Geometry and Trigonometry - Properties of 2D and 3D shapes

Grade 9IB

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

🔑Concepts

•

n

•

r

•

Prisms and Cylinders: A prism is a 3D shape with a uniform cross-section throughout its length. For example, a triangular prism looks like a tent with two identical triangular faces joined by three rectangular faces. A cylinder is a special prism with a circular cross-section, resembling a common soda can.

•

Pyramids and Cones: These 3D shapes taper to a single point called the apex. A square-based pyramid has four triangular faces meeting at the top, while a cone has a circular base and a curved surface that narrows to a point. The volume of these shapes is exactly one-third the volume of a prism with the same base and height.

•

Surface Area of 3D Shapes: The total surface area is the sum of the areas of all faces. For curved shapes like spheres, visualize the surface as a wrapper tightly covering the entire object. A sphere is perfectly round in 3D space, like a basketball, where every point on the surface is equidistant from the center.

•

a^2 + b^2 = c^2

•

\theta

📐Formulae

(n-2) \times 180^{\circ}

A = \pi r^2

C = 2 \pi r

A_{sector} = \frac{\theta}{360} \times \pi r^2

V = A_{base} \times h

V = \pi r^2 h

V = \frac{1}{3} \times A_{base} \times h

V = \frac{4}{3} \pi r^3

SA = 4 \pi r^2

d^2 = l^2 + w^2 + h^2

\sin(\theta) = \frac{O}{H}, \cos(\theta) = \frac{A}{H}, \tan(\theta) = \frac{O}{A}

💡Examples

Problem 1:

5

Solution:

2 \times \pi r^2 = 2 \times 3.14 \times 5^2 = 2 \times 3.14 \times 25 = 157

Explanation:

To find the total surface area, we must sum the areas of the flat top and bottom faces with the area of the rectangular 'wrapper' (curved surface) that goes around the cylinder.

Problem 2:

3

Solution:

d^2 = l^2 + w^2 + h^2

Explanation:

The space diagonal represents the longest possible straight line that can fit inside the box, spanning from one bottom corner to the diagonally opposite top corner.