Ganit Prakash - Class-X - Let us work out 4
Let us work out 4 Solutions Cube and Cuboid
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📚 Learning Resources
Study GuideFor a Cuboid (Length \(l\), Breadth \(b\), Height \(h\)):
1. Volume = \(l \times b \times h\)
2. Total Surface Area = \(2(lb + bh + lh)\)
3. Length of Diagonal = \(\sqrt{l^2 + b^2 + h^2}\)
For a Cube (Side \(a\)):
1. Volume = \(a^3\)
2. Total Surface Area = \(6a^2\)
3. Length of Diagonal = \(\sqrt{3}a\)
Cuboid: A three-dimensional solid object which has six rectangular faces. It is also known as a rectangular parallelepiped. All angles are right angles.
Cube: A special case of a cuboid where the length, breadth, and height are all equal (\(l = b = h\)). All six faces are squares.
Dimensions: We measure these solids using three parameters: Length, Breadth, and Height. The space occupied by the object is its Volume, and the total area of all its faces is the Surface Area.
1. Unit Mismatch: Always ensure all dimensions (m, cm, dcm) are converted to the same unit before starting calculations.
2. Area vs Volume: Remember that Area is measured in square units (sq.m.) while Volume is in cubic units (cubic m. or litres).
3. Diagonal Calculation: Students often confuse the surface diagonal (\(\sqrt{l^2 + b^2}\)) with the space diagonal (\(\sqrt{l^2 + b^2 + h^2}\)). Always use the space diagonal for the "longest rod."
1. Draw Diagrams: For word problems involving holes, canals, or roads, draw a rough 3D sketch to visualize the dimensions.
2. Write Given Data: Always start your answer by listing: "Given: Length = ..., Breadth = ..., etc." to get step-marking.
3. Check Density: In problems involving weight (like tea or brass), first find the volume and then multiply by weight per unit volume.
4. Volume Displacement: For problems like digging a hole and spreading soil, remember: Volume of soil removed = Volume of soil spread on land.
📗 Let us work out 4
Q1 to Q234 Cuboid shaped things: 1. Brick, 2. Text-book, 3. Matchbox, 4. Almirah.
4 Cube shaped things: 1. Ludo dice, 2. Rubik's cube, 3. Sugar cube, 4. Ice cube (square shaped).
Referring to the cuboidal figure (ABCDEFGH) provided in the textbook image:
Surfaces (6): ABCD, EFGH, ABFE, DCGH, ADEH, BCGF.
Edges (12): AB, BC, CD, DA, EF, FG, GH, HE, AE, BF, CG, DH.
Vertices (8): A, B, C, D, E, F, G, H.
Given, Length (l) = 5m, Breadth (b) = 4m, and Height (h) = 3m.
The length of the longest rod is equal to the length of the diagonal of the room.
\(\text{Length of diagonal} = \sqrt{l^2 + b^2 + h^2}\)
\(= \sqrt{5^2 + 4^2 + 3^2} = \sqrt{25 + 16 + 9} = \sqrt{50}\) m.
\(= \sqrt{25 \times 2} = 5\sqrt{2}\) m.
Answer: The length of the longest rod is \(5\sqrt{2}\) m.
Let the length of each edge of the cube be \(a\) m.
Area of one surface = \(a^2 = 64\)
\(a = \sqrt{64} = 8\) m.
\(\text{Volume of the cube} = a^3 = 8^3 = 512\) cubic m.
Answer: The volume of the cube is 512 cubic m.
Breadth of the canal = 2 m.
Depth of the canal = 8 dcm. = \(\frac{8}{10}\) m. = 0.8 m.
Total quantity of soil extracted (Volume) = 240 cubic m.
Let the length of the canal be \(L\) m.
\(\text{Volume} = \text{Length} \times \text{Breadth} \times \text{Depth}\)
\(240 = L \times 2 \times 0.8\)
\(240 = 1.6L\)
\(L = \frac{240}{1.6} = \frac{2400}{16} = 150\) m.
