Measuring Space: Perimeter and Area (Class 9): Notes और MCQ Practice

आसान भाषा में

किसी भी 2D shape की बाहरी boundary की कुल लंबाई उसका Perimeter होती है और उसके अंदर घिरी हुई जगह उसका Area होती है। किसी भी circle के लिए Circumference और Diameter का ratio हमेशा constant (π) रहता है। Parallelogram का area base × height होता है, और triangle का area उसका आधा (1/2 × base × height) होता है।

Important Formulas & Points

• Perimeter:
• Square = 4a
• Rectangle = 2(a + b)
• Circle (Circumference) = 2πr = πd
• π (Pi): Circumference/Diameter का ratio है। यह एक irrational number है (decimal कभी खत्म या repeat नहीं होता)। Common approximations: 22/7, 3.1416, 355/113.
• Arc Length: Arc length = 2πr × (θ ÷ 360°)
• Area:
• Rectangle = a × b
• Parallelogram = b × h (base × height)
• Triangle = 1/2 × b × h
• Median Theorem: किसी triangle का median उसे equal area वाले दो triangles में divide करता है।

Memory Tricks

1. Digits of π याद रखने की ट्रिक: "How I wish I could recollect pi" — शब्दों के letters गिनें: 3, 1, 4, 1, 5, 9, 2 → π ≈ 3.141592.
2. Arc Length: पूरा circle 360° होता है, इसलिए arc सिर्फ full circle (2πr) का θ/360° fraction है → 2πr × (θ ÷ 360°).

Common Mistakes

1. Perimeter of a Sector: बच्चे अक्सर सिर्फ arc length निकालकर छोड़ देते हैं। याद रखें, sector के perimeter में दोनों straight radii भी जुड़ेंगे: Arc length + 2r.
2. π = 22/7 समझना: π rational नहीं है; 22/7 सिर्फ एक practical approximation है (π ≈ 22/7, परंतु π ≠ 22/7).

Measuring Space: Perimeter and Area

1. Chapter in Short

This chapter introduces the fundamental concepts of measuring two-dimensional space through perimeter (the total boundary length) and area (the surface space covered). It explores how perimeters of polygons compare to their dimensions, delves into the fascinating history and geometry of the circle's circumference-to-diameter ratio (the irrational number π), explains how to calculate arc lengths and lane staggers in running tracks, and demonstrates how areas of rectangles, parallelograms, and triangles are connected—culminating in the theorem that a median divides a triangle into two triangles of equal area.

2. Key Concepts Explained Simply

Perimeter: The Boundary Walk

Imagine a tiny insect walking along the outer edge of a closed shape without turning back until it returns to where it started. The total distance walked is the perimeter.

• For a square of side a, perimeter = 4a (Ratio of perimeter to side is always 4:1).
• For an equilateral triangle of side a, perimeter = 3a (Ratio of perimeter to side is always 3:1).
• For a rectangle of length a and width b, perimeter = 2(a + b).

The C/D Ratio and the Story of π (Pi)

For every circle, no matter how small or large, dividing its circumference (C) by its diameter (D) always gives the exact same constant ratio:

• C ÷ D = π (read as pi, like 'pie').
• Therefore, Circumference C = π × D = 2πr, where r is the radius.
• π is an irrational number: Its decimal expansion never terminates and never repeats in a pattern. Johann Lambert proved that π is irrational in 1761.
• Fractions like 22 ÷ 7 and 355 ÷ 113 are useful approximations, but π is not exactly equal to them.

Length of an Arc of a Circle

An arc is a portion of a circle's circumference. If an arc subtends an angle θ° at the centre of a circle of radius r:

• A full circle is 360° with length 2πr.
• A semicircle is 180°, so length = (180° ÷ 360°) × 2πr = πr.
• A quarter circle is 90°, so length = (90° ÷ 360°) × 2πr = (πr) ÷ 2.
• For any angle θ°, Arc Length = (θ° ÷ 360°) × 2πr.

Running Tracks and Lane Staggers

On a standard 400 m running track, runners in the outer lanes run curves with larger radii than those in the inner lanes. Because a larger radius means a longer semicircle, race organisers place starting marks for outer lanes ahead of the inner lanes. This distance between adjacent starting points is called the stagger, ensuring every runner runs the exact same total distance.

Area of Polygons

Area is the measure of flat space enclosed by a boundary, measured using a 1 × 1 square as 1 square unit (unit²).

• Rectangle: Area = base × height = a × b.
• Parallelogram: A parallelogram can be cut and rearranged into a rectangle of the same base and height. Therefore, Area = base × height = bh.
• Triangle: Two congruent copies of any triangle fit together to form a parallelogram. Therefore, the triangle covers half the area of that parallelogram: Area = 1/2 × base × height = 1/2 bh.

The Median Property of a Triangle

A median is a line drawn from any vertex of a triangle to the midpoint of the opposite side.

• Theorem: A median of a triangle divides it into two triangles of equal area.
• Why? Both smaller triangles share the exact same perpendicular height and have bases of equal length (since the side was divided into two equal halves).

