NCERT Solutions for Class 9 Maths Chapter 13

Download here the NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.1, 13.2, 13.3, 13.4, 13.5, 13.6, 13.7, 13.8 and 13.9 in English Medium as well as Hindi Medium updated for session 2020-2021. UP Board Students of Class 9 (High School) are using NCERT Book in session 2020-2021, so they can also using these textbooks solutions as UP Board Solutions for Class 9 Maths Chapter 13. Here, they can download Prashnavali 13.1, Prashnavali 13.2, Prashnavali 13.3, Prashnavali 13.4, Prashnavali 13.5, Prashnavali 13.6, Prashnavali 13.7, Prashnavali 13.8 and Prashnavali 13.9

9 in Hindi Medium to use offline for session 2020-21. NCERT Solutions for class 9 chapter 13 is applicable for CBSE Delhi Board, MP Board, UP Board – High School, UK Board, Gujrat Board and other board using NCERT Books 2020-21. NCERT Solutions 2020-2021 for other are also given in downloadable format.

NCERT Solutions for Class 9 Maths Chapter 13

Class: 9Maths (English and Hindi Medium)
Chapter 13:Surface Areas and Volumes

9th Maths Chapter 13 Solutions in English & Hindi Medium

Download free NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes all exercises which are given below in PDF format to download for new academic session 2020-21. NCERT Solutions as well as NCERT Solutions Offline Apps are updated for new academic session 2020-2021 based on latest CBSE Syllabus 2020-21. NCERT Books are now implemented in Uttar Pradesh also. So, students can download UP Board solutions for class 9 Maths chapter 13 all exercises from here.




Class 9 Maths Exercise 13.1 and 13.2 Solutions in Video

Class 9 Maths Exercise 13.1 Solutions in Video
Class 9 Maths Exercise 13.2 Solutions in Video

Class 9 Maths Exercise 13.3 and 13.4 Solutions in Video

Class 9 Maths Exercise 13.3 Solutions in Video
Class 9 Maths Exercise 13.4 Solutions in Video

Class 9 Maths Exercise 13.5 and 13.6 Solutions in Video

Class 9 Maths Exercise 13.5 Solutions in Video
Class 9 Maths Exercise 13.6 Solutions in Video

Class 9 Maths Exercise 13.7 and 13.8 Solutions in Video

Class 9 Maths Exercise 13.7 Solutions in Video
Class 9 Maths Exercise 13.8 Solutions in Video



Class 9 Maths Exercise 13.9 Solutions in Video

What are the formulae for Cuboid?

Cuboid: A cuboid is a solid bounded by six rectangular plane surfaces for example match box, brick, books, etc. are cuboid.
1. Surface area (or total surface area) of cuboid = 2 (lb + bh + hl) square units
2. Lateral surface area of cuboid = 2(l + b)h square units
3. Diagonal of a cuboid = √[l² + b² + h²] units
4. Total length of a edges of a cuboid = 4 (l + b + h) units
5. Volume of cuboid = lbh cubic units

What are the formulae for Cube?

Cube: A cuboid whose length, breadth and height are equal, is called a cube.
1. Surface area (or total surface area) of cuboid = 6a² square units
2. Lateral surface area of cuboid = 4a² square units
3. Diagonal of a cuboid = √3a units
4. Total length of a edges of a cuboid = 12a units
5. Volume of cuboid = a³ cubic units

What is meant by a Cylinder? Write its formulae?

Cylinder: A solid generated by the revolution of a rectangle about one of its sides which is kept fixed is called right circular cylinder.
Curved Surface Area (CSA) = 2πrh square units
Total Surface Area (TSA) = 2πr(r + h) square units
Volume = πr²h cubic units

What do you understand by a Cone? Write its Formulae?

Cone: A right circular cone is solid generated by revolving a line segment which passes through a fixed point and which makes a constant angle with a fixed line.
Slant Height = √[r² + h²]
Curved Surface Area (CSA) = πrl square units
Total Surface Area (TSA) = πr(r + l) square units
Volume = 1/3 πr²h cubic units

What are the basic formulae of Sphare?

