📚 Questions 71-80
Most Expected Maths Questions for Class 10
Circle – Segment Area
Find the area of the segment of a circle given that the angle of the sector is 120° and the radius of the circle is 21 cm (Take π = 22/7).
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✅ Solution
Given: Radius r = 21 cm, θ = 120°, π = 22/7
Step 1: Area of sector
Area of sector = \(\frac{θ}{360} × πr^2\)
= \(\frac{120}{360} × \frac{22}{7} × 21 × 21\)
= \(\frac{1}{3} × 22 × 3 × 21\)
= \(22 × 21 = 462 \text{ cm}^2\)
Step 2: Area of triangle
Area of triangle = \(\frac{1}{2}r^2 \sin θ\)
= \(\frac{1}{2} × 21 × 21 × \sin 120°\)
= \(\frac{1}{2} × 441 × \frac{\sqrt{3}}{2}\)
= \(\frac{441\sqrt{3}}{4} \text{ cm}^2\)
Step 3: Area of segment
Area of segment = Area of sector – Area of triangle
= \(462 – \frac{441\sqrt{3}}{4}\)
Using √3 ≈ 1.732:
= \(462 – \frac{441 × 1.732}{4}\)
= \(462 – \frac{763.812}{4}\)
= \(462 – 190.953 = 271.047 \text{ cm}^2\)
Answer: Area ≈ 271.05 cm² (exact: \(462 – \frac{441\sqrt{3}}{4}\) cm²)
Sector Perimeter Problem
In figure, is shown a sector OAP of a circle with centre O, containing ∠θ. AB is perpendicular to the radius OA and meets OP produced at B. Prove that the perimeter of shaded region is \(\tan θ + \sec θ + \frac{πθ}{180} – 1\)
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✅ Proof
Step 1: Understand the figure
Let radius OA = 1 unit (for simplicity)
∠AOP = θ
AB ⊥ OA, so ΔOAB is right-angled at A
Shaded region consists of arc AP and lines AB and BP
Step 2: Find lengths
In ΔOAB:
OA = 1
tan θ = AB/OA ⇒ AB = tan θ
sec θ = OB/OA ⇒ OB = sec θ
Arc AP length = \(\frac{θ}{180} × π × 1 = \frac{πθ}{180}\)
BP = OB – OP = sec θ – 1
Step 3: Perimeter of shaded region
Perimeter = Arc AP + AB + BP
= \(\frac{πθ}{180} + \tan θ + (\sec θ – 1)\)
= \(\tan θ + \sec θ + \frac{πθ}{180} – 1\)
Hence proved.
Volume – Spherical Marbles in Cylinder
150 spherical marbles, each of diameter 1.4 cm, are dropped in a cylindrical vessel of diameter 7 cm containing some water, which are completely immersed in water. Find the rise in the level of water in the vessel.
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✅ Solution
Step 1: Volume of one marble
Diameter of marble = 1.4 cm
Radius r = 0.7 cm
Volume of one marble = \(\frac{4}{3}πr^3\)
= \(\frac{4}{3} × \frac{22}{7} × (0.7)^3\)
= \(\frac{4}{3} × \frac{22}{7} × 0.343\)
= \(\frac{4 × 22 × 0.343}{21} = \frac{30.184}{21} = 1.437 \text{ cm}^3\)
Step 2: Total volume of 150 marbles
Total volume = 150 × 1.437 = 215.55 cm³
Step 3: Rise in water level
Cylinder diameter = 7 cm ⇒ radius R = 3.5 cm
Let rise in water level = h cm
Volume displaced = πR²h = \(\frac{22}{7} × 3.5 × 3.5 × h\)
= \(38.5 × h \text{ cm}^3\)
Volume displaced = Volume of marbles
38.5h = 215.55
h = \(\frac{215.55}{38.5} = 5.6 \text{ cm}\)
Answer: Rise in water level = 5.6 cm
Surface Area – Hemisphere & Cylinder
A vessel is in the form of a hemispherical bowl, surmounted by a hollow cylinder of same diameter. The diameter of the hemispherical bowl is 14 cm and the total height of the vessel is 13 cm. Find the surface area of the vessel. [Use π = 22/7]
