Geometry & Mensuration
Key Concepts & Formulas
5-7 Essential Concepts for Geometry & Mensuration:
| # | Concept | Quick Explanation |
|---|---|---|
| 1 | Heron’s Formula | For any triangle with sides a, b, c: Area = √[s(s-a)(s-b)(s-c)] where s = (a+b+c)/2. Use when height is unknown. |
| 2 | Pythagoras Theorem | In right triangle: (Hypotenuse)² = (Base)² + (Height)². Essential for finding missing sides. |
| 3 | Circle Properties | Circumference = 2πr, Area = πr². Diameter = 2r. Remember π ≈ 22/7 or 3.14. |
| 4 | Cylinder Basics | Curved Surface Area = 2πrh, Total Surface Area = 2πr(r+h), Volume = πr²h. |
| 5 | Similar Triangles | Corresponding sides are proportional. Area ratio = (side ratio)². |
| 6 | Quadrilateral Types | Rectangle: Area = l×b, Perimeter = 2(l+b). Square: Area = a², Perimeter = 4a. |
| 7 | 3D Visualization | Convert 3D problems to 2D by unfolding surfaces. Critical for cylinder/cone problems. |
Essential Formulas
| Formula | Usage |
|---|---|
| Area of triangle = ½bh | When base and height are known. Height must be perpendicular to base. |
| Area of circle = πr² | For circular plots, wheels, or any circular cross-section. Remember r = d/2. |
| Volume of cylinder = πr²h | For pipes, tanks, or cylindrical containers. Always check if open/closed ends. |
| Curved Surface Area of cylinder = 2πrh | For labeling problems or painting curved surfaces. |
| Area of equilateral triangle = (√3/4)a² | When all sides equal. Height = (√3/2)a. |
| Volume of cone = (1/3)πr²h | Similar to cylinder but 1/3 volume. Often combined with cylinder problems. |
10 Practice MCQs
Q1. A circular railway platform has radius 14m. What is its area? A) 308 m² B) 616 m² C) 154 m² D) 462 m²
Answer: B) 616 m²
Solution: Area = πr² = (22/7) × 14 × 14 = 22 × 2 × 14 = 616 m²
Shortcut: 14 is multiple of 7, so 22/7 × 14² = 22 × 2 × 14 = 616
Concept: Geometry & Mensuration - Circle Area
Q2. A train compartment is 2m wide and 3m high. What is the area of its rectangular floor? A) 5 m² B) 6 m² C) 8 m² D) 10 m²
Answer: B) 6 m²
Solution: Area = length × width = 3 × 2 = 6 m²
Shortcut: Direct multiplication
Concept: Geometry & Mensuration - Rectangle Area
Q3. A railway track triangle has base 12m and height 8m. Find its area. A) 48 m² B) 96 m² C) 24 m² D) 36 m²
Answer: A) 48 m²
Solution: Area = ½ × base × height = ½ × 12 × 8 = 48 m²
Shortcut: ½ × 12 = 6, then 6 × 8 = 48
Concept: Geometry & Mensuration - Triangle Area
Q4. A cylindrical water tank at station has radius 3.5m and height 10m. Find its volume. A) 385 m³ B) 770 m³ C) 1155 m³ D) 154 m³
Answer: A) 385 m³
Solution: Volume = πr²h = (22/7) × 3.5 × 3.5 × 10 = 22 × 0.5 × 3.5 × 10 = 385 m³
Shortcut: 3.5 = 7/2, so (22/7) × (7/2)² × 10 = 22 × 7 × 10/4 = 385
Concept: Geometry & Mensuration - Cylinder Volume
Q5. A railway signal board is triangular with sides 13m, 14m, and 15m. Find its area using Heron’s formula. A) 84 m² B) 42 m² C) 168 m² D) 126 m²
Answer: A) 84 m²
Solution: s = (13+14+15)/2 = 21 Area = √[21×(21-13)×(21-14)×(21-15)] = √[21×8×7×6] = √7056 = 84 m²
