Ex 14.3 Class 9 Maths Ganita Manjari Solutions | Sid Classes
Welcome to Sid Classes, your trusted destination for comprehensive academic resources! In this detailed study guide, we present complete, step-by-step solutions for the NCERT Class 9 Maths Ganita Manjari Chapter 14 Exercise 14.3 Surface Areas and Volumes of Right Circular Cones, Slant Height Relations, and Conical Applications. Designed to reflect how top-scoring students write their solutions and reasoning in school examinations, this walkthrough covers curved surface area, total surface area, cone volume, tarpaulin tent requirements, rotated right triangles, and fractional water volume calculations entirely clear.
🔑 Key Formulas & Principles:
- Slant Height (\(l\)): Calculated using radius (\(r\)) and height (\(h\)) via Pythagoras' theorem: \(l = \sqrt{r^2 + h^2}\).
- Curved Surface Area (CSA): \(\text{CSA} = \pi rl\).
- Total Surface Area (TSA): \(\text{TSA} = \pi rl + \pi r^2 = \pi r(l + r)\).
- Volume of a Cone: \(V = \frac{1}{3}\pi r^2 h\).
NCERT Class 9 Maths Ganita Manjari Chapter 14 Exercise 14.3 Solutions
Question 1
Find the total surface area of a cone, if its slant height is \(21\text{ m}\) and diameter of its base is \(24\text{ m}\).
Solution:
1. Given:
- Slant height (\(l\)) = \(21\text{ m}\)
- Base diameter (\(d\)) = \(24\text{ m} \implies \text{Radius } (r) = \frac{24}{2} = 12\text{ m}\)
2. Total Surface Area (\(\text{TSA}\)) of a cone = \(\pi r(l + r)\)
3. Substituting the values (\(\pi \approx \frac{22}{7}\)):
\(\text{TSA} = \frac{22}{7} \times 12 \times (21 + 12)\)
\(\text{TSA} = \frac{22}{7} \times 12 \times 33 = \frac{8712}{7} = \mathbf{1244.57\text{ m}^2}\) (or \(1244.57\text{ m}^2\)).
Question 2
Find the curved surface area of a right circular cone whose slant height is \(10\text{ cm}\) and base radius is \(7\text{ cm}\).
Solution:
1. Given:
- Slant height (\(l\)) = \(10\text{ cm}\)
- Base radius (\(r\)) = \(7\text{ cm}\)
2. Curved Surface Area (\(\text{CSA}\)) = \(\pi rl\)
3. Substituting the values (\(\pi = \frac{22}{7}\)):
\(\text{CSA} = \frac{22}{7} \times 7 \times 10 = \mathbf{220\text{ cm}^2}\).
Question 3
The height of a cone is \(16\text{ cm}\), and its base radius is \(12\text{ cm}\). Find the curved surface area and the total surface area of the cone.
Solution:
1. Given: Height (\(h\)) = \(16\text{ cm}\), Radius (\(r\)) = \(12\text{ cm}\).
2. Find slant height (\(l\)):
\(l = \sqrt{r^2 + h^2} = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20\text{ cm}\).
3. Curved Surface Area (\(\text{CSA}\)):
\(\text{CSA} = \pi rl = 3.14 \times 12 \times 20 = \mathbf{753.6\text{ cm}^2}\) (or \(\frac{5280}{7} \approx 754.29\text{ cm}^2\)).
4. Total Surface Area (\(\text{TSA}\)):
\(\text{TSA} = \pi r(l + r) = 3.14 \times 12 \times (20 + 12) = 3.14 \times 12 \times 32 = \mathbf{1205.76\text{ cm}^2}\).
Question 4
A cone has a height of \(15\text{ cm}\). If its volume is \(1570\text{ cm}^3\), find the radius of the base. (\(\pi = 3.14\))
Solution:
1. Given: Height (\(h\)) = \(15\text{ cm}\), Volume (\(V\)) = \(1570\text{ cm}^3\), \(\pi = 3.14\).
2. Volume formula: \(V = \frac{1}{3}\pi r^2 h\)
3. Substituting the values:
\(1570 = \frac{1}{3} \times 3.14 \times r^2 \times 15\)
\(1570 = 3.14 \times 5 \times r^2\)
\(1570 = 15.7 \times r^2\)
4. Solving for \(r^2\):
\(r^2 = \frac{1570}{15.7} = 100 \implies r = \mathbf{10\text{ cm}}\).
Question 5
The curved surface area of a cone is \(308\text{ cm}^2\) and its slant height is \(14\text{ cm}\). Find (i) the radius of the base, (ii) the total surface area of the cone.