Sphere: A sphere is three dimensional figure which is made up of all points in the space, which lie at a constant distance, form a fixed point called the centre of the sphere and the constant distant is called its radius.
Curved Surface Area (CSA) = Total Surface Area (TSA) = 4πr² square units
Volume = 4/3 πr³ cubic units

Important Notes on 9th Maths Chapter 13

Surface Area:
Surface area of a solid body is the area of all of its surface together and it is always measured in square units. Surface area is also known as total surface area (TSA).
Volume:
Space occupied by an object (solid body) is called the volume of the object. Volume is always measured in cubic units.



Important Questions on 9th Maths Chapter 13

उस लंब वृत्तीय शंकु का आयतन ज्ञात कीजिए जिसकी त्रिज्या 6 cm और ऊँचाई 7 cm है।
शंकु के आधार की त्रिज्या r = 6 cm और
ऊँचाई h = 7 cm है।
शंकु का आयतन = 1/3 πr^2 h
= 1/3 ×22/7 × 6 × 6 ×7
= 264 cm^3
अतः, लंब वृत्तीय शंकु का आयतन 264 cm^3 है।
The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm3 of wood has a mass of 0.6 g.
Inner radius of cylindrical pipe r = 24/2 = 12 cm,
outer radius R = 28/2 = 14 cm and length h = 35 m
Volume of cylindrical wooden pipe = π(R^2-r^2 )h
= 22/7 × (〖14〗^2-〖12〗^2 )×35
= 22 × (196 – 144) × 5
= 22 × 52 × 5 = 5720 cm^3
Mass of cylindrical wooden pipe
= 5720 × 0.6 g = 3432 g =3.432 kg
[∵1 cm3 of wood has a mass of 0.6 g]
Hence, the volume of cylindrical wooden pipe is 3.432 kg.
A soft drink is available in two packs – (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?
Length of tin can l = 5 cm, breadth b = 4 cm and height h = 15 cm
Volume of tin can
= lbh = 5 × 4 × 15
= 300 cm^3
Radius of plastic cylinder r
= 7/2 = 3.5 cm and height H = 10 cm
Volume of plastic cylinder = πr^2 H
= 22/7 × 3.5 × 3.5 × 10
= 22 × 0.5 × 3.5 × 10
= 385 cm^3
Difference between capacities of two packs
= 385 – 300 = 85 cm^3
Hence, the capacity of plastic cylindrical pack is greater than tin can by 85 cm^3.
एक शंकु की ऊँचाई 15 cm है। यदि इसका आयतन 1570 cm3 है, तो इसके आधार की त्रिज्या ज्ञात कीजिए।
शंकु का आयतन V = 1570 cm3 और ऊँचाई h = 15 cm है।
माना, शंकु के आधार की त्रिज्या = r cm
शंकु का आयतन = 1/3 πr^2 h
⇒ 1570 = 1/3 × 3.14 × r^2 × 15
⇒ 1570 = 3.14 × r^2 × 5
⇒ r^2 = 1570/(3.14 × 5)
= 100
⇒ r = √100 = 10 cm
अतः, शंकु के आधार की त्रिज्या 10 cm है।
ऊपरी व्यास 3.5 m वाले शंकु के आकार का एक गढ्ढा 12 m गहरा है। इसकी धारिता किलोलीटरों में कितनी है?
गढ्ढे के आधार की त्रिज्या r = 3.5/2 = 1.75 m और ऊँचाई h = 12 m है।
गढ्ढे की धारिता = 1/3 πr^2 h
= 1/3 × 22/7 × 1.75 × 1.75 × 12
= 38.5 m^3
= 38.5 किलोलीटर
[∵1 m^3=1 किलोलीटर]

अतः, गढ्ढे की धारिता 38.5 किलोलीटर है।

If the lateral surface of a cylinder is 94.2 cm2 and its height is 5 cm, then find radius of its base.
Lateral surface area of cylinder C = 94.2 cm^2
and height h = 5 cm.
Let, the radius of cylinder = r cm
Lateral surface area of cylinder C = 2πrh
⇒ 94.2 = 2 × 3.14 × r × 5
⇒ r = 94.2/(3.14×10) = 3 cm

Hence, the radius of base is 3 cm.

