Overview
Introduction: Mensuration is the branch of mathematics that deals with measurement of lengths, areas and volumes of geometric shapes. In Class 8 (Mathematics – VIII, Chapter: Mensuration) students extend prior knowledge of perimeter and area to a wider range of plane figures and learn surface area and volume of simple solids. Importance: Mensuration connects geometry with real-life measurements — from finding the area of a plot, the carpet needed for a room, to the volume of containers — and builds strong problem-solving and modelling skills. Key themes: perimeter and area of standard plane figures (square, rectangle, triangle, parallelogram, trapezium), area of polygons and composite figures, circumference and area of a circle (use of π), unit conversions, and surface area and volume of solids (cube, cuboid, cylinder). What the student will learn: derive and apply standard area and perimeter formulas, compute areas and circumferences involving π, break complex shapes into simpler parts, convert between units of length/area/volume, and calculate lateral/total surface area and volumes of cuboids, cubes and cylinders; solve related word problems and practical applications.
Learning Objectives
- Define perimeter, area, surface area and volume with correct units.
- Explain and state formulas for area of rectangle, square, parallelogram, triangle and circle.
- Apply appropriate formulas to calculate perimeter and area of plane figures in numerical problems.
- Derive the area formula of a triangle and a parallelogram using geometric reasoning.
- Calculate curved surface area, total surface area and lateral area of cubes, cuboids and right circular cylinders.
- Compute volumes of cubes, cuboids and right circular cylinders and relate volume to capacity (litres and cubic metres).
- Convert units of length, area and volume (mm, cm, m, km; cm², m²; cm³, m³, litres) correctly in problem solving.
- Solve numerical problems on surface area and volume of composite solids formed by combining or subtracting basic solids.
Topics in this chapter
8 topics · tap a topic title to jump straight to it.
Introduction to Mensuration
What is Mensuration?
Mensuration is the branch of mathematics that deals with measuring geometric quantities such as length, area, surface area and volume. It helps us calculate how much space a shape covers (area), how far its boundary runs (perimeter), how much material is needed to cover a surface (surface area), and how much a solid can contain (volume).
Key Concepts
- Perimeter: Total length around a 2‑D figure (sum of its sides).
- Area: Measure of the region enclosed by a plane figure, usually in square units (e.g., cm²).
- Surface area: Total area of the outer surfaces of a 3‑D object, in square units.
- Volume: Amount of space occupied by a 3‑D object, in cubic units (e.g., cm³).
- Units and conversion: Always use compatible units; convert (e.g., m to cm) before applying formulas. Area units are squared, volume units are cubed.
How formulas arise (intuition)
Most area formulas come from counting unit squares or decomposing a shape into simpler shapes. For example, area of a rectangle A = length × breadth counts how many unit squares fit inside. For triangles, area = 1/2 × base × height because a rectangle can be split into two congruent right triangles.
Practical tips
- Draw and label diagrams: base, height, radius, slant height clearly.
- Decompose complex shapes into rectangles, triangles, sectors, then add/subtract areas.
- Check units: convert, then compute; append correct units to final answer.
- For solids, visualize nets to find surface area and use cross‑sections for volume reasoning.
Why it matters (real life)
Mensuration is used in construction (flooring, painting, fencing), packing and shipping (box volumes), manufacturing (material required), landscaping, and many daily measurements.
- Flooring: To buy tiles for a room of 5 m by 4 m, compute area = 5 × 4 = 20 m². Buy tiles for slightly more than 20 m² to allow wastage.
- Fencing a rectangular field: For a field 30 m by 20 m, perimeter = 2(30 + 20) = 100 m of fence required.
- Painting a wall: For a rectangular wall 3 m × 4 m, area = 12 m²; multiply by paint coverage (e.g., 1 litre/10 m²) to find paint needed.
- Water in a tank: A cuboid tank 2 m × 1.5 m × 1 m has volume = 2 × 1.5 × 1 = 3 m³ (3000 litres).
- Wrapping a gift box: For a cuboid of 40 cm × 30 cm × 20 cm, total surface area = 2(lb + bh + hl) gives amount of wrapping paper needed.