Answer: The length of the canal is 150 m.
Let the length of each edge of the cube be \(a\) cm.
Length of the diagonal of a cube = \(\sqrt{3} \times a\)
According to the condition, \(\sqrt{3} \times a = 4\sqrt{3} \implies a = 4\) cm.
\(\text{Total surface area} = 6a^2 = 6 \times (4)^2 = 6 \times 16 = 96\) sq.cm.
Answer: The total surface area of the cube is 96 sq.cm.
A cube has 12 edges.
Let the length of each edge be \(a\) cm.
Sum of edges = \(12a = 60 \implies a = 5\) cm.
\(\text{Volume of the cube} = a^3 = 5^3 = 125\) cubic cm.
Answer: The volume of the cube is 125 cubic cm.
Let the length of each edge of the cube be \(a\) cm.
Total surface area (sum of 6 surfaces) = \(6a^2 = 216\)
\(a^2 = \frac{216}{6} = 36 \implies a = \sqrt{36} = 6\) cm.
\(\text{Volume} = a^3 = 6^3 = 216\) cubic cm.
Answer: The volume of the cube is 216 cubic cm.
Total volume of the parallelopiped = 432 cubic cm.
Since it is converted into two equal cubes, volume of each cube = \(\frac{432}{2} = 216\) cubic cm.
Let the length of each edge of the cube be \(a\) cm.
\(a^3 = 216 \implies a = \sqrt[3]{216} = 6\) cm.
Answer: The length of each edge of each cube is 6 cm.
Let the side of the original cube be \(a\) units.
Volume of original cube (\(V_1\)) = \(a^3\).
New side after 50% decrease = \(a - 50\% \text{ of } a = a - 0.5a = 0.5a = \frac{a}{2}\).
Volume of changed cube (\(V_2\)) = \((\frac{a}{2})^3 = \frac{a^3}{8}\).
Ratio of volumes = \(V_1 : V_2 = a^3 : \frac{a^3}{8} = 1 : \frac{1}{8} = 8 : 1\).
Answer: The ratio of the volumes of the original and changed cube is 8 : 1.
Let length (\(l\)) = \(3x\) cm, breadth (\(b\)) = \(2x\) cm, and height (\(h\)) = \(x\) cm.
Volume = \(l \times b \times h = (3x)(2x)(x) = 6x^3\).
According to condition: \(6x^3 = 384 \implies x^3 = \frac{384}{6} = 64 \implies x = \sqrt[3]{64} = 4\).
So, \(l = 3 \times 4 = 12\) cm, \(b = 2 \times 4 = 8\) cm, \(h = 4\) cm.
Total Surface Area = \(2(lb + bh + lh) = 2(12 \times 8 + 8 \times 4 + 12 \times 4)\)
\(= 2(96 + 32 + 48) = 2(176) = 352\) sq.cm.
Answer: 352 sq.cm.
Volume of the box = \(7.5 \times 6 \times 5.4 = 243\) cubic dcm.
Weight of tea = (Total weight) - (Weight of Empty box)
\(= 52.350 \text{ kg} - 3.75 \text{ kg} = 48.600 \text{ kg}\).
Weight of 1 cubic dcm tea = \(\frac{48.6}{243} = 0.2 \text{ kg} = 200 \text{ gm}\).
Answer: 200 gm.
Base is square, so Length = Breadth = \(x\) cm.
Thickness (Height) = \(1\) mm = \(0.1\) cm.
Volume of plate = \(x \times x \times 0.1 = 0.1x^2\) cubic cm.
Total weight = Volume \(\times\) Density \(\implies 4725 = 0.1x^2 \times 8.4\)
\(4725 = 0.84x^2 \implies x^2 = \frac{4725}{0.84} = 5625\)
\(x = \sqrt{5625} = 75\).
Answer: 75.
Let the depth of each hole be \(h\) m.
Volume of 1 hole = \(14 \times 8 \times h = 112h\) cubic m.
