(a) A company makes a chicken run in the shape of a triangular prism, as shown below. The uniform cross-section of the chicken run is an isosceles triangle.
The run covers a rectangular ground area of 5·46 m2.
The vertical height of the run is 1·5 m.
Diagram not drawn to scale
Each face of the chicken run is to be covered in wire mesh, apart from the base. The wire mesh costs £5.60 per m2.
Calculate the cost of the wire mesh that is needed for the chicken run. [7]
Cost of the wire mesh = £ …………………………
(b) The company also makes chicken coops that are mathematically similar.
Medium coop
Capacity = 8 m3
Area of wire mesh = 3 m2
Large coop
Capacity = 27 m3
Use the above information to calculate the area of wire mesh in the large coop. [4]
Mark scheme (a)
Scheme
Marks
Comments
(Width of rectangle/Base of triangle =) 1.3 (m) (Sloping length2 =) 1.52 + 0.652 Sloping length2 = 2.6725 or
(Sloping length =) √2.6725
B1 M1 A1
May be implied by use of 0.65 in further work FT ‘their 1.3/2’
(Sloping length =) 1.63(…) (m)
A1
FT from M1 for the correctly evaluated square root of ‘their 2.6725’ provided their answer > 1.5 If trigonometry used to calculate the sloping length, M1 for base angle = tan-1(1.5/(1.3÷2)) (= 66.5(71…º)) or apex angle = tan-1((1.3÷2)/1.5) (= 23.4(28…º)) m1 for correct rearrangement of a sin or cos equation using their base or apex angle with 1.5 or 1.3÷2 A1 for (Sloping length =) 1.63(…) (m)
(Cost of wire mesh =) ( 2×0.5×1.3×1.5 + 2×1.63(…)×4.2 ) (× 5.6(0)) 2×0.5×1.3×1.5: 1.95 m2, (£) 10.92 2×1.63(…)×4.2: 13.69 to 13.73(…) m2, (£) 76.66(…) to (£) 76.91
M2
FT ‘their 1.63(…)’ provided Pythagoras or trigonometry attempted and ‘their 1.3’ M1 for
0.5×1.3×1.5 + 1.63(…)×4.2 or
…… + 2×1.63(…)×4.2 or
2×1.63(…)×4.2 × 5.6(0)
OR M1 for
2×0.5×1.3×1.5 + …… or
2×0.5×1.3×1.5 × 5.6(0)
where Pythagoras or trigonometry may not have been attempted
= (£) 87.58 to (£) 87.83
A1
FT from previous M2 only Note: A sloping length of 1.6 (m) leads to an answer of (£)86.18(4)
Mark scheme (b)
Scheme
Marks
Comments
\(\sqrt[3]{\dfrac{27}{8}}\) or \(\sqrt[3]{\dfrac{8}{27}}\) OR \(\left(\dfrac{27}{8}\right)^{2}\) or \(\left(\dfrac{8}{27}\right)^{2}\) or 3:2 or 2:3 or equivalent
B1
(= 1.5 or 0.666… OR 11.3(9…) or 0.08(7…))
\(\left(\sqrt[3]{\dfrac{27}{8}}\right)^{2}\) OR \(\left(\sqrt[3]{\dfrac{8}{27}}\right)^{2}\) or 32:22 or 22:32 or equivalent
B1
(= 2.25 OR 0.444…)
(Area of wire mesh = ) 3 × \(\left(\sqrt[3]{\dfrac{27}{8}}\right)^{2}\) or equivalent
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