Mathematics Concept Notes for Class 8: Chapter 11 Mensuration
Explore reliable concept summaries and revision materials through the CBSE Class 8 Mathematics Mensuration Notes Set 10. Tailored for Class 8 learners, utilizing these notes for Mathematics Chapter 11 Mensuration ensures thorough preparation and strengthens core academic definitions before final evaluations.
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Access the complete concept summary PDF for Chapter 11 Mensuration below. Regular review of these targeted notes builds familiarity with complex Class 8 Mathematics themes and helps secure higher marks in final school evaluations.
Plane Figures: Perimeter and Area
Area is the measure of the surface enclosed by a closed figure and perimeter is the length of its boundary. A figure bounded by straight lines is called a rectilinear figure. Area is always in square units and perimeter in units of length.
| Figure | Perimeter | Area |
|---|---|---|
| Rectangle (length l, breadth b) | 2(l + b) | l × b; diagonal d = √(l² + b²) |
| Square (side a) | 4a | a² = ½ × (diagonal)² |
| Parallelogram (base b, height h) | 2(sum of adjacent sides) | b × h |
| Rhombus (side a, diagonals d₁, d₂) | 4a | ½ × d₁ × d₂ |
| Trapezium (parallel sides a, b, height h) | sum of the four sides | ½ × (a + b) × h |
| General quadrilateral (diagonal d, perpendiculars h₁, h₂ on it) | sum of the four sides | ½ × d × (h₁ + h₂) |
| Triangle (base b, height h) | a + b + c | ½ × b × h, or Heron: √(s(s − a)(s − b)(s − c)) with s = half the perimeter |
| Equilateral triangle (side a) | 3a | \( \frac{\sqrt{3}}{4}a^2 \); height = \( \frac{\sqrt{3}}{2}a \) |
| Isosceles triangle (equal sides a, base c) | 2a + c | \( \frac{c}{4}\sqrt{4a^2 - c^2} \) |
| Circle (radius r) | 2πr | πr² |
| Semicircle | πr + 2r | \( \frac{1}{2}\pi r^2 \) |
| Ring (radii R and r) | 2π(R + r) for both edges | π(R² − r²) |
| Sector (angle θ) | arc + 2r, arc = \( \frac{\theta}{360}\times 2\pi r \) | \( \frac{\theta}{360}\times \pi r^2 \) |
Why These Formulas Work
1. A diagonal cuts a parallelogram into two congruent triangles, so its area is twice that of one triangle, which gives base × height.
2. The diagonals of a rhombus cut each other at right angles. Splitting it into four right triangles gives half the product of the diagonals.
3. A trapezium can be split into two triangles and a rectangle. Adding their areas gives half of the sum of the parallel sides times the distance between them.
4. A quadrilateral is split by one diagonal into two triangles with the diagonal as common base. Their heights are the perpendiculars from the other two corners.
Irregular Rectilinear Fields
To find the area of an irregular field, join the two farthest corners with a base line and drop perpendiculars (offsets) from the other corners onto it. Note the distances along the base line and the offset lengths. The field then divides into right triangles at the ends and trapezia in between. Add the areas of all these pieces. In the usual example a field with base line 250 m and the offsets noted at 30, 60, 100 and 180 m comes to 16050 m².
Solids: Surface Area and Volume
| Solid | Curved / lateral surface area | Total surface area | Volume |
|---|---|---|---|
| Cuboid (l, b, h) | 2h(l + b) | 2(lb + bh + hl) | lbh |
| Cube (edge a) | 4a² | 6a² | a³ |
| Right circular cylinder (r, h) | 2πrh | 2πr(r + h) | πr²h |
| Right circular cone (r, h, slant height l) | πrl, with l = √(r² + h²) | πr(r + l) | \( \frac{1}{3}\pi r^2 h \) |
Diagonal of a cuboid = √(l² + b² + h²). Diagonal of a cube = a√3.
Handy Results and Methods
1. If every length of a figure is multiplied by k, the area is multiplied by k² and the volume by k³. Doubling the edge of a cube therefore makes its volume 8 times and its surface area 4 times.
2. A length changed by x% changes the area by (1 + x/100)² − 1. A 25% longer side gives 56.25% more area, not 25%.
3. When a solid is melted and recast, the volume stays the same. Equate volumes to find the new dimensions.
4. For a wheel, distance travelled in one revolution is the circumference 2πr. Total distance = number of revolutions × circumference.
5. A rectangle rolled along its width to form a cylinder has the width as the height and the other side as the circumference of the base.
6. For paths inside a rectangle of width w, area of the path = outer area − inner area, and the inner rectangle has each side reduced by 2w.
Worked Examples
1. Area of a square with diagonal 2.5 cm = ½ × 2.5² = 3.125 cm².
2. A square has area 42.25 m². Side = √42.25 = 6.5 m = 650 cm. Tiles of 13 cm × 13 cm needed = 422500 ÷ 169 = 2500.
3. A room 5 m × 4 m × 3 m: area of four walls = 2h(l + b) = 2 × 3 × 9 = 54 m².
4. A rectangle's sides are in the ratio 4 : 3 and its area is 3072 m². Then 12x² = 3072, x = 16, sides 64 m and 48 m, and the perimeter is 224 m.
5. A parallelogram of area 64 cm² has sides 16 cm and 5 cm. The heights on these sides are 64 ÷ 16 = 4 cm and 64 ÷ 5 = 12.8 cm.
6. A rhombus with diagonals 10 cm and 24 cm has area 120 cm² and side √(5² + 12²) = 13 cm.
7. In a trapezium with parallel sides 13 cm and 23 cm and equal slant sides 13 cm, the foot of the perpendicular from the top corner cuts 5 cm off the longer base. Height = √(13² − 5²) = 12 cm and area = ½ × 36 × 12 = 216 cm².
8. A quadrilateral with a diagonal of 15 cm and perpendiculars 8.2 cm and 9.1 cm on it has area ½ × 15 × 17.3 = 129.75 cm².
9. Cuboid 10 × 8 × 6 cm: volume = 480 cm³, surface area = 2(80 + 48 + 60) = 376 cm².
10. A cylinder of radius 35 cm and height 1 m: volume = 22/7 × 0.35 × 0.35 × 1 = 0.385 m³, curved area = 2.2 m², total area = 2.97 m².
11. A metal pipe 154 cm long with outer radius 5.5 cm and inner radius 4.5 cm contains 22/7 × 154 × (5.5² − 4.5²) = 4840 cm³ of metal.
12. A rectangle 44 cm × 20 cm rolled along its width of 20 cm: 2πr = 44 gives r = 7 cm, height 20 cm, volume = 22/7 × 49 × 20 = 3080 cm³.
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Study Material and Revision Notes for Class 8 Mathematics Chapter 11 Mensuration
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