Syllabus Note: In older editions of the NCERT textbook, Surface Areas and Volumes was listed as Chapter 13. Under the newly rationalized CBSE syllabus (2023-2024 onwards), it has been renumbered to Chapter 11. Furthermore, earlier exercises covering cubes and cuboids have been removed, making the Cone exercise the new Exercise 11.1.
NCERT Solutions for Class 9 Maths Chapter 11: Surface Areas and Volumes (Exercise 11.1)
Welcome to the complete guide for NCERT Solutions for Class 9 Maths Chapter 11 – Surface Areas and Volumes. This specific exercise, Exercise 11.1, focuses purely on calculating the Curved Surface Area (CSA) and Total Surface Area (TSA) of a Right Circular Cone.
Exercise 11.1 Complete Solutions
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. (Assume π = 22/7)
Solution:
Given parameters of the cone:
- Diameter (d) = 10.5 cm
- Radius (r) = d/2 = 10.5 / 2 = 5.25 cm
- Slant height (l) = 10 cm
Curved Surface Area (CSA) of a cone = πrl
CSA = (22/7) × 5.25 × 10
CSA = (22/7) × 52.5
CSA = 22 × 7.5 = 165 cm²
Final Answer: The curved surface area of the cone is 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:
Given parameters:
- Slant height (l) = 21 m
- Diameter = 24 m ⇒ Radius (r) = 12 m
Total Surface Area (TSA) of a cone = πr(l + r)
TSA = (22/7) × 12 × (21 + 12)
TSA = (22/7) × 12 × 33
TSA = 8712 / 7 ≈ 1244.57 m²
Final Answer: The total surface area of the cone is 1244.57 m².
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:
Given parameters:
- CSA = 308 cm²
- Slant height (l) = 14 cm
(i) Radius of the base:
Curved Surface Area = πrl
308 = (22/7) × r × 14
308 = 44 × r
r = 308 / 44 = 7 cm
(ii) Total surface area:
TSA = CSA + Area of base = CSA + πr²
TSA = 308 + (22/7) × (7)²
TSA = 308 + (22 × 7)
TSA = 308 + 154 = 462 cm²
Final Answer: (i) The radius is 7 cm. (ii) The total surface area is 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 ₹70.
Solution:
Given parameters:
- Height (h) = 10 m
- Radius (r) = 24 m
(i) Slant height of the tent (l):
l = √(r² + h²)
l = √(24² + 10²)
l = √(576 + 100) = √676 = 26 m
(ii) Cost of the canvas:
The canvas required equals the Curved Surface Area (CSA) of the tent.
CSA = πrl = (22/7) × 24 × 26 m²
Cost of 1 m² canvas = ₹70
Total Cost = CSA × 70
Total Cost = (22/7) × 24 × 26 × 70
Total Cost = 22 × 24 × 26 × 10 = ₹1,37,280
Final Answer: (i) Slant height is 26 m. (ii) Total cost of the canvas is ₹1,37,280.
Question 5
What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm (Use π = 3.14).
Solution:
Given parameters for the tent:
- Height (h) = 8 m
- Radius (r) = 6 m
First, find the slant height (l):
l = √(r² + h²) = √(6² + 8²)
l = √(36 + 64) = √100 = 10 m
Curved Surface Area (CSA) of the tent = πrl
CSA = 3.14 × 6 × 10 = 188.4 m²
Area of tarpaulin required = CSA = 188.4 m².
Width of tarpaulin = 3 m. Let the length be L.
Length × Width = Area
L × 3 = 188.4
L = 188.4 / 3 = 62.8 m
Extra margin required = 20 cm = 0.2 m.
Total length of tarpaulin required = 62.8 + 0.2 = 63 m.
Final Answer: The total length of tarpaulin required is 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 ₹210 per 100 m².
Solution:
Given parameters:
- Slant height (l) = 25 m
- Base diameter = 14 m ⇒ Radius (r) = 7 m
Curved Surface Area (CSA) to be whitewashed = πrl
CSA = (22/7) × 7 × 25
CSA = 22 × 25 = 550 m²
Rate of white-washing = ₹210 per 100 m² = ₹2.10 per m².
Total Cost = CSA × Rate
Total Cost = 550 × (210 / 100) = 550 × 2.10 = ₹1155
Final Answer: The cost of white-washing the tomb is ₹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:
Given parameters for one cap:
- Radius (r) = 7 cm
- Height (h) = 24 cm
First, calculate the slant height (l):
l = √(r² + h²) = √(7² + 24²)
l = √(49 + 576) = √625 = 25 cm
Sheet required for 1 cap = CSA of cone = πrl
CSA = (22/7) × 7 × 25 = 550 cm²
Total area of sheet required for 10 caps = 10 × CSA
Total Area = 10 × 550 = 5500 cm²
Final Answer: The area of the sheet required is 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 ₹12 per m², what will be the cost of painting all these cones? (Use π = 3.14 and take √1.04 = 1.02)
Solution:
Given parameters for one cone:
- Diameter = 40 cm ⇒ Radius (r) = 20 cm = 0.2 m
- Height (h) = 1 m
First, calculate the slant height (l):
l = √(r² + h²) = √(0.2² + 1²)
l = √(0.04 + 1) = √1.04 = 1.02 m
Curved Surface Area (CSA) of 1 cone = πrl
CSA = 3.14 × 0.2 × 1.02 = 0.64056 m²
Total CSA for 50 cones:
Total CSA = 50 × 0.64056 = 32.028 m²
Cost of painting per m² = ₹12.
Total Cost = Total CSA × 12
Total Cost = 32.028 × 12 = 384.336
Final Answer: The approximate cost of painting all the cones will be ₹384.34.
