Class 9 Maths Ganita Manjari Chapter 14 Exercise 14.3 Solutions Math of Space: Surface Area and Volume
NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 14 Exercise 14.3 Math of Space: Surface Area and Volume provides step-by-step solutions to help students understand and solve questions related to the surface area and volume of three-dimensional shapes. The solutions are explained in simple language, making it easier for students to understand the concepts and apply the correct formulas.
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NCERT Class 9 Maths Ganita Manjari Chapter 14 Exercise Set 14.3
1. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Answer:
Given:
- Slant height, l = 21 m
- Diameter = 24 m, so radius r = 12 m
Total surface area of a cone:
TSA = πr(r + l)
Substituting the values:
TSA = π × 12 × (12 + 21)
= 12π × 33
= 396π m²
Using π ≈ 22/7:
TSA ≈ 1244.57 m²
Therefore, the total surface area of the cone is 396π m² or approximately 1244.57 m².
2. Find the curved surface area of a right circular cone whose slant height is 10 cm and base radius is 7 cm.
Answer:
Given:
- Slant height, l = 10 cm
- Radius, r = 7 cm
Curved surface area of a cone:
CSA = πrl
Substituting the values:
CSA = π × 7 × 10
= 70π cm²
Using π = 22/7:
CSA = 70 × 22/7 = 220 cm²
Therefore, the curved surface area of the cone is 220 cm².
3. The height of a cone is 16 cm, and its base radius is 12 cm. Find the curved surface area and the total surface area of the cone.
Given:
- Height, h = 16 cm
- Radius, r = 12 cm
First, find the slant height:
l = √(r² + h²)
= √(12² + 16²)
= √(144 + 256)
= √400
= 20 cm
Curved Surface Area
CSA = πrl
= π × 12 × 20
= 240π cm²
Using π = 22/7:
CSA ≈ 754.29 cm²
Total Surface Area
TSA = πr(l + r)
= π × 12 × (20 + 12)
= 384π cm²
Using π = 22/7:
TSA ≈ 1206.86 cm²
Therefore:
- Curved surface area = 240π cm² ≈ 754.29 cm²
- Total surface area = 384π cm² ≈ 1206.86 cm².
4. A cone has a height of 15 cm. If its volume is 1570 cm³, find the radius of the base.
Answer:
Given:
- Height, h = 15 cm
- Volume, V = 1570 cm³
Volume of a cone:
V = (1/3)πr²h
Substituting the values:
1570 = (1/3) × π × r² × 15
1570 = 5πr²
Using π = 3.14:
1570 = 15.7r²
r² = 100
r = 10 cm
Therefore, the radius of the base is 10 cm
5. The curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find (i) the radius of the base, (ii) the total surface area of the cone.
Answer:
Given:
- Curved surface area, CSA = 308 cm²
- Slant height, l = 14 cm
(i) Radius of the base
Curved surface area of a cone:
CSA = πrl
Substituting the values:
308 = (22/7) × r × 14
308 = 44r
r = 7 cm
Therefore, the radius of the base is 7 cm.
(ii) Total surface area of the cone
TSA = πr(l + r)
= (22/7) × 7 × (14 + 7)
= 22 × 21
= 462 cm²
Therefore:
(i) Radius = 7 cm
(ii) Total surface area = 462 cm²
6. 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.
Answer:
Given:
- Radius, r = 7 cm
- Height, h = 24 cm
First, find the slant height:
l = √(r² + h²)
= √(7² + 24²)
= √(49 + 576)
= √625
= 25 cm
Since a joker’s cap is open at the base, only the curved surface area is required.
Area of one cap:
CSA = πrl
= (22/7) × 7 × 25
= 550 cm²
For 10 caps:
10 × 550 = 5500 cm²
Therefore, the area of sheet required to make 10 caps is 5500 cm².
7. What length of tarpaulin 3 m wide is required to make a conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that is required for stitching margins and wastage in cutting is 20 cm.
Answer:
Given:
- Height of tent, h = 8 m
- Base radius, r = 6 m
- Width of tarpaulin = 3 m
First, find the slant height:
l = √(r² + h²)
= √(6² + 8²)
= √(36 + 64)
= 10 m
Since the tent is open at the bottom, only the curved surface area is required.
CSA = πrl
= π × 6 × 10
= 60π m²
Using π = 22/7:
CSA = 60 × 22/7 = 1320/7 m²
Let the required length of tarpaulin be L.
3 × L = 1320/7
L = 440/7 ≈ 62.86 m
Extra material required = 20 cm = 0.20 m
Therefore,
Total length = 62.86 + 0.20 = 63.06 m
Hence, approximately 63.06 m of tarpaulin is required.
8. A right triangle with sides 6 cm, 8 cm and 10 cm is rotated through 360° about the side of 8 cm. Find the volume and the curved surface area of the solid so formed.
When the right triangle with sides 6 cm, 8 cm and 10 cm is rotated about the side of 8 cm, it forms a right circular cone.
For the cone:
- Height, h = 8 cm
- Radius, r = 6 cm
- Slant height, l = 10 cm
Volume of the cone
V = (1/3)πr²h
= (1/3) × π × 6² × 8
= 96π cm³
Using π = 22/7:
V ≈ 301.71 cm³
Curved surface area
CSA = πrl
= π × 6 × 10
= 60π cm²
Using π = 22/7:
CSA ≈ 188.57 cm²
Therefore:
Volume = 96π cm³ ≈ 301.71 cm³
Curved surface area = 60π cm² ≈ 188.57 cm²
9. Suppose you have a cup in the shape of a right circular cone. Fill it with water to half the depth of the cone. What fraction of the volume of the cup is occupied by the water?
Let the full cone have height h and radius r.
If the water is filled to half the depth, then the water forms a smaller similar cone with:
Height = h/2
Since the cones are similar:
Radius = r/2
Volume of full cup:
V = (1/3)πr²h
Volume of water:
V₁ = (1/3)π(r/2)²(h/2)
= (1/3)πr²h × 1/8
So,
V₁ = V/8
Therefore, the water occupies:
1/8 of the volume of the cup.