EXERCISE SET 14.1
Question 1
The volume of a cube is 64 cm^3. What is its total surface area?
[fig]
Solution:
Given:
Volume of the cube = 64 cm^3
Let the edge length of the cube be a cm.
Formula for volume of a cube = a^3
a^3 = 64
a = 64^(1/3) = 4 cm
Formula for total surface area of a cube = 6 * a^2
Total Surface Area = 6 * (4)^2 = 6 * 16 = 96 cm^2
Answer: 96 cm^2
Question 2
How many small cubes with side 20 cm can be packed tight in a cubical box with side 2 m?
[fig]
Solution:
Side of the cubical box (A) = 2 m = 200 cm
Side of each small cube (a) = 20 cm
Volume of the cubical box = A^3 = 200 * 200 * 200 = 8,000,000 cm^3
Volume of one small cube = a^3 = 20 * 20 * 20 = 8,000 cm^3
Number of small cubes = (Volume of cubical box) / (Volume of one small cube)
Number of small cubes = 8,000,000 / 8,000 = 1000
Answer: 1000 cubes
Question 3
The dimensions of a godown are 40 m x 25 m x 10 m. If it is filled with cuboidal boxes, each of dimensions 2 m x 1.25 m x 1 m, then find the number of boxes.
[fig]
Solution:
Dimensions of the godown:
Length (L) = 40 m
Breadth (B) = 25 m
Height (H) = 10 m
Volume of the godown = L * B * H = 40 * 25 * 10 = 10,000 m^3
Dimensions of one cuboidal box:
length (l) = 2 m
breadth (b) = 1.25 m
height (h) = 1 m
Volume of one cuboidal box = l * b * h = 2 * 1.25 * 1 = 2.5 m^3
Number of boxes = (Volume of godown) / (Volume of one box)
Number of boxes = 10,000 / 2.5 = 4000
Answer: 4000 boxes
Question 4
Two cubes each of volume 125 cm^3 are joined end to end. Find the surface area of the resulting cuboid.
[fig]
Solution:
Volume of each cube = 125 cm^3
Let the side of each cube be a cm.
a^3 = 125
a = 5 cm
When two cubes are joined end to end:
Length of the resulting cuboid (l) = 5 + 5 = 10 cm
Breadth of the resulting cuboid (b) = 5 cm
Height of the resulting cuboid (h) = 5 cm
Total surface area of the resulting cuboid = 2 * (lb + bh + hl)
= 2 * (105 + 55 + 510)
= 2 * (50 + 25 + 50)
= 2 * 125
= 250 cm^2
Answer: 250 cm^2
Question 5
A cube of side 4 cm is cut into cubes of side 1 cm. What is the ratio of the surface areas of the original cube and all the cut-out cubes?
[fig]
Solution:
Side of original cube (A) = 4 cm
Surface area of original cube = 6 * A^2 = 6 * (4)^2 = 6 * 16 = 96 cm^2
Volume of original cube = A^3 = 4^3 = 64 cm^3
Side of small cube (a) = 1 cm
Volume of one small cube = a^3 = 1^3 = 1 cm^3
Number of small cut-out cubes = (Volume of original cube) / (Volume of one small cube) = 64 / 1 = 64
Surface area of one small cube = 6 * a^2 = 6 * (1)^2 = 6 cm^2
Total surface area of all 64 cut-out cubes = 64 * 6 = 384 cm^2
Ratio of surface area of original cube to total surface area of all cut-out cubes:
Ratio = 96 / 384 = 1 / 4
Answer: 1 : 4
Question 6
The surface areas of the three faces of a cuboid that meet at one of the corners of the cuboid are 6 cm^2, 15 cm^2, and 10 cm^2 respectively. What is the volume of the cuboid?
[fig]
Solution:
Let the dimensions of the cuboid be length l, breadth b, and height h.
Area of face 1 (l * b) = 6 cm^2
Area of face 2 (b * h) = 15 cm^2
Area of face 3 (h * l) = 10 cm^2
Multiplying the three equations:
(l * b) * (b * h) * (h * l) = 6 * 15 * 10
(l * b * h)^2 = 900
Taking square root on both sides:
l * b * h = sqrt(900) = 30 cm^3
Volume of the cuboid = l * b * h = 30 cm^3
Answer: 30 cm^3
Question 7
A cube of side 5 cm is painted on all its faces. If it is sliced into 1 cm^3 cubes, how many of these 1 cm^3 cubes have:
(i) exactly three faces painted?
(ii) exactly two faces painted?
(iii) exactly one face painted?
(iv) no face painted?
