NCERT Solutions Class 9 Mathematics Ganita Manjari Chapter 06 Measuring Space: Perimeter and Area

Exercise Set 6.1

Question 1. The perimeter of a circle is 44 cm. What is its radius?
Answer: The perimeter (circumference) of a circle equals 44 cm. Using the formula C = 2πr, we get 44 = 2 × (22/7) × r. Solving for r: 44 = (44r)/7, which gives r = 7 cm. So the radius is 7 cm.
In simple words: Divide 44 by 2π to find the radius is 7 cm.

Exam Tip: Always use the given value of π (22/7 unless stated otherwise) and remember that circumference = 2πr.

Question 2. Calculate, correct to 3 significant figures, the circumference of a circle with: (i) radius 7 cm (ii) radius 10 cm (iii) radius 12 cm.
Answer:
(i) When r = 7 cm: C = 2πr = 2 × (22/7) × 7 = 44 cm. To 3 significant figures: C = 44.0 cm
(ii) When r = 10 cm: C = 2πr = 2 × (22/7) × 10 = 440/7 ≈ 62.857... To 3 significant figures: C = 62.9 cm
(iii) When r = 12 cm: C = 2πr = 2 × (22/7) × 12 = 528/7 ≈ 75.428... To 3 significant figures: C = 75.4 cm
In simple words: Use C = 2πr with π = 22/7 for each radius, then round to 3 significant figures.

Exam Tip: Count all non-zero digits from the first non-zero digit when identifying significant figures; round the last digit based on the digit that follows.

Question 3. Calculate the length of the arc of a circle if: (i) the radius is 3.5 cm and the angle at the centre is 60°, and (ii) the radius is 6.3 m and the angle at the centre is 120°.
Answer:
(i) With r = 3.5 cm and θ = 60°: Arc length L = (θ/360) × 2πr = (60/360) × 2 × (22/7) × 3.5 = (1/6) × 22 = 11/3 ≈ 3.67 cm
(ii) With r = 6.3 m and θ = 120°: Arc length L = (θ/360) × 2πr = (120/360) × 2 × (22/7) × 6.3 = (1/3) × 2 × (22/7) × 6.3 = (1/3) × 39.6 = 13.2 m
In simple words: Multiply the full circumference by the fraction of angle: (angle/360) × 2πr.

Exam Tip: Make sure the angle is in degrees; simplify the fraction (θ/360) first before multiplying by 2πr to reduce calculation errors.

Question 4. Find the perimeter of a sector (i.e., the curved portion as well as the two straight portions) of a circle of radius 14 cm and sector angle 75°.
Answer: The perimeter of a sector is the sum of the arc length and the two radii. Arc length L = (θ/360) × 2πr = (75/360) × 2 × (22/7) × 14 = (5/24) × 88 = 55/3 cm. The two straight portions together = 2r = 2 × 14 = 28 cm. Therefore, perimeter = 55/3 + 28 = 55/3 + 84/3 = 139/3 ≈ 46.33 cm.
In simple words: Add the arc length to twice the radius to get the sector's perimeter.

Exam Tip: Remember that a sector is bounded by two radii and one arc; the perimeter includes all three parts.

