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Geometry - Three-Dimensional Shapes: Cubes, Cuboids, Cylinders, Cones, Spheres

Grade 6ICSE

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

🔑Concepts

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A cube is a three-dimensional solid with six equal square faces. For a cube with edge length aa, it has 8 vertices, 12 equal edges, and 6 faces.

A 3D cube showing edge length a
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A cuboid (rectangular prism) has 6 rectangular faces. The dimensions are identified as length (ll), breadth (bb), and height (hh). Opposing faces of a cuboid are identical.

A 3D cuboid showing length, breadth, and height
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A cylinder consists of two parallel circular bases and a curved surface. The distance between the circular bases is the height (hh) and the radius of the circular base is rr.

Cylinder with radius r and height h
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A sphere is a perfectly round 3D object where every point on the surface is at an equal distance (radius rr) from the center.

Sphere with radius r

📐Formulae

Volume of a Cuboid = l×b×hl \times b \times h

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

Volume of a Cube = a3a^3 (where aa is the side)

Total Surface Area (TSA) of a Cube = 6a26a^2

Euler's Formula for Polyhedra: F+V−E=2F + V - E = 2

Diagonal of a Cuboid = l2+b2+h2\sqrt{l^2 + b^2 + h^2}

💡Examples

Problem 1:

Find the volume and 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}.

  1. Volume: V=l×b×h=12×8×5=480 cm3V = l \times b \times h = 12 \times 8 \times 5 = 480\text{ cm}^3
  2. Total Surface Area: 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)=2(196)=392 cm2TSA = 2(96 + 40 + 60) = 2(196) = 392\text{ cm}^2

Explanation:

To find the volume, multiply all three dimensions. For the total surface area, calculate the area of the three pairs of opposite rectangular faces and sum them up.

Problem 2:

A cube has an edge length of 6 cm6\text{ cm}. Verify Euler's formula for this shape and calculate its volume.

Solution:

  1. Identification: A cube has F=6F = 6 (faces), V=8V = 8 (vertices), and E=12E = 12 (edges).
  2. Euler's Formula Verification: F+V−E=6+8−12=14−12=2F + V - E = 6 + 8 - 12 = 14 - 12 = 2 Since the result is 22, Euler's formula is verified.
  3. Volume Calculation: V=a3=63=6×6×6=216 cm3V = a^3 = 6^3 = 6 \times 6 \times 6 = 216\text{ cm}^3

Explanation:

First, identify the number of faces, vertices, and edges for the cube to plug into Euler's formula (F+V−E=2F + V - E = 2). Then, use the side length to find the volume by cubing the side value.

Problem 3:

Calculate the Total Surface Area (TSA) of a cube whose edge is 4 cm4\text{ cm}. Also, find the area of its four lateral faces.

Cube with 4 cm edges

Solution:

a=4 cma = 4\text{ cm} Total Surface Area=6a2\text{Total Surface Area} = 6a^2 TSA=6×(4)2=6×16=96 cm2\text{TSA} = 6 \times (4)^2 = 6 \times 16 = 96\text{ cm}^2 Lateral Surface Area (LSA)=4a2\text{Lateral Surface Area (LSA)} = 4a^2 LSA=4×(4)2=4×16=64 cm2\text{LSA} = 4 \times (4)^2 = 4 \times 16 = 64\text{ cm}^2

Explanation:

To find the total surface area, we multiply the area of one square face (a2a^2) by the 6 faces. For lateral surface area (excluding top and bottom), we multiply by 4.

Problem 4:

A wooden box is in the shape of a cuboid with length 10 cm10\text{ cm}, breadth 6 cm6\text{ cm}, and height 4 cm4\text{ cm}. Find the length of the longest rod that can be placed inside this box.

Cuboid with diagonal line representing the longest rod

Solution:

l=10,b=6,h=4l = 10, b = 6, h = 4 Diagonal=l2+b2+h2\text{Diagonal} = \sqrt{l^2 + b^2 + h^2} Diagonal=102+62+42\text{Diagonal} = \sqrt{10^2 + 6^2 + 4^2} Diagonal=100+36+16=152\text{Diagonal} = \sqrt{100 + 36 + 16} = \sqrt{152} Diagonal≈12.33 cm\text{Diagonal} \approx 12.33\text{ cm}

Explanation:

The longest rod that can fit inside a cuboid is equal to the length of its space diagonal, connecting two opposite vertices.