NCERT Solutions Class 10 Maths Chapter 12: Surface Areas and Volumes

Q: Two cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.

Given volume of one cube = 64 cm³, the edge is 4 cm. When joined, the resulting cuboid has Length = 8 cm, Breadth = 4 cm, Height = 4 cm. The Surface Area = 2(lb + bh + hl) = 160 cm².

Q: 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.

Radius = 3.5 cm. Height of cone = 15.5 - 3.5 = 12 cm. Slant height = 12.5 cm. Total Surface Area = CSA of cone + CSA of hemisphere = 214.5 cm².

Q: 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.

The greatest diameter is equal to the side of the cube, which is 7 cm. The total surface area of the solid is 332.5 cm².

Q: A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

Radius = 2.5 mm. Height of cylindrical part = 14 - 5 = 9 mm. Surface Area = CSA of cylinder + 2 × CSA of hemisphere = 220 mm².

Q: A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent.

Radius = 2 m. Area of canvas = CSA of cylinder + CSA of cone = 44 m². The cost of canvas at ₹ 500 per m² is ₹ 22,000.

Q: From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm².

Total Surface Area = CSA of cylinder + CSA of cone + Area of base = 17.6 cm². Rounded to the nearest cm², it is 18 cm².

Detailed solutions for Exercise 12.1 (Surface Area of Combination of Solids).

Ex 12.1 Ex 12.2

#### **Q1. Two cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.** #### **A1. Solution:** Given: Volume of one cube = $64$ cm³. Let edge of the cube be $a$. $a^3 = 64 \Rightarrow a = \sqrt[3]{64} = 4$ cm. When two cubes are joined end to end, a cuboid is formed. Dimensions of the cuboid: Length ($l$) = $4 + 4 = 8$ cm. Breadth ($b$) = $4$ cm. Height ($h$) = $4$ cm. Surface Area of cuboid = $2(lb + bh + hl)$ $= 2(8 \times 4 + 4 \times 4 + 4 \times 8)$ $= 2(32 + 16 + 32)$ $= 2(80) = 160$ cm². #### **Q2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.** #### **A2. Solution:** Given: Diameter of hemisphere = 14 cm $\Rightarrow$ Radius ($r$) = 7 cm. Total height of vessel = 13 cm. Height of cylinder ($h$) = Total height $-$ Radius of hemisphere $h = 13 - 7 = 6$ cm. Radius of cylinder ($r$) = 7 cm. Inner Surface Area = CSA of cylinder + CSA of hemisphere $= 2\pi rh + 2\pi r^2$ $= 2\pi r(h + r)$ $= 2 \times \frac{22}{7} \times 7 \times (6 + 7)$ $= 44 \times 13 = 572$ 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.** #### **A3. Solution:** Given: Radius ($r$) = 3.5 cm = $\frac{7}{2}$ cm. Total height of toy = 15.5 cm. Height of cone ($h$) = Total height $-$ Radius of hemisphere $h = 15.5 - 3.5 = 12$ cm. Slant height of cone ($l$) = $\sqrt{r^2 + h^2}$ $= \sqrt{(3.5)^2 + (12)^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5$ cm. Total Surface Area = CSA of cone + CSA of hemisphere $= \pi rl + 2\pi r^2 = \pi r(l + 2r)$ $= \frac{22}{7} \times 3.5 \times (12.5 + 2 \times 3.5)$ $= \frac{22}{7} \times \frac{7}{2} \times (12.5 + 7)$ $= 11 \times 19.5 = 214.5$ 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.** #### **A4. Solution:** Given: Side of cubical block ($l$) = 7 cm. The greatest diameter the hemisphere can have is equal to the side of the cube. Diameter ($d$) = 7 cm $\Rightarrow$ Radius ($r$) = $\frac{7}{2}$ cm. Total Surface Area (TSA) of the solid = TSA of cube $-$ Area of base of hemisphere + CSA of hemisphere. $= 6l^2 - \pi r^2 + 2\pi r^2$ $= 6l^2 + \pi r^2$ $= 6(7)^2 + \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}$ $= 6(49) + \frac{77}{2}$ $= 294 + 38.5 = 332.5$ cm². #### **Q5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter $l$ of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.** #### **A5. Solution:** Given: Edge of cube = $l$. Diameter of hemisphere = $l$ $\Rightarrow$ Radius ($r$) = $\frac{l}{2}$. Surface Area of remaining solid = TSA of cube $-$ Area of circular base of hemisphere + CSA of hemisphere. $= 6l^2 - \pi r^2 + 2\pi r^2$ $= 6l^2 + \pi r^2$ $= 6l^2 + \pi \left(\frac{l}{2}\right)^2$ $= 6l^2 + \frac{\pi l^2}{4}$ $= \frac{l^2}{4} (24 + \pi)$ units². #### **Q6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.** #### **A6. Solution:** Given: Total length = 14 mm. Diameter = 5 mm $\Rightarrow$ Radius ($r$) = 2.5 mm. Height of cylinder ($h$) = Total length $-$ $2 \times$ Radius of hemisphere $h = 14 - (2.5 + 2.5) = 14 - 5 = 9$ mm. Surface Area = CSA of cylinder + $2 \times$ CSA of hemisphere $= 2\pi rh + 2(2\pi r^2)$ $= 2\pi r (h + 2r)$ $= 2 \times \frac{22}{7} \times 2.5 \times (9 + 2 \times 2.5)$ $= \frac{44}{7} \times 2.5 \times (9 + 5)$ $= \frac{110}{7} \times 14 = 220$ mm². #### **Q7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of ₹ 500 per m². (Note that the base of the tent will not be covered with canvas.)** #### **A7. Solution:** Given: Cylinder: Radius ($r$) = 2 m, Height ($h$) = 2.1 m. Cone: Radius ($r$) = 2 m, Slant height ($l$) = 2.8 m. Area of canvas = CSA of cylinder + CSA of cone $= 2\pi rh + \pi rl$ $= \pi r (2h + l)$ $= \frac{22}{7} \times 2 \times (2 \times 2.1 + 2.8)$ $= \frac{44}{7} \times (4.2 + 2.8)$ $= \frac{44}{7} \times 7 = 44$ m². Cost of canvas = Area $\times$ Rate $= 44 \times 500 = \text{₹ } 22,000$. #### **Q8. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm².** #### **A8. Solution:** Given: Height ($h$) = 2.4 cm, Diameter = 1.4 cm $\Rightarrow$ Radius ($r$) = 0.7 cm. Slant height ($l$) = $\sqrt{r^2 + h^2} = \sqrt{(0.7)^2 + (2.4)^2} = \sqrt{0.49 + 5.76} = \sqrt{6.25} = 2.5$ cm. TSA of remaining solid = CSA of cylinder + CSA of cone + Area of cylindrical base (top is hollow). $= 2\pi rh + \pi rl + \pi r^2$ $= \pi r (2h + l + r)$ $= \frac{22}{7} \times 0.7 \times (2 \times 2.4 + 2.5 + 0.7)$ $= 2.2 \times (4.8 + 2.5 + 0.7)$ $= 2.2 \times 8 = 17.6$ cm². Nearest cm² $\approx 18$ cm². #### **Q9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure. 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.** #### **A9. Solution:** Given: Height of cylinder ($h$) = 10 cm, Radius ($r$) = 3.5 cm. 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.