**NCERT Solutions Class 10 Maths Chapter 13 Surface Areas And Volumes** – Here are all the NCERT solutions for Class 10 Maths Chapter 13. This solution contains questions, answers, images, explanations of the complete Chapter 13 titled Surface Areas And Volumes of Maths taught in Class 10. If you are a student of Class 10 who is using NCERT Textbook to study Maths, then you must come across Chapter 13 Surface Areas And Volumes. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas And Volumes in one place.

## NCERT Solutions Class 10 Maths Chapter 13 Surface Areas And Volumes

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For a better understanding of this chapter, you should also see summary of Chapter 13 Surface Areas And Volumes , Maths, Class 10.

Class | 10 |

Subject | Maths |

Book | Mathematics |

Chapter Number | 13 |

Chapter Name |
Surface Areas And Volumes |

### NCERT Solutions Class 10 Maths chapter 13 Surface Areas And Volumes

Class 10, Maths chapter 13, Surface Areas And Volumes solutions are given below in PDF format. You can view them online or download PDF file for future use.

### Surface Areas And Volumes Download

**NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes**

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### Question & Answer

Q.1:2 cubes each of volume \( 64 \mathrm{cm}^{3}= \) are joined end to end. Find the surface area of the resulting cuboid.

Ans :Given that , Volume of cubes= \( (\text { Edge })^{3}=64\) \( (\text { Edge })^{3}=64\) Edge \( =4 \mathrm{cm} \) If cubes are joined end to end, the dimensions of the resulting cuboid will be 4 cm, 4 cm, 8 cm. Surface area of cuboids \( \begin{array}{l}{=2(l b+b h+l h)} \\ {=2(4 \times 4+4 \times 8+4 \times 8)} \\ {=2(16+32+32)} \\ {=2(16+64)} \\ {=2 \times 80=160 \mathrm{cm}^{2}}\end{array} \)

Q.2:A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

Ans :It can be observed that radius (r) of the cylindrical pa t and the hemispherical part is the same (i.e., 7 cm). Height of hemispherical part = Radius = 7 cm Height of cylindrical part (h) = 13-7 = 6 cm Inner surface area of the vessel = CSA of cylindrical part + CSA of hemispherical part \( =2 \pi r h+2 \pi r^{2} \) Inner Surface area of vessel \( \begin{array}{l}{=2 \times \frac{22}{7} \times 7 \times 6+2 \times \frac{22}{7} \times 7 \times 7} \\ {=44(6+7)=44 \times 13} \\ {=572 \mathrm{cm}^{2}}\end{array} \)

Q.3:A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

Ans :It can be observed that the radius of the conical part and the hemispherica part is same (i.e., 3.5cm). Height of hemispherical part=Radius(r) =3.5 =7/2cm Height Of conical part (h) = 15.5 -3.5 = 12 cm Slant Height \( \begin{array}{l}{=\sqrt{r^{2}+h^{2}}} \\ {=\sqrt{\left(\frac{7}{2}\right)^{2}+(12)^{2}}=\sqrt{\frac{49}{4}+144}=\sqrt{\frac{49+576}{4}}} \\ {=\sqrt{\frac{625}{4}}=\frac{25}{2}}\end{array} \) of conical part + CSA of hemispherical part Total surface area of toy CSA of conica part + CSA of hemispherical part \( \begin{array}{l}{=\pi r l+2 \pi r^{2}} \\ {=\frac{22}{7} \times \frac{7}{2} \times \frac{25}{2}+2 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}} \\ {=137.5+77=214.5 \mathrm{cm}^{2}}\end{array} \)

Q.4:A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid .

Ans :From the figure, it can be observed that the greatest diameter possible for such hemisphere is equal to the cube's edge, i.e., 7cm. Radius (r) of hemispherical part=7/2=3.5cm Total surface area of solid = Surface area of cubical part + CSA of hemispherical part -Area of base of hemispherical part \( =6(\mathrm{Edge})^{2}+2 \pi r^{2}-\pi r^{2}=6(\mathrm{Edge})^{2}+\pi v^{2} \) Total surface area of solid \( \begin{array}{l}{=6(7)^{2}+\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}} \\ {=294+38.5=332.5 \mathrm{cm}^{2}}\end{array} \)

Q.5:A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

Ans :Diameter of hemisphere =Edge Of cube = l Radius of hemisphere = l/2 Total surface area of solid = Surface area of cubical part + CSA of hemispherical part - Area of base of hemispherical part =\( 6(\text { Edge })^{2}+2 \pi r^{2}-\pi r^{2}=6(\text { Edge })^{2}+\pi r^{2} \) Total surface area of solid \( \begin{array}{l}{=6 l^{2}+\pi \times\left(\frac{l}{2}\right)^{2}} \\ {=6 l^{2}+\frac{\pi I^{2}}{4}} \\ {=\frac{1}{4}(24+\pi) l^{2} \text { unit }^{2}}\end{array} \)

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