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IB Mathematics AA Distance , volume and Surface Area Study Notes | New Syllabus

IB Mathematics AA Distance , volume and Surface Area Study Notes

IB Mathematics AA Distance , volume and Surface Area Study Notes

IB Mathematics AA Distance , volume and Surface Area Study Notes Offer a clear explanation of Distance , volume and Surface Area , including various formula, rules, exam style questions as example to explain the topics. Worked Out  examples and common problem types provided here will be sufficient to cover for topic Distance , volume and Surface Area

Distance, Volume, and Surface Area Study

Distance, volume, and surface area calculations are fundamental for analyzing three-dimensional geometric objects. This section delves into measuring distances in 3D space, calculating volumes and surface areas, and determining angles in 3D shapes.

Guidance, Clarifications, and Syllabus Links:

  • Distance in 3D Space:
    • Distance Formula:

      The distance between two points \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \) is calculated as:

      \( d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2} \)

    • Midpoint Formula:

      The midpoint of the line segment joining two points \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \) is given by:

      \( \text{Midpoint} = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2}\right) \)

  • Volume of 3D Solids:
    • Right pyramid: \( V = \frac{1}{3} \times \text{Base Area} \times \text{Height} \)
      • Right cone: \( V = \frac{1}{3} \pi r^2 h \)
      • Sphere: \( V = \frac{4}{3} \pi r^3 \)
      • Hemisphere: \( V = \frac{2}{3} \pi r^3 \)
      • Combinations of solids: Add or subtract volumes as required based on the configuration.
  • Surface Area of 3D Solids:
    • Right cone: \( A = \pi r (r + l) \), where \( l \) is the slant height.
    • Sphere: \( A = 4 \pi r^2 \)
    • Hemisphere: \( A = 2 \pi r^2 + \pi r^2 \)
    • Composite solids: Add or subtract surface areas based on the shape configuration.

Examples:

  • Example 1: Find the distance between the points \( (1, 2, 3) \) and \( (4, 6, 8) \).

    \( d = \sqrt{(4 – 1)^2 + (6 – 2)^2 + (8 – 3)^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2} \)

  • Example 2: Calculate the volume of a composite solid formed by placing a hemisphere with radius 3 cm on top of a cylinder with the same radius and height 7 cm.
    • Volume of hemisphere: \( V_h = \frac{2}{3} \pi 3^3 = 18\pi \)
    • Volume of cylinder: \( V_c = \pi r^2 h = \pi 3^2 7 = 63\pi \)
    • Total volume: \( V = V_h + V_c = 18\pi + 63\pi = 81\pi \)
  • Example 3: Find the surface area of a sphere with radius 5 cm.

    \( A = 4 \pi r^2 = 4 \pi 5^2 = 100\pi \)

IB Mathematics AA SL Distance , volume and Surface Area Exam Style Worked Out Questions

Question

A company is designing a new logo. The logo is created by removing two equal segments from a rectangle, as shown in the following diagram.

The rectangle measures 5cm by 4cm. The points A and B lie on a circle, with centre O and radius 2cm, such that AÔB = θ, where 0 < θ < π. This information is shown in the following diagram.

(a) Find the area of one of the shaded segments in terms of θ.
(b) Given that the area of the logo is 13.4cm2, find the value of θ.

▶️Answer/Explanation

Ans:

(a) valid approach to find area of segment by finding area of sector – area of triangle

(b) EITHER
area of logo = area of rectangle – area of segments

5 × 4 – 2 × (2θ – 2sin θ) = 13.4

OR

area of one segment = \(\frac{20-13.4}{2}(=3.3)\)

2θ – 2sinθ = 3.3

THEN

θ = 2.35672….
θ = 2.36 (do not accept an answer in degrees)

Note: Award (M1)(A1)A0 if there is more than one solution.
Award (M1)(A1FT)A0 if the candidate works in degrees and obtains a final answer of 135.030…

Question

For a right pyramid of square base of slide 8 and vertical height 3 find
(a) the volume
(b) the surface are

▶️Answer/Explanation

Ans
(a) \(V=\frac{1}{3}8^23=64\)
(b) \(AM^2=4^2+3^2 \Rightarrow AM =5\)
\(S=8^2+4 \times (\frac{1}{2} \times 8 \times 5)=64+80=144\)

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