[fig]
Solution:
Number of divisions along each edge, n = 5 cm / 1 cm = 5
(i) Cubes with exactly three faces painted:
These cubes are located at the 8 vertices of the large cube.
Number of cubes = 8
(ii) Cubes with exactly two faces painted:
These cubes lie along the edges, excluding the vertices.
Formula = 12 * (n – 2)
Number of cubes = 12 * (5 – 2) = 12 * 3 = 36
(iii) Cubes with exactly one face painted:
These cubes are located on the interior of each face.
Formula = 6 * (n – 2)^2
Number of cubes = 6 * (5 – 2)^2 = 6 * 3^2 = 6 * 9 = 54
(iv) Cubes with no face painted:
These form an inner cube of side (n – 2).
Formula = (n – 2)^3
Number of cubes = (5 – 2)^3 = 3^3 = 27
Answer:
(i) 8
(ii) 36
(iii) 54
(iv) 27
Question 8
Find a cuboid with edges whose lengths are integers (in cm), given that it has a total surface area of exactly 100 cm^2.
(i) Is there more than one such cuboid?
(ii) Can you find them all?
(iii) Show that you have found them all.
[fig]
Solution:
Let the integer dimensions of the cuboid be l, b, and h with 1 <= l <= b <= h.
Total Surface Area = 2 * (lb + bh + hl) = 100
So, lb + bh + hl = 50.
We test possible integer values for l:
Case 1: l = 1
1b + bh + h1 = 50
b + bh + h = 50
(b + 1)(h + 1) = 51
Factors of 51 are (1, 51) and (3, 17).
Since b >= l = 1, b+1 >= 2.
- b + 1 = 3 => b = 2
- h + 1 = 17 => h = 16
Dimensions: (1, 2, 16)
Case 2: l = 2
2b + bh + 2*h = 50
(b + 2)(h + 2) = 50 + 4 = 54
Factors of 54 where b+2 >= 4:
- b + 2 = 6 => b = 4
- h + 2 = 9 => h = 7
Dimensions: (2, 4, 7)
Case 3: l = 3
3b + bh + 3*h = 50
(b + 3)(h + 3) = 50 + 9 = 59
Since 59 is prime, no integer solutions with b >= 3.
Case 4: l = 4
(b + 4)(h + 4) = 50 + 16 = 66
Factors of 66 with b+4 >= 8:
- b + 4 = 11 => b = 7, h + 4 = 6 (Impossible as b <= h)
Testing higher values of l proves no further solutions exist.
Answer:
(i) Yes, there are two such cuboids.
(ii) The dimensions are 1 cm x 2 cm x 16 cm and 2 cm x 4 cm x 7 cm.
(iii) Systematic testing of l from 1 to 4 shows all potential integer factor pairs for (l+b)(l+h) are exhausted.
EXERCISE SET 14.2
Question 1
Two cylinders, A and B, are given. The radius of cylinder B is twice that of cylinder A, and the height of cylinder B is half that of cylinder A. Find the ratio of the curved surface area of A to the curved surface area of B. Also find the ratio of the volume of A to the volume of B.
[fig]
Solution:
Let radius of cylinder A = r, height of cylinder A = h.
Then radius of cylinder B (r_B) = 2r.
Height of cylinder B (h_B) = h / 2.
- Ratio of Curved Surface Area (CSA):
CSA_A = 2 * pi * r * h
CSA_B = 2 * pi * (2r) * (h / 2) = 2 * pi * r * h
Ratio = 1 : 1 - Ratio of Volume:
V_A = pi * r^2 * h
V_B = pi * (2r)^2 * (h / 2) = pi * 4r^2 * h / 2 = 2 * pi * r^2 * h
Ratio = 1 : 2
Answer:
Ratio of CSA = 1 : 1
Ratio of Volume = 1 : 2
Question 2
The radii of two cylinders are in the ratio 2:3, and their heights are in the ratio 3:2. Find (a) the ratio of their volumes, and (b) the ratio of their curved surface areas.
[fig]
Solution:
Let r1/r2 = 2/3 and h1/h2 = 3/2.
(a) Volume Ratio:
V1/V2 = (pi * r1^2 * h1) / (pi * r2^2 * h2)
= (r1/r2)^2 * (h1/h2)
= (2/3)^2 * (3/2) = (4/9) * (3/2) = 12/18 = 2/3
(b) CSA Ratio:
CSA1/CSA2 = (2 * pi * r1 * h1) / (2 * pi * r2 * h2)
= (r1/r2) * (h1/h2)
= (2/3) * (3/2) = 1/1
Answer:
(a) Ratio of volumes = 2 : 3
(b) Ratio of CSAs = 1 : 1
Question 3
The edge of a cube measures r cm. The largest possible right circular cylinder is cut out of the cube. What do you think is the volume of the cylinder (in cm^3)?
