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

Grade 9IB

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

🔑Concepts

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The sum of the interior angles of a polygon with nn sides is given by the formula (n−2)×180∘(n-2) \times 180^{\circ}. For a regular polygon, each interior angle is (n−2)×180∘n\frac{(n-2) \times 180^{\circ}}{n}.

Regular pentagon showing interior angles
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A sector is a portion of a circle enclosed by two radii and an arc. The area of a sector depends on the central angle θ\theta and the radius rr, calculated as θ360×πr2\frac{\theta}{360} \times \pi r^2.

A circle sector with radius r and central angle theta
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The 3D Pythagorean theorem relates the space diagonal dd of a rectangular cuboid to its length ll, width ww, and height hh using the relationship d2=l2+w2+h2d^2 = l^2 + w^2 + h^2.

Cuboid showing the space diagonal
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Right-angled trigonometry (SOH CAH TOA) is used to find missing side lengths and angles in 3D problems by identifying 2D right-angled triangles within the 3D shape.

📐Formulae

Sum of interior angles: (n−2)×180∘(n-2) \times 180^{\circ}

Area of a Circle: A=πr2A = \pi r^2

Circumference of a Circle: C=2πrC = 2 \pi r

Area of a Sector: Asector=θ360×πr2A_{sector} = \frac{\theta}{360} \times \pi r^2

Volume of a Prism: V=Abase×hV = A_{base} \times h

Volume of a Cylinder: V=πr2hV = \pi r^2 h

Volume of a Pyramid or Cone: V=13×Abase×hV = \frac{1}{3} \times A_{base} \times h

Volume of a Sphere: V=43πr3V = \frac{4}{3} \pi r^3

Surface Area of a Sphere: SA=4πr2SA = 4 \pi r^2

3D Pythagoras Theorem: d2=l2+w2+h2d^2 = l^2 + w^2 + h^2

SOH CAH TOA: sin⁡(θ)=OH,cos⁡(θ)=AH,tan⁡(θ)=OA\sin(\theta) = \frac{O}{H}, \cos(\theta) = \frac{A}{H}, \tan(\theta) = \frac{O}{A}

💡Examples

Problem 1:

A cylinder has a radius of 55 cm and a height of 1212 cm. Calculate its total surface area. (Take π≈3.14\pi \approx 3.14)

Solution:

Step 1: Identify the components of the surface area. A cylinder has two circular bases and one curved surface area. Step 2: Calculate the area of the two circular bases: 2×πr2=2×3.14×52=2×3.14×25=1572 \times \pi r^2 = 2 \times 3.14 \times 5^2 = 2 \times 3.14 \times 25 = 157 cm2^2. Step 3: Calculate the curved surface area (circumference ×\times height): 2πrh=2×3.14×5×12=376.82 \pi r h = 2 \times 3.14 \times 5 \times 12 = 376.8 cm2^2. Step 4: Add the areas together: Total SA =157+376.8=533.8= 157 + 376.8 = 533.8 cm2^2.

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:

Find the length of the space diagonal of a rectangular cuboid with dimensions 33 cm by 44 cm by 1212 cm.

Solution:

Step 1: Use the 3D version of Pythagoras' Theorem: d2=l2+w2+h2d^2 = l^2 + w^2 + h^2. Step 2: Substitute the known values: d2=32+42+122d^2 = 3^2 + 4^2 + 12^2. Step 3: Calculate the squares: d2=9+16+144d^2 = 9 + 16 + 144. Step 4: Sum the values: d2=169d^2 = 169. Step 5: Take the square root: d=169=13d = \sqrt{169} = 13 cm.

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.

Problem 3:

Calculate the volume of a right-angled triangular prism with a base length of 66 cm, a base height of 88 cm, and a prism length of 1515 cm.

Triangular prism with dimensions 6, 8, and 15 cm

Solution:

  1. Find the area of the triangular base: Abase=12×base×heightA_{base} = \frac{1}{2} \times \text{base} \times \text{height} Abase=12×6×8=24 cm2A_{base} = \frac{1}{2} \times 6 \times 8 = 24 \text{ cm}^2
  2. Multiply the base area by the length of the prism: V=Abase×hV = A_{base} \times h V=24×15=360 cm3V = 24 \times 15 = 360 \text{ cm}^3

Explanation:

To find the volume of any prism, you must first calculate the area of the uniform cross-section (the base) and then multiply it by the length or depth of the prism.

Problem 4:

A square-based pyramid has a base side length of 1010 m and a vertical height of 1212 m. Determine its volume.

Square based pyramid with side 10m and height 12m

Solution:

  1. Calculate the area of the square base: Abase=s2=102=100 m2A_{base} = s^2 = 10^2 = 100 \text{ m}^2
  2. Apply the volume formula for a pyramid: V=13×Abase×hV = \frac{1}{3} \times A_{base} \times h V=13×100×12V = \frac{1}{3} \times 100 \times 12 V=100×4=400 m3V = 100 \times 4 = 400 \text{ m}^3

Explanation:

The volume of a pyramid is exactly one-third of the volume of a prism with the same base and height. Start by finding the area of the base, then multiply by height and divide by three.