Chapter 12 — Surface Areas and Volumes

Solution:

Radius of cone and hemisphere:

\[ r = 3.5\ \text{cm} \]

Total height of toy = \(15.5\ \text{cm}\)

Height of cone:

\[ \begin{aligned} h &= 15.5 - 3.5 \\ h &= 12\ \text{cm} \end{aligned} \]

Slant height of cone:

\[ \begin{aligned} l &= \sqrt{r^2 + h^2} \\ &= \sqrt{(3.5)^2 + (12)^2} \\ &= \sqrt{12.25 + 144} \\ &= \sqrt{156.25} \\ &= 12.5\ \text{cm} \end{aligned} \]

Total surface area = CSA of cone + CSA of hemisphere

\[ \begin{aligned} \text{TSA} &= \pi r l + 2\pi r^2 \end{aligned} \] \[ \begin{aligned} &= \pi r (l + 2r) \end{aligned} \] \[ \begin{aligned} &= \frac{22}{7} \times 3.5 \times (12.5 + 7) \end{aligned} \] \[ \begin{aligned} &= \frac{22}{7} \times 3.5 \times 19.5 \end{aligned} \] \[ \begin{aligned} &= 22 \times 0.5 \times 19.5 \ &= 11 \times 19.5 \ &= 214.5\ \text{cm}^2 \end{aligned} \]

Final Answer: \(214.5\ \text{cm}^2\)

Solution:

Side of cube:

\[ a = 7\ \text{cm} \]

The largest hemisphere that can fit on the cube will have diameter equal to the side of cube:

\[ \text{Diameter} = 7\ \text{cm} \] \[ \begin{aligned} r &= \frac{7}{2} \\ r &= 3.5\ \text{cm} \end{aligned} \]

Surface area of solid:

= Surface area of cube + Curved surface area of hemisphere − Area of circular base

\[ \begin{aligned} \text{SA} &= 6a^2 + 2\pi r^2 - \pi r^2 \end{aligned} \] \[ \begin{aligned} &= 6a^2 + \pi r^2 \end{aligned} \]

Substituting values:

\[ \begin{aligned} &= 6 \times (7)^2 + \frac{22}{7} \times (3.5)^2 \end{aligned} \] \[ \begin{aligned} &= 6 \times 49 + \frac{22}{7} \times 12.25 \end{aligned} \] \[ \begin{aligned} &= 294 + 22 \times 1.75 \end{aligned} \] \[ \begin{aligned} &= 294 + 38.5 \\ &= 332.5\ \text{cm}^2 \end{aligned} \]

Final Answer: \(332.5\ \text{cm}^2\)

Solution:

Edge of cube:

\[ l \]

Diameter of hemisphere = \(l\)

\[ \begin{aligned} r &= \frac{l}{2} \end{aligned} \]

Surface area of remaining solid:

= Surface area of cube − Area of circular portion removed + Curved surface area of hemisphere

\[ \begin{aligned} \text{SA} &= 6l^2 - \pi r^2 + 2\pi r^2 \end{aligned} \] \[ \begin{aligned} &= 6l^2 + \pi r^2 \end{aligned} \]

Substituting \(r = \frac{l}{2}\):

\[ \begin{aligned} \text{SA} &= 6l^2 + \pi \left(\frac{l}{2}\right)^2 \end{aligned} \] \[ \begin{aligned} &= 6l^2 + \pi \cdot \frac{l^2}{4} \end{aligned} \]

Taking LCM:

\[ \begin{aligned} &= \frac{24l^2}{4} + \frac{\pi l^2}{4} \end{aligned} \] \[ \begin{aligned} &= \frac{l^2}{4}(24 + \pi) \end{aligned} \]

Final Answer: \(\frac{l^2}{4}(24 + \pi)\)

Solution:

Diameter of capsule = \(5\ \text{mm}\)

\[ \begin{aligned} r &= \frac{5}{2} \\ r &= 2.5\ \text{mm} \end{aligned} \]

Total length of capsule = \(14\ \text{mm}\)

Height of cylindrical part:

\[ \begin{aligned} h &= 14 - (2.5 + 2.5) \\ h &= 14 - 5 \\ h &= 9\ \text{mm} \end{aligned} \]

