Submit your solutions on separate paper. Indicate your team name on each sheet.
One solution set only is required from each team.
Part marks are awarded for partial or incorrect solutions.
The steps to a solution can be omitted if you are confident your final answer is correct,
however the steps will be used for part marks for incomplete or incorrect solutions.
Marks may be deducted for poor mathematical form.
Group Problem Set #2
1.The new volume of a cube is 512 cubic units after its sides have been doubled in length. What was the original volume of the cube?
2.A solid sphere with outside radius 8.4 cm is placed tightly inside a cylinder with inside radius 8.4 cm and height 21.6 cm. Find the volume of air left inside the cylinder (to the nearest 0.1 cm3).
3.The Great Pyramid in Egypt (when it was constructed) had a square base measuring 234 m by 234 m and a height of 156 m. Following years of weathering effects and neglect, the government of Egypt is undertaking a face-lift for this landmark.
a)If one can of paint is needed to cover 120 m2 of its exposed faces, find the number of cans of paint needed to give two coats of paint to the exposed faces of the Great Pyramid.
b)On average, one cubic metre of the material used to construct the Great Pyramid (including the inside) weighs 172 kg. Find the weight of the stones and other inside materials used to build the Great Pyramid (to the nearest 1000 kg).
4. A part for a machine is the constructed in the shape of a right-circular cone with its base hollowed out to form a cavity which is also in the shape of a right circular cone. The points and lengths stated refer to the cross-section of the cone shown below. The vertices of these two cones (points A and D) lie in a line directly above the centre of the circular base. The diameter of the base of the original cone and the hollowed out cone (line segment BC) is 80 cm. the slant height of the large cone (AB and AC) is 85 cm. The slant height of the hollowed out cone (BD and CD) is 50 cm.
4.continued
a)Find the volume of the hollowed out object (after the bottom cone has been removed) correct to the nearest 0.1 cm3.
b)Find the total surface area of the hollowed out object (after the bottom cone has been removed) correct to nearest 100 cm2.
5.A large circle has a diameter AB of 40 cm. Two identical smaller circles are drawn inside the larger circle so that the diameter of the larger circle passes through both centres of the smaller circles, and the smaller circles touch each other and the inside of the larger circle (see diagram below).
a)Find the ratio of the area of one of the smaller circles to the area of the larger circle.
b)Find the area that lies inside the large circle but outside the two smaller circles (to the nearest 0.1 cm2).
c)Find the length of line segment QP shown if P is the center of one of the smaller circles, Q is a point on the larger circle so that PQ is perpendicular to the line joining the centres of the three circles. (Hint:AQB = 90.)
Solutions to Group Problem Set #2
1.(8 marks)
Let the original side length of the cube be x (or any suitable variable).(1 mark)
Hence the new (doubled) side length of the cube is 2x(1 mark)
The volume of a cube (in terms of its side length x) is x3(1 mark)
Hence:(2x)3 = 512(1 mark)
8x3 = 512
x3 = 64
x = 4(2 marks: 1 mark for any failed solution)
Original volume is given by:V = x3 = 43 = 64(1 mark)
Thus the original volume of the cube is 64 cubic units.(1 mark for units/statement)
Award 8 Marksimmediately if the answer is correct without examining the solution.
As with #1 above and all subsequent questions (unless otherwise stated), award full marks immediately if the final answer is correct without regard to the details of their solution.
2.(9 marks)
The diagram is shown above.(1 mark)
The radius of the sphere and the cylinder is 8.4 cm(1 mark)
Volume of sphere is given by the formula:(1 mark)
Volume of cylinder is given by the formula:(1 mark)
For the sphere:V = (8.4)3 = 2482.71 cm3 (nearest 0.01 cm3) (1 mark)
For the cylinder:V = (8.4)2(21.6) = 4788.09 cm3 (nearest 0.01 cm3) (1 mark)
Volume of the air inside is given by the difference between these values. (1 mark)
Volume (air) = 4788.09 – 2482.71 = 2305.37 (1 mark)
Thus the volume of the air left inside (to nearest 0.1 cm3) is 2305.4 cm3. (1 mark)
(9 marks for question 2.)
3.(13 marks)
a)Lateral Surface Area = 4 (1 mark)
(where the base is 234 m but we must find the (slant) height using the Pythagorean Theorem)
Let x represent the slant height (see diagram below)(1 mark)
x = = 195 (which is the h in above) (2 marks;
1 mark for any failed attempt)
Thus the lateral surface area (in m2) is given by:
SA = 4 0.5 234 195 = 91260(1 mark)
The number of cans of paint required for two coats of paint is given by:
2 91260 120 = 1521(2 marks
1 mark for any failed attempt) 1521 cans of paint are required for the task.
(7 marks for part a)
b)The volume of a square-based pyramid is given by: (1 mark)
(where B is the area of the base and h is the perpendicular height)
V = 2342 156 = 2847312(2 marks; 1 mark for any failed attempt)
Weight = 172 2847312 = 489737664(1 mark)
Thus the weight of the material used the build the Great Pyramid (inside and outside) was
489738000 kg.