It costs ₹ 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of ₹ 20 per m2, find inner curved surface area of the vessel.
Cost of painting the inner curved surface of cylindrical vessel = ₹ 2200
and height h = 10 m.
Let, the inner radius of cylindrical vessel = r m
The inner curved surface area of cylindrical vessel = 2πrh
The cost of painting is at the rate of ₹ 20 per m2 = ₹ 20 × 2πrh
According to question,
₹ 20 × 2πrh = ₹ 2200
⇒ 2πrh = 2200/20 = 110

Hence, the inner curved surface area is 110 m^2.

A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?
Radius of cylindrical bowl r = 7/2 = 3.5 and
height of soup inside the cylindrical bowl h = 4 cm
Volume of cylindrical bowl = πr^2 h
= 22/7 × (3.5)^2 × 4
= 22/7× 3.5 × 3.5 × 4
= 22 × 0.5 × 3.5 × 4
= 154 cm^3
Therefore, the volume of soup per day for 250 patient
= 250 × 154 = 38500 cm^3
Hence, hospital has to prepare 38500 cm^3 soup daily to serve 250 patients.
माचिस की डिब्बी के माप 4 cm × 2.5 cm × 1.5 cm हैं। ऐसी 12 डिब्बियों के एक पैकिट का आयतन क्या होगा?
माचिस की डिब्बी की लंबाई l = 4 cm,
चौड़ाई b = 2.5 cm और
ऊँचाई h = 1.5 cm है।
माचिस की डिब्बी का आयतन
= lbh
= 4 × 2.5 × 1.5
= 15 cm^3
इसलिए, माचिस की 12 डिब्बियों का आयतन
= 12 × 15
= 180 cm^3

अतः, माचिस की 12 डिब्बियों का आयतन 180 cm^3 है।

एक घनाभाकार पानी की टंकी 6 m लंबी, 5 m चौड़ी और 4.5 m गहरी है। इसमें कितने लीटर पानी आ सकता है?
पानी की टंकी की लंबाई l = 6 m,
चौड़ाई b = 5 m और
गहराई h = 4.5 m है।

पानी की टंकी का आयतन
= lbh
= 6 × 5 × 4.5
= 135 cm^3
= 135 × 1000 l
= 135000 l

अतः, पानी की टंकी में 135000 लीटर पानी आ सकता है।

The curved surface area of a right circular cylinder of height 14 cm is 88 cm^2. Find the diameter of the base of the cylinder.
Curved surface area of cylinder 88 cm^2
and height h = 14 cm
Let, the radius of base of cylinder = r cm
Curved surface area of cylinder
= 2πrh
⇒ 88 = 2 × 22/7 × r ×14
⇒ 88 = 88r
⇒ r = 1 cm

Hence, the diameter of base of cylinder
= 2r = 2×1 = 2 cm

The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m^2.
Radius of roller r = 84/2 = 42 cm = 0.42 m
and length h = 120 cm = 1.2 m
Outer curved surface area of roller
= 2πrh
= 2 × 22/7 × 0.42 × 1.2
= 2 × 22 × 0.06 × 1.2
= 3.168 m^2
Area of ground levelled in on revolution
= 3.168 m^2
Therefore, area of ground levelled in 500 revolutions
= 500 × 3.168
= 1584 m^2

Hence, the area of playground is 1584 m^2.

एक कमरे की लंबाई, चौड़ाई और ऊँचाई क्रमशः 5 m, 4 m और 3 m हैं। ₹ 7.50 प्रति m^2 की दर से इस कमरे की दीवारों और छत पर सफेदी कराने का व्यय ज्ञात कीजिए।
कमरे की लंबाई l = 5 m,
चौड़ाई b = 4 m
और ऊँचाई h = 3 m है।
कमरे की दीवारों और छत का क्षेत्रफल
= कमरे का कुल पृष्ठीय क्षेत्रफल – फर्श का क्षेत्रफल
= 2(lb + bh + hl) – lb
= 2(5×4 + 4×3 + 3×5) – 5×4 m^2
= 2(20 + 12 + 15) – 20 m^2
= 2(47) – 20 m^2
= 94 – 20
= 74 m^2

