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

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

IB DP Maths AI SL Paper 2 - All Topics

Question

AirFlow Industries is developing a prototype for a heavy-duty doorstop modeled as a triangular prism, \(ABCDEF\). In a 3D coordinate system where units are in centimeters, the vertices of one triangular base are given by \(A(13, 1, 0)\), \(B(13, 25, 0)\), and \(C(13, 25, 7)\). Another vertex is located at \(D(4, 25, 7)\), as illustrated in the following diagram.
3D diagram of triangular prism doorstop
(a) State the 3D coordinates of vertex \(F\).
(b) Calculate the length of the edge \(AC\).
(c) Use the side lengths of triangle \(ABC\) to verify that it is a right-angled triangle.
(d) Determine the total volume of the material required to manufacture the original doorstop.
To optimize production costs, the company decides to scale down the product to a volume of \(625 \text{ cm}^3\) by removing a portion of the material (indicated by the shaded region in the second diagram). The geometric properties of the triangular faces remain right-angled.
Diagram showing the shaded region to be removed
The length of the new base edge, \(AX\), is determined by the relation \(AX = AB \times \sqrt{\frac{V_s}{V_o}}\), where \(V_s\) represents the volume of the smaller doorstop and \(V_o\) represents the original volume.
(e) Calculate the length of \(AX\).
The raw material costs \(0.025 \text{ USD}\) per \(\text{cm}^3\). The manufacturing process requires a \(10\%\) surplus of material beyond the final product volume. The company sets the retail price to achieve a \(20\%\) profit margin over the total material cost.
(f) Determine the final retail price for the new doorstop, rounding your answer to two decimal places.

Most-appropriate topic codes (IB Mathematics AI SL 2025):

• SL 3.1: 3D Coordinate geometry, distance formula, and volume of prisms — parts (a), (b), (d)
• SL 3.2: Geometry of right-angled triangles — part (c)
• SL 1.6 & 1.7: Percentage calculations and financial applications — part (f)
▶️ Answer/Explanation

(a)
\(F\) is directly below \(D\) in the \(z\)-direction, sharing \(x, y\) with \(A\). From \(A(13,1,0)\) and \(D(4,25,7)\), prism is right, so \(F\) has \(x=4, y=1, z=0\).
\(\boxed{(4, 1, 0)}\)

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

(c)
\(AB = 25 – 1 = 24\), \(BC = 7 – 0 = 7\), \(AC = 25\).
Check Pythagoras: \(24^2 + 7^2 = 576 + 49 = 625 = 25^2\).
Hence triangle \(ABC\) is right-angled at \(B\).
\(\boxed{AB^2 + BC^2 = AC^2}\)

(d)
Triangular base area = \(\frac{1}{2} \times AB \times BC = \frac{1}{2} \times 24 \times 7 = 84 \text{ cm}^2\).
Depth (width) = \(AD_x = 13 – 4 = 9 \text{ cm}\).
Volume = base area × width = \(84 \times 9 = 756 \text{ cm}^3\).
\(\boxed{756 \text{ cm}^3}\)

(e)
Given \(V_s = 625\), \(V_o = 756\), \(AB = 24\):
\(AX = AB \times \sqrt{\frac{V_s}{V_o}} = 24 \times \sqrt{\frac{625}{756}} \approx 24 \times 0.90925 \approx 21.82 \text{ cm}\).
\(\boxed{21.8 \text{ cm}}\)

(f)
Material needed = \(625 \times 1.10 = 687.5 \text{ cm}^3\).
Cost = \(687.5 \times 0.025 = 17.1875 \text{ USD}\).
Selling price = \(17.1875 \times 1.20 = 20.625 \text{ USD}\).
\(\boxed{20.63 \text{ USD}}\)

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