2020 H2 Maths Paper 2 Q4 — Worked Solution

Part (i)

2 marks

Approach

We need two relations: how $a$ and $h$ are tied together by the $30$ cm cardboard (Fig. 1), and how $h$ , $H$ and $a$ are tied together inside the folded pyramid (Fig. 2).

From Fig. 1, the net’s overall width is $30$ cm, made up of the central square of side $a$ together with two triangular flaps of height $h$ on either side, so

a + 2h = 30

h = \frac{30 - a}{2}.

In the folded pyramid, $h$ is the slant height from the apex to the midpoint of a base edge, $H$ is the vertical height, and the horizontal distance from the centre of the square base to the midpoint of an edge is $\frac{a}{2}$ . These three lengths form a right-angled triangle, so by Pythagoras’ theorem,

h^{2} = H^{2} + \left( \frac{a}{2} \right)^{2}

H^{2} = h^{2} - \frac{a^{2}}{4}.

Substituting $h = \frac{30 - a}{2}$ ,

H^{2} = \left( \frac{30 - a}{2} \right)^{2} - \frac{a^{2}}{4}

= \frac{(30 - a)^{2} - a^{2}}{4}

= \frac{900 - 60a}{4}

= 225 - 15a.\ \ \left( \text{shown} \right)

Part (ii)

5 marks

Approach

With $H = \sqrt{225 - 15a}$ from (i), the volume is a function of $a$ alone; we differentiate and solve $\frac{dV}{da} = 0$ .

Since the volume of the pyramid is $\frac{1}{3} \times$ base area $\text{×}$ height,

V = \frac{1}{3}a^{2}\sqrt{225 - 15a},\quad\quad 0 < a < 15

(the upper bound comes from requiring $H^{2} > 0$ ).

Differentiating using the product rule,

\frac{dV}{da} = \frac{1}{3}\left\lbrack 2a\sqrt{225 - 15a} + a^{2} \cdot \frac{- 15}{2\sqrt{225 - 15a}} \right\rbrack

= \frac{a}{3\sqrt{225 - 15a}}\left\lbrack 2(225 - 15a) - \frac{15a}{2} \right\rbrack

= \frac{a}{3\sqrt{225 - 15a}}\left( 450 - \frac{75a}{2} \right).

Setting $\frac{dV}{da} = 0$ : since $0 < a < 15$ , $a \neq 0$ and $\sqrt{225 - 15a} \neq 0$ , so

450 - \frac{75a}{2} = 0

a = 12.

At $a = 12$ ,

H^{2} = 225 - 15(12) = 45,\quad\quad H = 3\sqrt{5},

V = \frac{1}{3}(12)^{2}\left( 3\sqrt{5} \right) = 144\sqrt{5}.

$\therefore$ the maximum possible volume of the pyramid is $144\sqrt{5} \approx 322\ cm^{3}$ (3 s.f.).

Part (iii)

Part (a)

3 marks

Approach

The total surface area of the four triangular faces is $4 \times \frac{1}{2}ah = 2ah$ ; substitute $h = \frac{30 - a}{2}$ from (i) to write it in terms of $a$ alone, then maximise.

S = 2ah = 2a \cdot \frac{30 - a}{2}

= 30a - a^{2}.

Differentiating,

\frac{dS}{da} = 30 - 2a.

Setting $\frac{dS}{da} = 0$ ,

30 - 2a = 0

a = 15.

$\therefore$ the total surface area of the four triangular faces is a maximum when $a = 15$ .

Note: at $a = 15$ , $H^{2} = 225 - 15(15) = 0$ , so this is exactly the boundary value at which the pyramid degenerates — see (b).

Part (b)

1 mark

Since $H = 0$ when $a = 15$ , the solid has zero height: each triangular flap folds all the way down onto the base square instead of standing up. At $a = 15$ , $h = \frac{30 - 15}{2} = 7.5 = \frac{a}{2}$ , which is exactly the distance from the centre of the square base to the midpoint of an edge — so each folded-down triangle reaches exactly to the centre, and all four meet there.

$\therefore$ the net folds completely flat: no pyramid is formed at all, only a flat square shape (of diagonal $30$ cm, whose corners are the midpoints of the sides of the original $30$ cm cardboard square).

2020 H2 Maths Paper 2, Question 4

Part (i)

Part (ii)

Part (iii)

Part (a)

Part (b)

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