NCERT Class 9 Maths Ch. 11 Surface Areas and Volumes Other Exercises:-
- Exercise 11.1
- Exercise 11.2
- Exercise 11.3
- Exercise 11.4
Q.1 Construct an angle of 90° at the initial point of a given ray and justify the construction.
Ans
Given:
A ray OA with initial point O is given.
To construct:
An angle of 90°
Steps of Construction:
(i) Draw a ray OA with initial point O.
(ii) Mark an arc of any radius with centre O.
(iii) Mark two arcs with same radius and centre at the points M1 and M2 respectively.
(iv) Now mark an arc with centre M2 and another arc with centre M3 with any radius. These arcs intersect at N.
(v) Join O and N by a ray OB.
(vi) ∠BOA is required angle of 90°.
Justification:
(i) Since, ∠M2OA = ∠M3OM2 = 60°
(ii) Ray OB is bisector of ∠M3OM2 i.e.
∠BOM2 = (1/2) ∠M3OM2
= (1/2) 60°
= 30°
(iii) ∠BOA = ∠BOM2 + ∠M2OA
= 30° + 60°
= 90°
Hence, it is justified that ∠BOA = 90°.
Q.2 Construct an angle of 45° at the initial point of a given ray and justify the construction.
Ans
Given:
A ray with initial point O is given.
To construct:
An angle of 45°
Steps of Construction:
(i) Draw a ray OA with initial point O.
(ii) Mark an arc of any radius with centre O.
(iii) Mark two arcs with same radius and centre at the points M1 and M2 respectively.
(iv) Now mark an arc with centre M2 and another arc with centre M3 with any radius. These arcs intersect at N.
(v) Join O and N by a ray OB.
(vi) ∠BOA is an angle of 90°.
(vii) Now, mark arcs of any radius with centre M1 and Q respectively, which intersect at P.
(viii) Draw ray OC through P, we get ∠COA which is equal to 45°.
(ix) Thus, ∠COA is required angle.
Justification:
∠COA = (1/2) ∠BOA
= (1/2) 90°
= 45°
Q.3 Construct the angles of the following measurements: (i) 30° (ii) 22.5° (iii) 15°
Ans
(i)
Given:
A ray with initial point O is given.
To construct:
An angle of 30°
Steps of Construction:
(i) Draw a ray OA with initial point O.
(ii) Mark an arc of any radius with centre O, which intersects ray OA at the point M1.
(iii) Mark another arc with centre M1 and same radius, which intersects earlier arc at M2.
(iv) Now, we draw arcs with any radius and centres M1 and M2 respectively. These arcs intersect at N.
(v) Join O and N with the help of ray OB.
(vi) ∠BOA is required angle of 30°.
(ii)
Given:
A ray with initial point O
To construct:
An angle of 22.5°
Steps of Construction:
(i) Draw a ray OA with initial point O.
(ii) Mark an arc of any radius with centre O.
(iii) Mark two arcs with same radius and centre at the points M1 and M2 respectively.
(iv) Now mark an arc with centre M2 and another arc with centre M3 with any radius. These arcs intersect at N.
(v) Join O and N by a ray OB.
(vi) ∠BOA is an angle of 90°.
(vii) Now, mark arcs of any radius with centre M1 and X respectively, which intersect at P.
(viii) Draw ray OC through P, we get ∠COA which is equal to 45°.
(ix) Now we draw bisector of ∠COA and we get ∠DOA, which is equal to 22.5°.
(x) Thus, ∠DOA is required angle.
(iii)
Given:
A ray with initial point O is given.
To construct:
An angle of 15°.
Steps of Construction:
(i) Draw a ray OA with initial point O.
(ii) Mark an arc of any radius with centre O, which intersects ray OA at the point M1.
(iii) Mark another arc with centre M1 and same radius, which intersects earlier arc at M2.
(iv) Now, we draw arcs with any radius and centres M1 and M2 respectively. These arcs intersect at N.
(v) Join O and N with the help of ray OB.
(vi) ∠BOA is angle of 30°.
(vii) Now, draw bisector OC of ∠BOA.
(viii) ∠COA is required angle of 15°.
Q.4 Construct the following angles and verify by measuring them by a protractor:
(i) 75° (ii) 105° (iii) 135°
Ans
(i)
Given:
A ray with initial point O is given.
To construct:
An angle of 75°
Steps of Construction:
(i) Draw ray OP and construct angle of 90°.
(ii) Mark arcs with centre P and R respectively with any radius. These arcs intersect at Q.
(iii) Draw ray OC throw point Q.
(iv) ∠COA is required angle.
(ii)
Given:
A ray with initial point O is given.
To construct:
An angle of 105°
Steps of Construction:
(v) Draw ray OP and construct angle of 90°.
(vi) Mark arcs with centre P and Q respectively with any radius. These arcs intersect at R.
(vii) Draw ray OC throw point R.
(viii) ∠COA is required angle.
(iii)
Given:
A ray with initial point O is given.
To construct:
An angle of 135°
Steps of Construction:
(i) Draw ray OP and construct angle of 90°.
(ii) Mark arcs with centre P and Q respectively with any radius. These arcs intersect at R.
(iii) Draw ray OC throw point R.
(iv) ∠COA is required angle.
Q.5 Construct an equilateral triangle, given its side and justify the construction.
Ans
Given:
A side of triangle is given.
To construct:
An equilateral triangle ABC
Steps of Construction:
(i) Draw a side of given length.
(ii) Draw two arcs of radius AB with centres A and B respectively. These arcs intersect at C.
(iii) Join AC and BC.
(iv) ΔABC is required triangle.
FAQs (Frequently Asked Questions)
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