Volume of 30 holes = \(30 \times 112h = 3360h\).
Given total volume = \(2520 \implies 3360h = 2520\)
\(h = \frac{2520}{3360} = \frac{252}{336} = 0.75 \text{ m} = 7.5 \text{ dcm}\).
Answer: 0.75 m or 7.5 dcm.
Edge = \(1.2\) m = \(12\) dcm.
Total volume of tank = \((12)^3 = 1728\) cubic dcm (litres).
Water remaining = \(\frac{1}{3} \times 1728 = 576\) litres.
Water taken out = \(1728 - 576 = 1152\) litres.
Capacity of each bucket = \(\frac{1152}{64} = 18\) litres.
Answer: 18 litres.
Volume of packet = \(2.8 \times 1.5 \times 0.9 = 3.78\) cubic dcm = \(3780\) cubic cm.
One gross = \(12 \times 12 = 144\) matchboxes.
Volume of 1 matchbox = \(\frac{3780}{144} = 26.25\) cubic cm.
For one matchbox: \(5 \times 3.5 \times \text{height} = 26.25\)
\(17.5 \times \text{height} = 26.25 \implies \text{height} = \frac{26.25}{17.5} = 1.5\) cm.
Answer: Volume = 26.25 cc, Height = 1.5 cm.
Length (\(l\)) = \(2.1\) m = \(21\) dcm.
Breadth (\(b\)) = \(1.5\) m = \(15\) dcm.
Volume of poured water = \(630\) litres = \(630\) cubic dcm.
Let the increased depth be \(h\) dcm.
According to condition: \(21 \times 15 \times h = 630\)
\(315h = 630 \implies h = \frac{630}{315} = 2\) dcm.
Answer: 2 dcm or 0.2 metre.
Total area of field = \(20 \times 15 = 300\) sq.m.
Area of 4 holes = \(4 \times (4 \times 4) = 64\) sq.m.
Remaining area for spreading soil = \(300 - 64 = 236\) sq.m.
Volume of soil from 4 cubic holes = \(4 \times (4 \times 4 \times 4) = 256\) cubic m.
Let increased height be \(h\) m.
\(236 \times h = 256 \implies h = \frac{256}{236} = \frac{64}{59} = 1\frac{5}{59}\) m.
Answer: \(1\frac{5}{59}\) metre.
Volume of soil required for low land = \(48 \times 31.5 \times 0.65\) cubic m (since \(6.5\) dcm = \(0.65\) m).
Volume of soil = \(982.8\) cubic m.
Let depth of hole be \(d\) m.
Volume of hole = \(27 \times 18.2 \times d = 491.4d\).
\(491.4d = 982.8 \implies d = \frac{982.8}{491.4} = 2\) m.
Answer: 2 metre.
If the depth of the cuboidal pot would be 5 dcm, then let us calculate whether 1620 lit. oil can be kept or not in that pot.
Total oil = \(800 + 725 + 575 = 2100\) litres = \(2100\) cubic dcm.
Depth (\(h\)) = \(7\) cm = \(0.7\) dcm.
Let length (\(l\)) = \(4x\) dcm, breadth (\(b\)) = \(3x\) dcm.
\(4x \times 3x \times 0.7 = 2100 \implies 8.4x^2 = 2100 \implies x^2 = 250 \implies x = \sqrt{250} = 5\sqrt{10}\).
Wait, checking units: 2100 L = 2100 cubic dcm. \(h = 0.7\) dcm. Correct.
Length = \(20\sqrt{10}\) dcm, Breadth = \(15\sqrt{10}\) dcm.
Part 2: If depth = \(5\) dcm, Volume = Area \(\times\) Depth = \(3000 \times 5\) (if area was \(3000\)).
Using calculated Area = \(\frac{2100}{0.7} = 3000\) sq.dcm.
New Volume = \(3000 \times 5 = 15000\) litres. Yes, 1620 litres can be kept.
Answer: Length = \(20\sqrt{10}\) dcm, Breadth = \(15\sqrt{10}\) dcm. Yes, it can be kept.