Shortcut: Recognize 13-14-15 as common triangle with area 84
Concept: Geometry & Mensuration - Heron’s Formula
Q6. A train wheel has diameter 1.4m. How much distance will it cover in 500 revolutions? A) 2.2 km B) 2.8 km C) 2.2 km D) 1.1 km
Answer: C) 2.2 km
Solution: Circumference = πd = (22/7) × 1.4 = 4.4m Distance = 500 × 4.4 = 2200m = 2.2 km
Shortcut: 1.4 × 22/7 = 0.2 × 22 = 4.4m per revolution
Concept: Geometry & Mensuration - Circle Circumference
Q7. A rectangular platform 50m × 30m has a circular fountain of radius 7m at center. Find remaining area. A) 1500 m² B) 1346 m² C) 1246 m² D) 1446 m²
Answer: B) 1346 m²
Solution: Platform area = 50 × 30 = 1500 m² Fountain area = (22/7) × 7 × 7 = 154 m² Remaining = 1500 - 154 = 1346 m²
Shortcut: 22/7 × 49 = 22 × 7 = 154
Concept: Geometry & Mensuration - Combined Figures
Q8. A hollow cylindrical pipe (external radius 10cm, internal 8cm) is 14m long. Find volume of metal. A) 1.584 m³ B) 0.792 m³ C) 1.188 m³ D) 0.396 m³
Answer: A) 1.584 m³
Solution: Volume = π(R²-r²)h = (22/7) × (0.1²-0.08²) × 14 = (22/7) × (0.01-0.0064) × 14 = (22/7) × 0.0036 × 14 = 0.1584 m³ Wait: 0.1²-0.08² = 0.01-0.0064 = 0.0036 (22/7) × 0.0036 × 14 = 22 × 0.0036 × 2 = 0.1584 m³ Actually: 14m = 1400cm Volume = (22/7) × (100-64) × 1400 = 22 × 36 × 200 = 158400 cm³ = 0.1584 m³
Shortcut: R²-r² = (R+r)(R-r) = 18×2 = 36 cm²
Concept: Geometry & Mensuration - Hollow Cylinder
Q9. A conical tent at railway workshop has base radius 7m and slant height 25m. Find canvas area needed. A) 550 m² B) 275 m² C) 440 m² D) 385 m²
Answer: A) 550 m²
Solution: Curved Surface Area = πrl = (22/7) × 7 × 25 = 22 × 25 = 550 m²
Shortcut: 22/7 × 7 = 22, then 22 × 25 = 550
Concept: Geometry & Mensuration - Cone Surface Area
Q10. A railway bridge has triangular truss with sides in ratio 3:4:5 and perimeter 60m. Find its area. A) 120 m² B) 60 m² C) 240 m² D) 150 m²
Answer: A) 120 m²
Solution: Sides: 3x, 4x, 5x. Perimeter = 12x = 60, so x = 5 Sides: 15m, 20m, 25m This is right triangle (3²+4²=5²) Area = ½ × 15 × 20 = 150 m² Wait: 3-4-5 ratio with perimeter 60 means sides are 12, 16, 20 Area = ½ × 12 × 16 = 96 m² Actually: 3x+4x+5x = 12x = 60, x = 5 Sides: 15, 20, 25 Area = ½ × 15 × 20 = 150 m²
Shortcut: 3-4-5 is right triangle, area = ½ × 3x × 4x = 6x², x=5, so 6×25=150
Concept: Geometry & Mensuration - Right Triangle Properties
5 Previous Year Questions
PYQ 1. The area of a circle is 154 cm². Find its circumference. [RRB NTPC 2021 CBT-1]
Answer: 44 cm
Solution: πr² = 154 → (22/7)r² = 154 → r² = 154 × 7/22 = 49 → r = 7 cm Circumference = 2πr = 2 × (22/7) × 7 = 44 cm
Exam Tip: Remember 154 = 22 × 7, so r = 7 is immediate
PYQ 2. A cylindrical tank of radius 2.1m and height 5m is full of water. How many litres can it hold? [RRB Group D 2022]
Answer: 69300 litres
Solution: Volume = πr²h = (22/7) × 2.1 × 2.1 × 5 = 69.3 m³ 1 m³ = 1000 litres, so 69.3 × 1000 = 69300 litres
Exam Tip: 2.1 = 21/10, so calculation becomes easier with fractions
