Surface Area of Combinations of Solids

The surface area of combinations of solids refers to the total area covering the outer surface of objects formed by combining two or more basic 3D shapes. These basic shapes include cubes, cuboids, spheres, hemispheres, cylinders, and cones. Calculating the surface area of these combined solids requires careful consideration of which surfaces are exposed and which are hidden due to the joining of the shapes. Often, portions of the surface areas of the individual solids will be removed or overlap when the solids are combined.

Formulae

The following formulae are essential for calculating the surface area of the individual shapes, which will be used in calculating surface area of combinations:

• Cube: Surface Area = $6a^2$ (where $a$ is the side length)
• Cuboid: Surface Area = $2(lw + lh + wh)$ (where $l$ is length, $w$ is width, and $h$ is height)
• Sphere: Surface Area = $4\pi r^2$ (where $r$ is the radius)
• Hemisphere: Curved Surface Area = $2\pi r^2$; Total Surface Area = $3\pi r^2$ (where $r$ is the radius)
• Cylinder: Curved Surface Area = $2\pi rh$; Total Surface Area = $2\pi r(h+r)$ (where $r$ is the radius and $h$ is the height)
• Cone: Curved Surface Area = $\pi r l$; Total Surface Area = $\pi r(l+r)$ (where $r$ is the radius and $l$ is the slant height) and $l = \sqrt{r^2 + h^2}$ (where $h$ is the height)

Examples

Example-1: A solid is in the shape of a cone mounted on a hemisphere. The radius of each is 3.5 cm and the total height of the solid is 15.5 cm. Find the surface area of the solid.

Solution:

1. Radius of the cone, $r = 3.5$ cm
2. Height of the cone, $h = 15.5 – 3.5 = 12$ cm
3. Slant height of the cone, $l = \sqrt{r^2 + h^2} = \sqrt{3.5^2 + 12^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5$ cm
4. Surface area of the cone = $\pi r l = \pi \times 3.5 \times 12.5 = 43.75\pi$ cm²
5. Surface area of the hemisphere = $2\pi r^2 = 2 \pi (3.5)^2 = 24.5\pi$ cm²
6. Total surface area of the solid = Surface area of the cone + Surface area of the hemisphere = $43.75\pi + 24.5\pi = 68.25\pi \approx 214.2$ cm²

Example-2: Two cubes each of volume $64 \text{ cm}^3$ are joined end to end. Find the surface area of the resulting cuboid.

Solution:

1. Volume of a cube = $a^3$. Therefore, $a^3 = 64$. Thus, $a = 4$ cm.
2. The cuboid formed has dimensions: length $l = 4 + 4 = 8$ cm, width $w = 4$ cm, and height $h = 4$ cm.
3. Surface Area of the cuboid = $2(lw + lh + wh) = 2((8 \times 4) + (8 \times 4) + (4 \times 4)) = 2(32 + 32 + 16) = 2(80) = 160$ cm²

Common mistakes by students

• Forgetting to subtract overlapping areas: When two solids are joined, the area where they are joined is not exposed and should not be included in the total surface area.
• Using incorrect formulas: Students often mix up formulas for surface area and volume, or use the wrong formula for a particular shape (e.g., using the total surface area of a hemisphere when only the curved surface is needed).
• Incorrect units: Always remember to include the correct units (e.g., cm², m²) in the final answer.
• Not reading the question carefully Students often fail to properly understand how shapes are combined, leading to errors in calculations.

Real Life Application

Understanding the surface area of combinations of solids is critical in many real-world scenarios:

• Painting and Construction: Calculating the amount of paint needed to cover a complex structure like a house or a building that combines various shapes (cuboids, cylinders for pipes, spheres for decorative elements).
• Packaging: Designing packaging, such as boxes, that use combinations of shapes and calculating the amount of material needed.
• Manufacturing: Determining the surface area of objects produced by manufacturing companies, for coating, insulation or determining materials usage.
• Engineering: Calculating surface areas of components and structures (e.g., the surface area of a silo which might combine a cylinder and a hemisphere).

Fun Fact

Archimedes, one of the greatest mathematicians of all time, was fascinated by the geometry of spheres and cylinders. He requested that a sphere be inscribed in a cylinder on his tombstone to represent his greatest mathematical achievement: the relationship between the volume and surface area of a sphere and its circumscribing cylinder.

Recommended YouTube Videos for Deeper Understanding

Q.1 A solid is in the shape of a cone mounted on a hemisphere. The radius of each is 7 cm and the height of the cone is also 24 cm. Find the total surface area of the solid.

Check Solution

Ans: B

The total surface area is the sum of the curved surface area of the cone and the curved surface area of the hemisphere. Slant height of cone $l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25$ Curved surface area of cone = $\pi r l = \frac{22}{7} \times 7 \times 25 = 550 cm^2$ Curved surface area of hemisphere = $2 \pi r^2 = 2 \times \frac{22}{7} \times 7^2 = 308 cm^2$ Total surface area = $550 + 308 = 858 cm^2$

Q.2 Two cubes, each of volume $64 cm^3$, are joined end to end. Find the surface area of the resulting cuboid.