Total surface area:

\[ \begin{aligned} \text{TSA} &= 2\pi rh + 4\pi r^2 \end{aligned} \] \[ \begin{aligned} &= 2\pi r (h + 2r) \end{aligned} \] \[ \begin{aligned} &= 2 \times \frac{22}{7} \times 2.5 \times (9 + 5) \end{aligned} \] \[ \begin{aligned} &= 2 \times \frac{22}{7} \times 2.5 \times 14 \end{aligned} \] \[ \begin{aligned} &= 2 \times 22 \times 2.5 \times 2 \end{aligned} \] \[ \begin{aligned} &= 44 \times 5 \\ &= 220\ \text{mm}^2 \end{aligned} \]

Final Answer: \(220\ \text{mm}^2\)

Solution:

Diameter of cylindrical part = \(4\ \text{m}\)

\[ \begin{aligned} r &= \frac{4}{2} \\ r &= 2\ \text{m} \end{aligned} \]

Height of cylindrical part:

\[ h = 2.1\ \text{m} \]

Slant height of cone:

\[ l = 2.8\ \text{m} \]

Area of canvas required:

= Curved surface area of cylinder + Curved surface area of cone

\[ \begin{aligned} \text{Area} &= 2\pi rh + \pi rl \end{aligned} \] \[ \begin{aligned} &= \pi r (2h + l) \end{aligned} \] \[ \begin{aligned} &= \frac{22}{7} \times 2 \times (2 \times 2.1 + 2.8) \end{aligned} \] \[ \begin{aligned} &= \frac{22}{7} \times 2 \times (4.2 + 2.8) \end{aligned} \] \[ \begin{aligned} &= \frac{22}{7} \times 2 \times 7 \end{aligned} \] \[ \begin{aligned} &= 44\ \text{m}^2 \end{aligned} \]

Cost of canvas:

\[ \begin{aligned} \text{Cost} &= 44 \times 500 \\ &= 22000 \end{aligned} \]

Final Answer: Area = \(44\ \text{m}^2\), Cost = \(\text{₹ }22000\)

Solution:

Diameter of cylinder = \(1.4\ \text{cm}\)

\[ \begin{aligned} r &= \frac{1.4}{2} \\ r &= 0.7\ \text{cm} \end{aligned} \]

Height:

\[ h = 2.4\ \text{cm} \]

Slant height of cone:

\[ \begin{aligned} l &= \sqrt{r^2 + h^2} \\ &= \sqrt{(0.7)^2 + (2.4)^2} \\ &= \sqrt{0.49 + 5.76} \\ &= \sqrt{6.25} \\ &= 2.5\ \text{cm} \end{aligned} \]

Total surface area:

\[ \begin{aligned} \text{TSA} &= 2\pi rh + \pi rl + \pi r^2 \end{aligned} \] \[ \begin{aligned} &= \pi r (2h + l + r) \end{aligned} \] \[ \begin{aligned} &= \frac{22}{7} \times 0.7 \times (2 \times 2.4 + 2.5 + 0.7) \end{aligned} \] \[ \begin{aligned} &= \frac{22}{7} \times 0.7 \times (4.8 + 2.5 + 0.7) \end{aligned} \] \[ \begin{aligned} &= \frac{22}{7} \times 0.7 \times 8 \end{aligned} \] \[ \begin{aligned} &= 2.2 \times 8 \\ &= 17.6\ \text{cm}^2 \end{aligned} \]

Rounding to nearest \(\text{cm}^2\):

\[ \approx 18\ \text{cm}^2 \]

Final Answer: \(\approx 18\ \text{cm}^2\)

Solution:

Radius of cylinder:

\[ r = 3.5\ \text{cm} \]

Height:

\[ h = 10\ \text{cm} \]

Total surface area:

= Curved surface area of cylinder + Curved surface area of two hemispheres

\[ \begin{aligned} \text{TSA} &= 2\pi rh + 2 \times (2\pi r^2) \end{aligned} \] \[ \begin{aligned} &= 2\pi rh + 4\pi r^2 \end{aligned} \] \[ \begin{aligned} &= 2\pi r (h + 2r) \end{aligned} \]

Substituting values:

\[ \begin{aligned} &= 2 \times \frac{22}{7} \times 3.5 \times (10 + 2 \times 3.5) \end{aligned} \] \[ \begin{aligned} &= 2 \times \frac{22}{7} \times 3.5 \times (10 + 7) \end{aligned} \] \[ \begin{aligned} &= 2 \times 22 \times 0.5 \times 17 \end{aligned} \] \[ \begin{aligned} &= 22 \times 17 \\ &= 374\ \text{cm}^2 \end{aligned} \]

Final Answer: \(374\ \text{cm}^2\)