अतः, कमरे की दीवारों और छत का क्षेत्रफल 74 m^2 है।
कमरे की दीवारों और छत पर सफेदी कराने का व्यय
= ₹ 7.50 ×74
= ₹ 555.00

एक घनाकार डिब्बे का एक किनारा 10 cm लंबाई का है तथा एक अन्य घनाभाकार डिब्बे की लंबाई, चौड़ाई और ऊँचाई क्रमशः 12.5 cm, 10 cm और 8 cm हैं। किस डिब्बे का पार्श्व पृष्ठीय क्षेत्रफल अधिक है और कितना अधिक है?
घनाकार डिब्बे की लंबाई l =10 cm है।
घनाकार डिब्बे का पार्श्व पृष्ठीय क्षेत्रफल
= 4l^2
= 4(10)^2 cm^2
= 4(100) cm^2
= 400 cm^2
घनाभाकार डिब्बे की लंबाई l = 12.5 cm,
चौड़ाई b = 10 cm और ऊँचाई h = 8 cm है।

घनाभाकार डिब्बे का पार्श्व पृष्ठीय क्षेत्रफल
= 2(l + b)h
= 2(12.5 + 10) × 8 cm^2
= 2(22.5) × 8 cm^2
= 360 cm^2

पार्श्व पृष्ठीय क्षेत्रफलों में अंतर
= 400 cm^2 – 360 cm^2
= 40 cm^2

अतः, घनाकार डिब्बे का पार्श्व पृष्ठीय क्षेत्रफल घनाभाकार डिब्बे के पार्श्व पृष्ठीय क्षेत्रफल से 40 cm^2 अधिक है।

एक छोटा पौधा घर (Green house) सम्पूर्ण रूप से शीशे की पट्टियों से (आधार भी सम्मिलित है) घर के अंदर ही बनाया गया है और शीशे की पट्टियों को टेप द्वारा चिपका कर रोका गया है। यह पौधा घर 30 cm लंबा, 25 cm चौड़ा और 25 cm ऊँचा है। सभी 12 किनारों के लिए कितने टेप की आवश्यकता है?
पौधा घर की लंबाई l = 30 cm,
चौड़ाई b = 25 cm और
ऊँचाई h = 25 cm है।

सभी 12 किनारों (4 लंबाई, 4 चौड़ाई और 4 ऊँचाई) के लिए टेप
= 4(l + b + h)
= 4(30 + 25 + 25) cm
= 4(80) cm
= 320 cm

अतः, सभी 12 किनारों के लिए 320 cm टेप की आवश्यकता है।

Curved surface area of a right circular cylinder is 4.4 m^2. If the radius of the base of the cylinder is 0.7 m, find its height.
Curved surface area of cylinder 4.4 m^2
and radius r = 0.7 m
Let, the height of cylinder = h m
Curved surface area of cylinder
= 2πrh
⇒ 4.4 = 2 × 22/7 × 0.7 × h
⇒ 4.4 = 4.4h
⇒ h = 1 m

Hence, the height of the cylinder is 1 m.

Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)
Radius of hemisphere r = 10 cm
Surface area of hemisphere = 3πr^2
= 3 × 3.14 × 10 × 10
= 942 cm^2

Hence, the total surface area of hemisphere is 942 cm^2.

एक घनाभाकार टंकी की धारिता 50000 लीटर पानी की है। यदि इस टंकी की लंबाई और गहराई क्रमशः 2.5 m और 10 m हैं, तो इसकी चौड़ाई ज्ञात कीजिए।
घनाभाकार टंकी की लंबाई l = 2.5 m,
गहराई h = 10 m
और आयतन V = 50000 लीटर = 50 m3 है।

माना, घनाभाकार टंकी की चौड़ाई = b m
घनाभाकार टंकी का आयतन
= lbh
⇒ 50 = 2.5 × b × 10
⇒ b = 50/(2.5 × 10) = 2 m

अतः, इस घनाभाकार टंकी की चौड़ाई 2 m है।

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.1 in PDF form
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