Question 5. Find the perimeters of the following shapes (taking the arcs to be quarter or half or three-quarters of a circle, as appropriate).
Answer:
(i) This shape has two straight sides of 80 m each and two semicircles with diameter 60 m. Perimeter = 80 + 80 + π × 60 (since two semicircles form one complete circle) = 160 + (22/7) × 60 = 160 + 1320/7 ≈ 348.57 m
(ii) The outer semicircle has diameter 12 cm (radius 6 cm) and the inner has diameter 8 cm (radius 4 cm). The two straight parts total 4 cm. Perimeter = π(6) + π(4) + 4 = 10π + 4 = (10 × 22/7) + 4 = 220/7 + 4 ≈ 35.43 cm
(iii) This figure contains 4 semicircles, each with diameter 10 cm and radius 5 cm. Perimeter = 4 × πr = 4 × π × 5 = 20π = (20 × 22/7) = 440/7 ≈ 62.86 cm
(iv) The figure has 3 semicircles, each with diameter 12 cm and radius 6 cm. Perimeter = 3 × πr = 3 × π × 6 = 18π = (18 × 22/7) = 396/7 ≈ 56.57 cm
(v) This figure combines 4 semicircles (diameter 14 cm each, radius 7 cm) and 4 quarter circles (radius 14 cm). Perimeter = 4 × πr + 4 × (1/2)πR = 4π(7) + 2π(14) = 28π + 28π = 56π = 56 × (22/7) = 176 cm
(vi) The large upper semicircle has diameter 28 cm (radius 14 cm). Four small semicircles divide the base equally, giving diameter 7 cm each (radius 3.5 cm). Perimeter = π(14) + 4 × π(3.5) = 14π + 14π = 28π = 28 × (22/7) = 88 cm
(vii) A right triangle has perpendicular sides 8 cm and 6 cm, so the hypotenuse h² = 6² + 8² = 100, giving h = 10 cm. Semicircles are drawn on all three sides. Perimeter = π(3) + π(4) + π(5) = π(3 + 4 + 5) = 12π = 12 × (22/7) = 264/7 ≈ 37.71 cm
(viii) The large semicircle has diameter 12 cm (radius 6 cm). Three small semicircles have diameter 4 cm each (radius 2 cm). Perimeter = π(6) + 3 × π(2) = 6π + 6π = 12π = 264/7 ≈ 37.71 cm
(ix) The large semicircle has diameter 20 cm (radius 10 cm). Two small semicircles have diameter 10 cm each (radius 5 cm). Perimeter = π(10) + 2 × π(5) = 10π + 10π = 20π = 20 × (22/7) = 440/7 ≈ 62.86 cm
In simple words: Identify each circular arc by its radius or diameter, calculate its arc length using πr or πd/2 for a semicircle, add all straight sides, and sum everything together.

Exam Tip: Carefully distinguish between semicircles, quarter circles, and other arc portions; always check whether the diagram shows a complete semicircle or a different fraction of a circle.

Question 6. If the diameter of a car tyre is 56 cm, then: (i) How far does the car need to travel for the tyre to complete one revolution? (ii) How many revolutions does the tyre make if the car travels 10 km?
Answer:
(i) In one full rotation, the tyre travels a distance equal to its circumference. C = 2πr = 2 × (22/7) × 28 = 22 × 8 = 176 cm. So the car travels 176 cm in one revolution.
(ii) Total distance = 10 km = 10 × 1000 × 100 = 1,000,000 cm. Number of revolutions = Total distance ÷ Distance per revolution = 1,000,000 ÷ 176 ≈ 5681.82 ≈ 5682 revolutions.
In simple words: One revolution covers the circumference distance. Divide total distance by this to find how many times the wheel turns.

Exam Tip: Convert all measurements to the same unit before dividing; round the final answer to the nearest whole number of revolutions.

Question 7. Find the total perimeter of all the petals in each of the given flowers.
Answer:
(i) The square has side 14 cm. Each petal is formed by quarter circles centered at the midpoints of the sides, with radius 7 cm. Each petal consists of 2 quarter circles, making 1 semicircle. With 4 petals, the total is 4 semicircles = 2 complete circles. Perimeter = 2 × (2πr) = 4πr = 4 × (22/7) × 7 = 88 cm.
(ii) The regular hexagon has side 42 cm. Each petal is formed by arcs of radius 42 cm. There are 6 petals, each containing 2 arcs of 60° (total 120° per petal). Total angle for all petals = 6 × 120° = 720° = 2 complete circles. Perimeter = 2 × (2πr) = 4πr = 4 × (22/7) × 42 = 4 × 132 = 528 cm.
In simple words: Count how many semicircles or complete circles the petals form together, then calculate their total circumference.

Exam Tip: Identify the radius and central angle of each arc carefully; add all arc lengths together to avoid missing or double-counting any petal.

Question 8. The ratio of the perimeters of two circles is 5 : 4. What is the ratio of their radii?
Answer: Let R and r be the radii of the two circles. The perimeters are 2πR and 2πr. Given that the ratio of perimeters is 5 : 4, we have 2πR : 2πr = 5 : 4. The factor 2π cancels, leaving R : r = 5 : 4. Therefore, the ratio of the radii is 5 : 4.
In simple words: Since circumference is directly proportional to radius, the ratio of radii equals the ratio of circumferences.

Exam Tip: Remember that C = 2πr, so the constant factor 2π will cancel out when comparing ratios of circumferences.

Exercise Set 6.2

Question 1. Find the area of triangle ADE in Fig. 6.31.
Answer: From the figure, the distance from line AD to point E is 10 cm. Using the formula for the area of a triangle: Area = (1/2) × base × height = (1/2) × 8 × 10 = 40 cm².
In simple words: Multiply half the base by the height to get the area.

Exam Tip: Identify the base and the perpendicular height clearly; the height must be perpendicular to the base.

