Mensuration II
Lesson Objectives
- Recall and apply area and volume formulas of advanced geometric shapes.
- Calculate surface areas and volumes of combined solids.
- Solve word problems involving composite shapes and real-life applications.
- Interpret and solve past examination problems on difficult mensuration tasks.
Lesson Introduction
Mensuration at this level involves computing areas and volumes of complex shapes including frustums, hollow cylinders, and composite solids. This lesson will walk you through difficult-level mensuration problems that integrate multiple shapes and real-life problem-solving strategies.
Core Lesson Content
Formulas Recap
- Volume of a cone: \( V = \frac{1}{3} \pi r^2 h \)
- Surface area of a cone: \( \pi r (r + l) \)
- Volume of a sphere: \( V = \frac{4}{3} \pi r^3 \)
- Volume of a frustum: \( V = \frac{1}{3} \pi h (R^2 + r^2 + Rr) \)
- Volume of a hollow cylinder: \( V = \pi h (R^2 - r^2) \)
Worked Example
Example 1: Calculate the volume of a cone with radius 7 cm and height 12 cm.
\( V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (7^2)(12) = \frac{1}{3} \pi (588) = 196\pi \approx 615.75\ \text{cm}^3 \)
\( V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (7^2)(12) = \frac{1}{3} \pi (588) = 196\pi \approx 615.75\ \text{cm}^3 \)
Example 2: A sphere has a radius of 6 cm. Find its volume.
\( V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (6)^3 = \frac{4}{3} \pi (216) = 288\pi \approx 904.32\ \text{cm}^3 \)
\( V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (6)^3 = \frac{4}{3} \pi (216) = 288\pi \approx 904.32\ \text{cm}^3 \)
Example 3: Find the surface area of a cone with radius 5 cm and slant height 13 cm.
\( SA = \pi r (r + l) = \pi (5)(5 + 13) = \pi (5)(18) = 90\pi \approx 282.74\ \text{cm}^2 \)
\( SA = \pi r (r + l) = \pi (5)(5 + 13) = \pi (5)(18) = 90\pi \approx 282.74\ \text{cm}^2 \)
Example 4: A frustum of a cone has radii 10 cm and 6 cm, and height 8 cm. Find its volume.
\( V = \frac{1}{3} \pi h (R^2 + r^2 + Rr) \)
\( = \frac{1}{3} \pi (8)(100 + 36 + 60) = \frac{1}{3} \pi (8)(196) = \frac{1}{3} \pi (1568) = 522.67\pi \approx 1641.86\ \text{cm}^3 \)
\( V = \frac{1}{3} \pi h (R^2 + r^2 + Rr) \)
\( = \frac{1}{3} \pi (8)(100 + 36 + 60) = \frac{1}{3} \pi (8)(196) = \frac{1}{3} \pi (1568) = 522.67\pi \approx 1641.86\ \text{cm}^3 \)
Example 5: Calculate the volume of a hollow cylinder with outer radius 10 cm, inner radius 6 cm, and height 20 cm.
\( V = \pi h (R^2 - r^2) = \pi (20)(100 - 36) = \pi (20)(64) = 1280\pi \approx 4021.24\ \text{cm}^3 \)
\( V = \pi h (R^2 - r^2) = \pi (20)(100 - 36) = \pi (20)(64) = 1280\pi \approx 4021.24\ \text{cm}^3 \)
Example 6: [WAEC] A cylindrical pipe 3 m long has an external radius of 14 cm and thickness 2 cm. Find its volume.
Convert 3 m to 300 cm.
Outer radius \( R = 14 \) cm, inner radius \( r = 12 \) cm.
\( V = \pi h (R^2 - r^2) = \pi (300)(196 - 144) = \pi (300)(52) = 15600\pi \approx 49008\ \text{cm}^3 \)
Convert 3 m to 300 cm.
Outer radius \( R = 14 \) cm, inner radius \( r = 12 \) cm.
