Table of Contents
Introduction
Surface Areas and Volume are vital concepts in Geometry, focusing on the measurement of the areas and volumes of 3D shapes. These are often used in real-world applications, such as in construction, design, and engineering. The formulas and concepts in this chapter will equip you with the knowledge needed to calculate these properties of solid objects.
Exercise 9.1: Surface Area of a Cube
Q1. A cube has an edge of 5 cm. Find its surface area.
Solution:
The formula for the surface area of a cube is:Surface Area=6a2\text{Surface Area} = 6a^2Surface Area=6a2
Where:
- aaa is the length of the edge of the cube.
Given:
Edge length a=5a = 5a=5 cm.
Now, substitute this into the formula:Surface Area=6×(5)2=6×25=150 cm2\text{Surface Area} = 6 \times (5)^2 = 6 \times 25 = 150 \text{ cm}^2Surface Area=6×(5)2=6×25=150 cm2
Answer: The surface area of the cube is 150 cm².
Exercise 9.2: Surface Area of a Cuboid
Q1. A cuboid has dimensions 10 cm, 5 cm, and 2 cm. Find its surface area.
Solution:
The formula for the surface area of a cuboid is:Surface Area=2(lb+bh+hl)\text{Surface Area} = 2(lb + bh + hl)Surface Area=2(lb+bh+hl)
Where:
- lll is the length,
- bbb is the breadth,
- hhh is the height.
Given:
- l=10l = 10l=10 cm,
- b=5b = 5b=5 cm,
- h=2h = 2h=2 cm.
Now, substitute these values into the formula:Surface Area=2×(10×5+5×2+2×10)\text{Surface Area} = 2 \times \left( 10 \times 5 + 5 \times 2 + 2 \times 10 \right)Surface Area=2×(10×5+5×2+2×10) =2×(50+10+20)= 2 \times (50 + 10 + 20)=2×(50+10+20) =2×80=160 cm2= 2 \times 80 = 160 \text{ cm}^2=2×80=160 cm2
Answer: The surface area of the cuboid is 160 cm².
Exercise 9.3: Surface Area of a Cylinder
Q1. A cylinder has a radius of 7 cm and a height of 14 cm. Find its surface area.
Solution:
The formula for the surface area of a cylinder is:Surface Area=2πr(h+r)\text{Surface Area} = 2\pi r(h + r)Surface Area=2πr(h+r)
Where:
- rrr is the radius of the base,
- hhh is the height of the cylinder.
Given:
- r=7r = 7r=7 cm,
- h=14h = 14h=14 cm.
Now, substitute these values into the formula:Surface Area=2×π×7×(14+7)\text{Surface Area} = 2 \times \pi \times 7 \times (14 + 7)Surface Area=2×π×7×(14+7) =2×π×7×21=294π cm2= 2 \times \pi \times 7 \times 21 = 294 \pi \, \text{cm}^2=2×π×7×21=294πcm2
Approximating π=3.14\pi = 3.14π=3.14:Surface Area=294×3.14=922.44 cm2\text{Surface Area} = 294 \times 3.14 = 922.44 \, \text{cm}^2Surface Area=294×3.14=922.44cm2
Answer: The surface area of the cylinder is approximately 922.44 cm².
Exercise 9.4: Volume of a Cube
Q1. A cube has an edge length of 4 cm. Find its volume.
Solution:
The formula for the volume of a cube is:Volume=a3\text{Volume} = a^3Volume=a3
Where aaa is the length of the edge of the cube.
Given:
- a=4a = 4a=4 cm.
Now, substitute this into the formula:Volume=43=64 cm3\text{Volume} = 4^3 = 64 \, \text{cm}^3Volume=43=64cm3
Answer: The volume of the cube is 64 cm³.
Exercise 9.5: Volume of a Cuboid
Q1. A cuboid has dimensions 8 cm, 6 cm, and 3 cm. Find its volume.
Solution:
The formula for the volume of a cuboid is:Volume=l×b×h\text{Volume} = l \times b \times hVolume=l×b×h
Where:
- lll is the length,
- bbb is the breadth,
- hhh is the height.
Given:
- l=8l = 8l=8 cm,
- b=6b = 6b=6 cm,
- h=3h = 3h=3 cm.
Now, substitute these values into the formula:Volume=8×6×3=144 cm3\text{Volume} = 8 \times 6 \times 3 = 144 \, \text{cm}^3Volume=8×6×3=144cm3
Answer: The volume of the cuboid is 144 cm³.
Exercise 9.6: Volume of a Cylinder
Q1. A cylinder has a radius of 3 cm and a height of 10 cm. Find its volume.
Solution:
The formula for the volume of a cylinder is:Volume=πr2h\text{Volume} = \pi r^2 hVolume=πr2h
Where:
- rrr is the radius,
- hhh is the height.
Given:
- r=3r = 3r=3 cm,
- h=10h = 10h=10 cm.
