# NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.4

## NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes (Ex 13.4) Exercise 13.4 Free PDF download of NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.4 (Ex 13.4) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.4 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all NCERT Solutions in your emails. You can also download NCERT Solutions Class 9 Maths to help you to revise complete syllabus and score more marks in your examinations. Students can also avail of NCERT Solutions for Class 9 Science from our website.

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Exercise (13.4)

1. Find the surface area of a sphere of radius:

i. $\text{10}\text{.5 cm}$

Ans:

Given radius of the sphere $\text{r = 10}\text{.5 cm}$

The surface area of the sphere $\text{A = 4 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{4 }\!\!\times\!\!\text{ }\frac{\text{22}}{\text{7}}\text{ }\!\!\times\!\!\text{ }{{\left( \text{10}\text{.5} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = }\left( \text{88 }\!\!\times\!\!\text{ 1}\text{.5 }\!\!\times\!\!\text{ 10}\text{.5} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 1386 c}{{\text{m}}^{\text{2}}}$

Hence, the surface area of the sphere is $\text{1386 c}{{\text{m}}^{\text{2}}}$.

ii. $\text{5}\text{.6 cm}$

Ans:

Given radius of the sphere $\text{r = 5}\text{.6 cm}$

The surface area of the sphere $\text{A = 4 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{4 }\!\!\times\!\!\text{ }\frac{\text{22}}{\text{7}}\text{ }\!\!\times\!\!\text{ }{{\left( \text{5}\text{.6} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = }\left( \text{88 }\!\!\times\!\!\text{ 0}\text{.8 }\!\!\times\!\!\text{ 5}\text{.6} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 394}\text{.24 c}{{\text{m}}^{\text{2}}}$

Hence, the surface area of the sphere is $\text{394}\text{.24 c}{{\text{m}}^{\text{2}}}$.

iii. $\text{14 cm}$ $\left[ \text{Assume }\!\!\pi\!\!\text{ = }\frac{\text{22}}{\text{7}} \right]$

Ans:

Given radius of the sphere $\text{r = 14 cm}$

The surface area of the sphere $\text{A = 4 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{4 }\!\!\times\!\!\text{ }\frac{\text{22}}{\text{7}}\text{ }\!\!\times\!\!\text{ }{{\left( \text{14} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = }\left( \text{4 }\!\!\times\!\!\text{ 44 }\!\!\times\!\!\text{ 14} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 2464 c}{{\text{m}}^{\text{2}}}$

Hence, the surface area of the sphere is $\text{2464 c}{{\text{m}}^{\text{2}}}$.

2. Find the surface area of a sphere of diameter:

i. $\text{14 cm}$

Ans:

Given diameter of the sphere $\text{= 14 cm}$

So, the radius of the sphere $\text{r = }\frac{\text{14}}{\text{2}}\text{ = 7 cm}$

The surface area of the sphere $\text{A = 4 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{4 }\!\!\times\!\!\text{ }\frac{\text{22}}{\text{7}}\text{ }\!\!\times\!\!\text{ }{{\left( \text{7} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = }\left( \text{88 }\!\!\times\!\!\text{ 7} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 616 c}{{\text{m}}^{\text{2}}}$

Hence, the surface area of the sphere is $\text{616 c}{{\text{m}}^{\text{2}}}$.

ii. $\text{21 cm}$

Ans:

Given diameter of the sphere $\text{= 21 cm}$

So, the radius of the sphere $\text{r = }\frac{\text{21}}{\text{2}}\text{ = 10}\text{.5 cm}$

The surface area of the sphere $\text{A = 4 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{4 }\!\!\times\!\!\text{ }\frac{\text{22}}{\text{7}}\text{ }\!\!\times\!\!\text{ }{{\left( \text{10}\text{.5} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 1386 c}{{\text{m}}^{\text{2}}}$

Hence, the surface area of the sphere is $\text{1386 c}{{\text{m}}^{\text{2}}}$.

iii. $\text{3}\text{.5 m}$ $\left[ \text{Assume }\!\!\pi\!\!\text{ = }\frac{\text{22}}{\text{7}} \right]$

Ans:

Given diameter of the sphere $\text{= 3}\text{.5 m}$

So, the radius of the sphere $\text{r = }\frac{\text{3}\text{.5}}{\text{2}}\text{ = 1}\text{.75 m}$

The surface area of the sphere $\text{A = 4 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{4 }\!\!\times\!\!\text{ }\frac{\text{22}}{\text{7}}\text{ }\!\!\times\!\!\text{ }{{\left( \text{1}\text{.75} \right)}^{\text{2}}} \right]\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 38}\text{.5 }{{\text{m}}^{\text{2}}}$

Hence, the surface area of the sphere is $\text{38}\text{.5 }{{\text{m}}^{\text{2}}}$.

3. Find the total surface area of a hemisphere of radius $\text{10 cm}$. $\left[ \text{Use }\!\!\pi\!\!\text{ = 3}\text{.14} \right]$

Ans:

Given the radius of hemisphere $\text{r = 10 cm}$

The total surface area of the hemisphere is the sum of its curved surface area and the circular base.