If the breadth of the land would be more by 4 dcm, then let us calculate the depth of the tank to be made.
Total requirement = \(1200 + 1050 + 950 = 3200\) litres.
25% extra = \(3200 \times 1.25 = 4000\) litres = \(4000\) cubic dcm.
Land area = \(2.5 \text{m} \times 1.6 \text{m} = 25 \text{dcm} \times 16 \text{dcm} = 400\) sq.dcm.
Depth \(d = \frac{4000}{400} = 10\) dcm = \(1\) m.
If breadth increases by 4dcm: New breadth = \(16 + 4 = 20\) dcm.
New depth \(d' = \frac{4000}{25 \times 20} = \frac{4000}{500} = 8\) dcm = \(0.8\) m.
Answer: Original depth = 1 m; New depth = 0.8 m.
Weight of rice = \(880.5 - 115.5 = 765\) kg.
Inner volume = \(\frac{765}{1.5} = 510\) cubic dcm.
Inner height (\(h\)) = \(\frac{510}{12 \times 8.5} = \frac{510}{102} = 5\) dcm.
Outside dimensions (Thickness 5cm = 0.5 dcm):
\(L = 12 + 2(0.5) = 13\) dcm, \(B = 8.5 + 2(0.5) = 9.5\) dcm, \(H = 5 + 2(0.5) = 6\) dcm.
Outside Area = \(2(LB + BH + LH) = 2(13 \times 9.5 + 9.5 \times 6 + 13 \times 6)\)
\(= 2(123.5 + 57 + 78) = 2(258.5) = 517\) sq.dcm.
Cost = \(517 \times 1.50 = \text{₹ } 775.50\).
Answer: Inner height = 5 dcm; Cost = ₹ 775.50.
Volume of pond = \(20 \times 18.5 \times 3.2 = 1184\) cubic m = \(1184\) kilo litres.
Time required = \(\frac{1184}{160} = 7.4\) hours = \(7\) hrs \(24\) mins.
Depth in paddy field (\(h\)):
\(59.2 \times 40 \times h = 1184\)
\(2368 \times h = 1184 \implies h = \frac{1184}{2368} = 0.5\) m.
Answer: Time = 7 hrs 24 mins; Paddy field depth = 0.5 metre.
📗 24. (A) M.C.Q.
Multiple Choice(a) 4 cm.
(b) 5 cm.
(c) 3 cm.
(d) 6 cm.
\(\text{Volume} = \text{Base Area} \times \text{Height}\)
\(440 = 88 \times h \implies h = \frac{440}{88} = 5\) cm.
Answer: (b) 5 cm.
(a) 190
(b) 192
(c) 184
(d) 180
\(\text{Number of planks} = \frac{\text{Volume of hole}}{\text{Volume of one plank}}\)
\(= \frac{40 \times 12 \times 16}{5 \times 4 \times 2} = \frac{7680}{40} = 192\).
Answer: (b) 192
(a) 64 cubic metre
(b) 216 cubic metre
(c) 256 cubic metre
(d) 512 cubic metre
Let side be \(a\). Total surface area = \(6a^2 = 256 \implies a^2 = \frac{256}{6}\).
Note: Based on the "Hints: The number of surfaces is 6", let's re-examine. If the problem meant area of each surface is 256, then \(a = 16\). If it meant total surface area is 256, the numbers are not perfect cubes. Let's assume it meant each surface or there is a typo in text (often 216 is used). Using 216: \(6a^2 = 216 \implies a=6 \implies V=216\). Using 256 as total surface area doesn't match standard MCQ options. If side area is 256: \(a=16, V=4096\). Let's check option (d) 512: \(a^3=512 \implies a=8 \implies 6a^2 = 384\). Standard textbook correction: if surface area of 4 walls is 256, \(4a^2=256 \implies a=8 \implies a^3=512\).
Answer: (d) 512 cubic metre (Assuming area of 4 side surfaces is 256).