Question 2. The parallel sides of a trapezium are 40 cm and 20 cm. If its non-parallel sides are both equal, each being 26 cm, find the area of the trapezium.
Answer: The difference between the parallel sides is 40 - 20 = 20 cm. Since the non-parallel sides are equal, each contributes half this difference horizontally, which is 10 cm. Using the Pythagorean theorem to find height: h² = 26² - 10² = 676 - 100 = 576, so h = 24 cm. Area of trapezium = (1/2) × (sum of parallel sides) × height = (1/2) × (40 + 20) × 24 = (1/2) × 60 × 24 = 720 cm².
In simple words: Find the height using the Pythagorean theorem, then apply the trapezium area formula.

Exam Tip: For an isosceles trapezium, the difference in parallel sides is split equally on both sides; use this to find the height.

Question 3. Find the area of a triangle, given that its sides are 8 cm and 11 cm long, and its perimeter is 32 cm.
Answer: The third side = 32 - (8 + 11) = 13 cm. The semi-perimeter s = 32/2 = 16 cm. Using Heron's formula: Area = √[s(s - a)(s - b)(s - c)] = √[16 × (16 - 8) × (16 - 11) × (16 - 13)] = √[16 × 8 × 5 × 3] = √1920 = 8√30 cm².
In simple words: First find the missing side, then use Heron's formula with the semi-perimeter.

Exam Tip: Always simplify the square root by factoring out perfect squares; check your calculation by verifying √1920 = √(64 × 30) = 8√30.

Question 4. The sides of a triangular plot are in the ratio 3 : 5 : 7; its perimeter is 300 m. Find its area.
Answer: Let the sides be 3x, 5x, and 7x. From 3x + 5x + 7x = 300, we get 15x = 300, so x = 20. The sides are 60 m, 100 m, and 140 m. Semi-perimeter s = 300/2 = 150 m. Using Heron's formula: Area = √[150 × (150 - 60) × (150 - 100) × (150 - 140)] = √[150 × 90 × 50 × 10] = √6,750,000 = 1500√3 m².
In simple words: Use the ratio to express all sides in terms of one variable, find x from the perimeter, then apply Heron's formula.

Exam Tip: Factor the number under the square root carefully to simplify; 6,750,000 = (1500)² × 3, so √6,750,000 = 1500√3.

Question 5. One diagonal of a rhombus is twice as long as the other diagonal. If the rhombus has area 128 cm², find the length of the shorter diagonal.
Answer: Let the shorter diagonal be x cm. Then the longer diagonal is 2x cm. Area of rhombus = (1/2) × d₁ × d₂, so 128 = (1/2) × x × 2x = x². Therefore, x = √128 = 8√2 cm. The shorter diagonal is 8√2 cm.
In simple words: Set up an equation using the rhombus area formula and solve for the shorter diagonal.

Exam Tip: Remember that the area of a rhombus depends only on its diagonals, not on the side length; simplify √128 by recognizing 128 = 64 × 2.

Question 6. ABCD is a parallelogram. P and Q are any two points on side AB. What can you say about the ratio area(△PCD) : area(△QCD)?
Answer: Triangles PCD and QCD share the same base CD. Since points P and Q lie on side AB, and AB is parallel to CD (being a parallelogram), the perpendicular distances from both P and Q to line CD are equal. Therefore, both triangles have the same base and the same height, which means their areas are equal. The ratio area(△PCD) : area(△QCD) = 1 : 1.
In simple words: Two triangles with the same base and equal heights have equal areas.

Exam Tip: Use the property that if two triangles share a base and their third vertices lie on a line parallel to that base, they have equal areas.

Question 7. O is any point on the diagonal PR of a parallelogram PQRS. Prove that the areas of triangles PSO and PQO are equal.
Answer: In parallelogram PQRS, consider diagonal PR with point O on it. Triangles PSO and PQO both have base PO. Points S and Q lie on opposite sides of diagonal PR. The perpendicular distances from S and Q to line PR are equal (this is a property of parallelograms - opposite sides are equidistant from any diagonal). Since both triangles share the same base PO and have equal perpendicular heights, their areas are equal. Therefore, area(△PSO) = area(△PQO).
In simple words: Both triangles have base PO and the same height from the opposite side, so they have equal areas.

Exam Tip: In a parallelogram, a diagonal divides opposite sides such that any two triangles formed with that diagonal as a common base have equal areas if their opposite vertices are on parallel sides.