\( V = \pi h (R^2 - r^2) = \pi (300)(196 - 144) = \pi (300)(52) = 15600\pi \approx 49008\ \text{cm}^3 \)
Example 7: A hemisphere has a radius of 5 cm. Calculate its volume.
Volume of a hemisphere = \( \frac{2}{3} \pi r^3 = \frac{2}{3} \pi (125) = \frac{250}{3} \pi \approx 261.80\ \text{cm}^3 \)
Volume of a hemisphere = \( \frac{2}{3} \pi r^3 = \frac{2}{3} \pi (125) = \frac{250}{3} \pi \approx 261.80\ \text{cm}^3 \)
Example 8: A cone is exactly fitted into a hemisphere. If the radius is 6 cm, find the ratio of their volumes.
Volume of cone = \( \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (36)(6) = 72\pi \)
Volume of hemisphere = \( \frac{2}{3} \pi r^3 = \frac{2}{3} \pi (216) = 144\pi \)
Ratio = \( \frac{72\pi}{144\pi} = \frac{1}{2} \)
Volume of cone = \( \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (36)(6) = 72\pi \)
Volume of hemisphere = \( \frac{2}{3} \pi r^3 = \frac{2}{3} \pi (216) = 144\pi \)
Ratio = \( \frac{72\pi}{144\pi} = \frac{1}{2} \)
Example 9: [NECO] A cylindrical tank has a height of 4 m and radius 2 m. Find the surface area.
SA = \( 2\pi r^2 + 2\pi rh = 2\pi (4) + 2\pi (2)(4) = 8\pi + 16\pi = 24\pi \approx 75.40\ \text{m}^2 \)
SA = \( 2\pi r^2 + 2\pi rh = 2\pi (4) + 2\pi (2)(4) = 8\pi + 16\pi = 24\pi \approx 75.40\ \text{m}^2 \)
Example 10: A frustum has a top radius of 4 cm, bottom radius of 6 cm, and height of 10 cm. Find its volume.
\( V = \frac{1}{3} \pi h (R^2 + r^2 + Rr) = \frac{1}{3} \pi (10)(36 + 16 + 24) = \frac{1}{3} \pi (10)(76) = 253.33\pi \approx 795.87\ \text{cm}^3 \)
\( V = \frac{1}{3} \pi h (R^2 + r^2 + Rr) = \frac{1}{3} \pi (10)(36 + 16 + 24) = \frac{1}{3} \pi (10)(76) = 253.33\pi \approx 795.87\ \text{cm}^3 \)
Exercises
- Find the volume of a cone with base radius 9 cm and height 15 cm.
- [NECO] A frustum has upper radius 5 cm, lower radius 9 cm, and height 7 cm. Find its volume. (Past Question)
- Calculate the volume of a hollow metal tube with outer radius 10 cm, inner radius 7 cm, and height 30 cm.
- A sphere has diameter 10 cm. Find its volume and surface area.
- Find the surface area of a cone with radius 6 cm and slant height 10 cm.
- [WAEC] A bucket is in the shape of a frustum with radii 8 cm and 12 cm, and height 18 cm. Calculate its volume. (Past Question)
- Find the volume of a hemisphere of radius 4 cm.
- [JAMB] A cylindrical tank is 1.5 m high and has a diameter of 2 m. Find its volume. (Past Question)
- Calculate the surface area of a hollow cylinder with radii 9 cm and 6 cm, and height 20 cm.
- [WAEC] A solid metal cone has a radius of 4 cm and height 9 cm. How much material was used to make it? (Past Question)
Conclusion/Recap
We explored challenging mensuration problems, including volumes and surface areas of frustums, hollow cylinders, and composite solids. Mastery of these formulas and problem-solving techniques is crucial for tackling advanced-level mathematics and real-world engineering tasks.
Clip It!
Share your ANSWER in the Chat. Indicate TITLE e.g Linear Equation 1. .....2. e.t.c