(2 marks; 1 for units and 1 for rounding correctly)
(6 marks for part b)
4.(18 marks)
The length of the radius for both cones is 40 cm. (2 marks; 1 for each cone)
The formula for the volume of a cone is V = πr2h(1 mark)
Let H be the altitude of the full cone (ABC).
The length of the altitude H for the full cone is H2 = 852 – 402 = 752 = 5625
Thus H = 75(2 marks; 1 for any failed effort)
Let h be the altitude of the hollowed out cone at the bottom (DBC).
The length of the altitude h for the hollowed out cone is h2 = 502 – 402 = 302 = 900
Thus h = 30(2 marks; 1 for any failed effort)
a)Thus the volume of the solid object is given by
V = πr2H – πr2h=π(40)2[75 – 30] = 75398.223
(3 marks for arriving at V = 75398.22; 2 marks if correct method is used but arithmetic error is made; 1 mark for any other failed effort)
Thus the volume of the hollowed out object is approximately 75398.2 cm3
(2 marks; 1 mark for units and 1 mark for correct rounding)
(12 marks for part a)
b)The formula for the lateral surface area of a cone is SA = πrs(1 mark)
Let the slant heights of the full and hollowed out cones be S and s respectively(1 mark)
The total surface area of the solid object is given by πrS + πrs(where S = 85 and s = 50)
TSA = πrS + πrs= π(40)[85 + 50] = 16964.6057 (2 marks; 1 mark for any failed effort)
The total surface area if 17000 (or 1.70 104) cm2(2 marks; 1 for rounding, 1 for units)
(6 marks for part b)
5.(18 marks)
a)The radius of the larger circle is 20 cm.(1 mark)
The radius of each smaller circle is 10 cm.(1 mark)
The area of a circle is given by A = r2(1 mark)
The ratio of the area of the smaller circle to the area of the larger circle is given by:
(10)2 : (20)2 = 100 : 400 = 1 : 4(2 marks; 1 mark for any failed effort)
(5 marks for part a)
b)Solution One
Since the ratio of the areas of a smaller circle to the larger circle is 1:4, the area outside the two smaller circles is one-half of the area of the larger circle and is 200 cm2. (4 marks)
b)Solution Two
The area inside the larger circle but outside the smaller circles is given by:
A = (20)2 2 (10)2 = 400 200 = 200628.318
(2 marks; 1 mark for any failed effort)
The required area is 628.3 cm2.(2 marks; 1 for rounding, 1 for units)
(4 marks for part a)
c)Solution One:
Join AQ and QB (see diagram below)(1 mark)
APQ is similar to AQB because they each have a right angle and a common angle at A. QPB is similar to AQB because they each have a right angle and a common angle at B.
Thus APQ QPB (they are both similar to the same triangle)
(1 mark for any reasonable attempt to explain why APQ QPB)
Thus we have:(1 mark)
Let the length of PQ (or QP) be x(or simply use PQ in the following equation).
AP is 10 cm (established earlier) and PB is 30 cm.(1 mark)
Hence:
And we get:x2 = 300(1 mark)
Which gives:x = (1 mark)
(But the negative answer is inadmissible.)(1 markfor this concept and the )
Thus the length of PQ is:cm(1 mark)
Thus the length of PQ is:17.3 or 17.32 cm(1 mark)
Thus the length of PQ is:cm(2 marks)
(Award 2 or 1 or 0 marks for the conclusion depending on the answer selected and whether or not units were stated)
c)Solution two:
AQB = 90 (which is given) and QPA = 90 and QPB = 90
(since QP is perpendicular to AB)
Let QP be represented by x, AQ by y and QB by z(1 mark)
(Using the Pythagorean Theorem in each case here)
In APQ:y2 = x2 + 102thus y2 = x2 + 100(1)(1 mark)
In QPB:z2 = x2 + 302thusz2 = x2 + 900(2)(1 mark)
In AQB:402 = y2 + z2thus1600 = y2 + z2(3)(1 mark)
By substituting (1) and (2) into (3) (i.e. substitute for y2 and z2 in the right side of (3))
we get:1600 = x2 + 100 + x2 + 900
which gives:600 = 2x2 or 300 = x2(1 mark)
This gives:x = (1 mark)
(But the negative answer is inadmissible.)(1 markfor this concept and the )
Thus the length of PQ is:cm(1 mark)
Thus the length of PQ is:17.3 or 17.32 cm(1 mark)
Thus the length of PQ is:cm(2 marks)
(Award 2 or 1 or 0 marks for the conclusion depending on the answer selected and whether or not units were stated)
(9 marks possible for part b)
Note that you can also solve this problem using special right triangles or QP as the geometric mean between AP and PB(and obtain 9 marks if arriving at).
You can also use trigonometric ratios and solving the two right triangles (i.e. SOHCAHTOA) to solve it and get the approximate answer (for 8 marks only).
Quick Mark Summary For This problem Set:
- 8 Marks
- 9 Marks
- 13 Marks
- 18 Marks
- 18 Marks
Total Possible Marks for Problems Set #2 is 66
Indicate Total Team Score Only on Blackboard Summary (Never give percentages)