- Perimeter of rectangle: P = 2(l + b)
- Area of rectangle: A = l × b
- Perimeter of square: P = 4s
- Area of square: A = s²
- Area of triangle: A = 1/2 × base × height
- Area of parallelogram: A = base × height
Perimeter of Plane Figures
What is Perimeter? The perimeter of a plane figure is the total length of its boundary — i.e., the sum of the lengths of all its sides. It is a linear measure and is expressed in units such as mm, cm, m, km.
Basic idea and steps to find perimeter
- Identify all boundary segments (straight or curved) of the figure.
- Measure or use given lengths for each boundary segment.
- Add the lengths (make sure units are the same).
- Report the result with the appropriate unit.
Perimeter vs Area: Perimeter measures the boundary length (1-dimensional), while area measures the region covered (2-dimensional). They are different properties and use different formulae.
Common shapes and how to compute their perimeters
- Square: add four equal sides.
- Rectangle / Parallelogram: add two pairs of equal opposite sides.
- Triangle: sum of its three side lengths.
- Rhombus: four equal sides.
- Circle: the perimeter is called the circumference (a curved boundary).
- Regular polygon: number of sides × length of one side.
- Composite figures: break the figure into simple parts, add outer boundary lengths (do not count internal shared edges).
Practical tips
- Always check and convert units before adding (e.g., cm to m).
- For curved edges (circle), use π (pi). Use π ≈ 3.14 or 22/7 as instructed.
- When figures are composite, sketch and label every side to avoid double counting.
- Rectangle: Find the perimeter of a rectangle with length 12 cm and breadth 5 cm. P = 2(l + b) = 2(12 + 5) = 2 × 17 = 34 cm.
- Square: A square has side 7 m. P = 4 × side = 4 × 7 = 28 m.
- Triangle: A triangle has sides 8 cm, 6 cm and 5 cm. P = 8 + 6 + 5 = 19 cm.
- Circle: A circle has radius 3.5 cm. Circumference C = 2πr = 2 × π × 3.5 = 7π cm. Using π = 22/7 gives C = 22 cm (exact form: 7π cm).
- Composite figure (L-shape): An L-shaped figure made of two rectangles: A (6 cm × 4 cm) attached to B (4 cm × 3 cm) sharing a 4 cm side. Outline the outer boundary and add those side lengths only: P = 6 + 4 + 3 + 4 + (remaining verticals) etc. (Sketch, label and sum outer edges — do not include the shared internal edge.)
- Square: P = 4a, where a = side
- Rectangle: P = 2(l + b), where l = length, b = breadth
- Triangle: P = a + b + c (sum of three sides)
- Parallelogram: P = 2(a + b), where a and b are adjacent sides
- Rhombus: P = 4a, where a = side
- Regular n-sided polygon: P = n × s, where n = number of sides, s = side length
Area: Basic Concepts and Formulae
What is Area? Area is the measure of the region occupied by a flat (two-dimensional) figure. It tells how much surface the figure covers and is measured in square units (for example, cm², m², mm²).
Units and conversion: Common units: mm², cm², m², km². To convert, use the square of the linear conversion factor. Example: 1 m = 100 cm, so 1 m² = (100 cm)² = 10,000 cm².
Basic ideas and methods:
- Unit-square method: Cover the figure with unit squares (exact or fractional) to estimate area.
- Decomposition / rearrangement: Break a complex figure into simple shapes (rectangles, triangles, circles) whose areas are known and add or subtract them.
- Use of height (altitude): For parallelograms and triangles the area depends on the base and the corresponding height (perpendicular distance).
How formulae are connected (short derivations):
- Rectangle: area = base × height because a rectangle is exactly base rows of unit squares of height 1.
- Square: a special rectangle with all sides equal, so area = side².
- Triangle: two copies of the same triangle can form a parallelogram (or rectangle when rearranged), so area = 1/2 × base × height.
- Parallelogram: it can be rearranged into a rectangle with same base and height, so area = base × height.
- Trapezium (trapezoid): average of parallel sides times height → area = 1/2 × (sum of parallel sides) × height.