Total surface area of hemisphere $\text{A = 2 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}\text{ + }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = 3 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{3 }\!\!\times\!\!\text{ 3}\text{.14 }\!\!\times\!\!\text{ }{{\left( \text{10} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 942 c}{{\text{m}}^{\text{2}}}$

Hence, the total surface area of the hemisphere is $\text{942 c}{{\text{m}}^{\text{2}}}$.

4. The radius of a spherical balloon increases from $\text{7 cm}$ to $\text{14 cm}$ as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Ans:

Given the initial radius of the balloon ${{\text{r}}_{1}}\text{ = 7 cm}$

The final radius of the balloon ${{\text{r}}_{2}}\text{ = 14 cm}$

We have to find the ratio of surface areas of the balloon in the two cases.

The required ratio $\text{R = }\frac{\text{4 }\!\!\pi\!\!\text{ }{{\text{r}}_{\text{1}}}^{\text{2}}}{\text{4 }\!\!\pi\!\!\text{ }{{\text{r}}_{\text{2}}}^{\text{2}}}$

$\Rightarrow \text{R = }{{\left( \frac{{{\text{r}}_{\text{1}}}}{{{\text{r}}_{\text{2}}}} \right)}^{\text{2}}}$

$\Rightarrow \text{R = }{{\left( \frac{\text{7}}{\text{14}} \right)}^{\text{2}}}$

$\Rightarrow \text{R = }\frac{\text{1}}{\text{4}}$

Hence, the ratio of the surface areas of the balloon in both case is $\text{1 : 4}$.

5. A hemispherical bowl made of brass has inner diameter $\text{10}\text{.5 cm}$. Find the cost of tin plating it on the inside at the rate of $\text{Rs}\text{. 16}$ per $\text{100 c}{{\text{m}}^{\text{2}}}$. $\left[ \text{Assume }\!\!\pi\!\!\text{ = }\frac{\text{22}}{\text{7}} \right]$

Ans:

Given the radius of inner hemispherical bowl $\text{r = }\frac{\text{10}\text{.5}}{\text{2}}\text{ = 5}\text{.25 cm}$

The surface area of the hemispherical bowl $\text{A = 2 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{2 }\!\!\times\!\!\text{ }\frac{\text{22}}{\text{7}}\text{ }\!\!\times\!\!\text{ }{{\left( \text{5}\text{.25} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 173}\text{.25 c}{{\text{m}}^{\text{2}}}$

It is given that the cost of tin-plating $\text{100 c}{{\text{m}}^{\text{2}}}$ area $\text{= Rs}\text{. 16}$

So, the cost of tin-plating $173.25\text{ c}{{\text{m}}^{\text{2}}}$ area $\text{= Rs}\text{. }\left( \frac{\text{16}}{\text{100}}\text{ }\!\!\times\!\!\text{ 173}\text{.25} \right)\text{ = Rs}\text{. 27}\text{.72}$

Hence, the cost of tin-plating the hemispherical bowl is $\text{Rs}\text{. 27}\text{.72}$.

6. Find the radius of a sphere whose surface area is $\text{154 c}{{\text{m}}^{\text{2}}}$. $\left[ \text{Assume }\!\!\pi\!\!\text{ = }\frac{\text{22}}{\text{7}} \right]$

Ans:

Let us assume the radius of the sphere be $\text{r}$.

We are given the surface area of the sphere, $\text{A = 154 c}{{\text{m}}^{\text{2}}}$.

$\therefore \text{4 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}\text{ = 154 c}{{\text{m}}^{\text{2}}}$

$\Rightarrow r^{2}=\left(\frac{154 \times 7}{2 \times 22}\right) \mathrm{cm}^{2}$

$\Rightarrow \text{r = }\left( \frac{\text{7}}{\text{2}} \right)\text{ cm}$

$\Rightarrow \text{r = 3}\text{.5 cm}$

Therefore, the radius of the sphere is $\text{3}\text{.5 cm}$.

7. The diameter of the moon is approximately one-fourth of the diameter of the earth. Find the ratio of their surface area.

Ans:

Let us assume the diameter of earth is $\text{d}$.

So, the diameter of the moon will be $\frac{\text{d}}{\text{4}}$.

The radius of the earth ${{\text{r}}_{\text{1}}}\text{ = }\frac{\text{d}}{\text{2}}$

The radius of the moon ${{\text{r}}_{\text{2}}}\text{ = }\frac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{2}}\text{ = }\frac{\text{d}}{\text{8}}$

The ratio of surface area of moon and earth $\text{R = }\frac{\text{4 }\!\!\pi\!\!\text{ }{{\text{r}}_{\text{2}}}^{\text{2}}}{\text{4 }\!\!\pi\!\!\text{ }{{\text{r}}_{\text{1}}}^{\text{2}}}$

$\Rightarrow \text{R = }\frac{\text{4 }\!\!\pi\!\!\text{ }{{\left( \frac{\text{d}}{\text{8}} \right)}^{\text{2}}}}{\text{4 }\!\!\pi\!\!\text{ }{{\left( \frac{\text{d}}{\text{2}} \right)}^{\text{2}}}}$

$\Rightarrow \text{R = }\frac{\text{4}}{\text{64}}$

$\Rightarrow \text{R = }\frac{\text{1}}{\text{16}}$

Therefore, the ratio of surface area of the moon and earth is $\text{1 : 16}$.