Now, substitute these values into the formula:Volume=π×32×10=π×9×10=90π cm3\text{Volume} = \pi \times 3^2 \times 10 = \pi \times 9 \times 10 = 90 \pi \, \text{cm}^3Volume=π×32×10=π×9×10=90πcm3
Approximating π=3.14\pi = 3.14π=3.14:Volume=90×3.14=282.6 cm3\text{Volume} = 90 \times 3.14 = 282.6 \, \text{cm}^3Volume=90×3.14=282.6cm3
Answer: The volume of the cylinder is approximately 282.6 cm³.
Exercise 9.7: Volume of a Cone
Q1. A cone has a radius of 5 cm and a height of 12 cm. Find its volume.
Solution:
The formula for the volume of a cone is:Volume=13πr2h\text{Volume} = \frac{1}{3} \pi r^2 hVolume=31πr2h
Where:
- rrr is the radius of the base,
- hhh is the height.
Given:
- r=5r = 5r=5 cm,
- h=12h = 12h=12 cm.
Now, substitute these values into the formula:Volume=13π×52×12=13π×25×12=100π cm3\text{Volume} = \frac{1}{3} \pi \times 5^2 \times 12 = \frac{1}{3} \pi \times 25 \times 12 = 100 \pi \, \text{cm}^3Volume=31π×52×12=31π×25×12=100πcm3
Approximating π=3.14\pi = 3.14π=3.14:Volume=100×3.14=314 cm3\text{Volume} = 100 \times 3.14 = 314 \, \text{cm}^3Volume=100×3.14=314cm3
Answer: The volume of the cone is approximately 314 cm³.
Exercise 9.8: Volume of a Sphere
Q1. A sphere has a radius of 7 cm. Find its volume.
Solution:
The formula for the volume of a sphere is:Volume=43πr3\text{Volume} = \frac{4}{3} \pi r^3Volume=34πr3
Where rrr is the radius.
Given:
- r=7r = 7r=7 cm.
Now, substitute these values into the formula:Volume=43π×73=43π×343=13723π\text{Volume} = \frac{4}{3} \pi \times 7^3 = \frac{4}{3} \pi \times 343 = \frac{1372}{3} \piVolume=34π×73=34π×343=31372π
Approximating π=3.14\pi = 3.14π=3.14:Volume=13723×3.14=1436.08 cm3\text{Volume} = \frac{1372}{3} \times 3.14 = 1436.08 \, \text{cm}^3Volume=31372×3.14=1436.08cm3
Answer: The volume of the sphere is approximately 1436.08 cm³.
Additional Questions for Chapter 9: Surface Areas and Volumes
1. Find the surface area of a cube whose edge length is 12 cm.
Solution: To find the surface area of a cube, use the formula:Surface Area of Cube=6a2\text{Surface Area of Cube} = 6a^2Surface Area of Cube=6a2
Where:
- aaa is the edge length of the cube.
2. A cuboid has a length of 8 cm, breadth of 5 cm, and height of 6 cm. Find its total surface area.
Solution: Use the formula for the surface area of a cuboid:Surface Area of Cuboid=2(lb+bh+hl)\text{Surface Area of Cuboid} = 2(lb + bh + hl)Surface Area of Cuboid=2(lb+bh+hl)
Where:
- l=8l = 8l=8 cm,
- b=5b = 5b=5 cm,
- h=6h = 6h=6 cm.
3. A cylindrical tank has a radius of 7 m and height of 14 m. Find the total surface area of the tank.
Solution: Use the formula for the surface area of a cylinder:Surface Area of Cylinder=2πr(h+r)\text{Surface Area of Cylinder} = 2\pi r(h + r)Surface Area of Cylinder=2πr(h+r)
Where:
- r=7r = 7r=7 m,
- h=14h = 14h=14 m.
4. Find the volume of a cone with a radius of 4 cm and a height of 9 cm.
Solution: Use the formula for the volume of a cone:Volume of Cone=13πr2h\text{Volume of Cone} = \frac{1}{3} \pi r^2 hVolume of Cone=31πr2h
Where:
- r=4r = 4r=4 cm,
- h=9h = 9h=9 cm.
5. The radius of a sphere is 10 cm. Find its surface area and volume.
Solution: For surface area:Surface Area of Sphere=4πr2\text{Surface Area of Sphere} = 4\pi r^2Surface Area of Sphere=4πr2
For volume:Volume of Sphere=43πr3\text{Volume of Sphere} = \frac{4}{3} \pi r^3Volume of Sphere=34πr3
Where:
- r=10r = 10r=10 cm.
6. A cylindrical pipe has an inner radius of 5 cm and outer radius of 7 cm. The height of the pipe is 20 cm. Find the surface area of the pipe.
Solution: To find the surface area of a cylindrical pipe, use the formula:Surface Area of Pipe=2πh(r1+r2)+2π(r22−r12)\text{Surface Area of Pipe} = 2\pi h \left( r_1 + r_2 \right) + 2\pi \left( r_2^2 – r_1^2 \right)Surface Area of Pipe=2πh(r1+r2)+2π(r22−r12)
Where:
- r1=5r_1 = 5r1=5 cm (inner radius),
- r2=7r_2 = 7r2=7 cm (outer radius),
- h=20h = 20h=20 cm.
