NCERT Solutions Class 9 Maths Chapter 11 Surface Area and Volume Exercise 11.1
Understanding the world around us often begins with shapes and sizes, and that’s exactly what Class 9 Maths Chapter 11, Surface Area and Volume, explores. Exercise 11.1 focuses on calculating the surface area of a cuboid and a cube, two of the most fundamental 3-dimensional solids. This exercise helps students grasp how to find the total area covering the outer surfaces of these solids using their length, breadth, and height.
Our NCERT Solutions for Class 9 Maths Chapter 11 Surface Area and Volume Exercise 11.1 focuses on applying formulas to real-life problems, enhancing problem-solving skills and spatial understanding. With clear step-by-step solutions, students can easily grasp concepts and build confidence for exams. These solutions follow the latest NCERT guidelines, making them a reliable resource for thorough preparation and practice.
1.0Download NCERT Solutions of Class 9 Maths Chapter 11 Surface Area and Volume Exercise 11.1 Free PDF
Download the up-to-date NCERT Solutions for Class 9 Maths Chapter 11 Exercise 11.1 as a PDF file. You will get all the clear explanations and step-by-step methods to solve surface areas of cuboids and cubes.
Key Concepts of Exercise 11.1
- Understanding the structure of a cuboid and cube.
- Understanding lateral surface area vs total surface area.
- Knowing the surface area formulas:
Cuboid: 2(lb + bh + hl)
Cube: 6a²
- Real life applications problems dealing with example boxes, bricks, containers, etc.
- Unit conversion procedures (if needed).
Also Read: 2026 Class 10 Solved Question Papers
2.0NCERT Exercise Solutions Class 9 Chapter 11 Surface Area and Volume: All Exercises
Here is a quick overview and access to solutions for all exercises from Chapter 11, Surface Area and Volume.
3.0NCERT Class 9 Maths Chapter 11 Exercise 11.1 : Detailed Solutions
1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.
Sol.
Diameter of the base = 10.5 cm
Radius of the base (r) = 10.5/2 cm = 5.25 cm
Slant height (ℓ) = 10 cm
Curved surface area (CSA) of the cone = πrℓ
= (22/7) × 5.25 × 10
= 165 cm².
2. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Sol.
Slant height (ℓ) = 21 m
Diameter of base = 24 m
Radius (r) = 24/2 = 12 m
Total surface area (TSA) of the cone = πr(r+ℓ)
= (22/7) × 12 × (12+21) m²
= (22/7) × 12 × 33 m²
= 8712/7 m²
= 1244.57 m² (approximately).
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.
Sol.
(i) Slant height (ℓ) = 14 cm
Curved surface area (CSA) = 308 cm²
We know CSA = πrℓ
(22/7) × r × 14 = 308
22 × r × 2 = 308
44r = 308
r = 308/44
r = 7 cm.
Hence, the radius of the base is 7 cm.
(ii) Total surface area (TSA) of the cone = πr(ℓ+r)
= (22/7) × 7 × (14+7)
= (22/7) × 7 × 21
= 22 × 21
= 462 cm².
Hence, the total surface area of the cone is 462 cm².
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.
Sol.
Height of the tent (h) = 10 m
Radius of the base (r) = 24 m
(i) The slant height (ℓ) can be found using the Pythagorean theorem: ℓ² = h² + r²
ℓ = √(h² + r²)
ℓ = √((10)² + (24)²) m
ℓ = √(100 + 576) m
ℓ = √676 m
ℓ = 26 m.
Thus, the required slant height of the tent is 26 m.
(ii) Curved surface area (CSA) of the cone = πrℓ
Area of the canvas required = (22/7) × 24 × 26 m²
= (22 × 24 × 26)/7 m²
= 13728/7 m²
Cost of canvas = Area of canvas × Cost per m²
Cost of (13728/7) m² canvas = Rs. 70 × (13728/7)
= Rs. 10 × 13728
= Rs. 137280.
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).
Sol.
Height of conical tent (h) = 8 m
Base radius (r) = 6 m
First, find the slant height (ℓ):
ℓ = √(h² + r²) = √(8² + 6²) = √(64 + 36) = √100 = 10 m.
Curved surface area (CSA) of the conical tent = πrℓ
Area of tarpaulin required = 3.14 × 6 × 10 m² = 188.4 m².
The area of the tarpaulin is given by (width × length).
Given width of tarpaulin = 3 m.
So, 3 m × length = 188.4 m²
Length = 188.4 / 3 m = 62.8 m.
Extra length required for stitching margins and wastage = 20 cm = 0.2 m.
Total length required = 62.8 m + 0.2 m = 63 m.
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.
Sol.
Slant height (ℓ) = 25 m
Base diameter = 14 m
Radius (r) = 14/2 = 7 m
Curved surface area (CSA) of the conical tomb = πrℓ
= (22/7) × 7 × 25 m²
= 22 × 25 m² = 550 m².
Cost of white washing = (Total Area / 100 m²) × Rate per 100 m²
Cost of white washing = (550 / 100) × Rs. 210
= 5.5 × Rs. 210
= Rs. 1155.
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.
Sol.
Base radius (r) = 7 cm
Height (h) = 24 cm
First, find the slant height (ℓ):
ℓ² = h² + r² = (24)² + (7)² = 576 + 49 = 625
ℓ = √625 = 25 cm.
The area of the sheet required for one cap is its curved surface area (since the base is open as it's a cap).
Area of sheet for one cap = πrℓ
= (22/7) × 7 × 25 cm²
= 22 × 25 cm² = 550 cm².
Sheet required for 10 such caps = 10 × 550 cm² = 5500 cm².
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)
Sol.
Base diameter = 40 cm
Radius (r) = 40/2 cm = 20 cm = 20/100 m = 0.2 m.
Height (h) = 1 m.
First, find the slant height (ℓ):
ℓ = √(r² + h²) = √((0.2)² + (1)²) m
ℓ = √(0.04 + 1) m = √1.04 m.
Given to take √1.04 = 1.02.
So, ℓ = 1.02 m.
Curved surface area (CSA) of 1 cone = πrℓ
= 3.14 × 0.2 × 1.02 m².
CSA of 50 cones = 50 × (3.14 × 0.2 × 1.02) m²
= 50 × 0.628 × 1.02 m² (Incorrect calculation in original, 3.140.2 = 0.628)
Let's recalculate:
= 50 × 3.14 × (2/10) × (102/100) m²
= 50 × (314/100) × (2/10) × (102/100) m²
= (50 × 314 × 2 × 102) / (100 × 10 × 100) m²
= (314 × 102) / (10 × 100) m² (after canceling 502 from numerator/denominator)
= 32028 / 1000 m² = 32.028 m².
Cost of painting per m² = Rs. 12.
Total cost of painting 50 cones = 32.028 m² × Rs. 12/m²
= Rs. 384.336 (approximately Rs. 384.34).
4.0Key Features and Benefits: Class 9 Maths Chapter 11 Surface Area and Volume : Exercise 11.1
- Covers Basic Geometry: Develops strong foundational knowledge about 3D objects.
- Real-Life Applications: Help students see how maths relates to real-life objects like boxes and walls.
- Detailed solutions: To each question in a step-wise logical process. Clearly worked examples and clear use of any required formula.
- Ideal for doing CBSE exam preparation: Aligned with the latest CBSE syllabus and exam requirements
- Promotes visualization: Encourages students to imagine and relate to 3D shapes.