Ch 12: Surface Areas and Volumes (Part 1)
Based on the latest CBSE rationalized syllabus. Master the combinations of solids (SA & Volumes).
Exercise 12.1 Solutions
💡 Trick: TSA of Combined Solid = CSA of Part 1 + CSA of Part 2. Do NOT add their base areas if they are joined!
Q1. 2 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.
Solution:
Volume of one cube = $a^3 = 64 \implies a = 4 \text{ cm}$.
When joined end to end, a cuboid is formed with:
Length ($l$) = $4 + 4 = 8 \text{ cm}$
Breadth ($b$) = $4 \text{ cm}$
Height ($h$) = $4 \text{ cm}$
Surface Area = $2(lb + bh + hl) = 2(8\times4 + 4\times4 + 4\times8)$
= $2(32 + 16 + 32) = 2(80) =$ 160 cm².
Q3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Solution:
Radius ($r$) = $3.5 \text{ cm} = 7/2 \text{ cm}$. Total height = $15.5 \text{ cm}$.
Height of cone ($h$) = Total height - radius of hemisphere = $15.5 - 3.5 = 12 \text{ cm}$.
Slant height of cone ($l$) = $\sqrt{r^2 + h^2} = \sqrt{3.5^2 + 12^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = \mathbf{12.5 \text{ cm}}$.
TSA of toy = CSA of cone + CSA of hemisphere
= $\pi rl + 2\pi r^2 = \pi r (l + 2r)$
= $\frac{22}{7} \times \frac{7}{2} \times (12.5 + 2 \times 3.5)$
= $11 \times (12.5 + 7) = 11 \times 19.5 =$ 214.5 cm².
Q4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.
Solution:
Greatest diameter of hemisphere = side of cube = 7 cm.
Radius ($r$) = $7/2 = 3.5 \text{ cm}$.
TSA of solid = TSA of cube + CSA of hemisphere - Base area of hemisphere
= $6a^2 + 2\pi r^2 - \pi r^2 = 6a^2 + \pi r^2$
= $6(7)^2 + \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}$
= $6(49) + 11 \times 3.5$
= $294 + 38.5 =$ 332.5 cm².
Q9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.
Solution:
Radius ($r$) = $3.5 \text{ cm}$, Height ($h$) = $10 \text{ cm}$.
When hemispheres are scooped out, their inner curved surfaces become visible. So we ADD the areas, not subtract.
TSA of article = CSA of cylinder + 2 $\times$ CSA of hemisphere
= $2\pi rh + 2(2\pi r^2) = 2\pi r (h + 2r)$
= $2 \times \frac{22}{7} \times 3.5 \times (10 + 2 \times 3.5)$
= $22 \times (10 + 7) = 22 \times 17 =$ 374 cm².
Exercise 12.2 (Volumes)
💡 Trick: Volume is simple addition or subtraction. Total Volume = Vol 1 + Vol 2.
Q1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of $\pi$.
Solution:
Radius ($r$) = $1 \text{ cm}$, Height of cone ($h$) = $1 \text{ cm}$.
Total Volume = Volume of Cone + Volume of Hemisphere
= $\frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3$
= $\frac{1}{3}\pi (1)^2(1) + \frac{2}{3}\pi (1)^3$
= $\frac{1}{3}\pi + \frac{2}{3}\pi = \frac{3}{3}\pi =$ $\pi \text{ cm}^3$.
Q3. A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm.
Solution:
Diameter = $2.8 \text{ cm} \implies r = 1.4 \text{ cm}$. Total length = $5 \text{ cm}$.
Length of cylindrical part ($h$) = $5 - (1.4 + 1.4) = 5 - 2.8 = \mathbf{2.2 \text{ cm}}$.
Volume of 1 Gulab Jamun = Vol of cylinder + 2 $\times$ Vol of hemisphere
= $\pi r^2 h + 2(\frac{2}{3}\pi r^3) = \pi r^2 (h + \frac{4}{3}r)$
= $\frac{22}{7} \times 1.4 \times 1.4 \times (2.2 + \frac{4}{3} \times 1.4)$
= $6.16 \times (2.2 + 1.867) = 6.16 \times 4.067 \approx \mathbf{25.05 \text{ cm}^3}$.
Volume of 45 Gulab Jamuns = $45 \times 25.05 \approx \mathbf{1127.25 \text{ cm}^3}$.
Volume of syrup (30%) = $\frac{30}{100} \times 1127.25 \approx \mathbf{338.17 \text{ cm}^3}$.
Approximate syrup volume is 338 cm³.
Q4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood.
Solution:
Volume of Cuboid = $15 \times 10 \times 3.5 = \mathbf{525 \text{ cm}^3}$.
Volume of 1 conical depression = $\frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 0.5 \times 0.5 \times 1.4 = \mathbf{0.366 \text{ cm}^3}$.
Volume of 4 cones = $4 \times 0.366 = \mathbf{1.464 \text{ cm}^3}$.
Volume of wood = Volume of Cuboid - Volume of 4 cones
= $525 - 1.464 =$ 523.53 cm³ (approx).