[fig]
Solution:
For the largest cylinder:
Diameter = Side of cube = r cm => Radius (R) = r/2
Height (H) = Side of cube = r cm
Volume = pi * R^2 * H
= pi * (r/2)^2 * r
= pi * (r^2 / 4) * r
= (pi * r^3) / 4
Answer: (pi * r^3) / 4 cm^3
Question 4
The radius of the base of a cylinder is increased by 10%. At the same time, the height of the cylinder is decreased by x%. Given that the volume of the cylinder remains unchanged, find the value of x.
[fig]
Solution:
V1 = pi * r^2 * h
New radius r’ = 1.1r
New height h’ = h * (1 – x/100)
V2 = pi * (1.1r)^2 * h * (1 – x/100) = 1.21 * pi * r^2 * h * (1 – x/100)
Since V1 = V2:
1.21 * (1 – x/100) = 1
1 – x/100 = 1 / 1.21 = 100 / 121
x/100 = 1 – 100/121 = 21/121
x = 2100 / 121 approx 17.36
Answer: 17.36%
Question 5
A solid metallic cube of side 12 cm is melted and recast into solid cylindrical rods, each having radius 2 cm and height 12 cm. Find:
(i) the volume of the cube,
(ii) the volume of one cylindrical rod,
(iii) the approximate number of complete cylindrical rods that can be formed. (pi approx 22/7)
[fig]
Solution:
(i) Volume of cube = 12^3 = 1728 cm^3
(ii) Volume of rod = pi * 2^2 * 12 = (22/7) * 4 * 12 = 1056 / 7 approx 150.86 cm^3
(iii) Number of rods = 1728 / (1056/7) = (1728 * 7) / 1056 = 11.45
Complete rods = 11
Answer: (i) 1728 cm^3, (ii) 150.86 cm^3, (iii) 11
EXERCISE SET 14.3
Question 1
Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
[fig]
Solution:
Radius (r) = 24 / 2 = 12 m
Slant height (l) = 21 m
TSA = pi * r * (l + r)
= (22/7) * 12 * (21 + 12)
= (22/7) * 12 * 33 = 8712 / 7 approx 1244.57 m^2
Answer: 1244.57 m^2
Question 2
Find the curved surface area of a right circular cone whose slant height is 10 cm and base radius is 7 cm.
[fig]
Solution:
r = 7 cm, l = 10 cm
CSA = pi * r * l
= (22/7) * 7 * 10 = 220 cm^2
Answer: 220 cm^2
Question 3
The height of a cone is 16 cm, and its base radius is 12 cm. Find the curved surface area and the total surface area of the cone. (Use pi = 3.14)
[fig]
Solution:
h = 16 cm, r = 12 cm
l = sqrt(16^2 + 12^2) = sqrt(256 + 144) = sqrt(400) = 20 cm
CSA = 3.14 * 12 * 20 = 753.6 cm^2
TSA = 3.14 * 12 * (20 + 12) = 3.14 * 12 * 32 = 1205.76 cm^2
Answer: CSA = 753.6 cm^2, TSA = 1205.76 cm^2
Question 4
A cone has a height of 15 cm. If its volume is 1570 cm^3, find the radius of the base. (Use pi = 3.14)
[fig]
Solution:
Volume = (1/3) * pi * r^2 * h
1570 = (1/3) * 3.14 * r^2 * 15
1570 = 15.7 * r^2
r^2 = 100 => r = 10 cm
Answer: 10 cm
Question 5
The curved surface area of a cone is 308 cm^2 and its slant height is 14 cm. Find:
(i) the radius of the base,
(ii) the total surface area of the cone.
[fig]
Solution:
(i) CSA = pi * r * l
308 = (22/7) * r * 14
308 = 44 * r => r = 7 cm
(ii) TSA = pi * r * (l + r)
= (22/7) * 7 * (14 + 7) = 22 * 21 = 462 cm^2
Answer: (i) 7 cm, (ii) 462 cm^2
Question 6
A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.
