Ex 14.1 Class 9 Maths Ganita Manjari Solutions | Sid Classes
Welcome to Sid Classes, your trusted destination for comprehensive academic resources! In this detailed study guide, we present complete, step-by-step solutions for the NCERT Class 9 Maths Ganita Manjari Chapter 14 Exercise 14.1 Surface Areas and Volumes of Cubes and Cuboids. Designed to reflect how top-scoring students write their solutions and reasoning in school examinations, this walkthrough covers volume-to-surface-area conversions, packing densities, godown storage capacities, joined cube formations, surface-to-volume ratio changes, corner face products, painted cube face counts, and integer-dimension cuboid combinations entirely clear.
🔑 Key Formulas & Principles:
- Cube Formulas: Volume \(V = a^3\), Total Surface Area \(\text{TSA} = 6a^2\) (where \(a\) is the side length).
- Cuboid Formulas: Volume \(V = l \times b \times h\), Total Surface Area \(\text{TSA} = 2(lb + bh + hl)\) (where \(l, b, h\) are length, breadth, and height).
NCERT Class 9 Maths Ganita Manjari Chapter 14 Exercise 14.1 Solutions
Question 1
The volume of a cube is \(64\text{ cm}^3\). What is its total surface area?
Solution:
1. Let the side length of the cube be \(a\).
2. Given volume \(V = 64\text{ cm}^3\):
\(a^3 = 64 \implies a = \sqrt[3]{64} = 4\text{ cm}\).
3. Total surface area of the cube is given by \(6a^2\):
\(\text{TSA} = 6(4)^2 = 6(16) = \mathbf{96\text{ cm}^2}\).
Question 2
How many small cubes with side \(20\text{ cm}\) can be packed tight in a cubical box with side \(2\text{ m}\)?
Solution:
1. Convert the dimensions of the large cubical box into centimetres:
Side of the large box = \(2\text{ m} = 2 \times 100\text{ cm} = 200\text{ cm}\).
2. Side of each small cube = \(20\text{ cm}\).
3. Number of small cubes along each edge = \(\frac{200}{20} = 10\).
4. Total number of small cubes that can be packed = \(10 \times 10 \times 10 = \mathbf{1000}\).
Question 3
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.
Solution:
1. Volume of the godown = \(40 \times 25 \times 10 = 10000\text{ m}^3\).
2. Volume of one cuboidal box = \(2 \times 1.25 \times 1 = 2.5\text{ m}^3\).
3. Number of boxes that can fit in the godown = \(\frac{\text{Volume of godown}}{\text{Volume of one box}} = \frac{10000}{2.5} = \mathbf{4000}\).
Question 4
Two cubes each of volume \(125\text{ cm}^3\) are joined end to end. Find the surface area of the resulting cuboid.
Solution:
1. Volume of each cube = \(125\text{ cm}^3 \implies \text{side } a = \sqrt[3]{125} = 5\text{ cm}\).
2. When two such cubes are joined end to end, the dimensions of the resulting cuboid become:
- Length (\(l\)) = \(5 + 5 = 10\text{ cm}\)
- Breadth (\(b\)) = \(5\text{ cm}\)
- Height (\(h\)) = \(5\text{ cm}\)
3. Surface area of the resulting cuboid = \(2(lb + bh + hl)\):
\(\text{SA} = 2(10 \times 5 + 5 \times 5 + 5 \times 10) = 2(50 + 25 + 50) = 2(125) = \mathbf{250\text{ cm}^2}\).
Question 5
A cube of side \(4\text{ cm}\) is cut into cubes of side \(1\text{ cm}\). What is the ratio of the surface areas of the original cube and all the cut-out cubes?
Solution:
1. Surface area of the original cube (side \(4\text{ cm}\)) = \(6(4)^2 = 6(16) = 96\text{ cm}^2\).
2. Number of \(1\text{ cm}\) cubes formed = \(\frac{4^3}{1^3} = \frac{64}{1} = 64\text{ cubes}\).
3. Surface area of one \(1\text{ cm}\) cube = \(6(1)^2 = 6\text{ cm}^2\).
4. Total surface area of all 64 cut-out cubes = \(64 \times 6 = 384\text{ cm}^2\).
5. Ratio of surface area of the original cube to all cut-out cubes = \(\frac{96}{384} = \mathbf{1 : 4}\).
Question 6
The surface areas of the three faces of a cuboid that meet at one of the corners of the cuboid are \(6\text{ cm}^2\), \(15\text{ cm}^2\), and \(10\text{ cm}^2\) respectively. What is the volume of the cuboid?