- Rhombus (by diagonals): its diagonals bisect each other at right angles; area = 1/2 × d1 × d2.
- Circle: area found from calculus/geometric limit = π × r² (r = radius).
Practical tips:
- Always use the perpendicular height for triangles, parallelograms and trapeziums.
- Convert all dimensions to the same linear unit before applying formulae.
- For irregular shapes estimate by grid-counting or split into known shapes.
- Rectangle: A rectangle has length 12 cm and width 5 cm. Area = length × width = 12 × 5 = 60 cm².
- Triangle: A triangle has base 10 m and height 6 m. Area = (1/2) × base × height = 0.5 × 10 × 6 = 30 m².
- Trapezium: A trapezium has parallel sides 8 cm and 14 cm and height 5 cm. Area = (1/2) × (8 + 14) × 5 = 0.5 × 22 × 5 = 55 cm².
- Rhombus (diagonals): A rhombus has diagonals of lengths 10 cm and 24 cm. Area = (1/2) × d1 × d2 = 0.5 × 10 × 24 = 120 cm².
- Composite shape: A rectangle 10 m × 6 m has a triangular garden (base 4 m, height 3 m) removed from one corner. Area left = area rectangle − area triangle = (10×6) − (0.5×4×3) = 60 − 6 = 54 m².
- Square: A = a^2 (a = side)
- Rectangle: A = l × b (l = length, b = breadth/width)
- Parallelogram: A = base × height
- Triangle: A = (1/2) × base × height
- Trapezium: A = (1/2) × (sum of parallel sides) × height = (1/2) × (a + b) × h
- Rhombus (using diagonals): A = (1/2) × d1 × d2
Area of Trapezium and Polygons
Overview: Mensuration deals with measuring area of 2D shapes. A trapezium (trapezoid) has one pair of parallel sides. A polygon is a closed figure with straight sides — regular polygons have all sides and angles equal; irregular polygons do not.
Area of a trapezium (derivation and idea): Let the two parallel sides (bases) be a and b, and the perpendicular distance between them (height) be h. If you place two congruent trapezia together along their non-parallel sides, they form a parallelogram with base (a + b) and height h. Area of that parallelogram is (a + b) × h, so the area of one trapezium is half of that:
Area = ½(a + b) × h
Key points: h must be the perpendicular distance between the parallel sides. Units of area are square units (cm², m², etc.).
Area of polygons:
- Regular polygon: A regular n-sided polygon can be split into n congruent isosceles triangles, each with base equal to a side and height equal to the apothem (the perpendicular from centre to a side). Summing areas of the triangles gives:
Area = ½ × perimeter × apothem = ½ × P × a
- Perimeter P = n × side. Apothem a is the distance from the center to the midpoint of any side.
- Irregular polygon: Break it into familiar shapes (triangles, rectangles, trapezia). Find area of each and add. Alternatively, if coordinates of vertices are known, area can be found by coordinate methods (advanced).
Practical tips:
- Always use the perpendicular height for h or apothem.
- When shapes are irregular, draw auxiliary lines to divide them into triangles/rectangles/trapezia.
- Check units and convert lengths before calculating area.
- Example 1 (Trapezium): Bases a = 8 cm, b = 14 cm, height h = 5 cm. Area = 1/2 × (a + b) × h = 1/2 × (8 + 14) × 5 = 1/2 × 22 × 5 = 11 × 5 = 55 cm².
- Example 2 (Regular Hexagon): Side s = 6 cm. For a regular hexagon apothem a = s × √3/2 = 6 × √3/2 = 3√3 ≈ 5.196 cm. Perimeter P = 6 × 6 = 36 cm. Area = 1/2 × P × a = 1/2 × 36 × 5.196 ≈ 18 × 5.196 ≈ 93.53 cm².
- Example 3 (Irregular polygon by decomposition): Suppose a 5-sided shape can be split into a rectangle 8 cm × 4 cm and a right triangle with base 6 cm and height 3 cm. Area(rectangle) = 8 × 4 = 32 cm². Area(triangle) = 1/2 × 6 × 3 = 9 cm². Total area = 32 + 9 = 41 cm².
