Maharashtra Board Solutions Class 9-Maths (Part 2): Chapter 9.2- Surface Area and Volume
Maharashtra Board Solutions Class 9-Maths (Part 2): Chapter 9.2- Surface Area and Volume

Class 9: Maths Chapter 9.2 solutions. Complete Class 9 Maths Chapter 9.2 Notes.

Maharashtra Board Solutions Class 9-Maths (Part 2): Chapter 9.2- Surface Area and Volume

Maharashtra Board 9th Maths Chapter 9.2, Class 9 Maths Chapter 9.2 solutions

Question 1.
Perpendicular height of a cone is 12 cm and its slant height is 13 cm. Find the radius of the base of the cone.
Given: Height (h) = 12 cm, length (l) = 13 cm
To find: Radius of the base of the cone (r)
Solution:
l2 = r2 + h2
∴ 132 = r2 + 122
∴ 169 = r2 + 144
∴169 – 144 = r2
∴ r2 = 25
∴ r = √25 … [Taking square root on both sides]
= 5 cm
∴ The radius of base of the cone is 5 cm.

Question 2.
Find the volume of a cone, if its total surface area is 7128 sq.cm and radius of base is 28 cm. ( π = 227)
Given: Radius (r) = 28 cm,
Total surface area of cone = 7128 sq.cm
To find: Volume of the cone
Solution:
i. Total surface area of cone = πr (l + r)
∴ 7128= y x 28 x (l + 28)
∴ 7128 = 22 x 4 x(l +28)
∴ l + 28 = 712822×4
∴ l + 28 = 81
∴ l = 81 – 28
∴ l = 53cm

ii. Now, l2 = r2 + h2
∴ 532 = 282+ h2
∴ 2809 = 784 + h2
∴ 2809 – 784 = h2
∴ h2 = 2025
∴ h = 2025−−−−√ …… [Taking square root on both sides]
= 45 cm
Maharashtra Board Class 9 Maths Solutions Chapter 9 Surface Area and Volume Practice Set 9.2 1
= 22 x 4 x 28 x 15
= 36960 cubic.cm
∴ The volume of the cone is 36960 cubic.cm.

Question 3.
Curved surface area of a cone is 251.2 cm2 and radius of its base is 8 cm. Find its slant height and perpendicular height, (π = 3.14)
Given: Radius (r) = 8 cm, curved surface area
of cone = 251.2 cm2
To find: Slant height (l) and the perpendicular height (h) of the cone
Solution:
i. Curved surface area of cone = πrl
∴ 251.2 = 3.14 x 8 x l
Maharashtra Board Class 9 Maths Solutions Chapter 9 Surface Area and Volume Practice Set 9.2 2
∴ l= 10 cm

ii. Now, l2 = r2 + h2
∴ 102 = 82 + h2
∴ 100 = 64 + h2
∴ 100 – 64 = h2
∴ h2 = 36
∴ h = √36 … [Taking square root on both sides]
= 6 cm
∴ The slant height and the perpendicular height of the cone are 10 cm and 6 cm respectively.

Question 4.
What will be the cost of making a closed cone of tin sheet having radius of base 6 m and slant height 8 m if the rate of making is ₹ 10 per sq.m?
Given: Radius (r) = 6 m, length (l) = 8 m
To find: Total cost of making the cone
Solution:
i. To find the total cost of making the cone of tin sheet, first we need to find the total surface area of the cone.
Total surface area of the cone = πr (l + r)
= 227 x 6 x (8 + 6)
= 227 x 6 x 14
= 22 x 6 x 2 = 264 sq.m

ii. Rate of making the cone = ₹ 10 per sq.m
∴ Total cost = Total surface area x Rate of making the cone
= 264 x 10
= ₹ 2640
∴ A The total cost of making the cone of tin sheet is ₹ 2640.

Question 5.
Volume of a cone is 6280 cubic cm and base radius of the cone is 20 cm. Find its perpendicular height, (π = 3.14)
Given: Radius (r) = 20 cm,
Volume of cone = 6280 cubic cm
To find: Perpendicular height (h) of the cone
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 9 Surface Area and Volume Practice Set 9.2 3
∴ The perpendicular height of the cone is 15 cm.

Question 6.
Surface area of a cone is 188.4 sq.cm and its slant height is 10 cm. Find its perpendicular height (π = 3.14).
Given: Length (l) =10 cm, curved surface area of the cone = 188.4 sq.cm
To find: Perpendicular height (h) of the cone
Solution:
i. Curved surface area of the cone = πrl
∴ 188.4 = 3.14 x r x 10
Maharashtra Board Class 9 Maths Solutions Chapter 9 Surface Area and Volume Practice Set 9.2 4

ii. Now, l2 = r2 + h2
∴ 102 = 62 + h2
∴ 100 = 36 + h2
∴ 100 – 36 = h2
∴ h2 = 64
∴ h = 64−−√ … [Taking square root on both sides]
= 8 cm
∴ The perpendicular height of the cone is 8 cm.

Question 7.
Volume of a cone is 1232 cm3 and its height is 24 cm. Find the surface area of the cone. (π = 227)
Given: Height (h) = 24 cm,
Volume of cone = 1232 cm3
To find: Surface area of the cone
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 9 Surface Area and Volume Practice Set 9.2 5
∴ r2 = 49
∴ r = 49−−√ … [Taking square root on both sides]
= 7 cm

ii. Now, l2 = r2 + h2
∴ l2 = 72 + 242
= 49 + 576 = 625
∴ l = 625−−−√ … [Taking square root on both sides]
= 25

iii. Curved surface area of cone = πrl
= 227 x 7 x 25
= 22 x 25
= 550 sq.cm
∴The surface area of the cone is 550 sq.cm.

