Home / IB Mathematics SL 3.1 The distance between two points AI HL Paper 2- Exam Style Questions

IB Mathematics SL 3.1 The distance between two points AI HL Paper 2- Exam Style Questions- New Syllabus

IBDP Maths AI HL - Paper 2 - All Topics
Question

A large water reservoir is built in the form of part of an upside-down right pyramid with a horizontal square base of length \( 80 \) metres. The point \( C \) is the centre of the square base and point \( V \) is the vertex of the pyramid.

The bottom of the reservoir is a square of length \( 60 \) metres that is parallel to the base of the pyramid, such that the depth of the reservoir is \( 6 \) metres. A vertical cross section, \( MNOPC \), of the reservoir is shown.

(a) Find the angle of depression from \( M \) to \( N \).

(b) (i) Find \( CV \).
(ii) Hence or otherwise, show that the volume of the reservoir is \( 29\,600 \) m\(^3\).

(c) Every day \( 80 \) m\(^3\) of water from the reservoir is used for irrigation. Joshua states that, if no other water enters or leaves the reservoir, then when it is full there is enough irrigation water for at least one year. By finding an appropriate value, determine whether Joshua is correct.

(d) To avoid water leaking into the ground, the five interior sides of the reservoir have been painted with a watertight material. Find the area that was painted.

▶️ Answer/Explanation
Markscheme

(a)
\( \tan(\theta) = \frac{6}{10} \) (M1)
\( \theta = 31.0^\circ \) (30.9637…) or \( 0.540 \) rad (0.540419…) (A1)
Explanation:
The angle of depression is found using the right-angled triangle with opposite side \( 6 \) m and adjacent side \( 10 \) m.
Result:
Angle of depression is \( 31.0^\circ \) or \( 0.540 \) rad

(b)(i)
\( CV = 40 \tan(\theta) \) or \( CV = 4 \times 6 \) (M1)
\( CV = 24 \) m (A1)
Explanation:
Use trigonometry or similar triangles from the cross-section.
Result:
\( CV = 24 \) m

(b)(ii)
\( V = \frac{1}{3} \times 80^2 \times 24 – \frac{1}{3} \times 60^2 \times 18 \) (M1)(A1)(A1)
\( V = 29\,600 \) m\(^3\) (AG)
Explanation:
Volume is the difference between the large pyramid (base \( 80 \) m, height \( 24 \) m) and the smaller pyramid above (base \( 60 \) m, height \( 18 \) m).
Result:
Volume is \( 29\,600 \) m\(^3\)

(c)
METHOD 1
\( \frac{29\,600}{80} = 370 \) days (A1)
Since \( 370 > 366 \), Joshua is correct. (A1)
METHOD 2
\( 80 \times 366 = 29\,280 \) m\(^3\) or \( 80 \times 365 = 29\,200 \) m\(^3\) (A1)
Since \( 29\,280 < 29\,600 \), Joshua is correct. (A1)
Explanation:
Compare total water to one year’s usage (366 or 365 days).
Result:
Joshua is correct

(d)
Height of trapezium (slanted side) is \( \sqrt{10^2 + 6^2} \) (\( = 11.6619… \)) (M1)
Area of one trapezoidal side is \( \frac{80 + 60}{2} \times \sqrt{10^2 + 6^2} \) (\( = 816.333… \)) (M1)(A1)
Total painted area = \( 4 \times \left( \frac{80 + 60}{2} \times \sqrt{10^2 + 6^2} \right) + 60^2 \) (M1)
Total painted area = \( 6870 \) m\(^2\) (6865.33 m\(^2\)) (A1)
Explanation:
Painted area includes four trapezoidal sides and the square base (\( 60 \) m \( \times 60 \) m).
Result:
Painted area is \( 6870 \) m\(^2\)

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