Check Solution

Ans: A

Side of cube = $\sqrt[3]{64} = 4 cm$. Dimensions of cuboid: length = 8 cm, width = 4 cm, height = 4 cm. Surface area = $2(lb + bh + lh) = 2(8 \times 4 + 4 \times 4 + 8 \times 4) = 2(32 + 16 + 32) = 2(80) = 160 cm^2$

Q.3 A hemispherical bowl of internal radius 9 cm is filled with liquid. This liquid is to be filled into cylindrical bottles of radius 3 cm and height 6 cm. How many bottles are needed?

Check Solution

Ans: C

Volume of hemisphere = $\frac{2}{3}\pi r^3 = \frac{2}{3} \pi (9^3) = 486\pi$ Volume of cylinder = $\pi r^2 h = \pi (3^2) (6) = 54\pi$ Number of bottles = $\frac{486\pi}{54\pi} = 9$

Q.4 A solid sphere of radius 6 cm is melted into a hollow cylinder of uniform thickness. If the external radius of the cylinder is 8 cm and its height is 6 cm, find the thickness of the cylinder.

Check Solution

Ans: B

Volume of the sphere = $\frac{4}{3}\pi(6^3) = 288\pi$ Volume of the cylinder = $\pi h (R^2 – r^2) = \pi 6 (8^2 – r^2)$ Equating the volumes: $288\pi = 6\pi (64 – r^2)$ $48 = 64 – r^2$ $r^2 = 16$ $r = 4$ Thickness = $8 – 4 = 4 cm$

Q.5 A tent is in the shape of a cylinder surmounted by a cone. The diameter of the cylindrical part is 24 m and its height is 11 m. If the conical part is 5 m high, find the area of the canvas required to make the tent.

Check Solution

Ans: A

Radius of cylinder and cone = 12 m Height of cylinder = 11 m Height of cone = 5 m l = $\sqrt{12^2 + 5^2} = 13$ Surface area = $2 \pi r h + \pi r l$ $2 \times \frac{22}{7} \times 12 \times 11 + \frac{22}{7} \times 12 \times 13 = 829.7 + 489.4 = 1319.1 \approx 1318.8$

Q.1 A solid is in the shape of a cone mounted on a hemisphere. The radius of each is 7 cm and the height of the cone is also 24 cm. Find the total surface area of the solid.

Check Solution

Ans: B

The total surface area is the sum of the curved surface area of the cone and the curved surface area of the hemisphere. Slant height of cone $l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25$ Curved surface area of cone = $\pi r l = \frac{22}{7} \times 7 \times 25 = 550 cm^2$ Curved surface area of hemisphere = $2 \pi r^2 = 2 \times \frac{22}{7} \times 7^2 = 308 cm^2$ Total surface area = $550 + 308 = 858 cm^2$

Q.2 Two cubes, each of volume $64 cm^3$, are joined end to end. Find the surface area of the resulting cuboid.

Check Solution

Ans: A

Side of cube = $\sqrt[3]{64} = 4 cm$. Dimensions of cuboid: length = 8 cm, width = 4 cm, height = 4 cm. Surface area = $2(lb + bh + lh) = 2(8 \times 4 + 4 \times 4 + 8 \times 4) = 2(32 + 16 + 32) = 2(80) = 160 cm^2$

Q.3 A hemispherical bowl of internal radius 9 cm is filled with liquid. This liquid is to be filled into cylindrical bottles of radius 3 cm and height 6 cm. How many bottles are needed?

Check Solution

Ans: C

Volume of hemisphere = $\frac{2}{3}\pi r^3 = \frac{2}{3} \pi (9^3) = 486\pi$ Volume of cylinder = $\pi r^2 h = \pi (3^2) (6) = 54\pi$ Number of bottles = $\frac{486\pi}{54\pi} = 9$

Q.4 A solid sphere of radius 6 cm is melted into a hollow cylinder of uniform thickness. If the external radius of the cylinder is 8 cm and its height is 6 cm, find the thickness of the cylinder.

Check Solution

Ans: B

Volume of the sphere = $\frac{4}{3}\pi(6^3) = 288\pi$ Volume of the cylinder = $\pi h (R^2 – r^2) = \pi 6 (8^2 – r^2)$ Equating the volumes: $288\pi = 6\pi (64 – r^2)$ $48 = 64 – r^2$ $r^2 = 16$ $r = 4$ Thickness = $8 – 4 = 4 cm$

Q.5 A tent is in the shape of a cylinder surmounted by a cone. The diameter of the cylindrical part is 24 m and its height is 11 m. If the conical part is 5 m high, find the area of the canvas required to make the tent.

Check Solution

Ans: A

Radius of cylinder and cone = 12 m Height of cylinder = 11 m Height of cone = 5 m l = $\sqrt{12^2 + 5^2} = 13$ Surface area = $2 \pi r h + \pi r l$ $2 \times \frac{22}{7} \times 12 \times 11 + \frac{22}{7} \times 12 \times 13 = 829.7 + 489.4 = 1319.1 \approx 1318.8$

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