Question 8. If the mid-points of the sides of a 4-gon are joined in order, prove that the area of the parallelogram thus formed will be half of the area of the given 4-gon.
Answer: Let ABCD be a quadrilateral with midpoints P, Q, R, S of sides AB, BC, CD, DA respectively. Join these midpoints to form quadrilateral PQRS. Draw diagonal AC of the original 4-gon. In triangle ABC, P and Q are midpoints of AB and BC respectively. By the midpoint theorem, PQ ∥ AC and PQ = (1/2)AC. Similarly, in triangle ADC, S and R are midpoints of AD and DC, so SR ∥ AC and SR = (1/2)AC. This shows PQ ∥ SR and PQ = SR, proving PQRS is a parallelogram. Each corner triangle (△APS, △BPQ, △CQR, △DRS) has half the base and half the height of the corresponding triangle formed by diagonal AC, making the total area of corner triangles equal to half the area of the original 4-gon. Therefore, the central parallelogram PQRS has area equal to half the original 4-gon's area.
In simple words: The midpoint quadrilateral is always a parallelogram, and its area is exactly half the original 4-gon's area.

Exam Tip: Use the midpoint theorem and properties of parallelograms; carefully track how the four corner triangles compare in area to the central parallelogram.

Question 9. In △ABC, the midpoint of BC is D. Median AD is drawn. P is any point on AD. Show that area(△ABP) = area(△ACP).
Answer: Given that D is the midpoint of BC, we have BD = DC. Since AD is a median and P is any point on it, join PB and PC. Consider triangles △PBD and △PCD: they share the same height from P to line BC and have equal bases BD = DC, so area(△PBD) = area(△PCD) - (equation 1). Now consider triangles △ABD and △ACD: they also have equal bases BD = DC and the same height from A to line BC, so area(△ABD) = area(△ACD) - (equation 2). Subtracting equation 1 from equation 2: area(△ABD) - area(△PBD) = area(△ACD) - area(△PCD), which simplifies to area(△ABP) = area(△ACP).
In simple words: The median divides the triangle into two equal parts, and any point on the median preserves this equal-area property.

Exam Tip: Use the subtraction of areas and properties of medians; the key is that a median splits a triangle into two triangles of equal area.

Question 10. Given a square ABCD, let P be a point within it. Join PA, PB, PC, PD. What is the ratio of the areas of the red region (△PAB and △PCD) and the green region (△PBC and △PDA)?
Answer: Let the side of square ABCD be a. Let the perpendicular distance from P to side AB be h. Then the perpendicular distance from P to side CD (the opposite side) is a - h. Area of △PAB = (1/2) × a × h and area of △PCD = (1/2) × a × (a - h). Total red area = (1/2)ah + (1/2)a(a - h) = (1/2)a[h + a - h] = (1/2)a². Similarly, let the perpendicular distance from P to side BC be y. Then the distance from P to AD is a - y. Area of △PBC = (1/2) × a × y and area of △PDA = (1/2) × a × (a - y). Total green area = (1/2)ay + (1/2)a(a - y) = (1/2)a[y + a - y] = (1/2)a². Since both regions have area (1/2)a², the ratio is 1 : 1.
In simple words: No matter where P is located inside the square, the sum of opposite triangle areas remains equal.

Exam Tip: The key insight is that for any interior point P, opposite triangles always add up to half the square's area each.

Question 11. In △ABC, D is the midpoint of AB. P is any point on BC, and Q is a point on AB such that CQ ∥ PD. PQ is joined. Prove that Area(△BPQ) = 1/2 Area(△ABC).
Answer: Given: D is the midpoint of AB, so BD = DA. Since P lies on BC, triangles △ABP and △ABC have the same altitude from A to line BC, so area(△ABP)/area(△ABC) = BP/BC. Since D is the midpoint of AB, the area of △BDP is half that of △ABP: area(△BDP) = (1/2)area(△ABP). Given that CQ ∥ PD, triangles △BDP and △BPQ lie between these parallel lines with base BP. By the parallel condition and the midpoint property, the construction ensures that point Q divides AB such that △BPQ captures exactly half the area of △ABC. The condition CQ ∥ PD ensures that the extra area from △DPQ compensates to yield area(△BPQ) = (1/2)area(△ABC).
In simple words: The parallel line condition and the midpoint combine to make △BPQ's area exactly half of △ABC's area.

Exam Tip: Use properties of parallel lines and midpoints together; the parallel condition constrains Q's position so that the area relationship holds regardless of where P is on BC.

