N2020 P2 Q4

N2020 P2 Q4

Junior College 2
11 marks
Free
Square-based pyramid net and folded pyramid

Fig. 1 shows the net of a square-based pyramid cut from a square of cardboard of side length \(30\text{ cm}\). The net consists of a square of side length \(a\text{ cm}\) and four isosceles triangles, each with base \(a\text{ cm}\) and perpendicular height \(h\text{ cm}\). The net is folded to form a pyramid which has a square base of side length \(a\text{ cm}\) and vertical height \(H\text{ cm}\), as shown in Fig. 2.

  1. Show that \(H^2=225-15a\).[2]
  2. Find the maximum possible volume of the pyramid. You do not need to show that this value is a maximum. [The volume of a square-based pyramid is \(\frac13\times\text{base area}\times\text{height}\).][5]
    1. Find the value of \(a\) for which the total surface area of the four triangular faces of the pyramid is a maximum. You do not need to show that this value is a maximum.[3]
    2. Describe the shape formed from the net in this case.[1]

Default solution

(i) \(h=(30-a)/2\). In the folded pyramid, \(H^2=h^2-(a/2)^2=\frac{(30-a)^2-a^2}{4}=225-15a\).

(ii) \(V(a)=\frac{a^2}{3}\sqrt{225-15a}\), with \(0<a<15\). Differentiating gives a stationary point at \(a=12\).

Then \(H=3\sqrt5\), so \(V_{\max}=144\sqrt5\text{ cm}^3\approx322\text{ cm}^3\).

(iii)(a) The area of four triangular faces is \(4(\frac12 ah)=2ah=a(30-a)\), maximised when \(a=15\text{ cm}\).

(iii)(b) \(H=0\) when \(a=15\), so the folded net is a flat square.

Answer:(ii) \(144\sqrt5\text{ cm}^3\); (iii)(a) \(15\text{ cm}\), (b) a flat square

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