[fig]
Solution:
r = 7 cm, h = 24 cm
l = sqrt(7^2 + 24^2) = sqrt(49 + 576) = 25 cm
CSA for 1 cap = (22/7) * 7 * 25 = 550 cm^2
Total area for 10 caps = 550 * 10 = 5500 cm^2
Answer: 5500 cm^2
Question 7
What length of tarpaulin 3 m wide is required to make a conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is 20 cm. (Use pi = 3.14)
[fig]
Solution:
r = 6 m, h = 8 m
l = sqrt(6^2 + 8^2) = 10 m
CSA = 3.14 * 6 * 10 = 188.4 m^2
Length of tarpaulin = 188.4 / 3 = 62.8 m
Total length = 62.8 + 0.2 = 63 m
Answer: 63 m
Question 8
A right triangle with sides 6 cm, 8 cm and 10 cm is rotated through 360 degrees about the side of 8 cm. Find the volume and the curved surface area of the solid so formed.
[fig]
Solution:
Solid formed is a cone with r = 6 cm, h = 8 cm, l = 10 cm.
Volume = (1/3) * pi * 6^2 * 8 = 96 * pi approx 301.71 cm^3
CSA = pi * 6 * 10 = 60 * pi approx 188.57 cm^2
Answer: Volume = 96pi cm^3, CSA = 60pi cm^2
Question 9
Suppose you have a cup in the shape of a right circular cone. Fill it with water to half the depth of the cone. What fraction of the volume of the cup is occupied by the water?
[fig]
Solution:
Let total height = H, total radius = R.
Volume = (1/3) * pi * R^2 * H.
Water height h = H/2. By similarity, water radius r = R/2.
Water Volume = (1/3) * pi * (R/2)^2 * (H/2) = (1/3) * pi * R^2 * H * (1/8).
Fraction = 1/8.
Answer: 1/8
EXERCISE SET 14.4
Question 1
A ball bearing has a radius of 0.7 cm. Find its surface area.
[fig]
Solution:
r = 0.7 cm
SA = 4 * pi * r^2
= 4 * (22/7) * 0.7 * 0.7 = 6.16 cm^2
Answer: 6.16 cm^2
Question 2
Two solid spheres made of the same metal have weights 5920 g and 740 g. Determine the radius of the larger sphere, if the diameter of the smaller one is 5 cm.
[fig]
Solution:
Ratio of weights = Ratio of volumes
5920 / 740 = 8
8 = (R/r)^3 => R/r = 2
Smaller radius r = 5/2 = 2.5 cm
Larger radius R = 2 * 2.5 = 5 cm
Answer: 5 cm
Question 3
The diameter of the moon is approximately one fourth the diameter of the earth. Given that the moon and earth are both roughly spherical, find the ratio of their surface areas.
[fig]
Solution:
r_m = (1/4) * r_e
Ratio of SA = (r_m / r_e)^2 = (1/4)^2 = 1/16
Answer: 1 : 16
Question 4
Find (i) the curved surface area and (ii) the total surface area of a hemisphere of radius 21 cm.
[fig]
Solution:
r = 21 cm
(i) CSA = 2 * pi * r^2 = 2 * (22/7) * 21 * 21 = 2772 cm^2
(ii) TSA = 3 * pi * r^2 = 3 * (22/7) * 21 * 21 = 4158 cm^2
Answer: (i) 2772 cm^2, (ii) 4158 cm^2
Question 5
Metal spheres, each of radius 2 cm, are packed into a rectangular box of internal dimensions 16 cm x 8 cm x 8 cm. When 16 spheres are packed, the box is filled with preservative liquid. Find the volume of this liquid. Round your answer to the nearest integer.
[fig]
Solution:
Volume of box = 16 * 8 * 8 = 1024 cm^3
Volume of 1 sphere = (4/3) * (22/7) * 2^3 approx 33.51 cm^3
Volume of 16 spheres = 16 * 33.51 approx 536.16 cm^3
Volume of liquid = 1024 – 536.16 = 487.84 cm^3
Answer: 488 cm^3
Question 6
The radius of a sphere is increased by 10%. Show that the volume increases by approximately 33.1%.
[fig]
Solution:
New radius r’ = 1.1r
New volume V’ = (4/3) * pi * (1.1r)^3 = 1.331 * V
Increase = 1.331V – V = 0.331V
Percentage = 33.1%
Question 7
The radius of a sphere is increased by x%. The volume of the sphere increases by 72.8%. Find the value of x.
[fig]
Solution:
V’ = 1.728V
(r’/r)^3 = 1.728
r’/r = 1.2
x = 20%
Answer: 20
Question 8
The hemispherical dome of a building needs to be painted. If the circumference of the base of the dome is 35.2 m, find the cost of painting it, given the cost of painting is Rupees 10 per 100 cm^2.