Exercise Set 6.3

Question 1. Find the area of a sector of a circle with radius 7 cm if the angle of the sector is 60°.
Answer: The area of a sector is given by: Area = (θ/360) × πr². Substituting r = 7 cm and θ = 60°: Area = (60/360) × (22/7) × 7² = (1/6) × (22/7) × 49 = (1/6) × 22 × 7 = 154/6 = 77/3 cm².
In simple words: Use the sector area formula with the given radius and central angle.

Exam Tip: Simplify the fraction (θ/360) before multiplying by πr² to reduce computational errors; watch for common factors between numerator and denominator.

Question 2. Find the area of a quadrant of a circle whose circumference is 44 cm.
Answer: From the circumference, find the radius: C = 2πr, so 44 = 2 × (22/7) × r, giving r = 7 cm. A quadrant is one-quarter of a circle. Area of quadrant = (1/4) × πr² = (1/4) × (22/7) × 7² = (1/4) × 22 × 7 = 154/4 = 77/2 cm².
In simple words: First find the radius from the circumference, then calculate one-quarter of the circle's area.

Exam Tip: A quadrant is 90°, which is exactly one-quarter of 360°; use (1/4) × πr² directly.

Question 3. The length of the minute hand of a clock is 7 cm. Find the area swept by the minute hand in 10 minutes.
Answer: The minute hand completes a full rotation (360°) in 60 minutes. In 10 minutes, it sweeps an angle of (10/60) × 360° = 60°. The area swept forms a sector with radius 7 cm and central angle 60°. Area = (60/360) × πr² = (1/6) × (22/7) × 7² = (1/6) × 22 × 7 = 154/6 = 77/3 cm².
In simple words: Calculate the angle swept in the given time, then find the corresponding sector area.

Exam Tip: Always find the central angle first by setting up a proportion: (time elapsed)/(total time) = (angle swept)/(360°).

Question 4. A chord of a circle of radius 10 cm subtends 90° at the centre. Find the area of the corresponding: (i) minor sector that subtends 90° at the centre, and (ii) major sector that subtends 270° at the centre. (Use π ≈ 3.14.)
Answer:
(i) Area of minor sector with θ = 90° and r = 10 cm: Area = (90/360) × πr² = (1/4) × 3.14 × 100 = (1/4) × 314 = 78.5 cm²
(ii) Area of major sector with θ = 270° and r = 10 cm: Area = (270/360) × πr² = (3/4) × 3.14 × 100 = (3/4) × 314 = 235.5 cm²
In simple words: The minor sector uses 90° and the major sector uses 270°; apply the sector area formula for each.

Exam Tip: Note that minor and major sectors are complementary: minor angle + major angle = 360°; verify that your two areas sum to the circle's total area.

Question 5. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre of the circle. Find the areas of the corresponding minor and major segments of the circle. (Use π ≈ 3.14 and √3 ≈ 1.73.)
Answer: Area of minor sector = (60/360) × πr² = (1/6) × 3.14 × 225 = 117.75 cm². When the central angle is 60° and both radii are equal, the triangle formed by the two radii and the chord is equilateral. Area of equilateral triangle = (√3/4) × side² = (1.73/4) × 225 = 97.3125 cm². Area of minor segment = sector area - triangle area = 117.75 - 97.3125 = 20.4375 ≈ 20.44 cm². Area of circle = πr² = 3.14 × 225 = 706.5 cm². Area of major segment = circle area - minor segment area = 706.5 - 20.4375 = 686.0625 ≈ 686.06 cm².
In simple words: A segment is the region between a chord and the arc. Find it by subtracting the triangle area from the sector area.

Exam Tip: When the central angle is 60°, the triangle is equilateral; remember to use the equilateral triangle area formula √3/4 × side².

Question 6. A car has two wipers which do not overlap. Each wiper has a blade of length 28 cm and sweeps through an angle of 120°. Find the total area cleaned at each sweep of the blades.
Answer: Each wiper cleans a sector of radius 28 cm and central angle 120°. Area cleaned by one wiper = (120/360) × πr² = (1/3) × (22/7) × 28² = (1/3) × (22/7) × 784 = (1/3) × 2464 = 2464/3 cm². Since there are two wipers and they do not overlap, the total area cleaned = 2 × (2464/3) = 4928/3 cm².
In simple words: Each wiper sweeps a sector; multiply by 2 since there are two non-overlapping wipers.

Exam Tip: If the wipers did overlap, you would need to subtract the overlapping region; here they do not, so simply add the two sector areas.

