Surface Areas and Volumes - Exercise 12.1

00:00
Revise Chapter Notes

Overview

Comprehensive step-by-step NCERT solutions for CBSE Class 10 Maths Exercise 12.1 on Surface Areas of Combinations of Solids, covering joined cubes, hemispheres mounted on cylinders, conical toys, hollowed cavities, canvas tents, and medicine capsules.

NCERT Exercise Practice questions with detailed step-by-step solutions

Back to Chapters
Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9
Question 1: Cubes Joined End to End
Two cubes each of volume $125\text{ cm}^3$ are joined end to end. Find the surface area of the resulting cuboid.
Cube 1 Cube 2
Step 1: Finding Edge of the Cube
Volume of each cube $= 125\text{ cm}^3$.
Let the side of the cube be $a$. $$a^3 = 125 \implies a = \sqrt[3]{125} = 5\text{ cm}$$
Step 2: Dimensions of Resulting Cuboid
When two cubes are joined end to end: $$\text{Length } L = a + a = 5 + 5 = 10\text{ cm}$$ $$\text{Breadth } B = 5\text{ cm}, \quad \text{Height } H = 5\text{ cm}$$
Step 3: Total Surface Area
$$\text{Surface Area} = 2(LB + BH + HL)$$ $$= 2(10 \times 5 + 5 \times 5 + 5 \times 10)$$ $$= 2(50 + 25 + 50) = 2(125) = 250\text{ cm}^2$$
$$\therefore \text{The surface area of the resulting cuboid is } 250\text{ cm}^2.$$
Question 2: Hollow Vessel
A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is $28\text{ cm}$ and the total height of the vessel is $26\text{ cm}$. Find the inner surface area of the vessel. (Use $\pi = \frac{22}{7}$)
Cylinder Hemisphere
Step 1: Geometry & Dimensions
Diameter of hemisphere $= 28\text{ cm} \implies$ Radius $r = \frac{28}{2} = 14\text{ cm}$.
Radius of cylinder is the same: $r = 14\text{ cm}$.
Total height of vessel $= 26\text{ cm}$.
Height of cylindrical part, $h = \text{Total height} - \text{Radius of hemisphere}$: $$h = 26 - 14 = 12\text{ cm}$$
Step 2: Inner Surface Area Calculation
$$\text{Inner Surface Area} = \text{CSA of Cylinder} + \text{CSA of Hemisphere}$$ $$= 2\pi rh + 2\pi r^2 = 2\pi r(h + r)$$ $$= 2 \times \frac{22}{7} \times 14 \times (12 + 14)$$ $$= 2 \times 22 \times 2 \times 26 = 88 \times 26 = 2288\text{ cm}^2$$
$$\therefore \text{The inner surface area of the vessel is } 2288\text{ cm}^2.$$
Question 3: Conical Toy on Hemisphere
A toy is in the form of a cone of radius $7\text{ cm}$ mounted on a hemisphere of same radius. The total height of the toy is $31\text{ cm}$. Find the total surface area of the toy. (Use $\pi = \frac{22}{7}$)
Step 1: Dimensions of the Cone & Hemisphere
Common radius $r = 7\text{ cm}$.
Total height of the toy $= 31\text{ cm}$.
Height of the conical portion, $h = 31 - r = 31 - 7 = 24\text{ cm}$.
Step 2: Slant Height of the Cone
$$l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25\text{ cm}$$
Step 3: Total Surface Area
$$\text{Total Surface Area} = \text{CSA of Cone} + \text{CSA of Hemisphere}$$ $$= \pi r l + 2\pi r^2 = \pi r(l + 2r)$$ $$= \frac{22}{7} \times 7 \times (25 + 2 \times 7) = 22 \times (25 + 14) = 22 \times 39 = 858\text{ cm}^2$$
$$\therefore \text{The total surface area of the toy is } 858\text{ cm}^2.$$
Question 4: Cubical Block Surmounted by Hemisphere
A cubical block of side $14\text{ cm}$ is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid. (Use $\pi = \frac{22}{7}$)
Step 1: Greatest Diameter of Hemisphere
The base of the hemisphere sits entirely on the square top face of the cube.
Therefore, the greatest diameter the hemisphere can have equals the side of the cube: $$\text{Greatest Diameter } d = 14\text{ cm} \implies \text{Radius } r = 7\text{ cm}$$
Step 2: Surface Area of Solid
The total surface area consists of the 6 faces of the cube, minus the circular base where the hemisphere rests, plus the curved dome of the hemisphere: $$\text{Surface Area} = 6a^2 - \pi r^2 + 2\pi r^2 = 6a^2 + \pi r^2$$ $$= 6(14)^2 + \frac{22}{7} \times (7)^2$$ $$= 6(196) + 154 = 1176 + 154 = 1330\text{ cm}^2$$
$$\text{Greatest Diameter} = 14\text{ cm}, \quad \text{Total Surface Area} = 1330\text{ cm}^2.$$
Question 5: Hemispherical Depression in Cube
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.
Step 1: Geometric Analysis
Edge of the cubical block $= l$.
Diameter of hemisphere $= l \implies$ Radius $r = \frac{l}{2}$.
When the cavity is scooped out, the top flat circle is removed, but the inner curved surface of the hemisphere is added.