7. A sphere is inscribed in a cube whose edge length is 14 cm. Find the volume of the sphere.
Solution: The diameter of the sphere is equal to the edge length of the cube. So:Diameter of Sphere=Edge length of Cube=14 cm\text{Diameter of Sphere} = \text{Edge length of Cube} = 14 \, \text{cm}Diameter of Sphere=Edge length of Cube=14cm
Now, radius r=142=7 cmr = \frac{14}{2} = 7 \, \text{cm}r=214=7cm.
Use the formula for the volume of a sphere:Volume of Sphere=43πr3\text{Volume of Sphere} = \frac{4}{3} \pi r^3Volume of Sphere=34πr3
Where:
- r=7r = 7r=7 cm.
8. The height of a cone is 15 cm, and the radius of its base is 6 cm. A cylindrical hole is drilled through the center of the cone. The radius of the hole is 3 cm. Find the volume of the remaining solid.
Solution: First, calculate the volume of the cone, and then subtract the volume of the cylindrical hole.
Volume of cone:Volume of Cone=13πr2h\text{Volume of Cone} = \frac{1}{3} \pi r^2 hVolume of Cone=31πr2h
Volume of cylinder (hole):Volume of Cylinder=πr2h\text{Volume of Cylinder} = \pi r^2 hVolume of Cylinder=πr2h
Finally, subtract the volume of the cylinder from the volume of the cone.
9. A spherical balloon has a radius of 10 cm. How much air is needed to fill the balloon? (Find the volume of the balloon)
Solution: Use the formula for the volume of a sphere:Volume of Sphere=43πr3\text{Volume of Sphere} = \frac{4}{3} \pi r^3Volume of Sphere=34πr3
Where:
- r=10r = 10r=10 cm.
10. A cylinder has a radius of 4 cm and height of 6 cm. Find the total surface area and the volume of the cylinder.
Solution: For surface area:Surface Area of Cylinder=2πr(h+r)\text{Surface Area of Cylinder} = 2\pi r(h + r)Surface Area of Cylinder=2πr(h+r)
For volume:Volume of Cylinder=πr2h\text{Volume of Cylinder} = \pi r^2 hVolume of Cylinder=πr2h
Where:
- r=4r = 4r=4 cm,
- h=6h = 6h=6 cm.
11. A cube has a surface area of 384 cm². Find its edge length and volume.
Solution: Use the formula for the surface area of a cube:Surface Area of Cube=6a2\text{Surface Area of Cube} = 6a^2Surface Area of Cube=6a2
Where:
- Surface area = 384 cm².
Solve for aaa (edge length), then use the formula for the volume of the cube:Volume of Cube=a3\text{Volume of Cube} = a^3Volume of Cube=a3
12. The height of a cylinder is 12 cm, and its volume is 150 cm³. Find its radius.
Solution: Use the formula for the volume of a cylinder:Volume of Cylinder=πr2h\text{Volume of Cylinder} = \pi r^2 hVolume of Cylinder=πr2h
Given:
- Volume = 150 cm³,
- Height h=12h = 12h=12 cm.
Solve for rrr.
13. A cone and a cylinder have the same height and the same radius. If the volume of the cone is 240 cm³, find the volume of the cylinder.
Solution: Use the formula for the volume of a cone:Volume of Cone=13πr2h\text{Volume of Cone} = \frac{1}{3} \pi r^2 hVolume of Cone=31πr2h
Since the cylinder and the cone have the same radius and height, the volume of the cylinder is three times the volume of the cone.
So, Volume of cylinder = 3×Volume of cone3 \times \text{Volume of cone}3×Volume of cone.
14. A frustum of a cone has a radius of 7 cm and 4 cm at the top and bottom, and a height of 10 cm. Find its volume.
Solution: Use the formula for the volume of a frustum of a cone:Volume of Frustum=13πh(r12+r22+r1r2)\text{Volume of Frustum} = \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1r_2)Volume of Frustum=31πh(r12+r22+r1r2)
Where:
- r1=7r_1 = 7r1=7 cm (bottom radius),
- r2=4r_2 = 4r2=4 cm (top radius),
- h=10h = 10h=10 cm.
15. A cone has a height of 24 cm and a base radius of 3 cm. A small cone is cut off from the top of the cone. The height of the small cone is 8 cm. Find the volume of the frustum formed.
Solution: First, calculate the volume of the large cone and the small cone. Then, subtract the volume of the small cone from the volume of the large cone to find the volume of the frustum.
These additional questions provide a wide range of problems to practice various surface area and volume calculations for different 3D shapes. By solving them, you’ll enhance your skills and be well-prepared for exams. Make sure to apply the appropriate formulas and check your results carefully!
Also Read: Understanding and Solving Chapter 7: Motion – Class 9 Science Exercises