Solution:
1. Let the dimensions of the cuboid be \(l, b,\) and \(h\). The areas of the three adjacent faces meeting at a corner are:
\(lb = 6\), \(bh = 15\), and \(hl = 10\).
2. Multiplying the three equations together:
\((lb) \times (bh) \times (hl) = 6 \times 15 \times 10\)
\((lbh)^2 = 900\)
3. Volume (\(V = lbh\)) = \(\sqrt{900} = \mathbf{30\text{ cm}^3}\).
Question 7
A cube of side \(5\text{ cm}\) is painted on all its faces. If it is sliced into \(1\text{ cm}^3\) cubes, how many of these \(1\text{ cm}^3\) cubes have:
(i) exactly three faces painted?
(ii) exactly two faces painted?
(iii) exactly one face painted?
(iv) no face painted?
Solution:
* (i) Exactly three faces painted: These are the corner cubes of the large cube. A cube has 8 corners, so there are \(\mathbf{8}\) such cubes.
* (ii) Exactly two faces painted: These lie on the edges between the corners. Each of the 12 edges has \((5 - 2) = 3\) cubes with two faces painted. Total = \(12 \times 3 = \mathbf{36}\).
* (iii) Exactly one face painted: These lie in the center of each face. Each of the 6 faces has \((5 - 2)^2 = 3^2 = 9\) cubes. Total = \(6 \times 9 = \mathbf{54}\).
* (iv) No face painted: These form the inner core of the cube. The number of unpainted interior cubes is \((5 - 2)^3 = 3^3 = \mathbf{27}\).
Question 8
Find a cuboid with edges whose lengths are integers (in cm), given that it has a total surface area of exactly \(100\text{ cm}^2\).
(i) Is there more than one such cuboid?
(ii) Can you find them all?
(iii) Show that you have found them all.
Solution:
1. Total surface area of a cuboid is given by \(2(lb + bh + hl) = 100\), which simplifies to:
\(lb + bh + hl = 50\).
2. Rewriting this equation by completing terms with height \(h\):
\((l + h)(b + h) = 50 + h^2\).
3. Testing integer values for \(h\):
- For \(h = 1\): \((l + 1)(b + 1) = 50 + 1 = 51 = 3 \times 17 \implies l = 2, b = 16\) (Dimensions: \(\mathbf{16\text{ cm} \times 2\text{ cm} \times 1\text{ cm}}\)).
- For \(h = 2\): \((l + 2)(b + 2) = 50 + 4 = 54 = 6 \times 9 \implies l = 4, b = 7\) (Dimensions: \(\mathbf{7\text{ cm} \times 4\text{ cm} \times 2\text{ cm}}\)).
- For \(h = 4\): \((l + 4)(b + 4) = 50 + 16 = 66 = 6 \times 11 \implies l = 2, b = 7\) (Dimensions: \(\mathbf{7\text{ cm} \times 2\text{ cm} \times 4\text{ cm}}\)).
- For \(h = 7\): \((l + 7)(b + 7) = 50 + 49 = 99 = 9 \times 11 \implies l = 2, b = 4\) (Dimensions: \(\mathbf{4\text{ cm} \times 2\text{ cm} \times 7\text{ cm}}\)).
4. Conclusion: Yes, there is more than one cuboid. The unique set of integer-dimension cuboids satisfying the condition are permutations of \((16, 2, 1)\) and \((7, 4, 2)\).
Frequently Asked Questions (FAQs)
Q1: How does cutting a large cube into smaller cubes affect total surface area?
A: While the total volume remains constant, cutting a solid into smaller pieces exposes new interior surfaces, significantly increasing the total surface area.
Q2: What is the relation between the dimensions of adjacent faces meeting at a corner of a cuboid?
A: The product of the areas of the three adjacent faces meeting at a corner is equal to the square of the volume of the cuboid (\((lbh)^2 = (area_1)(area_2)(area_3)\)).
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