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✅ Solution
Given: Diameter = 14 cm ⇒ radius r = 7 cm
Total height = 13 cm
Height of hemisphere = radius = 7 cm
Step 1: Find height of cylinder
Height of cylinder = Total height – Height of hemisphere
= 13 – 7 = 6 cm
Step 2: Calculate surface area
Surface area = CSA of hemisphere + CSA of cylinder
(Note: Base of cylinder is not exposed as it’s attached to hemisphere)
CSA of hemisphere = \(2πr^2 = 2 × \frac{22}{7} × 7 × 7\)
= \(2 × 22 × 7 = 308 \text{ cm}^2\)
CSA of cylinder = \(2πrh = 2 × \frac{22}{7} × 7 × 6\)
= \(2 × 22 × 6 = 264 \text{ cm}^2\)
Step 3: Total surface area
Total SA = 308 + 264 = 572 cm²
Answer: Surface area = 572 cm²
Cubical Block with Hemisphere
A cubical block of side 10 cm is surmounted by a hemisphere. What is the largest diameter that the hemisphere can have? Find the cost of painting the total surface area of the solid so formed, at the rate of ₹5 per 100 sq. cm. [Use π = 3.14]
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✅ Solution
Step 1: Largest diameter of hemisphere
Largest possible diameter = side of cube = 10 cm
So radius r = 5 cm
Step 2: Total surface area
Total SA = Surface area of cube – Area of circle + CSA of hemisphere
Surface area of cube = 6a² = 6 × 10² = 600 cm²
Area of circular base covered by hemisphere = πr² = 3.14 × 25 = 78.5 cm²
CSA of hemisphere = 2πr² = 2 × 3.14 × 25 = 157 cm²
Total SA = 600 – 78.5 + 157 = 678.5 cm²
Step 3: Cost of painting
Rate = ₹5 per 100 cm²
Cost = \(\frac{678.5}{100} × 5\) = 6.785 × 5 = ₹33.925 ≈ ₹33.93
Answers:
1. Largest diameter = 10 cm
2. Total surface area = 678.5 cm²
3. Cost of painting = ₹33.93
Tent Surface Area Problem
A tent is in the shape of a cylinder surmounted by a conical top of same diameter. If the height and diameter of cylindrical part are 2.1 m and 3 m respectively and the slant height of conical part is 2.8 m, find the cost of canvas needed to make the tent if the canvas is available at the rate of ₹500/sq. metre. (Use π = 22/7)
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✅ Solution
Given: Cylinder: height h = 2.1 m, diameter = 3 m ⇒ radius r = 1.5 m
Cone: slant height l = 2.8 m, radius r = 1.5 m
Step 1: Canvas area needed
Canvas area = CSA of cylinder + CSA of cone
CSA of cylinder = 2πrh = \(2 × \frac{22}{7} × 1.5 × 2.1\)
= \(2 × 22 × 0.3 × 2.1 = 2 × 22 × 0.63 = 27.72 \text{ m}^2\)
CSA of cone = πrl = \(\frac{22}{7} × 1.5 × 2.8\)
= \(22 × 0.3 × 2.8 = 22 × 0.84 = 18.48 \text{ m}^2\)
Step 2: Total canvas area and cost
Total area = 27.72 + 18.48 = 46.2 m²
Cost = 46.2 × 500 = ₹23,100
Answer: Canvas area = 46.2 m², Cost = ₹23,100
Statistics – Mean Calculation
If the mean of the following distribution is 6, find the value of p:
x: 2 4 6 10 p+5
f: 3 2 3 1 2
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✅ Solution
Step 1: Create frequency table
| x | f | f×x |
|---|---|---|
| 2 | 3 | 6 |
| 4 | 2 | 8 |
| 6 | 3 | 18 |
| 10 | 1 | 10 |
| p+5 | 2 | 2(p+5) |
Step 2: Apply mean formula
Σf = 3 + 2 + 3 + 1 + 2 = 11
Σfx = 6 + 8 + 18 + 10 + 2(p+5) = 42 + 2p + 10 = 52 + 2p
Mean = Σfx/Σf
6 = (52 + 2p)/11