- Area of trapezium = 1/2 × (sum of parallel sides) × height = 1/2 × (a + b) × h
- Area of regular polygon = 1/2 × perimeter × apothem = 1/2 × P × a (where P = n × side)
- Area of triangle = 1/2 × base × height
- Area of rectangle = length × breadth
- For irregular polygons: Area = sum of areas of component shapes (triangles, rectangles, trapezia)
- Units: area expressed in square units, e.g., cm², m²
Circle: Circumference and Area
What is a circle? A circle is the set of all points in a plane at a fixed distance from a fixed point called the centre. The fixed distance is the radius (r). The line passing through the centre and touching the circle at two points is the diameter (d = 2r).
Circumference is the length of the boundary of the circle. The ratio of the circumference (C) of any circle to its diameter (d) is a constant called π (pi): C/d = π. Hence the circumference formulas are C = 2πr or C = πd. In practice π is taken as 22/7 or 3.14 (use 22/7 when d or r is a multiple of 7 for exactness).
Area of a circle is the amount of surface enclosed by the circle. The area (A) of a circle of radius r is A = πr². (A simple justification: cut the circle into many equal sectors and rearrange them to form a shape close to a rectangle whose base ≈ πr and height ≈ r, giving area ≈ πr × r = πr².)
Other useful results: Arc length of an angle θ (in degrees) = (θ/360) × 2πr. Area of a sector with angle θ = (θ/360) × πr². Area of an annulus (ring) with outer radius R and inner radius r = π(R² − r²).
Units and tips: Circumference has units of length (cm, m). Area has square units (cm², m²). Always use the same unit for r or d before applying formulas. Choose π = 22/7 when it simplifies calculation with 7, otherwise 3.14 or the calculator value of π is fine.
- Example 1 — Given radius r = 7 cm. Circumference C = 2πr = 2 × (22/7) × 7 = 44 cm. Area A = πr² = (22/7) × 7 × 7 = 154 cm².
- Example 2 — Given circumference C = 44 cm. Find radius. C = 2πr ⇒ r = C/(2π) = 44 / (2 × (22/7)) = 44 × 7 / 44 = 7 cm.
- Example 3 — Area of an annulus: outer radius R = 10 cm, inner radius r = 6 cm. Area = π(R² − r²) = (22/7) × (100 − 36) = (22/7) × 64 = 201.142... ≈ 201.14 cm² (or compute exactly as 1408/7 cm²).
- Example 4 — Sector and arc: For a circle with r = 14 cm and central angle θ = 90°. Arc length = (θ/360) × 2πr = (90/360) × 2 × π × 14 = (1/4) × 28π = 7π ≈ 21.99 cm. Sector area = (θ/360) × πr² = (1/4) × π × 14² = 49π ≈ 153.94 cm².
- Circumference: C = 2πr = πd
- Area: A = πr²
- Arc length (angle θ in degrees): L = (θ/360) × 2πr
- Sector area (angle θ in degrees): As = (θ/360) × πr²
- Annulus area (outer R, inner r): A = π(R² − r²)
- Common π values: π ≈ 3.14 or π = 22/7 (use 22/7 when r or d is multiple of 7 for exactness)
Area of Composite and Shaded Regions
What is a composite or shaded region? A composite region is a figure made up of two or more simple shapes (rectangles, triangles, circles, etc.). A shaded region is a part of a composite figure that is colored or removed; we find its area by combining areas of simple shapes.
Basic strategy (step-by-step):
- Identify the simple shapes that make up the whole figure (or the shapes removed).
- Write down known dimensions and decide whether you need radii, heights or bases.
- Calculate the area of each simple shape using the appropriate formula.
- Combine areas: add areas of parts that form the shaded region, or subtract areas of removed parts from the whole.
- Ensure units are consistent and give the final answer with proper units (cm², m², etc.).
Key tips: If a circular part is given by diameter, convert to radius (r = d/2). For semicircles and quarter circles, use a fractional part of πr². For annulus (ring) use difference of two circles. When parts overlap, avoid double counting by subtracting the overlap.