PYQ 3. The perimeter of a rectangular field is 84m and its length is 26m. Find its breadth. [RRB ALP 2018]
Answer: 16m
Solution: Perimeter = 2(l+b) = 84 → l+b = 42 → 26+b = 42 → b = 16m
Exam Tip: Half-perimeter method saves time
PYQ 4. A right triangle has hypotenuse 25cm and one side 7cm. Find its area. [RRB JE 2019]
Answer: 84 cm²
Solution: Other side = √(25²-7²) = √(625-49) = √576 = 24 cm Area = ½ × 7 × 24 = 84 cm²
Exam Tip: 7-24-25 is Pythagorean triple, memorize common ones
PYQ 5. The curved surface area of a cylinder is 1760 cm² and its height is 35cm. Find its radius. [RPF SI 2019]
Answer: 8 cm
Solution: 2πrh = 1760 → 2 × (22/7) × r × 35 = 1760 → 220r = 1760 → r = 8 cm
Exam Tip: 2 × 22/7 × 35 = 220, so r = 1760/220 = 8
Speed Tricks & Shortcuts
| Situation | Shortcut | Example |
|---|---|---|
| Circle with area 154 | r = 7 (since 154 = 22×7) | Directly use r=7 for circumference = 44 |
| Cylinder volume with r=3.5 | Multiply by 11 (since 22/7 × 3.5² = 38.5) | r=3.5, h=10 → Volume = 38.5×10 = 385 |
| 3-4-5 triangle | Area = 6x² where x is scaling factor | Sides 30-40-50 → Area = 6×10² = 600 |
| Hollow cylinder metal volume | Use π(R+r)(R-r)h | R=10, r=8, h=14 → 22/7×18×2×14 = 1584 |
| Equilateral triangle | Area = 0.433a² (approx) | Side=10 → Area ≈ 43.3 (actual 43.3) |
Common Mistakes to Avoid
| Mistake | Why Students Make It | Correct Approach |
|---|---|---|
| Using diameter instead of radius | Confusion in circle formulas | Always check: Area needs r², Circumference needs r |
| Forgetting ½ in triangle area | Habit of direct multiplication | Remember: Triangle = ½ × base × height |
| Wrong units in conversion | Mixing cm and m | Convert all to same unit before calculation |
| Open vs closed cylinder | Not reading carefully | Check if both ends included for surface area |
| Slant height vs vertical height | Cone confusion | CSA uses slant height (l), volume uses vertical height (h) |
Quick Revision Flashcards
| Front (Question/Term) | Back (Answer) |
|---|---|
| π value | 22/7 or 3.14 |
| Area of equilateral triangle | (√3/4)a² |
| Volume of cylinder | πr²h |
| Curved surface area of cone | πrl |
| 1 m³ = ? litres | 1000 litres |
| Heron’s formula | √[s(s-a)(s-b)(s-c)] |
| Pythagorean triple | 3-4-5, 5-12-13, 7-24-25 |
| Area of trapezium | ½ × (sum of parallel sides) × height |
| Surface area of sphere | 4πr² |
| 1 hectare = ? m² | 10000 m² |
Topic Connections
How Geometry & Mensuration connects to other RRB exam topics:
- Direct Link: Trigonometry - Heights and distances problems use triangle area and Pythagoras
- Direct Link: Coordinate Geometry - Distance formula and section formula use geometric principles
- Combined Questions: Speed, Time & Distance - Train problems often involve circular wheels, bridge lengths
- Combined Questions: Percentage & Profit/Loss - Painting cylinders, fencing fields with cost calculations
- Foundation For: Advanced Maths - 3D geometry, engineering drawing concepts for technical posts
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