Step 2: Algebraic Derivation
$$\text{Total Surface Area} = \text{TSA of Cube} - \text{Base Area of Hemisphere} + \text{CSA of Hemisphere}$$ $$= 6l^2 - \pi r^2 + 2\pi r^2 = 6l^2 + \pi r^2$$ Substitute $r = \frac{l}{2}$: $$= 6l^2 + \pi \left(\frac{l}{2}\right)^2 = 6l^2 + \frac{\pi l^2}{4}$$ Factoring out $\frac{l^2}{4}$: $$= \frac{l^2}{4}(24 + \pi)$$
$$\therefore \text{The surface area of the remaining solid is } \frac{l^2}{4}(24 + \pi)\text{ sq. units}.$$
Question 6: Medicine Capsule
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 $20\text{ mm}$ and the diameter of the capsule is $7\text{ mm}$. Find its surface area. (Use $\pi = \frac{22}{7}$)
Cylinder
Step 1: Breakdown of Components
Diameter $= 7\text{ mm} \implies$ Radius $r = \frac{7}{2} = 3.5\text{ mm}$.
Total length $= 20\text{ mm}$.
Length of cylindrical portion: $$h = 20 - 2r = 20 - (3.5 + 3.5) = 20 - 7 = 13\text{ mm}$$
Step 2: Surface Area Calculation
The surface area is the curved surface of the cylinder plus the two hemispherical caps: $$\text{Surface Area} = \text{CSA of Cylinder} + 2 \times (\text{CSA of Hemisphere})$$ $$= 2\pi rh + 2(2\pi r^2) = 2\pi r(h + 2r)$$ $$= 2 \times \frac{22}{7} \times 3.5 \times (13 + 7)$$ $$= 22 \times 20 = 440\text{ mm}^2$$
$$\therefore \text{The surface area of the capsule is } 440\text{ mm}^2.$$
Question 7: Canvas Tent
A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are $4\text{ m}$ and $6\text{ m}$ respectively, and the slant height of the top is $5\text{ m}$, find the area of the canvas used for making the tent. Also, find the cost of the canvas at the rate of $₹600\text{ per m}^2$. (Note that the base of the tent will not be covered with canvas).
Step 1: Dimensions of Tent Parts
Diameter of cylindrical part $= 6\text{ m} \implies$ Radius $r = 3\text{ m}$.
Height of cylindrical part, $h = 4\text{ m}$.
Slant height of conical top, $l = 5\text{ m}$.
Step 2: Area of Canvas
$$\text{Canvas Area} = \text{CSA of Cylinder} + \text{CSA of Cone}$$ $$= 2\pi rh + \pi rl = \pi r(2h + l)$$ $$= \frac{22}{7} \times 3 \times (2(4) + 5) = \frac{66}{7} \times (8 + 5) = \frac{66 \times 13}{7} = \frac{858}{7} \approx 122.57\text{ m}^2$$
Step 3: Total Cost
$$\text{Total Cost} = \text{Area} \times \text{Rate} = \frac{858}{7} \times 600 = \frac{514800}{7} \approx ₹73542.86$$
$$\text{Area of Canvas} = \frac{858}{7}\text{ m}^2 \approx 122.57\text{ m}^2, \quad \text{Total Cost} \approx ₹73543.$$
Question 8: Conical Cavity in Solid Cylinder
From a solid cylinder whose height is $12\text{ cm}$ and diameter $10\text{ cm}$, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid. (Use $\pi = \frac{22}{7}$)
Step 1: Geometry & Slant Height
Diameter $= 10\text{ cm} \implies$ Radius $r = 5\text{ cm}$.
Height of cylinder and cavity, $h = 12\text{ cm}$.
Slant height of conical cavity: $$l = \sqrt{r^2 + h^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\text{ cm}$$
Step 2: Surface Area of Remaining Solid
The surface area includes the curved surface of the cylinder, one flat circular base, and the inner conical curved surface: $$\text{TSA} = \text{CSA of Cylinder} + \text{Area of Base} + \text{CSA of Conical Cavity}$$ $$= 2\pi rh + \pi r^2 + \pi rl = \pi r(2h + r + l)$$ $$= \frac{22}{7} \times 5 \times (2(12) + 5 + 13)$$ $$= \frac{110}{7} \times (24 + 18) = \frac{110}{7} \times 42 = 110 \times 6 = 660\text{ cm}^2$$
$$\therefore \text{The total surface area of the remaining solid is } 660\text{ cm}^2.$$
Question 9: Wooden Article with Scooped Hemispheres
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder. If the height of the cylinder is $20\text{ cm}$, and its base is of radius $7\text{ cm}$, find the total surface area of the article. (Use $\pi = \frac{22}{7}$)
Step 1: Breakdown of Exposed Surfaces
Radius $r = 7\text{ cm}$, Height of cylinder $h = 20\text{ cm}$.
When two hemispheres are scooped out from the ends, the flat ends are gone, and two inner hemispherical surfaces are created.
Step 2: Total Surface Area
$$\text{Total Surface Area} = \text{CSA of Cylinder} + 2 \times (\text{CSA of Hemisphere})$$ $$= 2\pi rh + 2(2\pi r^2) = 2\pi r(h + 2r)$$ $$= 2 \times \frac{22}{7} \times 7 \times (20 + 2(7))$$ $$= 44 \times (20 + 14) = 44 \times 34 = 1496\text{ cm}^2$$
$$\therefore \text{The total surface area of the article is } 1496\text{ cm}^2.$$
Previous: Exercise 11.1 Next: Exercise 12.2