Example of combining areas (idea): If a rectangle has a semicircle cut out, shaded area = area(rectangle) − area(semicircle). If a figure is an L-shape formed by two rectangles, shaded area = area(rectangle1) + area(rectangle2) (or total rectangle − removed rectangle).
- Example 1 — Rectangle with a semicircle removed: A rectangle 20 cm by 10 cm has a semicircle (diameter 10 cm) removed from one short side. Find shaded area. Solution: Area(rectangle)=20×10=200 cm². Semicircle radius r=10/2=5 cm, area(semicircle)=½πr²=½×π×25=12.5π ≈ 39.27 cm² (using π≈3.1416). Shaded area = 200 − 12.5π ≈ 160.73 cm².
- Example 2 — Annulus (shaded ring): Two concentric circles have radii 10 cm and 6 cm. Area of shaded region between them = π(R² − r²) = π(100 − 36) = 64π ≈ 201.06 cm².
- Example 3 — L-shaped region by subtraction: A 12 cm by 8 cm rectangle has a 6 cm by 4 cm rectangle removed from one corner. Shaded area = 12×8 − 6×4 = 96 − 24 = 72 cm².
- Example 4 — Square with inscribed circle: Square side = 14 cm. Inscribed circle diameter = 14, radius = 7. Area(square) = 14² = 196 cm². Area(circle) = π×7² = 49π ≈ 153.94 cm². Shaded area (square minus circle) ≈ 196 − 153.94 = 42.06 cm².
- Real-life example 1 — Garden with a pond: A rectangular lawn 30 m by 20 m contains a semicircular pond of diameter 10 m along one edge. Lawn area = 30×20 = 600 m². Semicircle area = ½π(5)² = 12.5π ≈ 39.27 m². Grass area to maintain = 600 − 12.5π ≈ 560.73 m².
- Real-life example 2 — Painted wall with a window: A wall 5 m by 3 m has a rectangular window 1 m by 1.5 m to remain unpainted. Paint area = 5×3 − 1×1.5 = 15 − 1.5 = 13.5 m².
- Area of rectangle = length × breadth = l × b
- Area of square = side² = a²
- Area of triangle = 1/2 × base × height = (1/2) × b × h
- Area of parallelogram = base × height = b × h
- Area of trapezium = (1/2) × (sum of parallel sides) × height = (1/2) × (a + b) × h
- Area of circle = π × r² (use π = 22/7 or 3.1416 as required)
Applications and Word Problems
What this topic covers
Applications and Word Problems in Mensuration ask you to apply area, perimeter, surface area and volume formulas to real-life situations. Problems may involve single shapes, composite shapes (made of two or more simple shapes), solids (cube, cuboid, cylinder) and unit conversions (cm, m, litres).
General approach / steps to solve word problems
- Read the problem carefully and underline the quantities required and given.
- Visualize and draw a neat labelled diagram (show all dimensions). For composite shapes, split into simple parts.
- Convert all measurements to the same units (e.g. cm → m) before using formulas.
- Choose the correct formula (area, circumference, surface area, volume) and substitute values.
- Compute, give the answer with correct units, and check if the result is reasonable.
Common categories of problems
- Perimeter and fencing problems (use perimeter or circumference).
- Area problems — carpeting, tiling, painting of plane regions (use area of rectangles, triangles, circles, etc.).
- Composite figures — add or subtract areas of basic shapes.
- Volume problems — amount of water, number of bricks, capacity of tanks (use volume of cuboid, cylinder).
- Surface area problems — painting or wrapping solids (use total or curved surface area as required).
Tips
- When a part is removed (a hole or cut‑out), subtract its area/volume from the whole.
- For solids with circular bases use π (use 22/7 or 3.14 as instructed; state which you use).
- Always label units in the answer (m, cm, m2, m3, L).
- 1) Carpet for a rectangular room: A room is 6 m long and 4 m wide. How much carpet is needed? Solution: Area = length × breadth = 6 × 4 = 24 m². So 24 m² of carpet is needed.
