Master Surface Areas and Volumes: NCERT Ex 13.5 (Class 10 Maths)

Understanding the Frustum of a Cone

The frustum of a cone is a part of a cone, obtained when it is cut by a plane parallel to its base. Imagine a full cone, and then slice off its top portion parallel to the base – what remains is a frustum. This 3D shape has two circular bases of different radii, a vertical height, and a slant height. Understanding its components is the first step to mastering its calculations.

Let's define the key parameters for a frustum:

• R: Radius of the larger base.
• r: Radius of the smaller base.
• h: Vertical height of the frustum.
• l: Slant height of the frustum.

The formulas for the frustum are derived from those of a cone. It's essential to remember these:

1. Slant Height (l): $l = \sqrt{h^2 + (R-r)^2}$
2. Curved Surface Area (CSA): CSA $= \pi (R+r)l$
3. Total Surface Area (TSA): TSA $=$ CSA $+$ Area of larger base $+$ Area of smaller base

TSA $= \pi (R+r)l + \pi R^2 + \pi r^2$

1. Volume (V): $V = \frac{1}{3} \pi h (R^2 + r^2 + Rr)$

These formulas are your toolkit for solving problems related to Exercise 13.5. Pay close attention to units and ensure consistency throughout your calculations.

Key Definitions

Frustum of a Cone

A portion of a cone that remains when the top part is cut off by a plane parallel to its base. It has two parallel circular bases of different radii.

Slant Height of Frustum

The distance along the lateral surface from a point on the circumference of one base to a corresponding point on the circumference of the other base. It is calculated as $l = \sqrt{h^2 + (R-r)^2}$.

Curved Surface Area (CSA)

The area of the curved lateral surface of the frustum, excluding the areas of its two circular bases. Its formula is $\pi (R+r)l$.

Total Surface Area (TSA)

The sum of the curved surface area and the areas of both the top and bottom circular bases. TSA $= \pi (R+r)l + \pi R^2 + \pi r^2$.

Volume of Frustum

The amount of space occupied by the frustum. Its formula is $V = \frac{1}{3} \pi h (R^2 + r^2 + Rr)$.

Worked Example: Finding the Volume of a Frustum

1. Problem Statement — 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.
2. Step 1: Identify Given Values and Convert to Radii — Given: Height of frustum, $h = 14$ cm Diameter of larger end, $D_1 = 4$ cm $\Rightarrow$ Radius, $R = D_1/2 = 4/2 = 2$ cm Diameter of smaller end, $D_2 = 2$ cm $\Rightarrow$ Radius, $r = D_2/2 = 2/2 = 1$ cm We need to find the capacity of the glass, which is its volume.
3. Step 2: Recall the Volume Formula for a Frustum — The formula for the volume of a frustum of a cone is: $V = \frac{1}{3} \pi h (R^2 + r^2 + Rr)$
4. Step 3: Substitute the Values into the Formula — Substitute $h=14$, $R=2$, $r=1$, and $\pi = 22/7$ into the volume formula: $V = \frac{1}{3} \times \frac{22}{7} \times 14 (2^2 + 1^2 + 2 \times 1)$
5. Step 4: Perform the Calculations — $V = \frac{1}{3} \times \frac{22}{7} \times 14 (4 + 1 + 2)$ $V = \frac{1}{3} \times 22 \times 2 (7)$ $V = \frac{1}{3} \times 44 \times 7$ $V = \frac{308}{3}$ $V = 102.67$ (approximately)
6. Step 5: State the Final Answer with Units — The capacity of the glass (volume of the frustum) is approximately $102.67$ cm$^3$.

Exam Tip: Avoiding Common Mistakes

When solving problems involving frustums, students often make a few common errors. Be mindful of these to maximize your scores:

1. Radius vs. Diameter: Always check whether the given values are radii or diameters. Remember to divide the diameter by two to get the radius before using it in formulas.
2. Formula Confusion: Do not confuse frustum formulas with cone formulas. They are similar but distinct. For instance, the volume of a frustum has an extra term $(R^2 + r^2 + Rr)$ compared to a cone's $r^2$.
3. Slant Height Calculation: When finding the CSA or TSA, you first need the slant height, $l$. Use the correct formula $l = \sqrt{h^2 + (R-r)^2}$. A common mistake is using $(R+r)^2$ or just $R^2+r^2$.
4. Unit Consistency: Ensure all dimensions are in the same units (e.g., all cm or all m) before calculation. If not, convert them first.
5. Arithmetic Errors: Double-check your calculations, especially when dealing with squares and cubes. A small calculation error can lead to a completely wrong answer.