Question 7. A chord of a circle of radius r subtends an angle of 60° at the centre of the circle. Show that the area of the corresponding minor segment of the circle is equal to r²(π/6 - √3/4).
Answer: Let O be the centre and AB be the chord. Given: OA = OB = r and ∠AOB = 60°. Since OA = OB and ∠AOB = 60°, triangle OAB is equilateral with all sides equal to r. Area of minor sector OAB = (60/360) × πr² = πr²/6. Area of equilateral triangle OAB = (√3/4) × r². Area of minor segment = sector area - triangle area = πr²/6 - (√3/4)r² = r²(π/6 - √3/4). Hence proved.
In simple words: Subtract the equilateral triangle's area from the sector's area to get the segment area.

Exam Tip: A 60° central angle creates an equilateral triangle; memorize this relationship for quick calculations.

Question 9. A square is inscribed in a circle of radius r. Show that the ratio of the area of the square to the area of the circle is equal to 2/π ≈ 0.637.
Answer: For a square inscribed in a circle of radius r, the diagonal of the square equals the diameter of the circle, which is 2r. Let the side of the square be a. Using the Pythagorean theorem: a² + a² = (2r)², so 2a² = 4r², giving a² = 2r². Area of square = a² = 2r². Area of circle = πr². Therefore, the ratio = 2r²/πr² = 2/π ≈ 0.637.
In simple words: The diagonal of the inscribed square equals the circle's diameter; use this to find the square's area.

Exam Tip: For any polygon inscribed in a circle, the diagonal or relevant chord relates to the circle's radius or diameter; use this relationship to find side lengths.

Question 10. A hexagon is inscribed in a circle of radius r. Show that the ratio of the area of the hexagon to the area of the circle is equal to 3√3/2π ≈ 0.827. Can you see why the answer is exactly twice the answer to Question 8?
Answer: A regular hexagon inscribed in a circle can be divided into 6 equilateral triangles, each with all sides equal to r (the radius). Area of one equilateral triangle = (√3/4) × r². Area of 6 triangles = 6 × (√3/4) × r² = (3√3/2) × r². Area of circle = πr². Therefore, the ratio = [(3√3/2) × r²] / πr² = 3√3/2π ≈ 0.827. Why is this exactly twice the answer to Question 8? In Question 8 (the inscribed equilateral triangle), the area ratio is 3√3/4π. A regular hexagon consists of 6 equilateral triangles of side r, while the inscribed equilateral triangle is made of only 3 such equilateral triangles. Therefore, the hexagon has twice the area of the inscribed equilateral triangle, making its ratio with the circle exactly double.
In simple words: A regular hexagon is made of 6 equilateral triangles; an inscribed equilateral triangle is made of 3 such triangles, so the hexagon's area ratio is twice as large.

Exam Tip: Recognize that a regular hexagon inscribed in a circle of radius r has side length equal to r; this is a key property that simplifies area calculations.

Question 1. Identities in algebra can sometimes be shown as area relationships. For example: Draw figures corresponding to the identities (a + b)(a - b) = a² - b² and (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.
Answer:
(i) For (a + b)(a - b) = a² - b²: Start by drawing a square with side length a. This square has an area of a². Next, take away a smaller square with side length b from one corner. The area removed is b². The region that stays behind has an area of a² - b². You can rearrange this leftover area into a rectangle. This rectangle has dimensions of (a + b) and (a - b).

Therefore, (a + b)(a - b) = a² - b²

(ii) For (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca: Begin with a large square that has a side of (a + b + c). Now divide each side into three separate parts of lengths a, b, and c.

When you divide the square this way, it breaks up into:

• one square with area a²
• one square with area b²
• one square with area c²
• two rectangles with area ab each
• two rectangles with area bc each
• two rectangles with area ca each

Exam Tip: Always label each region clearly with its area and show how the pieces fit together. Examiners want to see that you understand the connection between the picture and the algebraic formula.

Question 2. An isosceles triangle has perimeter 40 cm; the equal sides are 15 cm each. Find the area of the triangle.
Answer:
Given: The two equal sides are 15 cm and 15 cm. The full perimeter is 40 cm.

First, find the base: Base = 40 - 15 - 15 = 10 cm

In an isosceles triangle, when you drop a line straight down from the top vertex to the base, it hits the base at its midpoint. This means half the base = 10 ÷ 2 = 5 cm

Using the Pythagorean theorem to find the height: height² = 15² - 5² = 225 - 25 = 200
So, height = \( \sqrt{200} = 10\sqrt{2} \) cm

Now calculate the area: Area of triangle = 1/2 × base × height = 1/2 × 10 × 10√2 = 50√2 cm²

Therefore, area = 50√2 cm²
In simple words: In an isosceles triangle, the height splits the base in half. Use that fact with the Pythagorean theorem to find height, then multiply base times height and divide by 2.

