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The volume of a cube is 64 cm³. What is its total surface area?
Answer: Let the side of the cube be $a$.
Volume = $a^3 = 64 \Rightarrow a = 4$ cm.
The Total Surface Area (TSA) of a cube is $6a^2$.
TSA = $6(4^2) = 6(16) = 96$ cm².
How many small cubes with side 20 cm can be packed tight in a cubical box with side 2 m?
Answer: First, convert the side of the large box to cm: $2$ m = $200$ cm.
Volume of the large box = $(200)^3 = 8,000,000$ cm³.
Volume of one small cube = $(20)^3 = 8,000$ cm³.
Number of small cubes = $\frac{8,000,000}{8,000} = 1000$ cubes.
The dimensions of a godown are $40 \text{ m} \times 25 \text{ m} \times 10 \text{ m}$. If it is filled with cuboidal boxes, each of dimensions $2 \text{ m} \times 1.25 \text{ m} \times 1 \text{ m}$, then find the number of boxes.
Answer: Volume of the godown = $40 \times 25 \times 10 = 10,000$ m³.
Volume of one box = $2 \times 1.25 \times 1 = 2.5$ m³.
Number of boxes = $\frac{10,000}{2.5} = 4000$ boxes.
Two cubes each of volume 125 cm³ are joined end to end. Find the surface area of the resulting cuboid.
Answer: Volume of one cube = $a^3 = 125 \Rightarrow a = 5$ cm.
When joined end to end, the resulting cuboid has:
Length ($l$) = $5 + 5 = 10$ cm
Width ($w$) = $5$ cm
Height ($h$) = $5$ cm
Total Surface Area = $2(lw + wh + hl) = 2((10)(5) + (5)(5) + (5)(10)) = 2(50 + 25 + 50) = 2(125) = 250$ cm².
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?
Answer: Surface area of the original cube = $6(4^2) = 6(16) = 96$ cm².
Volume of the original cube = $4^3 = 64$ cm³, which yields 64 smaller cubes of 1 cm³ each.
Surface area of one 1 cm cube = $6(1^2) = 6$ cm².
Total surface area of all 64 smaller cubes = $64 \times 6 = 384$ cm².
Ratio = $96 : 384 = 1 : 4$.
The surface areas of the three faces of a cuboid that meet at one of the corners of the cuboid are 6 cm², 15 cm², and 10 cm² respectively. What is the volume of the cuboid?
Answer: Let the dimensions be $l, w, h$.
Given: $lw = 6$, $wh = 15$, $lh = 10$.
Multiply all three equations: $(lw)(wh)(lh) = 6 \times 15 \times 10$
$(lwh)^2 = 900$
Volume $V = lwh = \sqrt{900} = 30$ cm³.
A cube of side 5 cm is painted on all its faces. If it is sliced into 1 cm³ cubes, how many of these 1 cm³ cubes have:
(i) exactly three faces painted?
(ii) exactly two faces painted?
(iii) exactly one face painted?
(iv) no face painted?
Answer: Total smaller cubes = $5 \times 5 \times 5 = 125$.
(i) Exactly three faces painted (corners): 8 cubes.
(ii) Exactly two faces painted (edges, excluding corners): 12 edges $\times$ (5 - 2) = $12 \times 3 = $ 36 cubes.
(iii) Exactly one face painted (centers of faces): 6 faces $\times$ (3 $\times$ 3) = $6 \times 9 = $ 54 cubes.
(iv) No face painted (interior cubes): $(5 - 2) \times (5 - 2) \times (5 - 2) = 3 \times 3 \times 3 = $ 27 cubes.
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.
Answer: Let cylinder A have radius $r$ and height $h$. Cylinder B has radius $2r$ and height $h/2$.
Curved Surface Area (CSA):
$CSA_A = 2\pi rh$
$CSA_B = 2\pi (2r)(h/2) = 2\pi rh$
Ratio of CSA = $1 : 1$.
Volume (V):
$V_A = \pi r^2 h$
$V_B = \pi (2r)^2 (h/2) = \pi (4r^2)(h/2) = 2\pi r^2 h$
Ratio of Volume = $1 : 2$.
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 cm³)?
Answer: The largest cylinder that fits in a cube of edge $r$ will have a diameter equal to the edge length ($2R = r \Rightarrow R = r/2$) and a height equal to the edge length ($h = r$).