(a) 1 : 3
(b) 1 : 8
(c) 1 : 9
(d) 1 : 18
\(\frac{V_1}{V_2} = \frac{a_1^3}{a_2^3} = \frac{1}{27} \implies \frac{a_1}{a_2} = \sqrt[3]{\frac{1}{27}} = \frac{1}{3}\).
\(\text{Ratio of Surface Areas} = \frac{6a_1^2}{6a_2^2} = \left(\frac{1}{3}\right)^2 = \frac{1}{9}\).
Answer: (c) 1 : 9
(a) \(s = 6d^2 \)
(b) \(3s = 7d \)
(c) \(s^3 = d^2 \)
(d) \(d^2 = \frac{s}{2} \)
\(s = 6a^2\) and \(d = \sqrt{3}a \implies d^2 = 3a^2 \implies a^2 = \frac{d^2}{3}\).
Substitute into \(s\): \(s = 6 \times \frac{d^2}{3} = 2d^2 \implies d^2 = \frac{s}{2}\).
Answer: (d) \(d^2 = \frac{s}{2}\)
📗 24. (B) True or False
Statement CheckIf side becomes \(2a\), volume becomes \((2a)^3 = 8a^3\). This is 8 times the original, not 4.
Answer: False
[Hints: 1 acre = 100 sq.m., 1 hectre = 100 acre]
Area = \(2 \text{ hectare} = 2 \times 10,000 \text{ sq.m} = 20,000 \text{ sq.m.}\)
Height = \(5 \text{ cm} = 0.05 \text{ m.}\)
\(\text{Volume} = 20,000 \times 0.05 = 1000 \text{ cubic metre.}\)
Answer: True
📗 24. (C) Fill in the blanks
Completion(i) The number of diagonals of a cuboid is 4.
(ii) The length of the diagonal on a surface of a cube = \(\sqrt{2}\) \(\times\) the length of one edge.
(iii) If the length, breadth and height of a rectengular parallelopiped are equal, then the special name of this solid is – Cube.
📗 25. S.A.
Short AnswerFor a cuboid:
- Number of surfaces (\(x\)) = 6
- Number of edges (\(y\)) = 12
- Number of vertices (\(z\)) = 8
- Number of diagonals (\(p\)) = 4
\(\text{Value} = x - y + z + p\)
\(= 6 - 12 + 8 + 4 = 6\).
Answer: 6
\(\text{Volume of 1st cuboid} = 4 \times 6 \times 4 = 96\) cubic units.
\(\text{Volume of 2nd cuboid} = 8 \times (2h - 1) \times 2 = 16(2h - 1)\) cubic units.
According to condition: \(16(2h - 1) = 96\)
\(2h - 1 = \frac{96}{16} = 6\)
\(2h = 6 + 1 = 7 \implies h = \frac{7}{2} = 3.5\).
Answer: 3.5
Let original edge be \(a\). Original Surface Area = \(6a^2\).
New edge = \(a + 50\% \text{ of } a = 1.5a\).
\(\text{New Surface Area} = 6(1.5a)^2 = 6(2.25a^2) = 13.5a^2\).
\(\text{Increase in Area} = 13.5a^2 - 6a^2 = 7.5a^2\).
\(\text{Percentage Increase} = \frac{7.5a^2}{6a^2} \times 100 = 1.25 \times 100 = 125\%\).
Answer: 125%
\(\text{Total Volume} = 3^3 + 4^3 + 5^3 = 27 + 64 + 125 = 216\) cubic cm.
Let edge of new cube be \(A\).
\(A^3 = 216 \implies A = \sqrt[3]{216} = 6\) cm.
Answer: 6 cm
The lengths of adjacent walls represent the length (\(l\)) and breadth (\(b\)) of the room.
\(l = 12\) m, \(b = 8\) m.
\(\text{Area of floor} = l \times b = 12 \times 8 = 96\) sq.m.
Note: Height is extra information here as floor area doesn't depend on it.
Answer: 96 sq.m.
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