Solution:
* (i) Radius of the base (\(r\)):
\(\text{CSA} = \pi rl \implies 308 = \frac{22}{7} \times r \times 14\)
\(308 = 44r \implies r = \frac{308}{44} = \mathbf{7\text{ cm}}\).
* (ii) Total surface area (\(\text{TSA}\)):
\(\text{TSA} = \text{CSA} + \pi r^2 = 308 + \frac{22}{7} \times (7)^2 = 308 + 154 = \mathbf{462\text{ cm}^2}\).
Question 6
A joker's cap is in the form of a right circular cone of base radius \(7\text{ cm}\) and height \(24\text{ cm}\). Find the area of the sheet required to make 10 such caps.
Solution:
1. Given: Radius (\(r\)) = \(7\text{ cm}\), Height (\(h\)) = \(24\text{ cm}\).
2. Find slant height (\(l\)):
\(l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25\text{ cm}\).
3. Curved surface area of one cap (since a cap is open at the base):
\(\text{CSA} = \pi rl = \frac{22}{7} \times 7 \times 25 = 550\text{ cm}^2\).
4. Area of sheet required for 10 such caps = \(10 \times 550 = \mathbf{5500\text{ cm}^2}\).
Question 7
What length of tarpaulin \(3\text{ m}\) wide is required to make a conical tent of height \(8\text{ m}\) and base radius \(6\text{ m}\)? Assume that the extra length of material that is required for stitching margins and wastage in cutting is \(20\text{ cm}\).
Solution:
1. Given: Height (\(h\)) = \(8\text{ m}\), Radius (\(r\)) = \(6\text{ m}\).
2. Slant height (\(l\)) = \(\sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\text{ m}\).
3. Curved surface area of tent (\(\text{CSA}\)):
\(\text{CSA} = \pi rl = 3.14 \times 6 \times 10 = 188.4\text{ m}^2\).
4. Area of tarpaulin = \(\text{Length} \times \text{Width}\):
\(\text{Length} \times 3 = 188.4 \implies \text{Length} = \frac{188.4}{3} = 62.8\text{ m}\).
5. Adding extra length for stitching margins and wastage (\(20\text{ cm} = 0.2\text{ m}\)):
Total length required = \(62.8 + 0.2 = \mathbf{63\text{ m}}\).
Question 8
A right triangle with sides \(6\text{ cm}\), \(8\text{ cm}\) and \(10\text{ cm}\) is rotated through \(360^\circ\) about the side of \(8\text{ cm}\). Find the volume and the curved surface area of the solid so formed.
Solution:
1. Rotating the triangle about its \(8\text{ cm}\) side forms a cone where:
- Height (\(h\)) = \(8\text{ cm}\)
- Radius (\(r\)) = \(6\text{ cm}\)
- Slant height (\(l\)) = \(10\text{ cm}\)
2. Volume of the cone:
\(V = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \pi \times 6^2 \times 8 = \frac{1}{3} \times \pi \times 36 \times 8 = 96\pi = \mathbf{301.44\text{ cm}^3}\) (or \(\frac{2112}{7}\)).
3. Curved Surface Area (\(\text{CSA}\)):
\(\text{CSA} = \pi rl = \pi \times 6 \times 10 = 60\pi = \mathbf{188.4\text{ cm}^2}\) (or \(\frac{1320}{7}\)).
Question 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?
Solution:
1. Let the total height of the cone be \(h\) and base radius be \(r\). Total volume \(V_{\text{total}} = \frac{1}{3}\pi r^2 h\).
2. When the water level is filled to half the depth, the height of the water cone is \(h' = \frac{h}{2}\).
3. By similar triangles, the radius of the water surface at half depth is also halved: \(r' = \frac{r}{2}\).
4. Volume of the water:
\(V_{\text{water}} = \frac{1}{3}\pi (r')^2 h' = \frac{1}{3}\pi \left(\frac{r}{2}\right)^2 \left(\frac{h}{2}\right) = \frac{1}{8} \left(\frac{1}{3}\pi r^2 h\right) = \frac{1}{8} V_{\text{total}}\).
5. Therefore, \(\mathbf{\frac{1}{8}}\) of the volume of the cup is occupied by the water.
Frequently Asked Questions (FAQs)
Q1: How do you find the slant height of a cone when radius and height are given?
A: Use Pythagoras' theorem: \(l = \sqrt{r^2 + h^2}\).
Q2: Why is the curved surface area formula of a cone \(\pi rl\) while total surface area includes \(\pi r^2\)?
A: \(\pi rl\) accounts only for the lateral sloping side of the cone, whereas \(\pi r^2\) adds the area of the circular base.
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