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Mensuration - Surface Area of Cube, Cuboid, and Cylinder

Grade 8CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Surface Area refers to the total area of all the faces of a 3D object. For a cuboid with length ll, breadth bb, and height hh, the Total Surface Area (TSA) is the sum of the areas of its 6 rectangular faces: 2(lb+bh+hl)2(lb + bh + hl). The Lateral Surface Area (LSA) excludes the top and bottom faces, given by 2h(l+b)2h(l + b).

3D cuboid showing length, breadth and height
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A cube is a special cuboid where all edges are of equal length aa. Since it has 6 identical square faces, its Total Surface Area is 6a26a^2. The Lateral Surface Area, which covers only the 4 side faces, is 4a24a^2.

3D cube with equal side lengths
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A cylinder consists of two congruent circular bases and a curved surface. The Curved Surface Area (CSA) is 2πrh2\pi rh, which represents the area of the side when 'unrolled' into a rectangle. The Total Surface Area (TSA) includes the CSA plus the area of the two circular ends: 2πrh+2πr2=2πr(r+h)2\pi rh + 2\pi r^2 = 2\pi r(r + h).

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When a cylindrical surface is opened along its height, it forms a rectangle where the length is the circumference of the base 2πr2\pi r and the breadth is the height hh. This explains why CSA=2πr×hCSA = 2\pi r \times h.

Unrolled curved surface of a cylinder forming a rectangle

📐Formulae

Total Surface Area of a Cuboid = 2(lb+bh+hl)2(lb + bh + hl)

Lateral Surface Area of a Cuboid = 2h(l+b)2h(l + b)

Total Surface Area of a Cube = 6a26a^2

Lateral Surface Area of a Cube = 4a24a^2

Curved Surface Area (CSA) of a Cylinder = 2πrh2\pi rh

Total Surface Area (TSA) of a Cylinder = 2πr(r+h)2\pi r(r + h)

Area of one circular base of a Cylinder = πr2\pi r^2

💡Examples

Problem 1:

Find the total surface area of a cuboid whose length is 12 cm12\text{ cm}, breadth is 8 cm8\text{ cm}, and height is 5 cm5\text{ cm}.

Solution:

Given: l=12 cml = 12\text{ cm}, b=8 cmb = 8\text{ cm}, h=5 cmh = 5\text{ cm}. Using the formula for Total Surface Area (TSA) of a cuboid: TSA=2(lb+bh+hl)TSA = 2(lb + bh + hl) TSA=2(12×8+8×5+5×12)TSA = 2(12 \times 8 + 8 \times 5 + 5 \times 12) TSA=2(96+40+60)TSA = 2(96 + 40 + 60) TSA=2(196)TSA = 2(196) TSA=392 cm2TSA = 392\text{ cm}^2.

Explanation:

To find the total surface area, we calculate the area of all six rectangular faces by summing the products of the dimensions and doubling the result because opposite faces are equal.

Problem 2:

A cylindrical tank has a radius of 7 m7\text{ m} and a height of 3 m3\text{ m}. Find its total surface area. (Take π=227\pi = \frac{22}{7})

Solution:

Given: r=7 mr = 7\text{ m}, h=3 mh = 3\text{ m}. Using the formula for Total Surface Area (TSA) of a cylinder: TSA=2πr(r+h)TSA = 2\pi r(r + h) TSA=2×227×7×(7+3)TSA = 2 \times \frac{22}{7} \times 7 \times (7 + 3) TSA=2×22×10TSA = 2 \times 22 \times 10 TSA=440 m2TSA = 440\text{ m}^2.

Explanation:

The total surface area includes the curved side of the tank plus the area of the circular top and bottom. We substitute the radius and height into the formula and simplify.

Problem 3:

A suitcase measures 80 cm×48 cm×24 cm80\text{ cm} \times 48\text{ cm} \times 24\text{ cm}. How many meters of tarpaulin of width 96 cm96\text{ cm} is required to cover 100100 such suitcases?

Diagram of a suitcase with dimensions 80x48x24 cm.

Solution:

  1. Find TSA of one suitcase: TSA=2(lb+bh+hl)TSA = 2(lb + bh + hl) TSA=2(80×48+48×24+24×80)TSA = 2(80 \times 48 + 48 \times 24 + 24 \times 80) TSA=2(3840+1152+1920)TSA = 2(3840 + 1152 + 1920) TSA=2(6912)=13824 cm2TSA = 2(6912) = 13824\text{ cm}^2

  2. TSA of 100100 suitcases: 100×13824=1382400 cm2100 \times 13824 = 1382400\text{ cm}^2

  3. Find length of tarpaulin: Area of tarpaulin = Length ×\times Width 1382400=L×961382400 = L \times 96 L=138240096=14400 cmL = \frac{1382400}{96} = 14400\text{ cm}

  4. Convert to meters: L=14400100=144 mL = \frac{14400}{100} = 144\text{ m}

Explanation:

We first calculate the total surface area of one suitcase to find the amount of fabric needed for one. Multiplying by 100 gives the total area required. Since the tarpaulin is rectangular, dividing its total area by its width gives the required length.

Problem 4:

The curved surface area of a hollow cylinder is 4224 cm24224\text{ cm}^2. It is cut along its height and forms a rectangular sheet of width 33 cm33\text{ cm}. Find the perimeter of the rectangular sheet.

A rectangular sheet representing the unfolded curved surface of the cylinder.

Solution:

  1. Area of rectangular sheet = Curved Surface Area of cylinder Area=4224 cm2Area = 4224\text{ cm}^2

  2. Find length (ll) of the sheet: Area=l×bArea = l \times b 4224=l×334224 = l \times 33 l=422433=128 cml = \frac{4224}{33} = 128\text{ cm}

  3. Find perimeter of the sheet: Perimeter=2(l+b)Perimeter = 2(l + b) Perimeter=2(128+33)Perimeter = 2(128 + 33) Perimeter=2(161)=322 cmPerimeter = 2(161) = 322\text{ cm}

Explanation:

When a cylinder is cut vertically, the curved surface unfolds into a rectangle. The height of the cylinder becomes one side of the rectangle, and the circumference becomes the other side. Here, the width is given, so we find the length and then calculate the perimeter.