Volume of cylinder = $\pi R^2 h = \pi \left(\frac{r}{2}\right)^2 (r) = \pi \left(\frac{r^2}{4}\right) (r) = \frac{\pi r^3}{4}$ cm³.
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.
Answer: Let the original radius be $r$ and height be $h$. Volume $V = \pi r^2 h$.
New radius = $1.1r$. New height = $h \left(1 - \frac{x}{100}\right)$.
New Volume = $\pi (1.1r)^2 \left(h \left(1 - \frac{x}{100}\right)\right) = 1.21 \pi r^2 h \left(1 - \frac{x}{100}\right)$.
Since the volume is unchanged: $1 = 1.21 \left(1 - \frac{x}{100}\right)$.
$1 - \frac{x}{100} = \frac{1}{1.21} = \frac{100}{121}$
$\frac{x}{100} = 1 - \frac{100}{121} = \frac{21}{121}$
$x = \frac{2100}{121} \approx 17.36\%$.
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. ($\pi \approx \frac{22}{7}$)
Answer:
(i) Volume of the cube = $a^3 = 12^3 = 1728$ cm³.
(ii) Volume of one cylindrical rod = $\pi r^2 h = \frac{22}{7} \times 2^2 \times 12 = \frac{22}{7} \times 48 = \frac{1056}{7} \approx 150.85$ cm³.
(iii) Number of complete rods = $\frac{\text{Volume of cube}}{\text{Volume of one rod}} = \frac{1728}{150.85} \approx 11.45$. So, 11 complete cylindrical rods can be formed.
Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Answer: Slant height ($l$) = 21 m. Radius ($r$) = $24/2 = 12$ m.
Total Surface Area (TSA) = $\pi r(l + r)$
TSA = $\frac{22}{7} \times 12 \times (21 + 12) = \frac{22}{7} \times 12 \times 33 = \frac{8712}{7} \approx 1244.57$ m².
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.
Answer: Height ($h$) = 16 cm, Radius ($r$) = 12 cm.
First, find slant height ($l$): $l = \sqrt{h^2 + r^2} = \sqrt{16^2 + 12^2} = \sqrt{256 + 144} = \sqrt{400} = 20$ cm.
Curved Surface Area (CSA) = $\pi rl = 3.14 \times 12 \times 20 = 753.6$ cm² (or exactly $240\pi$ cm²).
Total Surface Area (TSA) = $\pi r(l + r) = 3.14 \times 12 \times (20 + 12) = 3.14 \times 12 \times 32 = 1205.76$ cm² (or exactly $384\pi$ cm²).
A cone has a height of 15 cm. If its volume is 1570 cm³, find the radius of the base.
Answer: Volume ($V$) = $\frac{1}{3}\pi r^2 h$.
$1570 = \frac{1}{3} \times 3.14 \times r^2 \times 15$
$1570 = 5 \times 3.14 \times r^2$
$1570 = 15.7 \times r^2$
$r^2 = 100 \Rightarrow r = 10$ cm.
A right triangle with sides 6 cm, 8 cm and 10 cm is rotated through 360° about the side of 8 cm. Find the volume and the curved surface area of the solid so formed.
Answer: Rotating the triangle about the 8 cm side forms a right circular cone with height ($h$) = 8 cm, base radius ($r$) = 6 cm, and slant height ($l$) = 10 cm.
Volume ($V$) = $\frac{1}{3}\pi r^2 h = \frac{1}{3} \times 3.14 \times 36 \times 8 = 301.44$ cm³ (or exactly $96\pi$ cm³).
Curved Surface Area (CSA) = $\pi rl = 3.14 \times 6 \times 10 = 188.4$ cm² (or exactly $60\pi$ cm²).
A ball bearing has a radius of 0.7 cm. Find its surface area.
Answer: A ball bearing is spherical. Surface Area = $4\pi r^2$.
SA = $4 \times \frac{22}{7} \times (0.7)^2 = 4 \times \frac{22}{7} \times 0.49 = 4 \times 22 \times 0.07 = 6.16$ cm².
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.
Answer: Since they are made of the same metal, their weights are directly proportional to their volumes. Ratio of volumes = $5920 / 740 = 8$.
Volume ratio = $(\text{radius ratio})^3 = \left(\frac{R_{large}}{R_{small}}\right)^3 = 8 \Rightarrow \frac{R_{large}}{R_{small}} = 2$.
