Home / IB Mathematics SL 1.6 Approximation decimal places, significant figures AI HL Paper 2- Exam Style Questions

IB Mathematics SL 1.6 Approximation decimal places, significant figures AI HL Paper 2- Exam Style Questions- New Syllabus

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Question

A boutique furniture firm is prototyping a doorstop in the shape of a triangular prism, \(ABCDEF\). The vertices of one triangular face are given by the coordinates \( A(13, 1, 0) \), \( B(13, 25, 0) \), and \( C(13, 25, 7) \). Another vertex of the prism is located at \( D(4, 25, 7) \). All measurements are provided in centimetres.
3D diagram of a triangular prism doorstop
 
(a) Calculate the length of the edge segment \( AC \).
(b) Show that the triangular face \( ABC \) contains a right angle.
(c) Determine the volume of material required to construct the doorstop in its current form.
Design Update: To save on shipping costs, the volume is reduced to exactly \( 625 \, \text{cm}^3 \) by removing a section (indicated by the shaded region). The remaining triangular faces continue to be right-angled triangles.
3D diagram of modified prism with reduced volume
 
(d) (i) Find the value of the angle \( \theta \) at vertex \( A \).
  (ii) Hence, or otherwise, calculate the length of the new base segment \( AX \).
(e) The manufacturing material costs \( 0.025 \, \text{USD} \) per cubic centimetre. Due to the cutting process, \( 10\% \) more material than the final volume must be used. To ensure a profit, the retail price is set at \( 20\% \) above the total material cost. Determine the final selling price for the doorstop, rounded to two decimal places.

Most-appropriate topic codes:

• SL 3.1: Distance between two points in 3D space and volume of 3D solids — parts (a), (c)
• SL 3.2: Right-angled trigonometry and Pythagoras’ theorem — parts (b), (d)
• SL 1.6: Percentage calculations and financial applications — part (e)
▶️ Answer/Explanation

(a) Find \( AC \).

Using the distance formula in 3D:

\( AC = \sqrt{(13-13)^2 + (25-1)^2 + (7-0)^2} = \sqrt{0^2 + 24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25 \, \text{cm} \).

\( \boxed{25} \)

(b) Show that triangle \( ABC \) is right-angled.

Method 1 (vectors):
\( \overrightarrow{AB} = (0, 24, 0) \) and \( \overrightarrow{BC} = (0, 0, 7) \).
Since \( \overrightarrow{AB} \cdot \overrightarrow{BC} = 0 \), \( AB \perp BC \). Therefore, \( \triangle ABC \) is right-angled at \( B \).

Method 2 (Pythagoras):
\( AB = 24 \), \( BC = 7 \), \( AC = 25 \).
Check: \( 24^2 + 7^2 = 576 + 49 = 625 = 25^2 \).
Hence, by the converse of Pythagoras, \( \triangle ABC \) is right-angled at \( B \).

Triangle \( ABC \) is right-angled.

(c) Volume of the prism.

Cross-section is triangle \( ABC \), area = \( \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 24 \times 7 = 84 \, \text{cm}^2 \).
Width (depth) of prism = \( AD = 13 – 4 = 9 \, \text{cm} \).
Volume = area of cross-section × width = \( 84 \times 9 = 756 \, \text{cm}^3 \).

\( \boxed{756} \)

(d) (i) Find \( \theta \).

In triangle \( ABC \), \( \theta \) is the angle at \( A \).
Using right-angled trigonometry:
\( \tan \theta = \frac{BC}{AB} = \frac{7}{24} \Rightarrow \theta = \tan^{-1}\left(\frac{7}{24}\right) \approx 16.26^\circ \).

\( \boxed{16.3^\circ} \) (3 s.f.)

(d) (ii) Find \( AX \).

New volume = \( 625 \, \text{cm}^3 \), width remains \( 9 \, \text{cm} \).
Let new triangle be \( AXY \) similar to \( ABC \), with \( AX \) as base.
Area of new triangle = \( \frac{625}{9} \approx 69.444 \, \text{cm}^2 \).
For similar triangles, area ratio = \( k^2 \), where \( k = \frac{AX}{AB} \).
\( k^2 = \frac{625/9}{84} = \frac{625}{756} \Rightarrow k = \sqrt{\frac{625}{756}} \approx 0.90924 \).
Then \( AX = k \times AB = 0.90924 \times 24 \approx 21.82 \, \text{cm} \).

\( \boxed{21.8} \) (3 s.f.)

(e) Selling price.

Final volume = \( 625 \, \text{cm}^3 \).
Material needed (including 10% waste) = \( 625 \times 1.10 = 687.5 \, \text{cm}^3 \).
Cost of material = \( 687.5 \times 0.025 = 17.1875 \, \text{USD} \).
Selling price (including 20% profit) = \( 17.1875 \times 1.20 = 20.625 \, \text{USD} \).
Rounded to two decimal places = \( 20.63 \, \text{USD} \).

\( \boxed{20.63} \)

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