Practice Questions with Solutions

• Q: The radii of the top and bottom ends of a frustum of a cone are 14 cm and 7 cm respectively. If the height of the frustum is 24 cm, find its volume. A: Step 1: Identify given values. $R=14$ cm, $r=7$ cm, $h=24$ cm. Step 2: Apply the volume formula for a frustum: $V = \frac{1}{3} \pi h (R^2 + r^2 + Rr)$. Step 3: Substitute the values: $V = \frac{1}{3} \times \frac{22}{7} \times 24 (14^2 + 7^2 + 14 \times 7)$. Step 4: Calculate: $V = \frac{1}{3} \times \frac{22}{7} \times 24 (196 + 49 + 98) = \frac{1}{3} \times \frac{22}{7} \times 24 (343)$. $V = 22 \times 8 \times 49 = 8624$ cm$^3$. Final answer: The volume of the frustum is 8624 cm$^3$.
• Q: A bucket is in the form of a frustum of a cone with a height of 30 cm and radii of its lower and upper ends as 10 cm and 20 cm, respectively. Find the curved surface area of the bucket. (Use $\pi = 3.14$) A: Step 1: Identify given values. $h=30$ cm, $r=10$ cm, $R=20$ cm. Step 2: Calculate the slant height, $l = \sqrt{h^2 + (R-r)^2}$. $l = \sqrt{30^2 + (20-10)^2} = \sqrt{900 + 10^2} = \sqrt{900 + 100} = \sqrt{1000} = 10\sqrt{10}$ cm $\approx 10 \times 3.162 = 31.62$ cm. Step 3: Apply the CSA formula for a frustum: CSA $= \pi (R+r)l$. Step 4: Substitute and calculate: CSA $= 3.14 \times (20+10) \times 31.62 = 3.14 \times 30 \times 31.62 = 94.2 \times 31.62 \approx 2977.644$ cm$^2$. Final answer: The curved surface area of the bucket is approximately 2977.64 cm$^2$.
• Q: A tent is in the shape of a frustum of a cone surmounted by a cone. The radii of the bases of the frustum are 8 m and 10 m, and its height is 6 m. If the height of the conical part is 4 m, find the total canvas required for the tent (neglect the base of the frustum). (Use $\pi = 3.14$) A: Step 1: Frustum dimensions: $R=10$ m, $r=8$ m, $h_f=6$ m. Conical part dimensions: $r_{cone}=R=10$ m, $h_c=4$ m. Step 2: Calculate slant height of frustum ($l_f$) and conical part ($l_c$). $l_f = \sqrt{h_f^2 + (R-r)^2} = \sqrt{6^2 + (10-8)^2} = \sqrt{36 + 2^2} = \sqrt{36+4} = \sqrt{40} = 2\sqrt{10} \approx 6.32$ m. $l_c = \sqrt{r_{cone}^2 + h_c^2} = \sqrt{10^2 + 4^2} = \sqrt{100 + 16} = \sqrt{116} \approx 10.77$ m. Step 3: Calculate CSA of frustum: CSA$_f = \pi (R+r)l_f = 3.14 \times (10+8) \times 6.32 = 3.14 \times 18 \times 6.32 \approx 357.77$ m$^2$. Step 4: Calculate CSA of cone: CSA$_c = \pi r_{cone}l_c = 3.14 \times 10 \times 10.77 = 31.4 \times 10.77 \approx 338.298$ m$^2$. Step 5: Total canvas required = CSA$_f +$ CSA$_c = 357.77 + 338.298 \approx 696.068$ m$^2$. Final answer: The total canvas required for the tent is approximately 696.07 m$^2$.
• Q: A metallic right circular cone 20 cm high and whose vertical angle is $60^\circ$ is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained is drawn into a wire of diameter 1/16 cm, find the length of the wire. A: Step 1: Analyze the original cone. Height $H = 20$ cm, vertical angle $60^\circ$. This means the semi-vertical angle is $30^\circ$. Radius of base $R_1 = H \tan 30^\circ = 20 \times (1/\sqrt{3}) = 20/\sqrt{3}$ cm. Step 2: Analyze the smaller cone (top part). Height $h_s = 10$ cm. Radius of its base $R_2 = h_s \tan 30^\circ = 10 \times (1/\sqrt{3}) = 10/\sqrt{3}$ cm. Step 3: Frustum dimensions. $R = R_1 = 20/\sqrt{3}$ cm (larger radius), $r = R_2 = 10/\sqrt{3}$ cm (smaller radius), height of frustum $h = 10$ cm. Step 4: Calculate the volume of the frustum. $V_f = \frac{1}{3} \pi h (R^2 + r^2 + Rr)$. $V_f = \frac{1}{3} \pi (10) \left[ (\frac{20}{\sqrt{3}})^2 + (\frac{10}{\sqrt{3}})^2 + (\frac{20}{\sqrt{3}})(\frac{10}{\sqrt{3}}) \right]$ $V_f = \frac{10\pi}{3} \left[ \frac{400}{3} + \frac{100}{3} + \frac{200}{3} \right] = \frac{10\pi}{3} \left[ \frac{700}{3} \right] = \frac{7000\pi}{9}$ cm$^3$. Step 5: The frustum is drawn into a wire. Volume of wire = Volume of frustum. Wire is a cylinder. Diameter of wire $d = 1/16$ cm $\Rightarrow$ radius $r_w = 1/32$ cm. Let length be $L$. Volume of wire $V_w = \pi r_w^2 L = \pi (1/32)^2 L = \frac{\pi}{1024} L$. Step 6: Equate volumes and solve for $L$. $\frac{\pi}{1024} L = \frac{7000\pi}{9}$ $L = \frac{7000}{9} \times 1024 = \frac{7168000}{9} \approx 796444.44$ cm. $L = 796444.44 / 100 = 7964.44$ meters. Final answer: The length of the wire is approximately 7964.44 meters.

Frequently Asked Questions