)^{2}\)
\(=\left(\frac{7}{14}\right)^{2}=\frac{1}{4}\)
Therefore, the ratio between the surface areas is \(1:4\)
5. A hemispherical bowl made of brass has inner diameter 10.5 cm.
Find the cost of tin-plating it on the inside at the rate of Rs 16
per 100 \( {cm}^{2} \)
Answer
Inner radius (r) \( =\frac{10.5}{2}=5.25 {~cm} \)Surface area of hemispherical bowl \( =2 \pi {r}^{2}\)
\(=2 \times \frac{22}{7} \times 5.25 \times 5.25\)
\(=173.25 {~cm}^{2}\)
Cost of tin-plating \( 100 {~cm}^{2} \) area \( = \) Rs 16
Cost of tin-plating \( 173.25 {~cm}^{2} \) area \( =\left(\frac{16 \times 173.25}{100}\right) \)
\(=\text {Rs } 27.72\)
class 9 maths chapter 13 exercise 13.4 || class 9 surface area and volume exercise 13.4 || class 9 exercise 13.4 || ncert maths class 9 chapter 13 solutions || surface area and volume class 9 exercise 13.4
6. Find the radius of a sphere whose surface area is \( 154
{~cm}^{2} \).
Answer
Let the radius of the sphere be \(r\)Surface area of sphere \( =154 {~cm}^{2} \)
\(4 \pi r^{2}=154\)
\(r^{2}=\frac{154}{4 \pi}\)
\(\Rightarrow r^{2}=\frac{154 \times 7}{4 \times 22}\)
\(\Rightarrow r^{2}=\frac{49}{1}\)
\(\Rightarrow {r}^{2}=12.25\)
\(\Rightarrow {r}=3.5\)
Therefore, the radius of the sphere whose surface area is \( 154 {~cm}^{2} \) is 3.5 cm
7. The diameter of the moon is approximately one fourth of the
diameter of the earth. Find the ratio of their surface areas.
Answer
Let the diameter of earth be \( d \). Therefore, the diameter of
moon will be \( \frac{d}{4} \)Radius (Earth) \( =\frac{d}{2} \)
Radius (Moon) \( =\frac{1}{2} \times \frac{d}{4}=\frac{d}{8} \)
Surface Area of moon \( =4 \pi\left(\frac{d}{8}\right)^{2} \)
Surface Area of Earth \( =4 \pi\left(\frac{d}{2}\right)^{2} \)
Ratio \( =\frac{\text { Surface Area of moon }}{\text { Surface Area of Earth }} \)
\(=\frac{4 \pi \frac{d}{64}}{4 \pi \frac{d}{4}}\)
\(=\frac{4}{64}\)
\(=\frac{1}{16}\)
8. A hemispherical bowl is made of steel, 0.25 cm thick. The
inner radius of the bowl is 5 cm. Find the outer curved surface
area of the bowl.
Answer
The inner radius of hemispherical bowl \( =5 {~cm} \)The thickness of the bowl \( =0.25 {~cm} \)
Outer radius (r) of hemispherical bowl \( =(5+0.25) {cm} \)
\(=5.25 {~cm}\)
Outer CSA of hemispherical bowl \( =2 \pi {r}^{2} \)
\(=2 \times \frac{22}{7 }\times(5.25)^{2}\)
\(=173.25 {~cm}^{2}\)
9. A right circular cylinder just encloses a sphere of radius r
(see Fig. 13.22). Find
(i) Surface area of the sphere,
(ii) Curved surface area of the cylinder
(iii) Ratio of the areas obtained in (i) and (ii)
(i) Surface area of the sphere,
(ii) Curved surface area of the cylinder
(iii) Ratio of the areas obtained in (i) and (ii)
Answer
(i) Surface area of sphere \( =4 \pi {r}^{2} \)(ii) Height of cylinder \( ={r}+{r}=2 {r} \)
[As sphere touches both upper and lower surface of cylinder]
Radius of cylinder \( ={r} \)
CSA of cylinder \( =2 \pi {rh} \)
\( =2 \pi {r}(2 {r}) \)
\( =4 \pi r^{2} \)
(iii) Ratio \( =\frac{4 \pi r \times r}{4 \pi r \times r} \)
\( =\frac{1}{1}=1 \)
Ratio \( =1: 1 \)
