13. Surface Areas and Volumes Mathematics Exercise - 13.4 class 10 Maths in English - CBSE Study
NCERT Solutions for Class 10 Mathematics are carefully prepared according to the latest CBSE syllabus and NCERT textbooks to help students understand every concept clearly. These solutions cover all important 13. Surface Areas and Volumes with detailed explanations and step-by-step answers for better exam preparation. Each Exercise 13.4 is explained in simple language so that students can easily grasp the fundamentals and improve their academic performance. The study material is designed to support daily homework, revision practice, and final exam preparation for Class 10 students. With accurate answers, concept clarity, and structured content, these NCERT solutions help learners build confidence and score higher marks in their examinations. Whether you are revising a specific topic or preparing an entire chapter, this resource provides reliable and syllabus-based guidance for complete success in Mathematics.
Class 10 English Medium Mathematics All Chapters:
13. Surface Areas and Volumes
4. Exercise 13.4
13. Surface Areas and Volumes Exercise 13.4
Q1. 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:
Diameter of upper end D = 4 cm
Radius of upper end R = 2 cm
Diameter of lower end d = 2 cm
Radius of lower end r = 1 cm
Q2. 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:
Slant height (l ) of the frustum of the cone = 4 cm
Perimeter of upper end = 18 cm
Thus, the curved surface area of the frustum is 48 cm2.
Q3. A fez, the cap used by the Turks, is shaped like the frustum of a cone (see Fig. 13.24). 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:
Slant height (l) of the cap = 15 cm
Radius of open end (R) = 10 cm
Radius of upper end (r) = 4 cm
Area of material used for making it = CSA of Frustum + Area of upper part
Q4. 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. (Take π = 3.14)
Solution:
Height of the container (h) = 16 cm
Radius (R) of the top of the container = 20 cm
Radius of the lower end of the container (r) = 8 cm
Capacity in litre= 10.45 litre (Approx)
The cost of milk = 20 × 10.45 = ₹209.00
Area of used sheet = πl (R + r) + πr2
= 3.14 × 20 (20 + 8) + 3.14 × 8 × 8
= 3.14 × 20 (28) + 3.14 × 64
= 3.14 (560 + 64)
= 3.14 (624)
= 3.14 (624)
= 1959.36 cm2
The cost of sheet at the rate of ₹ 8 per 100 cm2
Hence, the cost of the metal sheet is ₹156.75.
Q5. 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:
AD = 20 cm
Then, AG = 10 cm (cut into two parts at the middle)
∠BAC = 60o
AD bisects ∠BAC
Thus, ∠CAD = 30o
In right angled triangle ΔAGF
Similarly, in right angled triangle DADC,
Lets the length of wire be H
Volume of the wire = Volume of the frustum obtained
H = 7964.44 m
Hence, the length of the wire is 7964.44 m.
Topic Lists: