REVIEW Volume Surface Area

The triangular faces of a square based pyramid, ABCDE, are all inclined at 70° to the base. The edges of the base ABCD are all 10 cm and M is the centre. G is the mid-point of CD. (a) Using the letters on the diagram draw a triangle showing the position of a 70° angle. (1) (b) Show that the height of the pyramid is 13.7 cm, to 3 significant figures. (2) (c) Calculate (i) the length of EG; (ii) the size of angle . (4) (d) Find the total surface area of the pyramid. (e) Find the volume of the pyramid.

(b)tan 70 = (M1) h = 5 tan 70 = 13.74 (A1) h = 13.7 cm (AG) 2 (c) EG2 = 52 + 13.72 OR 52 + (5 tan 70)2 (M1) EG = 14.6 cm (A1)(G2) (UP) (d) = 2 *tan–1 = (M1) = 37.8° (A1)(ft)(G2) 4 (e)Volume = (M1) = 457 cm3 (458 cm3) (A1)(G2) 2 (UP)

2. ABCDV is a solid glass pyramid 2.ABCDV is a solid glass pyramid. The base of the pyramid is a square of side 3.2 cm. The vertical height is 2.8 cm. The vertex V is directly above the centre O of the base. (a) Calculate the volume of the pyramid. (2) (b) The glass weighs 9.3 grams per cm3. Calculate the weight of the pyramid. (c) Show that the length of the sloping edge VC of the pyramid is 3.6 cm. (4) (d) Calculate the angle at the vertex, . (3) (e) Calculate the total surface area of the pyramid.

= 88.9 grams (A1)(ft)(G2) (UP) 2 Note: (M1) for substituting in correct formula = 9.56 cm3 (A1)(G2) 2 (UP) (b)9.56 * 9.3 (M1) = 88.9 grams (A1)(ft)(G2) (UP) 2 (c) base = 1.6 seen (M1) 4 Note: Award (M1) for halving base OC2 = 1.62 + 1.62 = 5.12 (A1) Note: Award (A1) for one correct use of Pythagoras 5.12 + 2.82 = 12.96 = VC2 (M1) Note: Award (M1) for using Pythagoras again to find VC2 VC = 3.6 AG (A1) Note: Award (A1) for 3.6 obtained from 12.96 only (not 12.95…) (d)3.22 = 3.62 + 3.62 – 2 * (3.6) (3.6)cos (M1)(A1) = 52.8° (no (ft) here) (A1)(G2) Note: Award (M1) for substituting in correct formula, (A1) for correct substitution OR sin = where M is the midpoint of BC (M1)(A1) = 52.8° (no (ft) here) (A1) (e) (UP) = 30.9 cm2 ((ft) from their (d)) (A1)(ft)(G2)4 (UP)

Jenny has a circular cylinder with a lid Jenny has a circular cylinder with a lid. The cylinder has height 39 cm and diameter 65 mm. (a) Calculate the volume of the cylinder in cm3. Give your answer correct to two decimal places. (3) The cylinder is used for storing tennis balls. Each ball has a radius of 3.25 cm. (b) Calculate how many balls Jenny can fit in the cylinder if it is filled to the top. (1) (c)(i) Jenny fills the cylinder with the number of balls found in part (b) and puts the lid on. Calculate the volume of air inside the cylinder in the spaces between the tennis balls. (ii) Convert your answer to (c) (i) into cubic metres. (4)

Answers (a) π x 3.252 x 39 (M1)(A1) (= 1294.1398) Answer 1294.14 (cm3)(2dp) (A1)(ft)(G2) 39/6.5 = 6 (i) Volume of one ball is Volume of air = π x 3.252 x 39 – 6 x x 3.253 = 431cm3 (M1)(A1)(ft) (ii) 0.000431m3 or 4.31 x 10–4 m3

7.The diagram below shows a child’s toy which is made up of a circular hoop, centre O, radius 7 cm. The hoop is suspended in a horizontal plane by three equal strings XA, XB, and XC. Each string is of length 25 cm. The points A, B and C are equally spaced round the circumference of the hoop and X is vertically above the point O. (a) Calculate the length of XO. (2) (b) Find the angle, in degrees, between any string and the horizontal plane. (c) Write down the size of angle (1) (d) Calculate the length of AB. (3) Find the angle between strings XA and XB.