Step 3: Solve for p
52 + 2p = 66
2p = 14
p = 7
Answer: p = 7
Probability – Card Problems
Two cards are drawn at random from a well shuffled pack of 52 cards. Find the probability that:
(a) Both are red cards
(b) One is a king and the other is a queen
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✅ Solution
Total possible outcomes:
Number of ways to draw 2 cards from 52:
\(^{52}C_2 = \frac{52 × 51}{2} = 1326\)
(a) Both are red cards:
Number of red cards = 26
Number of ways to choose 2 red cards:
\(^{26}C_2 = \frac{26 × 25}{2} = 325\)
P(both red) = \(\frac{325}{1326} = \frac{25}{102}\)
(b) One king and one queen:
Number of kings = 4
Number of queens = 4
Number of ways to choose 1 king and 1 queen:
\(^{4}C_1 × ^{4}C_1 = 4 × 4 = 16\)
P(one king and one queen) = \(\frac{16}{1326} = \frac{8}{663}\)
Answers:
(a) P(both red) = 25/102
(b) P(one king & one queen) = 8/663
Trigonometry – Height and Distance
From a point on the ground, the angles of elevation of the bottom and top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.
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✅ Solution
Step 1: Draw diagram and define variables
Let:
AB = Building height = 20 m
BC = Tower height = h m
AC = Total height = (20 + h) m
Distance from point to building = x m
Step 2: Apply trigonometric ratios
For building (45° angle):
tan 45° = AB/x = 20/x
Since tan 45° = 1, we get: x = 20 m
For total height (60° angle):
tan 60° = AC/x = (20 + h)/20
√3 = (20 + h)/20
Step 3: Solve for h
20 + h = 20√3
h = 20√3 – 20
h = 20(√3 – 1)
Using √3 ≈ 1.732:
h = 20(1.732 – 1) = 20 × 0.732 = 14.64 m
Answer: Height of tower = 20(√3 – 1) m ≈ 14.64 m
Arithmetic Progression
The first and last terms of an AP are 5 and 45 respectively. If the sum of all its terms is 400, find the common difference and the number of terms.
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✅ Solution
Given:
a = 5 (first term)
l = 45 (last term)
Sₙ = 400 (sum of n terms)
Step 1: Find number of terms (n)
Using sum formula: Sₙ = \(\frac{n}{2}\)(a + l)
400 = \(\frac{n}{2}\)(5 + 45)
400 = \(\frac{n}{2} × 50\)
400 = 25n
n = 16
Step 2: Find common difference (d)
Last term formula: l = a + (n-1)d
45 = 5 + (16-1)d
45 = 5 + 15d
40 = 15d
d = 40/15 = 8/3 ≈ 2.667
Answer:
Number of terms (n) = 16
Common difference (d) = 8/3 ≈ 2.667
Geometry – Triangles
In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC. Prove that 9AD² = 7AB².
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✅ Proof
Step 1: Draw figure and define variables
Let side of equilateral triangle ABC = a
BD = (1/3)BC = a/3
DC = BC – BD = a – a/3 = 2a/3
Draw AE ⊥ BC (E is midpoint of BC in equilateral triangle)
BE = EC = a/2
Step 2: Find DE and AE
DE = BE – BD = a/2 – a/3 = (3a – 2a)/6 = a/6
In right triangle AEB:
AE² = AB² – BE² (Pythagoras theorem)
AE² = a² – (a/2)² = a² – a²/4 = 3a²/4
So AE = √(3a²/4) = (a√3)/2
Step 3: Apply Pythagoras in triangle ADE
In right triangle ADE:
AD² = AE² + DE²
AD² = (3a²/4) + (a/6)²
AD² = (3a²/4) + (a²/36)
AD² = (27a² + a²)/36
AD² = (28a²)/36 = (7a²)/9
Multiply both sides by 9:
9AD² = 7a²
But a² = AB² (since side = a)
∴ 9AD² = 7AB²