3. Important Definitions, Formulas, Facts, and Dates

Formulas

• Perimeter of a Square = 4a
• Perimeter of a Rectangle = 2(a + b)
• Circumference of a Circle = 2πr = πD
• Length of an Arc = (θ° ÷ 360°) × 2πr
• Perimeter of a Sector = Arc Length + 2r = [(θ° ÷ 360°) × 2πr] + 2r
• Area of a Rectangle = length × breadth = ab
• Area of a Parallelogram = base × height = bh
• Area of a Triangle = 1/2 × base × height = 1/2 bh

Historical Values and Approximations of π

• Mesopotamia (c. 1900 BCE): Estimated π = 3 + 1/8 = 3.125 (comparing an inscribed hexagon to the circle).
• Archimedes (c. 250 BCE): Used inscribed and circumscribed polygons up to 96 sides to trap π: 3 10/71 < π < 3 1/7 (between approx. 3.1408 and 3.1428).
• Ptolemy of Alexandria (c. 150 CE): 377 ÷ 120 ≈ 3.14167.
• Liu Hui (263 CE): Developed the 'circle-cutting method'.
• Zu Chongzhi (480 CE): Used a 24,576-sided polygon to give the Yuelü (approximate ratio) 22 ÷ 7 ≈ 3.1428 and the Miü (close ratio) 355 ÷ 113 ≈ 3.1415929.
• Āryabhaṭa (499 CE): Stated π ≈ 62832 ÷ 20000 = 3.1416, calling it asanna (approaching or approximate).
• Brahmagupta (628 CE): Used √10 ≈ 3.1622 for algebraic ease.
• Mādhava of Sangamagrāma (c. 14th century): Discovered the first exact infinite series: π ÷ 4 = 1 − 1/3 + 1/5 − 1/7 + ... and calculated π to 11 decimal places (3.14159265358).
• William Jones (1706): First used the Greek letter π (from Greek perimetros).
• Leonhard Euler: Popularized the symbol π.
• Johann Lambert (1761): Proved that π is irrational.

Celebrations and Mnemonics

• Pi Day: March 14 (3-14).
• Pi Approximation Day: 22 July (22/7).
• Mnemonic for digits of π: "How(3) I(1) wish(4) I(1) could(5) recollect(9) pi(2)" = 3.141592...

4. Solved Examples

Example 1: Finding the Radius from Circumference

Question: The perimeter of a circle is 44 cm. Taking π = 22 ÷ 7, find its radius.

Solution:

• Circumference C = 2 × π × r
• 44 = 2 × (22 ÷ 7) × r
• 44 = (44 ÷ 7) × r
• r = (44 × 7) ÷ 44 = 7 cm

Answer: Radius = 7 cm.

Example 2: Length of an Arc and Perimeter of a Sector

Question: Find the length of the arc and the total perimeter of a sector of a circle with radius 14 cm and sector angle 75° (Use π = 22 ÷ 7).

Solution:

1. Length of the arc:

• Arc length = (θ° ÷ 360°) × 2πr
• Arc length = (75 ÷ 360) × 2 × (22 ÷ 7) × 14
• Arc length = (5 ÷ 24) × 2 × 22 × 2
• Arc length = (5 ÷ 24) × 88 = 440 ÷ 24 = 55 ÷ 3 = 18.33 cm

1. Perimeter of the sector:

• A sector is bounded by the arc and two radii.
• Perimeter = Arc length + 2r
• Perimeter = 18.33 + (2 × 14) = 18.33 + 28 = 46.33 cm (or 46 1/3 cm)

Answer: Arc length = 18.33 cm; Perimeter of sector = 46.33 cm.

Example 3: Tyre Rotations and Distance

Question: The diameter of a car tyre is 56 cm. (i) How far does the car travel in 1 revolution? (ii) How many revolutions does the tyre make if the car travels 10 km?

Solution:

1. Distance in one revolution = Circumference of the tyre

• D = 56 cm
• Distance = π × D = (22 ÷ 7) × 56 = 22 × 8 = 176 cm = 1.76 m

1. Number of revolutions in 10 km:

• Total distance = 10 km = 10 × 1000 m = 10,000 m = 1,000,000 cm
• Revolutions = Total distance ÷ Distance per revolution
• Revolutions = 1,000,000 ÷ 176 ≈ 5681.82 revolutions (approx. 5682 complete revolutions).

Answer: (i) 176 cm (1.76 m); (ii) approximately 5682 revolutions.

Example 4: Semicircular Paths Comparison

Question: Points P and Q lie on a straight line. Path 1 is a single semicircle on diameter PQ. Path 2 consists of three smaller semicircles whose diameters lie along PQ and add up to the total length PQ. Which path is longer?

Solution:

• Let the diameter of the large semicircle be D, so its radius is D ÷ 2.
• Length of Path 1 = π × (D ÷ 2) = (π × D) ÷ 2.
• Let the diameters of the three small semicircles be d1, d2, and d3, such that d1 + d2 + d3 = D.
• Length of Path 2 = (π × d1 ÷ 2) + (π × d2 ÷ 2) + (π × d3 ÷ 2) = [π × (d1 + d2 + d3)] ÷ 2 = (π × D) ÷ 2.
• Both expressions are identical.

Answer: Both paths have equal length.

5. Common Mistakes and Quick Revision Points

• Confusing π with 22/7: π is an irrational number; 22 ÷ 7 is a rational approximation. Do not write π = 22/7; write π ≈ 22/7.
• Perimeter of a Sector: Many students calculate only the arc length and forget to add the two radii forming the straight edges of the sector. Perimeter = Arc length + 2r.
• Perimeter vs. Area Units: Always check your units! Perimeter is a 1-D length (cm, m), while area is a 2-D surface measure (cm², m²).
• Parallelogram Height vs. Slant Side: The area of a parallelogram is base × perpendicular height, NOT base × adjacent side.
• Median divides area, not perimeter: A median splits a triangle into two parts of equal area, but the perimeters of those two parts are generally different.