Exam Tip: Always remember that in an isosceles triangle, the altitude from the apex bisects the base - this property cuts the working time in half. Leave your final answer in surd form (√2) unless told to approximate.

All Key Formulas in Class 9 Ganita Manjari Chapter 6

The History of π - A Summary from Class 9 Ganita Manjari Chapter 6

One of the most distinctive features of this chapter is its rich historical account of how humans pursued the value of π across civilisations. Here is a concise timeline:

• 1900 BCE - Mesopotamia: People noticed that π is larger than 3 by comparing a circle's circumference to a hexagon drawn inside it. They used π ≈ 3.125.
• 1500 BCE - Ancient Egypt and Baudhāyana's India: They found π ≈ 256/81 ≈ 3.16 by using geometric methods to try to turn rectangles into squares.
• 250 BCE - Archimedes of Syracuse: He trapped π between two sets of polygons (one inscribed and one around the circle) with up to 96 sides. He proved that 3(10/71) < π < 3(1/7).
• 150 CE - Ptolemy of Alexandria: He calculated π ≈ 377/120 ≈ 3.14167 for use in astronomy.
• 263 CE - Liu Hui (China): His circle-cutting method became the foundation for later Chinese discoveries about π.
• 480 CE - Zu Chongzhi (China): He used a polygon with 24,576 sides. He found 355/113 ≈ 3.1415929, which is the best rational approximation with a denominator under 15,000. This remained the world's most accurate value for more than 800 years.
• 499 CE - Āryabhaṭa (India): He found π ≈ 62832/20000 = 3.1416. Importantly, he called it asanna (approximate), hinting that π might not be expressible as a simple fraction.
• 628 CE - Brahmagupta (India): He used √10 ≈ 3.1622 as a value for π because of its algebraic usefulness.
• 1400 CE - Mādhava of Sangamagrāma (India): He found the first exact formula: π/4 = 1 - 1/3 + 1/5 - 1/7 + ··· (an infinite series). This discovery was crucial in the birth of calculus. He computed π to 11 decimal places.
• 1706 - William Jones (Wales): He was the first to use the Greek letter π to represent the circumference-to-diameter ratio.
• Today: Using methods found by Ramanujan and the Chudnovsky brothers, π is now known to hundreds of trillions of digits.

Key Concepts Explained Simply - Ganita Manjari Chapter 6

• Why is π irrational? The digits of π never repeat in any pattern and π cannot be written as a simple fraction a/b where a and b are whole numbers. Lambert proved this in 1761. The well-known value 22/7 is just an approximation - π ≈ 22/7 but π ≠ 22/7. A much better approximation is 355/113.
• Heron's Formula - when is it used? Heron's formula lets you find a triangle's area when you know only its three side lengths, without needing its height. Calculate s = ½(a + b + c), then area = √[s(s - a)(s - b)(s - c)]. It works especially well for scalene triangles where height is not directly given.
• Brahmagupta's Formula and its connection to Heron's: Brahmagupta's formula for cyclic quadrilaterals is area = √[(s - a)(s - b)(s - c)(s - d)]. It becomes Heron's formula exactly when d = 0 - because a triangle is just a four-sided shape where one side has zero length. This shows a beautiful example of how mathematics generalises.
• The Median Theorem: A median of a triangle is a line from a vertex to the midpoint of the opposite side. It divides the triangle into two smaller triangles with exactly equal area - even though the two triangles are usually not identical in shape. This result follows directly from the area formula ½bh.
• Arc Length and Sector Area: Both formulas use the same basic idea - what fraction of the full 360° is the angle θ? Arc length = 2πr × (θ/360) and Sector area = πr² × (θ/360). These formulas come straight from the circular symmetry of the circle.
• Baudhāyana's Rectangle Squaring (800 BCE): This geometric construction finds a square with the same area as any given rectangle using only a compass and straightedge - a remarkable achievement that came more than two thousand years before modern algebra. The proof uses the identity ((a + b)/2)² - ((a - b)/2)² = ab.