8. A hemispherical bowl is made of steel, $\text{0}\text{.25 cm}$ thick. The inner radius of the bowl is $\text{5 cm}$. Find the outer curved surface area of the bowl. $\left[ \text{Assume }\!\!\pi\!\!\text{ = }\frac{\text{22}}{\text{7}} \right]$

Ans:

Given the inner radius $\text{= 5 cm}$

The thickness of the bowl $\text{= 0}\text{.25 cm}$

So, the outer radius of the hemispherical bowl is $\text{r = }\left( \text{5 + 0}\text{.25} \right)\text{ cm = 5}\text{.25 cm}$

The outer curved surface area of the hemispherical bowl $\text{A = 2 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A =}\left[ \text{ 2 }\!\!\times\!\!\text{ }\frac{\text{2}}{\text{7}}\text{ }\!\!\times\!\!\text{ }{{\left( \text{5}\text{.25} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 173}\text{.25 c}{{\text{m}}^{\text{2}}}$

Therefore, the outer curved surface area of the hemispherical bowl is $\text{173}\text{.25 c}{{\text{m}}^{\text{2}}}$.

9. A right circular cylinder just encloses a sphere of radius $\text{r}$ (see figure). Find

(Image Will Be Updated Soon)

i. surface area of the sphere,

Ans:

The surface area of the sphere is $\text{4 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$.

ii. curved surface area of the cylinder,

Ans:

(Image Will Be Updated Soon)

Given the radius of cylinder $\text{= r}$

The height of cylinder $\text{= r + r = 2r}$

The curved surface area of cylinder $\text{A = 2 }\!\!\pi\!\!\text{ rh}$

$\Rightarrow \text{A = 2 }\!\!\pi\!\!\text{ r }\left( \text{2r} \right)$

$\Rightarrow \text{A = 4 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$

Therefore the curved surface area of the cylinder is  $\text{4 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$.

iii. ratio of the areas obtained in i. and ii.

The ratio of surface area of the sphere and curved surface area of cylinder  $\text{R = }\frac{\text{4 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}}{\text{4 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}}$

$\text{R = }\frac{\text{1}}{\text{1}}$

Therefore, the required ratio is $\text{1 : 1}$.

## NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.4

Opting for the NCERT solutions for Ex 13.4 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 13.4 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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Q1. What is meant by the term ‘surface area’?

Ans: The surface area of any solid or 3D shape can be defined as the sum of the areas of all its sides. It gives a measure of the area that a 3D figure occupies. There is a list of formulas that are commonly used to calculate the surface areas of 3D shapes, in maths. For example, there is a particular formula for calculating the surface area of a cube, whereas there is another formula for calculating the surface area of a sphere, and so on. Also, there are two types of surface areas for some 3D shapes, namely lateral surface area and total surface area.

Q2. What is meant by the ‘volume’ of a 3D shape?

Ans: The volume of a 3D shape or solid is the amount of space occupied by it. The volume of a 3D shape also gives the measure of the matter enclosed within the boundaries of the solid. If we measure the volume of a hollow or empty object, we can get the total volume of air or water that it can hold. Volume can be found only for 3-dimensional objects and nor for 2D objects, such as square or rectangle. In Maths, there are separate formulas to calculate the volume of various 3D objects. For example, the formula to find the volume of a cube is different from the formula to find the volume of a cuboid.

Q3. How many sums are there in the NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes, exercise 13.4?

Ans: There are a total of 9 sums in the NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes, exercise 13.4. Most of the sums in this exercise are based on the concept of surface areas and volumes of spheres and cylinders. When you refer to the solutions of these sums, you will be able to understand that you will have to write statements for solving these sums. All the 9 sums are solved and explained step by step in these NCERT solutions for Surface Areas and Volumes, exercise 13.4.

Q4. Are the NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes (Ex 13.4) Exercise 13.4 helpful?

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Q5. Where can I find the NCERT Solutions to Class 9 Maths Chapter 13 Exercise 13.4?

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Q6. Is Class 9 Maths Chapter 13 considered the toughest chapter?

Ans: No, not really, Chapter 13 of Class 9 Maths may be tricky so it requires good practice to understand the concepts covered in this chapter. If you practice regularly and with proper concentration then it will not be a tough chapter. Even if you don’t get the answer on your first try, solve it again, and eventually, you will be able to get the correct answer. You can verify your solutions from the link NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.4.

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Q9. Can I prepare for Class 9 in one month?

Ans: The time required for exam preparation entirely depends on the speed and accuracy of students. For some, one month can be a shorter period while for some it can be long enough. But the sooner the better, so it is advisable, to begin with the chapter as early as you can. SHARE TWEET SHARE SUBSCRIBE