Surface Areas and Volumes Exercise 13.4 Class 10 NCERT

Understanding the Frustum of a Cone

To understand a frustum, imagine cutting off the top portion of a standard cone. What remains is a solid with two parallel circular bases of different sizes. Let the radius of the larger circular base be $r_1$, the radius of the smaller circular base be $r_2$, the vertical height be $h$, and the slant height be $l$. The relationship between the slant height and the two radii is given by $l = \sqrt{h^2 + (r_1 - r_2)^2}$. Calculating the surface area and volume of a frustum involves finding the differences between the dimensions of the original large cone and the smaller cut-off cone. This similarity-of-triangles approach forms the basis of all frustum formulas.

Key Formulas for Frustum of a Cone

Slant Height (l)

l = \sqrt{h^2 + (r_1 - r_2)^2}, where r_1 is the radius of the larger circular base and r_2 is the radius of the smaller base.

Curved Surface Area (CSA)

CSA = \pi l (r_1 + r_2)

Total Surface Area (TSA)

TSA = \pi l (r_1 + r_2) + \pi r_1^2 + \pi r_2^2

Volume (Capacity)

V = \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 r_2)

Step-by-Step Problem-Solving Approach

1. Step 1: Identify and Label the Given Dimensions — Read the problem carefully to identify $r_1$, $r_2$, $h$, or $l$. Always assign $r_1$ to the larger radius ($r_1 > r_2$) to keep your calculations positive and standard.
2. Step 2: Align the Units — Ensure all dimensions are in the same units (e.g., all in cm or all in m). If capacity in litres is required, remember that $1000 \text{ cm}^3 = 1 \text{ litre}$ and $1 \text{ m}^3 = 1000 \text{ litres}$.
3. Step 3: Calculate the Slant Height if Needed — If the problem asks for the surface area but provides only the vertical height $h$ and radii, compute $l$ using $l = \sqrt{h^2 + (r_1 - r_2)^2}$ before moving forward.
4. Step 4: Substitute and Solve — Plug the values into the target formula. Keep $\pi$ as a symbol until the final steps of calculation to avoid premature decimal rounding errors.

Common Exam Mistakes to Avoid

• Open Bucket Pitfall: When a question asks for the metal sheet required to make a bucket (which is open at the top), do not use the full TSA formula. Use: $\text{Area of metal sheet} = \text{CSA} + \text{Area of base} = \pi l(r_1 + r_2) + \pi r_2^2$, where $r_2$ is the radius of the closed bottom end.
• Confusing Diameters with Radii: Board exam questions frequently provide diameters. Always divide them by 2 before substituting into the formulas.
• Sign Mistakes: In the slant height formula, remember it is $(r_1 - r_2)^2$, not $(r_1 + r_2)^2$. Double-check your arithmetic sign before squaring.

Practice Questions with Solutions

• Q: 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. (Use $\pi = \frac{22}{7}$) A: Step 1: Find the radii from the given diameters. $r_1 = \frac{4}{2} = 2$ cm, and $r_2 = \frac{2}{2} = 1$ cm. Height ($h$) = 14 cm. Step 2: Use the volume formula for a frustum: $V = \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 r_2)$ Step 3: Substitute the values: $V = \frac{1}{3} \times \frac{22}{7} \times 14 \times (2^2 + 1^2 + 2 \times 1)$ $V = \frac{1}{3} \times 22 \times 2 \times (4 + 1 + 2)$ $V = \frac{44}{3} \times 7 = \frac{308}{3} = 102 \frac{2}{3} \text{ cm}^3$ Final answer: The capacity of the glass is $102.67 \text{ cm}^3$ (or $102 \frac{2}{3} \text{ cm}^3$).
• Q: The slant height of a frustum of a cone is 4 cm and the perimeters (circumferences) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum. A: Step 1: Set up equations for the circumferences to find the radii. $2\pi r_1 = 18 \implies \pi r_1 = 9$ $2\pi r_2 = 6 \implies \pi r_2 = 3$ Adding these gives: $\pi(r_1 + r_2) = 12$ Step 2: Recall the Curved Surface Area (CSA) formula: $\text{CSA} = \pi l (r_1 + r_2)$ Step 3: Rearrange the formula as $(\pi(r_1 + r_2)) \times l$. Substitute the known values: $\text{CSA} = 12 \times 4 = 48 \text{ cm}^2$ Final answer: The curved surface area of the frustum is $48 \text{ cm}^2$.
• Q: A metal bucket open at the top is in the form of a frustum of a cone with height 16 cm. The radii of its lower and upper ends are 8 cm and 20 cm respectively. Find the cost of milk which can completely fill the bucket at the rate of Rs 40 per litre. (Take $\pi = 3.14$) A: Step 1: Identify given variables: $r_1 = 20$ cm, $r_2 = 8$ cm, $h = 16$ cm. Step 2: Calculate the volume of the bucket: $V = \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 r_2)$ $V = \frac{1}{3} \times 3.14 \times 16 \times (20^2 + 8^2 + 20 \times 8)$ $V = \frac{50.24}{3} \times (400 + 64 + 160)$ $V = \frac{50.24}{3} \times 624 = 50.24 \times 208 = 10449.92 \text{ cm}^3$ Step 3: Convert volume to litres: $\text{Capacity in litres} = \frac{10449.92}{1000} \approx 10.45 \text{ litres}$ Step 4: Find the total cost at Rs 40 per litre: $\text{Cost} = 10.45 \times 40 = \text{Rs } 418$ Final answer: The cost of the milk is Rs 418.
• Q: A Turkish cap, shaped like a frustum of a cone, has a radius of 10 cm at its open base and 4 cm at its upper base. If its slant height is 15 cm, find the area of material used for making it. A: Step 1: Identify the given values. $r_1 = 10$ cm (open base radius) $r_2 = 4$ cm (closed top radius) $l = 15$ cm Step 2: State the formula for the material used. Since the base is open, the area of material equals the CSA plus the area of the closed top circle: $\text{Area of material} = \pi l (r_1 + r_2) + \pi r_2^2$ Step 3: Substitute the values: $\text{Area} = \pi \times 15 \times (10 + 4) + \pi \times 4^2$ $\text{Area} = \pi \times 15 \times 14 + 16\pi$ $\text{Area} = 210\pi + 16\pi = 226\pi$ $\text{Area} = 226 \times \frac{22}{7} = \frac{4972}{7} \approx 710.28 \text{ cm}^2$ Final answer: The area of material used for making the cap is $710.28 \text{ cm}^2$.

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