Perimeter and Area — Important Questions Bank
Chapter 6 · Ganita Manjari · Grade 9 Part I · All questions strictly from chapter content
10 Long Answer (5 marks)
Theory + Numerical + Application
Complete Step-by-step Solutions
Part A — Short Answer Questions
15 questions · 2–3 marks each · Theory, Numerical, Application
Perimeter is the total length around the border of a shape — the total distance an insect would travel if it walked along the border and returned to the starting point.
(i) Square: Perimeter = 4a units
(ii) Rectangle: Perimeter = 2(a + b) units
(iii) Equilateral triangle: Perimeter = 3a units
Note: The square’s formula is a special case of the rectangle formula when a = b.
π (pi) is the ratio of the circumference (C) to the diameter (D) of a circle: π = C/D. This ratio is constant for all circles — a fundamental property of circles.
The ratio stays fixed because circles of different sizes are geometrically similar (same shape, different scale). Just as the ratio of perimeter to side is always 4:1 for all squares, the C/D ratio is always π for all circles.
Two common approximations: π ≈ 22/7 and π ≈ 3.14.
A much more accurate fraction is 355/113.
A number is irrational if it cannot be written as a fraction a/b where a and b are integers. The decimal expansion of an irrational number goes on forever with no repeating pattern.
π is irrational because its digits (3.14159265358…) continue indefinitely with no repeating cycle — unlike fractions such as 1/3 = 0.333… which have a pattern.
This means 22/7 is NOT exactly equal to π — it is only an approximation. We write π ≈ 22/7. Since π is irrational, there is no “best fraction” for π; we can always find a closer fraction, but never an exact one.
(ii) Parallelogram with base b and height h: Area = bh sq. units
(iii) Triangle with base b and height h: Area = ½ × b × h sq. unitsDerivation of parallelogram area: A parallelogram ABCD can be transformed into a rectangle by cutting the triangular portion from one side and attaching it to the other side. The resulting rectangle has the same base b and the same height h. Since they have equal area: Area of parallelogram = base × height = bh.
Theorem: A median of a triangle divides it into two triangles with equal area.
If AD is a median of △ABC (D is the midpoint of BC), then:
Area(△ABD) = Area(△ACD) = ½ × Area(△ABC)
Proof: BD = DC (D is midpoint), and both triangles △ABD and △ACD have the same height h from A. So Area = ½ × base × height is the same for both.
Why surprising: △ABD and △ACD are generally not congruent (different shapes), yet they have exactly equal areas. Equal area does not imply congruence!
where s = ½(a + b + c) is the semi-perimeter
The circle is cut into many thin pie-shaped slices. These slices are rearranged alternately to form a shape that looks like a parallelogram.
As the slices become thinner and thinner:
• The base of the parallelogram approaches half the circumference = πr
• The height approaches the radius = r
Area of circle = Area of parallelogram = base × height = πr × r = πr².
This beautifully connects the formula for circumference with the formula for area.
Brahmagupta’s Formula: For a cyclic quadrilateral with sides a, b, c, d and semi-perimeter s = ½(a + b + c + d):
Heron’s Formula as a Special Case: A triangle with sides a, b, c can be thought of as a cyclic quadrilateral with the fourth side d = 0 (two vertices coincide). Setting d = 0:
s = ½(a + b + c + 0) = ½(a + b + c) — same as Heron’s semi-perimeter.
Area = √[s(s−a)(s−b)(s−c)(s−0)] = √[s(s−a)(s−b)(s−c)] — this is exactly Heron’s formula!
Therefore, Brahmagupta’s formula is a generalisation of Heron’s formula.
Why staggers exist: On the straight sections, all lanes cover the same distance. But on the curved (semicircular) sections, outer lanes have a larger radius and therefore a larger circumference — so athletes in outer lanes run farther on the curves. To ensure everyone runs exactly 400 m, athletes in outer lanes start ahead.
Stagger calculation (Lane 1 vs Lane 2):
Part B — Long Answer Questions
10 questions · 5 marks each · Proof, Word Problems, Numerical
Using Heron’s Formula:
- For an equilateral triangle: all sides = a units.
- Semi-perimeter: s = ½(a + a + a) = 3a/2
- s − a = 3a/2 − a = a/2 (same for all three sides)
- Area = √[s(s−a)(s−b)(s−c)] = √[(3a/2)(a/2)(a/2)(a/2)]
- = √[(3a/2) × (a³/8)] = √[3a⁴/16] = (√3/4)a²
Verification using ½ × base × height:
Derivation from Archimedes’ Statement:
- Right-angled triangle with legs = radius (r) and circumference (2πr).
- Area of this triangle = ½ × base × height = ½ × r × 2πr
- = ½ × 2πr² = πr²
Calculation for r = 14 cm:
(i) Area of minor sector (θ = 60°, r = 15 cm):
(ii) Area of minor segment = Area of sector − Area of triangle:
Let the diameters of the three smaller semicircles b, c, d have values 2b’, 2c’, 2d’.
- Since all semicircles fit along PQ: PQ = 2b’ + 2c’ + 2d’
- Also, PQ = 2a’ (diameter of the large semicircle), so a’ = b’ + c’ + d’.
- Length of large semicircle path = πa’
- Length of three-semicircle path = πb’ + πc’ + πd’ = π(b’ + c’ + d’) = πa’
All rectangles are cyclic quadrilaterals (vertices lie on a circle).
- Sides of rectangle: a, b, a, b. Semi-perimeter s = ½(a + b + a + b) = a + b.
- s − a = b, s − b = a, s − a = b, s − b = a
- Area = √[(s−a)(s−b)(s−a)(s−b)] = √[(b)(a)(b)(a)] = √[a²b²]
- = ab
Setting up the geometry:
- Let the two circles have centres A and B, each with radius r. The circles intersect at C and D.
- Triangle ABC: AB = r, AC = r, BC = r → equilateral triangle. So ∠CAB = 60° = ∠CBA.
- Similarly, ∠DAB = 60° = ∠DBA. So ∠CAD = 120° = ∠CBD.
- Each dotted arc (inside the overlap) subtends 120° at its centre → it is 120/360 = 1/3 of circumference.
- Each outer (red) arc = remaining 2/3 of circumference.
- Total perimeter = 2 outer arcs = 2 × (2/3) × 2πr = (8/3)πr
For r = 6 cm:
(i) Finding the other leg:
(ii) Finding the hypotenuse:
(iii) Perimeter:
Leave a Reply