Question 8.
The curved surface area of a cone is 2200 sq.cm and its slant height is 50 cm. Find the total surface area of cone. (π = 227)
Given: Length (l) = 50 cm, curved surface area of cone = 2200 sq.cm
To find: Total surface area of the cone
Solution:
i. Curved surface area of cone = πrl
Maharashtra Board Class 9 Maths Solutions Chapter 9 Surface Area and Volume Practice Set 9.2 6

ii. Total surface area of cone = πr (l + r)
= 227 x 14 x (50 + 14)
= 227 x 14 x 64
= 22 x 2 x 64
= 2816 sq.cm
∴ The total surface area of the cone is 2816 sq.cm.

Question 9.
There are 25 persons in a tent which is conical in shape. Every person needs an area of 4 sq.m, of the ground inside the tent. If height of the tent is 18 m, find the volume of the tent.
Given: For the tent,
height (h) = 18m,
number of people in the tent = 25,
area required for each person = 4 sq.m
To find: Volume of the tent
Solution:
i. Every person needs an area of 4 sq.m, of the ground inside the tent.
Surface area of the base of the tent = number of people in the tent × area required for each person
= 25 × 4
= 100 sq.m

ii. Surface area of the base of the tent = πr2
∴ 100 = πr2
∴ πr2 = 100

iii. Volume of the tent= 13 πr2h
= 13 x 100 x 18 …….[∵ πr2 = 100]
= 100 x 6
= 600 cubic metre
∴ The volume of the tent is 600 cubic metre.

Question 10.
In a field, dry fodder for the cattle is heaped in a conical shape. The height of the cone is 2.1 m and diameter of base is 7.2 m. Find the volume of the heap of the fodder. If it is to be covered by polythene in rainy se&son then how much minimum polythene
sheet is needed? (π = 227 and 17.37−−−−√ = 4.17 ]
Given: Height of the heap (h) = 2.1 m.
diameter of the base (d) = 7.2 m
∴Radius of the base (r) = d2 = 7.22 = 3.6 m
To find: Volume of the heap of the fodder and polythene sheet required
Solution:
i. Volume of the heap of fodder = 13πr2h
= 13 x 227 x (3.6)2 x 2.1
= 13 x 227 x 3.6 x 3.6 x 2.1
= 1 x 22 x 1.2 x 3.6 x 0.3
= 28.51 cubic metre

ii. Now, l2 = r2 + h2
= (3.6)2 + (2.1)2
= 12.96 + 4.41
∴ l2 =17.37
∴ l2 = 17.37−−−−√ .. .[Taking square root on both sides]
= 4.17 m

iii. Area of the polythene sheet needed to cover the heap of the fodder = Curved surface area of the conical heap
= πrl
= 227 x 3.6 x 4.17
= 47.18 sq.m
∴ The volume of the heap of the fodder is 28.51 cubic metre and a polythene sheet of 47.18 sq.m will be required to cover it.

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Maharashtra Board Solutions Class 9-Maths (Part 2): Chapter 9.2- Surface Area and Volume

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Where do I get the Maharashtra State Board Books PDF For free download?

You can download the Maharashtra State Board Books from the eBalbharti official website, i.e. cart.ebalbharati.in or from this article.

How to Download Maharashtra State Board Books?

Students can get the Maharashtra Books for primary, secondary, and senior secondary classes from here.  You can view or download the Maharashtra State Board Books from this page or from the official website for free of cost. Students can follow the detailed steps below to visit the official website and download the e-books for all subjects or a specific subject in different mediums.
Step 1: Visit the official website ebalbharati.in
Step 2: On the top of the screen, select “Download PDF textbooks” 
Step 3: From the “Classes” section, select your class.
Step 4: From “Medium”, select the medium suitable to you.
Step 5: All Maharashtra board books for your class will now be displayed on the right side. 
Step 6: Click on the “Download” option to download the PDF book.

Who developed the Maharashtra State board books?

As of now, the MSCERT and Balbharti are responsible for the syllabus and textbooks of Classes 1 to 8, while Classes 9 and 10 are under the Maharashtra State Board of Secondary and Higher Secondary Education (MSBSHSE).

How many state boards are there in Maharashtra?

The Maharashtra State Board of Secondary & Higher Secondary Education, conducts the HSC and SSC Examinations in the state of Maharashtra through its nine Divisional Boards located at Pune, Mumbai, Aurangabad, Nasik, Kolhapur, Amravati, Latur, Nagpur and Ratnagiri.

About Maharashtra State Board (MSBSHSE)

The Maharashtra State Board of Secondary and Higher Secondary Education or MSBSHSE (Marathi: महाराष्ट्र राज्य माध्यमिक आणि उच्च माध्यमिक शिक्षण मंडळ), is an autonomous and statutory body established in 1965. The board was amended in the year 1977 under the provisions of the Maharashtra Act No. 41 of 1965.

The Maharashtra State Board of Secondary & Higher Secondary Education (MSBSHSE), Pune is an independent body of the Maharashtra Government. There are more than 1.4 million students that appear in the examination every year. The Maha State Board conducts the board examination twice a year. This board conducts the examination for SSC and HSC. 

The Maharashtra government established the Maharashtra State Bureau of Textbook Production and Curriculum Research, also commonly referred to as Ebalbharati, in 1967 to take up the responsibility of providing quality textbooks to students from all classes studying under the Maharashtra State Board. MSBHSE prepares and updates the curriculum to provide holistic development for students. It is designed to tackle the difficulty in understanding the concepts with simple language with simple illustrations. Every year around 10 lakh students are enrolled in schools that are affiliated with the Maharashtra State Board.