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BSEB Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 13 Surface Areas and Volumes Ex 13.3 Book Answers

BSEB Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 13 Surface Areas and Volumes Ex 13.3 Book Answers
BSEB Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 13 Surface Areas and Volumes Ex 13.3 Book Answers


BSEB Class 9th Maths Chapter 13 Surface Areas and Volumes Ex 13.3 Textbooks Solutions and answers for students are now available in pdf format. Bihar Board Class 9th Maths Chapter 13 Surface Areas and Volumes Ex 13.3 Book answers and solutions are one of the most important study materials for any student. The Bihar Board Class 9th Maths Chapter 13 Surface Areas and Volumes Ex 13.3 books are published by the Bihar Board Publishers. These Bihar Board Class 9th Maths Chapter 13 Surface Areas and Volumes Ex 13.3 textbooks are prepared by a group of expert faculty members. Students can download these BSEB STD 9th Maths Chapter 13 Surface Areas and Volumes Ex 13.3 book solutions pdf online from this page.

Bihar Board Class 9th Maths Chapter 13 Surface Areas and Volumes Ex 13.3 Textbooks Solutions PDF

Bihar Board STD 9th Maths Chapter 13 Surface Areas and Volumes Ex 13.3 Books Solutions with Answers are prepared and published by the Bihar Board Publishers. It is an autonomous organization to advise and assist qualitative improvements in school education. If you are in search of BSEB Class 9th Maths Chapter 13 Surface Areas and Volumes Ex 13.3 Books Answers Solutions, then you are in the right place. Here is a complete hub of Bihar Board Class 9th Maths Chapter 13 Surface Areas and Volumes Ex 13.3 solutions that are available here for free PDF downloads to help students for their adequate preparation. You can find all the subjects of Bihar Board STD 9th Maths Chapter 13 Surface Areas and Volumes Ex 13.3 Textbooks. These Bihar Board Class 9th Maths Chapter 13 Surface Areas and Volumes Ex 13.3 Textbooks Solutions English PDF will be helpful for effective education, and a maximum number of questions in exams are chosen from Bihar Board.

Bihar Board Class 9th Maths Chapter 13 Surface Areas and Volumes Ex 13.3 Books Solutions

Board BSEB
Materials Textbook Solutions/Guide
Format DOC/PDF
Class 9th
Subject Maths Chapter 13 Surface Areas and Volumes Ex 13.3
Chapters All
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BSEB Class 9th Maths Chapter 13 Surface Areas and Volumes Ex 13.3 Textbooks Solutions with Answer PDF Download

Find below the list of all BSEB Class 9th Maths Chapter 13 Surface Areas and Volumes Ex 13.3 Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:

BSEB Bihar Board Class 9th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3

Question 1.
Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.
Solution:
Here r = (10.52) cm = 5.25 cm and l = 10.
Curved surface area ofjthe cone = (πrl) cm²
= (227 x 5.25 x 10) cm²
= 165 cm².

Question 2.
Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Solution:
Here r = (242) m = 12 m and l = 21 m

Question 3.
Curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find (i) radius of the base and (ii) total surface area of the cone.
Solution:
(i) Curved surface of a cone = 308 cm²
Slant height, l = 14 cm.
∴ Let r be the radius of the base.
∴ πrl = 308
⇒ 227 x r x 14 = 308
⇒ r = 308×722×14
Thus, the radius of the base = 7 cm

(ii) Total surface area of the cone = πr(l + r) cm²
= 227 x 7 x (14 x 7) cm²
= (22 x 21) cm²
= 462 cm²

Question 4.
A conical tent is 10 m high and the radius of its base is 24 m. Find
(i) slant height of the tent.
(ii) cost of the canvas required to make the tent, if the cost of 1 m² canvas is Rs 70.
Solution:
(i) Here r = 24 m, h = 10 m.
Let l be the slant height of the cone. Then,
l² = h² + r² ⇒ l = ℎ2+𝑟2‾‾‾‾‾‾‾√
⇒ l = 242+102‾‾‾‾‾‾‾‾‾√ = 576+100‾‾‾‾‾‾‾‾‾‾√
= 676‾‾‾‾√ =26 cm

(ii) Canvas required to make the conical tent = Curved surface of the cone
= πrl x(227 x 24 x 26) cm²
Rate of canvas per 1 m² is Rs 70
∴ Total cost of canvas = Rs (227 x 24 x 26 x 70)
= Rs 137280

Question 5.
What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm. (Use n = 3.14)
Solution:
Let r m be the radius, Ambe the height and l m be the slant height of the tent. Then, r = 6m, h = 8 m
⇒ l = 𝑟2+ℎ2‾‾‾‾‾‾‾√
⇒ l = 62+82‾‾‾‾‾‾‾√
= 36+64‾‾‾‾‾‾‾‾√
= 100‾‾‾‾√ = 10 cm
Area of the canvas used for the tent = curved surface area of the tent = πrl = (3.14 x 6 x 10) m² = 188.4 m²
Now, this area is bought in the form of a rectangle of width 3 m.
∴ Length of tarpaulin required

The extra material required for stitching margins and cutting = 20 cm = 0.2 m.
So, the total length of tarpaulin required = (62.8 + 0.2) m = 63 m.

Question 6.
The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs 210 per 100 m².
Solution:
Here, l = 25 m and r = (142) m = 7 m.
Curved surface area = πrl m²
= (227 x 25 x 7) m²
= 550 m²
Rate of white-washing is Rs 210 per 100 m²
∴ Cost of white-washing the tomh = Rs ( 550 x 210100 )
= Rs 1155.

Question 7.
A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.
Solution:
Let r cm be the radius, h cm be the height and l cm be the slant height of the joker’s cap. Then, r = 7 cm, h = 24 cm

Sheet required for 10 such caps = 5500 cm².

Question 8.
A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is Rs 12 per m², what will be the cost of painting all these cones ? (Use π = 3.14 and take 1.04‾‾‾‾‾√ = 1.02)
Solution:
Let r. m be the radius, h m be the height and l m be the slant height of a cone. Then, r = (402) cm = 20 cm = 2 m, and h = 1 m
∴ l = 𝑟2+ℎ2‾‾‾‾‾‾‾√ = 0.04+1‾‾‾‾‾‾‾‾√
= 1.04‾‾‾‾√
= 1.02
Curved surface of 1 cone = πrl m²
= (227 x .2 x 1.02) m²
Curved surface of such 50 cones
= (50 x 227 x .2 x 1.02 x 12)
Cost of painting @ Rs 12 per m²
= Rs(50 x 227 x .2 x 1.02 x 12)
= Rs 384.68 (approx.)


BSEB Textbook Solutions PDF for Class 9th


Bihar Board Class 9th Maths Chapter 13 Surface Areas and Volumes Ex 13.3 Textbooks for Exam Preparations

Bihar Board Class 9th Maths Chapter 13 Surface Areas and Volumes Ex 13.3 Textbook Solutions can be of great help in your Bihar Board Class 9th Maths Chapter 13 Surface Areas and Volumes Ex 13.3 exam preparation. The BSEB STD 9th Maths Chapter 13 Surface Areas and Volumes Ex 13.3 Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 9th Maths Chapter 13 Surface Areas and Volumes Ex 13.3 Books State Board syllabus with maximum efficiency.

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