[fig]
Solution:
2 * pi * r = 35.2 => r = 5.6 m = 560 cm
SA = 2 * pi * r^2 = 2 * (22/7) * 560^2 = 1,971,200 cm^2
Cost = (1,971,200 / 100) * 10 = Rupees 1,97,120
Answer: Rupees 1,97,120
END-OF-CHAPTER EXERCISES
Question 1
Estimate how many scoops of ice cream can be obtained from a cuboidal container of dimensions 10 cm x 15 cm x 20 cm.
[fig]
Solution:
Container Volume = 3000 cm^3
Assume scoop radius = 2.5 cm => Volume approx 65.4 cm^3
Number of scoops = 3000 / 65.4 approx 45.8
Answer: Approx 46 scoops
Question 2
A cube of integer side length a is made of unit cubes. Write an expression giving the number of unit cubes to be added to make a cube of side length a + 1.
Solution:
Added cubes = (a + 1)^3 – a^3
= a^3 + 3a^2 + 3a + 1 – a^3 = 3a^2 + 3a + 1
Answer: 3a^2 + 3a + 1
Question 3
(i) Could a person drink enough water in a lifetime to fill an entire room? (ii) Estimate bricks in a classroom.
Solution:
(i) Classroom approx 126,000 litres. Lifetime consumption approx 73,000 litres. No.
(ii) Wall volume approx 14 m^3. Brick volume approx 0.002 m^3. Total approx 7000 bricks.
Question 4
Option A: 1 cube (50 mm side). Option B: 100 cubes (10 mm side). Which is more?
Solution:
V_A = 50^3 = 125,000 mm^3. V_B = 100 * 10^3 = 100,000 mm^3.
Answer: Option A
Question 5
If all people crowd in one location, how much area?
Solution:
8 billion people * 0.25 m^2/person = 2 * 10^9 m^2 = 2000 km^2.
Question 6
Cylindrical glasses (r=3.5, h=12). Litres for 1600 students?
Solution:
V_glass = pi * 3.5^2 * 12 = 462 cm^3
Total = 1600 * 462 = 739,200 cm^3 = 739.2 litres.
Answer: 739.2 litres
Question 7
SA of sphere (r=5) is 5 times CSA of cone (r=4). Find height and volume of cone.
Solution:
100*pi = 5 * (pi * 4 * l) => l = 5 cm
h = sqrt(5^2 – 4^2) = 3 cm
V = (1/3) * pi * 4^2 * 3 = 16*pi approx 50.24 cm^3
Question 8
Ratios of volumes of Jupiter/Earth and Sun/Earth.
Solution:
(i) (69900/6370)^3 approx 1321
(ii) (695700/6370)^3 approx 1,302,600
Question 9
Show sphere volume is 2/3 of enclosing cylinder.
Solution:
V_sphere = (4/3)pir^3. V_cyl = pir^2*(2r) = 2pi*r^3. Ratio = 4/6 = 2/3.
Question 10
String 1 m longer than Earth’s equator. Height gap?
Solution:
2pi(R+h) – 2piR = 1 => 2pih = 1 => h = 1/(2*pi) approx 15.9 cm. Constant for any sphere.
Question 11
Square pyramid in cylinder.
Solution:
Base area = 2r^2. V_pyr = (2/3)r^2h. V_cyl = pir^2h. Ratio = 2 : 3pi.
Question 12
Change in volume: (i) Cuboid (l+1) => wh. (ii) Cylinder (r+1) => 2pirh + pih. (iii) Sphere (r-1) => (4/3)pi(3r^2 – 3r + 1).
Question 13
S in terms of V.
Answer: S = 6 * V^(2/3)
Question 14
V in terms of S.
Answer: V = (S/6)^(3/2)
Question 15
Marbles (r=1) to raise water 4 cm in cylinder (R=4).
Solution:
V_needed = pi * 16 * 4 = 64*pi. V_marble = (4/3)pi. Number = 64 / (4/3) = 48.
Question 16
Ink in pen.
Answer: Approx 0.31 mL.
Question 17
Chapatis from 6 cm dough ball?
Answer: Approx 36 chapatis.
Question 18
Textbook pages vs floor.
Answer: Floor is larger. Book pages approx 8.8 m^2 vs floor 42 m^2.
Question 19
Giant cube for humanity.
Answer: Side approx 824 m.
Question 20
Water layer on Earth.
Answer: Approx 2.71 km.
Question 21
Cuboid dimensions (Volume halved, SA doubled).
Answer: Example: A 4 cm cube changing to approx 0.68 cm x 0.68 cm x 69.6 cm.
Question 22
Project: House volume.
Solution: Calculated by sum of section volumes (LWH). Example total: 294 m^3.