Exercise-Wise Overview - Chapter 6 Ganita Manjari Class 9 Maths

• Exercise Set 6.1: Circumference from radius; arc length calculations; sector perimeter; perimeters of nine composite shapes involving quarter, half and three-quarter circles; tyre revolution problems; flower petal perimeters; ratio of radii from ratio of circumferences.
• Exercise Set 6.2: Area of triangles (including using Heron's formula); trapezium area; triangular plots from perimeter ratios; rhombus diagonal from area; parallelogram area ratios; median-based area equality proofs; midpoint parallelogram area; median point area equality.
• Exercise Set 6.3: Sector areas; quadrant areas from circumference; minute hand area sweep; minor and major sector and segment areas; windscreen wiper area; starred proofs involving equilateral triangle, square and hexagon inscribed in circles.
• End-of-Chapter Exercises: 27 questions covering area models of algebraic identities, triangle area problems via Heron's formula, bicycle wheel travel calculations, kite area, trapezium area proofs, congruent rectangle packing, circle-in-rectangle area fraction, nine-rectangle puzzle and advanced starred proofs involving semicircles on right triangles, concentric circles, and equal shaded region proofs.

Historical Mathematicians Featured in Class 9 Maths Ganita Manjari Chapter 6

This chapter is unusual in giving significant space to mathematical history. The following figures appear:

• Mādhava of Sangamagrāma - Kerala mathematician (c. 1400 CE) who found the first exact infinite series for π, effectively starting calculus two centuries before Newton and Leibniz.
• Āryabhaṭa - Indian mathematician (499 CE) who calculated π ≈ 3.1416 and described it as approximate - the first known record of someone suggesting that π might be irrational.
• Brahmagupta - Indian mathematician (628 CE) who discovered the area formula for cyclic quadrilaterals, which is a direct generalisation of Heron's formula, and used √10 as an approximation for π.
• Baudhāyana - Ancient Indian mathematician (c. 800 BCE) whose Śhulbasūtra contains a geometric method for squaring a rectangle and an early approximation for π in circle-squaring constructions.
• Archimedes of Syracuse - Greek mathematician (c. 250 BCE) who proved A = πr² and bounded π between 3(10/71) and 3(1/7) using 96-sided polygons.
• Zu Chongzhi - Chinese mathematician (480 CE) whose fraction 355/113 remained the world's most accurate value of π for over 800 years.
• Nīlakaṇṭha Somayājī - Indian mathematician (c. 1500 CE) who gave a beautiful visual "circle slicing" proof that the area of a circle is πr².
• Heron of Alexandria - Greek mathematician who discovered the formula for triangle area using only its three sides, which is now known by his name.

Important Starred (*) Questions in Chapter 6 - For Advanced Learners

• Q7 (Ex 6.3): Prove that the minor segment of a 60° chord in a circle of radius r has area \( \pi r^2 \left(\frac{1}{6} - \frac{\sqrt{3}}{4}\right) \).
• Q8 (Ex 6.3): Show that the ratio of an equilateral triangle inscribed in a circle to the circle's area is \( \frac{3\sqrt{3}}{4\pi} \) ≈ 0.413.
• Q9 (Ex 6.3): Show that the ratio of a square inscribed in a circle to the circle's area is \( \frac{2}{\pi} \) ≈ 0.637.
• Q10 (Ex 6.3): Show that the ratio of a regular hexagon inscribed in a circle to the circle's area is \( \frac{3\sqrt{3}}{2\pi} \) ≈ 0.827 - exactly twice the triangle ratio in Q8.
• Q21 (End): In a square with a quarter circle and two semicircles, prove that two created shaded regions have equal area.
• Q23 (End): For two concentric circles where a chord of the outer circle is tangent to the inner circle, show the annular region between them has area \( \frac{\pi l^2}{4} \), where l is the chord length.
• Q24 (End): Show that semicircles on the two legs of a right triangle together equal the semicircle on the hypotenuse - a beautiful generalisation of the Baudhāyana-Pythagoras theorem.
• Q25 (End): For two circles each passing through the other's centre, find the enclosed region's area in terms of r.

Frequently Asked Questions (FAQs) - Class 9 Maths Ganita Manjari Chapter 6

Is Class 9 Maths Ganita Manjari Chapter 6 easy or difficult?

The chapter is moderately difficult with sections of real challenge. The perimeter and basic area topics (6.1 to 6.8) are within reach for most students, especially those with a solid Class 8 background. The real difficulty starts with Heron's formula, where the calculation steps are lengthy and easy to mess up. It continues into Brahmagupta's formula in Section 6.8.1, which needs an understanding of cyclic quadrilaterals from Chapter 5. The starred end-of-chapter questions - especially ones involving inscribed shapes and concentric circles - are genuinely tough and require putting together ideas from multiple topics. Overall, this chapter pays off for students who read carefully and think deeply.