- 2) Fencing a circular park: A circular park has radius 7 m. Find the length of the fence required. Solution: Circumference = 2πr = 2 × π × 7 = 14π ≈ 43.98 m (≈ 44.0 m).
- 3) Area of a composite figure: A rectangle of length 10 m and breadth 6 m has a semicircle attached to one of its shorter sides (diameter = 6 m). Find the total area. Solution: Area(rectangle) = 10 × 6 = 60 m². Radius of semicircle = 3 m. Area(semicircle) = (1/2)πr² = (1/2)π × 9 = 4.5π ≈ 14.14 m². Total ≈ 60 + 14.14 = 74.14 m².
- 4) Capacity of a cylindrical tank: A vertical cylindrical tank has radius 2 m and height 3 m. Find its capacity in litres and the number of 10‑litre buckets needed to fill it. Solution: Volume = πr²h = π × 4 × 3 = 12π m³ ≈ 37.699 m³. 1 m³ = 1000 L → capacity ≈ 37699 L. Number of 10 L buckets = 37699 / 10 ≈ 3770 buckets.
- 5) Painting a cuboidal box: A box measures 2 m × 1.5 m × 1 m. Find the total surface area to be painted and the paint required if 1 L covers 10 m². Solution: TSA = 2(lb + bh + hl) = 2(2×1.5 + 1.5×1 + 1×2) = 2(3 + 1.5 + 2) = 13 m². Paint needed = 13 / 10 = 1.3 L.
- Perimeter of rectangle = 2(l + b)
- Area of rectangle = l × b
- Area of square = a² ; Perimeter of square = 4a
- Area of triangle = (1/2) × base × height
- Area of parallelogram = base × height
- Area of trapezium = (1/2) × (sum of parallel sides) × height
Worked Examples and Exercises
What this topic covers
In Mensuration, 'Worked Examples and Exercises' helps students practise finding areas, surface areas and volumes of plane figures and solids (rectangles, squares, triangles, parallelograms, trapeziums, circles; cube, cuboid, cylinder, cone, sphere as taught). The focus is on choosing the correct formula, substituting dimensions with units, simplifying, and interpreting the result with correct units.
Step-by-step approach to solving problems
- Read and draw: Carefully read the question and draw a neat labelled diagram (show heights, radii, diagonals etc.).
- Identify shapes: Break composite figures into simple shapes (rectangles, triangles, semicircles, sectors) if needed.
- Pick formula(s): Choose the correct formula for area, surface area or volume.
- Convert units: Make sure all measurements are in the same unit (cm, m, etc.) before substituting.
- Substitute and calculate: Substitute numbers carefully, simplify step by step, keep π as 22/7 or 3.14 if required by the question, and give the final answer with the correct unit.
- Check: Check the answer for reasonableness (e.g., area positive, volume sensible compared to dimensions) and include units.
Common tips and mistakes
- Always label diagrams and show intermediate steps — marks are awarded for method.
- When dealing with composite shapes, add/subtract areas: area(composite)=sum of component areas − areas of holes.
- For curved surfaces, distinguish between curved (lateral) surface area and total surface area (which includes bases).
- Be consistent with π: if given 22/7 use it consistently; otherwise use 3.14 or π symbol as instructed.
- Watch for units: convert cm³ to m³ only when asked; for painting problems use area in m² and paint coverage in m² per litre.
Example (short worked illustration)
Find the area of a composite figure made of a rectangle 20 cm by 12 cm with a semicircle of diameter 12 cm attached along one 12 cm side. Solution outline: area = area(rectangle) + area(semicircle) = 20×12 + (1/2)πr², r=6 cm. So area = 240 + 0.5×π×36 = 240 + 18π ≈ 240 + 56.55 = 296.55 cm² (using π = 3.1416).
- Example 1 — Area of composite figure: A rectangle 20 cm by 12 cm has a semicircle of diameter 12 cm attached to one 12 cm side. Area = area(rectangle) + area(semicircle) = 20×12 + 1/2×π×6² = 240 + 18π ≈ 296.6 cm² (π ≈ 3.1416).
- Example 2 — Triangle area using height: Find area of triangle with base 10 m and height 6 m. Area = 1/2 × base × height = 1/2 × 10 × 6 = 30 m².
