Measuring Space: Perimeter and Area

1. Chapter at a glance

• Perimeter of any shape is the total length around its border.
• The ratio of circumference (C) to diameter (D) of any circle is a constant called π (pi), which is irrational.
• Circumference of a circle = 2πr (or πd); arc length of a sector with central angle θ° = (θ/360) × 2πr.
• Area of a rectangle = length × breadth; area of a parallelogram = base × height.
• Area of a triangle = (1/2) × base × height; a median divides a triangle into two triangles of equal area.
• Heron’s formula gives the area of a triangle when all three sides are known.
• Brahmagupta’s formula gives the area of a cyclic quadrilateral when all four sides are known.
• Area of a circle = πr²; area of a sector with central angle θ° = (θ/360) × πr².

2. Definitions, theorems and results

• Definition (Perimeter): Given any shape, its perimeter is the total length around its border.
• Definition (Circumference): The perimeter of a circle is called its circumference.
• Definition (C/D ratio / π): The ratio of the circumference to the diameter of a circle is the same for all circles; this constant is denoted by π.
• Definition (Irrational number): A number that cannot be written as a ratio of two integers (with non-zero denominator) is irrational. π is irrational.
• Theorem (Median divides triangle into equal areas): A median of a triangle divides it into two triangles with equal area. (Proof given in text; required in exam.)
• Result (Heron’s formula): For a triangle with sides a, b, c and semi-perimeter s = (a + b + c)/2, area = √[s(s − a)(s − b)(s − c)]. (Stated; proof in later grade.)
• Result (Brahmagupta’s formula): For a cyclic quadrilateral with sides a, b, c, d and semi-perimeter s = (a + b + c + d)/2, area = √[(s − a)(s − b)(s − c)(s − d)]. (Stated; proof not required.)
• Result (Sector and arc formulae): Derived from rotational symmetry of the circle; no separate named theorem.

3. Formula sheet

4. Solved-example patterns

• Perimeter/circumference problems: Identify shape(s); apply perimeter formula directly or convert given data (e.g., diameter to radius); use π ≈ 22/7 when specified.
• Arc-length problems: Use arc formula (θ/360) × 2πr after confirming central angle and radius.
• Area of triangle/parallelogram/rectangle: Choose base and corresponding height; apply (1/2)bh or bh; verify units.
• Heron’s formula problems: Compute semi-perimeter s first; substitute into √[s(s−a)(s−b)(s−c)]; verify result matches (1/2)bh when possible.
• Brahmagupta’s formula problems: Confirm quadrilateral is cyclic; compute s; substitute into √[(s−a)(s−b)(s−c)(s−d)].
• Circle area/sector problems: Use πr² or (θ/360)πr²; for segments combine sector and triangle areas when needed.
• Composite/perimeter-area figures: Decompose into standard shapes (rectangles, semicircles, sectors); add/subtract areas or lengths carefully; account for shared boundaries.

5. Common mistakes and exam pitfalls

• Using 22/7 for π without the problem stating “unless otherwise stated”; forgetting to write π ≠ 22/7.
• Confusing radius with diameter in circumference (2πr vs πd) or area (πr²).
• Omitting units or mixing cm with cm².
• For Heron’s formula: forgetting to halve the perimeter or using full perimeter as s.
• For sector/arc: using θ in radians instead of degrees or forgetting the 360 denominator.
• In median theorem: assuming the two triangles are congruent (they are equal in area only).
• In Brahmagupta problems: applying the formula to a non-cyclic quadrilateral.
• Sign errors when subtracting areas in composite figures or when finding segments.
• Missing “two semicircles make one full circle” when calculating track or flower perimeters/areas.