11. The diameter of the largest cylinder is 24cm, so the sum of the areas of the horizontal parts of the sculpture, excluding its base, is that of a circle of diameter 24cm, that is 144π cm2. The sum of the areas of the vertical parts of the sculpture is (2π × 1 × 2 + 2π × 2 × 2 + 2π × 3 × 2 + + 2π × 12 × 2)cm2, that … is 312π cm2. So, excluding the base, the total surface area of the sculpture is 456π cm2. is a 30-60-90 triangle, so 2OM. 12. OA = ΔAOM and are similar, and ΔAOM ΔAO′M′ = + (sum of two radii) OO′ OM O′M′ = : = 1 : 3 O’M’ OM (O’M’)2 : (OM)2 = 1 : 9   13. Join and AB, BC CA. Then is is equilateral, with length 3cm. The three medians meet at where O, = = = (2/3) × (33/2) = AO BO CO 3. = = = 3 OX OY OZ 3. − Hence the circle centre through has area (3 = (12 63) = 6π (2 3)2 3). O X, Y, Z π − π − − 14. The small circle can be enlarged to the size of the other circle by using a scale factor of / So if and and are R r. A B the centres of the circles = + + 2R r2 r R .
11. The diameter of the largest cylinder is 24cm, so the sum of the areas of the horizontal parts of the sculpture, excluding its base, is that of a circle of diameter 24cm, that is 144π cm2. The sum of the areas of the vertical parts of the sculpture is (2π × 1 × 2 + 2π × 2 × 2 + 2π × 3 × 2 + … + 2π × 12 × 2)cm2, that is 312π cm2. So, excluding the base, the total surface area of the sculpture is 456π cm2. 12. ΔAOM is a 30-60-90 triangle, so OA = 2OM. ΔAOM and ΔAO′M′ are similar, and OO′ = OM + O′M′ (sum of two radii) = O’M’ : OM = 1 : 3  (O’M’)2 :  (OM)2 = 1 : 9 13. Join AB, BC and CA. Then is is equilateral, with length 3cm. The three medians meet at O, where AO = BO = CO = (2/3) × (33/2) = 3. OX = OY = OZ = 3 − 3. Hence the circle centre O through X, Y, Z has area π (3 − 3)2 = π (12 − 63) = 6π (2 − 3). 14. The small circle can be enlarged to the size of the other circle by using a scale factor of R / r. So if A and B and are the centres of the circles 2R = r2 + r + R .
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