- Example 3 — Volume of a cylindrical water tank: A cylindrical tank has radius 1.5 m and height 4 m. Volume = πr²h = π × (1.5)² × 4 = π × 2.25 × 4 = 9π ≈ 28.27 m³ (π ≈ 3.1416).
- Example 4 — Paint required for room walls (lateral surface area): A rectangular room 6 m long, 4 m wide and 3 m high. Lateral surface area (walls only) = 2h(l + b) = 2×3×(6+4)=6×10=60 m². If 1 litre of paint covers 12 m², paint needed = 60/12 = 5 litres.
- Area of rectangle: A = length × breadth
- Area of square: A = side²
- Area of triangle: A = 1/2 × base × height
- Area of parallelogram: A = base × height
- Area of trapezium: A = 1/2 × (sum of parallel sides) × height
- Circumference of circle: C = 2πr = πd
Key Concepts
- Perimeter
- Total length around a closed plane figure; sum of all side lengths.
- Area
- Measure of the region enclosed by a plane figure, in square units.
- Rectangle
- A quadrilateral with opposite sides equal and four right angles; area = length × breadth.
- Square
- A rectangle with all four sides equal; area = side², perimeter = 4×side.
- Triangle
- A three-sided polygon; area = 1/2 × base × height (for a given base and corresponding height).
- Parallelogram
- A quadrilateral with opposite sides parallel and equal; area = base × height (height perpendicular to base).
- Trapezium (Trapezoid)
- A quadrilateral with one pair of parallel sides; area = 1/2×(sum of parallel sides)×height.
- Circle
- Set of all points in a plane at a fixed distance (radius) from a fixed center; area = πr².
- Radius
- Distance from the centre of a circle to any point on the circle.
- Diameter
- A chord that passes through the centre of a circle; equals twice the radius (d = 2r).
- Circumference
- Perimeter (boundary length) of a circle; C = 2πr = πd.
- Chord
- A line segment joining two points on a circle.
- Arc
- A continuous portion of the circumference between two points on a circle.
- Sector
- Region enclosed by two radii and the included arc; area = (θ/360)×πr² (θ in degrees).
- Segment (Circular Segment)
- Region between a chord and the corresponding arc of a circle (sector minus triangle portion).
- Total Surface Area (TSA)
- Sum of areas of all outer faces of a solid; expressed in square units.
- Lateral Surface Area (LSA)
- Surface area of the sides (excluding bases) of a solid like a cylinder or prism.
- Cuboid (includes Cube as special case)
- A rectangular box with length, breadth and height; Volume = l×b×h, TSA = 2(lb+bh+hl). Cube is a cuboid with l=b=h = side.
- Cylinder
- Solid with congruent circular bases parallel to each other and a curved lateral surface; Volume = πr²h, TSA = 2πr(h+r).
- Volume
- Amount of space occupied by a 3D object, measured in cubic units; computed by appropriate formula for each solid.
End-of-Chapter Trial Paper & Test Questions
Topic-wise questions to test your understanding of every concept in this chapter.