The radius of the smaller sphere = $5 / 2 = 2.5$ cm.
The radius of the larger sphere = $2 \times 2.5 = 5$ cm.
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.
Answer: Let the diameter of the Earth be $D_E$. Then the diameter of the moon is $D_M = \frac{1}{4} D_E$. Their radii hold the same ratio: $r_M = \frac{1}{4} r_E$.
Ratio of Surface Areas = $\frac{4\pi r_M^2}{4\pi r_E^2} = \left(\frac{r_M}{r_E}\right)^2 = \left(\frac{1}{4}\right)^2 = \frac{1}{16}$.
The ratio is $1 : 16$.
Metal spheres, each of radius 2 cm, are packed into a rectangular box of internal dimensions $16 \text{ cm} \times 8 \text{ cm} \times 8 \text{ cm}$. 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.
Answer: Volume of the box = $16 \times 8 \times 8 = 1024$ cm³.
Volume of one sphere = $\frac{4}{3}\pi r^3 = \frac{4}{3} \times 3.14 \times (2^3) = \frac{4}{3} \times 3.14 \times 8 \approx 33.51$ cm³.
Volume of 16 spheres = $16 \times 33.51 \approx 536.16$ cm³.
Volume of preservative liquid = Volume of box - Volume of 16 spheres = $1024 - 536.16 = 487.84$ cm³.
Rounded to the nearest integer, the volume is 488 cm³.
You are given two options: (A) one chocolate cube of side 50 mm, (B) hundred chocolate cubes of side 10 mm. Which option gives you more chocolate?
Answer:
Option A Volume = $a^3 = 50^3 = 125,000$ mm³.
Option B Volume = $100 \times (10^3) = 100 \times 1000 = 100,000$ mm³.
Option A gives you more chocolate.
A school provides milk to its students in cylindrical glasses, each with a diameter of 7 cm. If each glass is filled with milk to a height of 12 cm, how many litres of milk are needed to serve 1600 students? Use 1000 cm³ = 1 litre.
Answer: Radius of the glass ($r$) = $7 / 2 = 3.5$ cm. Height ($h$) = 12 cm.
Volume of milk per student = $\pi r^2 h = \frac{22}{7} \times 3.5^2 \times 12 = \frac{22}{7} \times 12.25 \times 12 = 462$ cm³.
Total volume for 1600 students = $1600 \times 462 = 739,200$ cm³.
Volume in litres = $\frac{739,200}{1000} = 739.2$ litres.
The surface area of a sphere of radius 5 cm is five times the area of the curved surface of a cone of radius 4 cm. Find the height and the volume of the cone.
Answer: Surface area of sphere ($r_s=5$) = $4\pi (5^2) = 100\pi$.
Curved surface area (CSA) of cone ($r_c=4$) = $\pi(4)l$.
Given: $100\pi = 5 \times (4\pi l) \Rightarrow 100\pi = 20\pi l \Rightarrow l = 5$ cm.
Height of cone ($h$) = $\sqrt{l^2 - r_c^2} = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = 3$ cm.
Volume of the cone = $\frac{1}{3}\pi r_c^2 h = \frac{1}{3} \times \pi \times (4^2) \times 3 = 16\pi \approx 50.24$ cm³.
We have a cylinder with a base radius of r cm and height h cm. A square pyramid is fitted inside it. The square base of the pyramid lies on the base of the cylinder, its corners on the boundary of the cylinder. The apex of the pyramid lies on the top of the cylinder. Find the ratio of the volume of this pyramid to the volume of the cylinder.
Answer: Let the square base of the pyramid have side $a$. Its diagonal connects opposite corners on the circle's boundary, so the diagonal = $2r$.
By Pythagoras, $a^2 + a^2 = (2r)^2 \Rightarrow 2a^2 = 4r^2 \Rightarrow a^2 = 2r^2$. Thus, the area of the square base = $2r^2$.
The height of the pyramid is $h$.
Volume of the pyramid = $\frac{1}{3} \times \text{Area of base} \times \text{height} = \frac{1}{3}(2r^2)h = \frac{2}{3}r^2 h$.
Volume of the cylinder = $\pi r^2 h$.
Ratio = $\frac{\frac{2}{3}r^2 h}{\pi r^2 h} = \frac{2}{3\pi}$. Thus, the ratio is $2 : 3\pi$.
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