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Shape and Space - Drawing 3D Shapes

Grade 6IB

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

🔑Concepts

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3D shapes are solid objects that have three dimensions: length, width, and height. Unlike 2D shapes, they occupy space (volume).

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Isometric Drawing: A technique to represent 3D objects on 2D paper using a grid of dots. In an isometric grid, the vertical lines remain vertical, and the horizontal axes are drawn at 30∘30^{\circ} angles to create a sense of depth.

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Orthographic Projections: These are 2D drawings used to describe a 3D object from different directions. The three main views are the 'Plan' (view from the top), the 'Front Elevation' (view from the front), and the 'Side Elevation' (view from the side).

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Nets: A net is a 2-dimensional flat shape that can be folded to make a 3-dimensional object. For example, a cube has 66 square faces, and its net consists of 66 connected squares.

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Properties of Polyhedra: 3D shapes with flat faces are called polyhedra. They are defined by their Faces (FF), Vertices (VV), and Edges (EE).

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Perspective: In perspective drawing, parallel lines appear to meet at a 'vanishing point' on the horizon to simulate how the human eye perceives distance.

📐Formulae

V−E+F=2 (Euler’s Formula for Polyhedra)V - E + F = 2 \text{ (Euler's Formula for Polyhedra)}

Vcube=s3V_{cube} = s^3

Vcuboid=l×w×hV_{cuboid} = l \times w \times h

SAcube=6s2SA_{cube} = 6s^2

💡Examples

Problem 1:

A hexagonal prism has 88 faces and 1212 vertices. Use Euler's Formula to calculate the number of edges (EE).

Solution:

Using the formula V−E+F=2V - E + F = 2, we substitute the given values: 12−E+8=212 - E + 8 = 2.

Explanation:

First, combine the constants: 20−E=220 - E = 2. To find EE, we calculate 20−2=E20 - 2 = E, which gives E=18E = 18. Therefore, a hexagonal prism has 1818 edges.

Problem 2:

Identify the 3D shape formed by a net consisting of one central square and four congruent triangles attached to each side of the square.

Solution:

The shape is a Square-based Pyramid.

Explanation:

Since the base is a square and it has four triangular faces that meet at a single vertex (apex) when folded, the resulting 3D shape is a square-based pyramid. It has F=5F = 5, V=5V = 5, and E=8E = 8.

Problem 3:

A cuboid has dimensions of length L=5 cmL = 5\text{ cm}, width W=3 cmW = 3\text{ cm}, and height H=4 cmH = 4\text{ cm}. Calculate its volume and surface area.

Solution:

Volume: V=5×3×4=60 cm3V = 5 \times 3 \times 4 = 60\text{ cm}^3. Surface Area: SA=2(5×3+3×4+5×4)=2(15+12+20)=2(47)=94 cm2SA = 2(5\times 3 + 3\times 4 + 5\times 4) = 2(15 + 12 + 20) = 2(47) = 94\text{ cm}^2.

Explanation:

Volume is found by multiplying the three dimensions. Surface area is the sum of the areas of all six rectangular faces (3 pairs of equal faces).

Drawing 3D Shapes Grade 6 Notes & Examples | IB Maths