ANSWERS α = 73.7° Note: Award (M1) for using any correct ratio. 120° (a) XO2 = 252 – 72 (M1) Note: Award (M1) for using Pythagoras’ Theorem with correct signs and values XO = 24 (A1)(G2) 2 α = 73.7° Note: Award (M1) for using any correct ratio. 120° (d) AB2 = 72 + 72 – 2 × 7 × 7 × cos 120° (M1)(A1) Note: Award (M1) for using cosine rule. Notes: Award (M1) for substituting values from the problem into the cosine rule, (A1) for correct values. Accept alternative, correct methods (e) (M1)(A1) Note: Award (M1) for substituting values from the problem into the cosine rule, (A1) for correct values. Θ = 28.1° (28.0°) (A1)(ft)(G2) 3 Note: Accept 28°. If using an isosceles triangle, award (M1) for angle, (A1) for answer, (A1) for doubling.

A child’s toy is made by combining a hemisphere of radius 3 cm and a right circular cone of slant height l as shown on the diagram below. (a) Show that the volume of the hemisphere is 18π cm3. The volume of the cone is two-thirds that of the hemisphere. (b) Show that the vertical height of the cone is 4 cm. (c) Calculate the slant height of the cone. (d) Calculate the angle between the slanting side of the cone and the flat surface of the hemisphere. (e) The toy is made of wood of density 0.6 g per cm3. Calculate the weight of the toy. (f) Calculate the total surface area of the toy.

(a)V = For using (with or without ) (a)V = For using (with or without ) = × π × 33 For using (their sphere formula) (M1) = 18π cm3 (AG) 2 (b V = × 18π For using × their answer to (a) (M1) = 12π (A1) 12π = π × 32 × h For equating the volumes (M1) (A1) h = 4 cm (c) l2 = 42 + 32 For using Pythagoras theorem (M1) l = 5 (A1) 2

(d) For identifying the correct angle (M1) tanΘ = or sin Θ = or cos Θ = (M1) Θ = 53.1° (0.927 radians) (A1) 3 (e) For summing volume of cone and hemisphere. (M1) Volume = 12π + 18π = 30π cm3 (94.2 cm3) For multiplying the volume by 0.6 (M1) Weight = 0.6 × 30π = 56.5 g (f) Surface area of cone = πrl = π × 3 × 5 = 15π (M1)(A1) Surface area of a hemisphere = × 4πr2 = × 4 × π × 32 = 18π (M1)(A1) Total surface area = 15π + 18π = 103.67 = 104 cm2 (A1) 5

4. An old tower (BT) leans at 10° away from the vertical (represented by line TG). The base of the tower is at B so that = 100°. Leonardo stands at L on flat ground 120 m away from B in the direction of the lean. He measures the angle between the ground and the top of the tower T to be = 26.5°. (a) (i) Find the value of angle BTL . (ii) Use triangle BTL to calculate the sloping distance BT from the base, B to the top, T of the tower. (5) (b) Calculate the vertical height TG of the top of the tower. (2) (c) Leonardo now walks to point M, a distance 200 m from B on the opposite side of the tower. Calculate the distance from M to the top of the tower at T. (3) (Total 10 marks)

(a)(i) Angle = 180 – 80 – 26.5 or 180 – 90 – 26.5 + 10 (M1) = 73.5° (A1)(G2) (ii) (M1)(A1)(ft) BT = 55.8 m (3s.f.) (A1)(ft) 5 (UP) (b) TG = 55.8 sin (80°) or 55.8 cos (10°) (M1) = 55.0 m (3s.f.) (A1)(ft)(G2) 2 Note: Apply (AP) if 0 missing (c) MT2 = 2002 + 55.82 – 2 x 200 x 55.8 cos (100°) (M1)(A1)(ft) MT = 217 m (3s.f.) (A1)(ft) 3 (UP)