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What is the area of a trapezium with parallel sides 8 cm and 14 cm and height 5 cm? / समानांतर भुजाओं 8 cm और 14 cm तथा ऊँचाई 5 cm वाले समलंब का क्षेत्रफल क्या है? (a) 55 cm² / 55 cm² (b) 110 cm² / 110 cm² (c) 44 cm² / 44 cm² (d) 70 cm² / 70 cm²
Show answer
(a) — Area of trapezium = ½ × (sum of parallel sides) × height = ½ × (8 + 14) × 5 = ½ × 22 × 5 = 55 cm². / समलंब का क्षेत्रफल = ½ × (समानांतर भुजाओं का योग) × ऊँचाई = ½ × 22 × 5 = 55 cm²।
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A cylindrical tank has radius 7 m and height 10 m. What is its volume? (Use π = 22/7) / एक बेलनाकार टंकी की त्रिज्या 7 m और ऊँचाई 10 m है। उसका आयतन क्या है? (π = 22/7 लें) (a) 1540 m³ / 1540 m³ (b) 1400 m³ / 1400 m³ (c) 2640 m³ / 2640 m³ (d) 4400 m³ / 4400 m³
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(a) — V = πr²h = (22/7) × 7² × 10 = (22/7) × 49 × 10 = 22 × 7 × 10 = 1540 m³. / V = πr²h = (22/7) × 49 × 10 = 1540 m³।
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The lateral surface area of a cuboid with dimensions 6 m × 4 m × 3 m is _____. / 6 m × 4 m × 3 m विमाओं वाले घनाभ का पार्श्व सतह क्षेत्रफल _____ है। (a) 60 m² / 60 m² (b) 120 m² / 120 m² (c) 108 m² / 108 m² (d) 72 m² / 72 m²
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(a) — Lateral Surface Area = 2h(l + b) = 2 × 3 × (6 + 4) = 6 × 10 = 60 m². The two rectangular base faces are not included. / पार्श्व सतह क्षेत्रफल = 2h(l + b) = 2 × 3 × 10 = 60 m²। दोनों आयताकार आधार फलक शामिल नहीं हैं।
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The formula for the area of a rhombus using its diagonals d₁ and d₂ is _____. / विकर्णों d₁ और d₂ वाले समचतुर्भुज के क्षेत्रफल का सूत्र _____ है।
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Area = ½ × d₁ × d₂ / क्षेत्रफल = ½ × d₁ × d₂ — The diagonals of a rhombus bisect each other at right angles, creating 4 right triangles; their combined area gives ½ × d₁ × d₂. / समचतुर्भुज के विकर्ण समकोण पर समद्विभाजित होते हैं, जिससे 4 समकोण त्रिभुज बनते हैं।
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1 m³ is equal to _____ litres. / 1 m³ _____ लीटर के बराबर है।
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1000 litres / 1000 लीटर — Since 1 m = 100 cm, 1 m³ = 100 × 100 × 100 cm³ = 1,000,000 cm³ = 1000 litres (because 1 litre = 1000 cm³). / 1 m = 100 cm, इसलिए 1 m³ = 1,000,000 cm³ = 1000 लीटर (1 लीटर = 1000 cm³)।
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True or False: The circumference of a circle with radius 7 cm is 44 cm when π = 22/7. / सत्य या असत्य: π = 22/7 लेने पर 7 cm त्रिज्या वाले वृत्त की परिधि 44 cm है।
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True / सत्य — C = 2πr = 2 × (22/7) × 7 = 2 × 22 = 44 cm. The factor 7 cancels neatly with the denominator of 22/7. / C = 2 × (22/7) × 7 = 44 cm। 7 और 22/7 के हर का गुणनफल ठीक आता है।
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A rectangular room is 8 m long and 5 m wide. What area of carpet (in m²) is needed to cover the floor? / एक आयताकार कमरा 8 m लंबा और 5 m चौड़ा है। फर्श ढकने के लिए कितने m² कालीन की आवश्यकता है?
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40 m² / 40 m² — Area = length × breadth = 8 × 5 = 40 m². Always use the same unit for both dimensions before computing area. / क्षेत्रफल = लंबाई × चौड़ाई = 8 × 5 = 40 m²। क्षेत्रफल निकालने से पहले दोनों मापों की इकाई एकसमान करें।
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Find the total surface area of a cube whose side is 5 cm. / 5 cm भुजा वाले घन का कुल सतह क्षेत्रफल ज्ञात कीजिए।
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TSA = 6a² = 6 × 5² = 6 × 25 = 150 cm² / कुल सतह क्षेत्रफल = 6a² = 6 × 25 = 150 cm² — A cube has 6 congruent square faces, each with area a² = 25 cm². Total = 6 × 25 = 150 cm². / घन के 6 सर्वांगसम वर्गाकार फलक होते हैं, प्रत्येक का क्षेत्रफल 25 cm²। कुल = 150 cm²।
Related Laws & Principles
Explore allFoundational laws & principles behind this chapter. Each one opens a full page — what it says, why it matters, five practice questions and the mistakes to avoid.