Class 10 Math Ch Surface Areas and Volumes – Ex 13.4 focuses on solving real-life problems involving the volume and surface area of a frustum of a cone. A frustum is formed when a cone is cut parallel to its base, and in this exercise, students learn how to apply the formulas for curved surface area, total surface area, and volume of a frustum.
These questions help develop spatial understanding and improve problem-solving skills in mensuration. The step-by-step NCERT solutions provided here make it easy for students to grasp the concept and apply it accurately in exams.
Class 10 Math Ch Surface Areas and Volumes Ex 13.4
Ex 13.4 Class 10 Maths Question 1.
A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.
Solution:
Given:
- Height of the frustum = 14 cm
- Diameter of the larger base = 4 cm → Radius (R) = 2 cm
- Diameter of the smaller base = 2 cm → Radius (r) = 1 cm
Formula for the volume of a frustum of a cone:
Volume = (1/3) × π × h × (R² + Rr + r²)
Volume = (1/3) × π × 14 × (2² + 2×1 + 1²)
Volume = (1/3) × π × 14 × (4 + 2 + 1)
Volume = (1/3) × π × 14 × 7
Volume = (1/3) × π × 98
Volume = (98π) / 3 cubic centimeters
Approximate value:
Volume ≈ 102.67 cm³
Ex 13.4 Class 10 Maths Question 2.
The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.
Solution:
Given:
- Slant height = 4 cm
- Perimeter (circumference) of the larger circular end = 18 cm
- Perimeter (circumference) of the smaller circular end = 6 cm
Formula for curved surface area (CSA) of a frustum of a cone:
CSA = (1/2) × (P₁ + P₂) × slant height
Where:
- P₁ and P₂ are the perimeters of the two circular ends
- Slant height = 4 cm
CSA = (1/2) × (18 + 6) × 4
CSA = (1/2) × 24 × 4
CSA = 12 × 4
CSA = 48 cm²
Ex 13.4 Class 10 Maths Question 3.
A fez, the cap used by the Turks, is shaped like the frustum of a cone. If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.
Solution:
Given:
- Radius of the larger (open) base = 10 cm
- Radius of the upper (smaller) base = 4 cm
- Slant height = 15 cm
Step 1: Formula for curved surface area (CSA) of a frustum:
CSA = π × (R + r) × l
Where:
- R = larger radius = 10 cm
- r = smaller radius = 4 cm
- l = slant height = 15 cm
CSA = π × (10 + 4) × 15
CSA = π × 14 × 15
CSA = 210π cm²
Step 2: Area of the top circular surface:
Area = π × r² = π × 4² = 16π cm²
Step 3: Total area of material used:
Total area = CSA + top area
= 210π + 16π = 226π cm²
Ex 13.4 Class 10 Maths Question 4.
A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of Rs 20 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs 8 per 100 cm2.
Solution:
Given:
- Height = 16 cm
- Lower radius (r) = 8 cm
- Upper radius (R) = 20 cm
- Milk cost = Rs 20 per litre
- Metal sheet cost = Rs 8 per 100 cm²
- Container is open at the top
Volume of frustum (for milk)
Volume = (1/3) × π × h × (R² + Rr + r²)
= (1/3) × π × 16 × (20² + 20×8 + 8²)
= (1/3) × π × 16 × (400 + 160 + 64)
= (1/3) × π × 16 × 624
= (1/3) × π × 9984
≈ 3.1416 × 3328 ≈ 10460.59 cm³
Convert to litres:
1 litre = 1000 cm³
So, Volume ≈ 10.46 litres
Milk cost = 10.46 × 20 = Rs 209.20
- Slant height (l):
l = √[(R – r)² + h²]
= √[(20 – 8)² + 16²] = √[144 + 256] = √400 = 20 cm
- CSA of frustum:
CSA = π × (R + r) × l
= π × (20 + 8) × 20 = π × 28 × 20 = 560π ≈ 1759.29 cm² - Bottom area:
= π × r² = π × 8² = 64π ≈ 201.06 cm² - Total metal area = CSA + bottom area:
= 1759.29 + 201.06 = 1960.35 cm²
Metal cost = (1960.35 / 100) × 8 = Rs 156.83
Ex 13.4 Class 10 Maths Question 5.
A metallic right circular cone 20 cm high and whose vertical angle is 60 is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1/16 cm, find the length of the wire.
Solution:
Given:
- Height of cone = 20 cm
- Vertical angle = 60 degrees
- The cone is cut halfway, so frustum height = 10 cm
- Diameter of wire = 1/16 cm, so radius = 1/32 cm
Step 1: Calculate radius of original cone base
tan 30° = radius / height
radius = 20 × (1/√3) ≈ 11.55 cm
Step 2: Radius of smaller cone (top part) = half of 11.55 = 5.775 cm
Step 3: Calculate volume of frustum
Volume = (1/3) × π × 10 × (11.55² + 11.55 × 5.775 + 5.775²)
= (1/3) × π × 10 × (133.4 + 66.7 + 33.35)
= (1/3) × π × 10 × 233.45
≈ 2446.4 cm³
Step 4: Calculate cross-sectional area of wire
Area = π × (1/32)² = π / 1024 ≈ 0.00307 cm²
Step 5: Find length of wire
Length = Volume / Area = 2446.4 / 0.00307 ≈ 796444 cm
Convert to meters: 796444 / 100 = 7964.4 meters
📘 Math & Science Solutions by Class
🔹 Class 10
🔹 Class 9
🔹 Class 8
🔹 Class 7
🔹 Class 6
🔹 Class 12
🔹 Class 11
- Class 11 Math Solutions
- Class 11 Physics Solutions
- Class 11 Chemistry Solutions
- Class 11 Biology Solutions
For the official Class 10 Mathematics Solutions, you can visit:
- NCERT Textbooks (for Class 10):