Surface Area of a Cuboid

A (rectangular) cuboid is a closed box which comprises of 3 pairs of rectangular faces that are parallel to each other, and joined at right angles. It is also known as a right rectangular prism. It has 8 vertices, 6 faces, and 12 edges. The cube is a cuboid whose faces are all squares.

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The figure above shows the developmental figure of a \(a\times b\times c\) cuboid. Notice that the developmental figure consists of two \(a\times b\) faces, two \(b\times c\) faces, and two \(c\times a\) faces. The surface area of a cuboid is the sum of the areas of these 6 faces. Hence the surface area of a \(a\times b\times c\) cuboid is \(2(ab+bc+ca)\).

What is the surface area of a \( 2 \times 3 \times 4 \) cuboid?

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The surface area is \( 2 ( 2 \times 3 + 3 \times 4 + 4 \times 2) = 52 \). \( _\square \)

hi

There is a cuboid whose base is a square and height is 10 cm. If the area of one side face is \(120\text{ cm}^2\), what is the total surface area of the cuboid?

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Since the area of one side face is \(120\text{ cm}^2\), the side length of the square base is \(\frac{120}{10} = 12\text{ cm}\). Thus, the cuboid is a \(12\text{ cm}\times 12\text{ cm}\times 10\text{ cm}\) cuboid. Hence, the surface area of the cuboid is

\[2 \times (12 \times 12 + 12 \times 10 + 12 \times 10) = 768\ (\text{cm}^2).\ _\square \]

A cuboid's base has an area of \(20\text{ cm}^2\), and a perimeter of 20 cm. If the cuboid's height is 6 cm, what is the surface area?

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The sum of the areas of the side faces is (perimeter of base) \(\times\) (height), and the surface area of the cuboid is (sum of areas of the side faces) + 2 \(\times\) (area of base). Thus, the surface area of the cuboid is \[20 \times 6 + 2\times 20 = 160\ (\text{cm}^2).\ _\square \]

Consider a cuboid whose base is a \(3\text{ cm}\times4\text{ cm}\) rectangle. If the volume of the cuboid is \(60\text{ cm}^3,\) what is the surface area?

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An \(a \times b \times c \) cuboid has a volume of \( abc \) and a surface area of \( 2 ( ab+bc+ca) \). Since the cuboid's volume is \(60\text{ cm}^3\) and the base is a \(3\text{ cm}\times4\text{ cm}\) rectangle, the height is \(\frac{60}{3 \times 4} = 5\text{ cm}.\) Then, the surface area of the cuboid is \[2 \times ( 3 \times 4 + 3 \times 5 + 4 \times 5) = 94\ (\text{cm}^2).\ _\square \]