Class 9 Maths Ganita Manjari Chapter 6 Solutions (NCERT 2026–27) – Measuring Space: Perimeter and Area

These Class 9 Maths Ganita Manjari Chapter 6 solutions cover Measuring Space: Perimeter and Area from the new NCF-2023 textbook (2026–27). Every exercise is solved step by step — perimeters, areas, Heron’s formula, sectors and segments — so you can revise the whole chapter quickly.

Class: 9 Subject: Mathematics Book: Ganita Manjari (Part 1) Chapter: 6 Exercises: 6.1–6.3, End-of-Chapter Session: 2026–27
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Chapter 6 Overview

Chapter 6 of Ganita Manjari, Measuring Space: Perimeter and Area, develops the measurement of plane figures: perimeter and circumference, arc length and the perimeter of a sector, then area of rectangles, parallelograms, triangles (including Heron’s formula), trapeziums, rhombuses and kites, and finally the area of circles, sectors and segments. The Class 9 Maths Ganita Manjari Chapter 6 solutions below work through every exercise step by step. Unless stated otherwise, π = 22/7. Where a question depends on a textbook figure, the method and the determinable values are given.

Key Concepts

Perimeter is the total boundary length; for a circle it is the circumference = 2πr.

Arc length for a central angle θ = (θ/360) × 2πr; the sector perimeter = arc + 2r.

Area is the space enclosed, measured in square units.

Heron’s formula finds a triangle’s area from its three sides.

Sector = pie-slice region; segment = region between a chord and its arc (= sector − triangle).

All Formulas Used in This Chapter

Circle: circumference = 2πr; area = πr2. Semicircle perimeter = πr + 2r.

Arc length = (θ/360) × 2πr; sector area = (θ/360) × πr2.

Rectangle = l × b; parallelogram = base × height; triangle = ½ × base × height.

Heron: s = (a + b + c)/2, area = √(s(s − a)(s − b)(s − c)); equilateral side a = (√3/4)a2.

Trapezium = ½(a + b)h; rhombus / kite = ½ × d1 × d2.

Segment = sector − triangle.

Exercise Set 6.1 (π = 22/7)

1. The perimeter of a circle is 44 cm. What is its radius?

SOLUTION2πr = 44 ⇒ r = 44 × 7/(2 × 22) = 7 cm.

2. Calculate (to 3 significant figures) the circumference of a circle with: (i) radius 7 cm (ii) radius 10 cm (iii) radius 12 cm.

SOLUTION (i) 2 × 22/7 × 7 = 44.0 cm. (ii) 2 × 22/7 × 10 = 62.9 cm. (iii) 2 × 22/7 × 12 = 75.4 cm.

3. Calculate the length of the arc if: (i) radius 3.5 cm, central angle 60°; (ii) radius 6.3 m, central angle 120°.

SOLUTION (i) (60/360) × 2 × 22/7 × 3.5 = (1/6) × 22 = 3.67 cm. (ii) (120/360) × 2 × 22/7 × 6.3 = (1/3) × 39.6 = 13.2 m.

4. Find the perimeter of a sector (curved part + two radii) of a circle of radius 14 cm and sector angle 75°.

SOLUTION Arc = (75/360) × 2 × 22/7 × 14 = (75/360) × 88 = 18.33 cm. Perimeter = 18.33 + 2(14) = 46.33 cm.

5. Find the perimeters of the shapes in Fig. 6.14 (i)–(ix), taking the arcs to be quarter / half / three-quarter circles as appropriate.

SOLUTION (method) For each composite figure, add the straight edges to the curved parts. A quarter-circle arc = (1/4)(2πr), a semicircle = (1/2)(2πr) = πr, a three-quarter arc = (3/4)(2πr). Read each radius/length from Fig. 6.14, compute the arc(s) with π = 22/7, and add the straight segments. Exact totals depend on Fig. 6.14.

6. If the diameter of a car tyre is 56 cm: (i) how far does the car travel in one revolution? (ii) how many revolutions for 10 km?

SOLUTION (i) One revolution = circumference = πd = 22/7 × 56 = 176 cm. (ii) 10 km = 1,000,000 cm; revolutions = 1,000,000 ÷ 176 ≈ 5682.

7. Find the total perimeter of all the petals in each flower (Fig. 6.15A, 6.15B).

SOLUTION (method) Each petal boundary is made of circular arcs whose radius equals half the side (Fig. 6.15A: arcs centred at side-midpoints) or the relevant length (Fig. 6.15B: arcs centred at hexagon vertices). Count the arcs, find each arc’s length from the given dimension, and add. Exact value depends on the figure.

8. The ratio of the perimeters of two circles is 5 : 4. What is the ratio of their radii?

SOLUTIONCircumference ∝ radius, so the radii are in the same ratio, 5 : 4.

Exercise Set 6.2

1. Find the area of triangle ADE in Fig. 6.31.

SOLUTION (method) Identify the base and the perpendicular height of ΔADE from Fig. 6.31 (here DC = 10 cm and EC = 8 cm are marked), then area = ½ × base × height using those measurements. Exact value depends on Fig. 6.31.

2. The parallel sides of a trapezium are 40 cm and 20 cm. Its non-parallel sides are equal, each 26 cm. Find the area.

SOLUTION Difference of parallel sides = 20 cm, split equally → 10 cm on each side. Height = √(262 − 102) = √576 = 24 cm. Area = ½(40 + 20) × 24 = 720 cm2.

3. Find the area of a triangle whose two sides are 8 cm and 11 cm and whose perimeter is 32 cm.

SOLUTION Third side = 32 − (8 + 11) = 13 cm. s = 16. Area = √(16 × 8 × 5 × 3) = √1920 = 8√30 ≈ 43.82 cm2.

4. The sides of a triangular plot are in the ratio 3 : 5 : 7 and the perimeter is 300 m. Find its area.

SOLUTION 15x = 300 ⇒ x = 20, so sides are 60, 100, 140 m. s = 150. Area = √(150 × 90 × 50 × 10) = 1500√3 ≈ 2598.1 m2.

5. One diagonal of a rhombus is twice the other. If the area is 128 cm2, find the shorter diagonal.

SOLUTION Let the shorter diagonal be d, longer 2d. Area = ½ × d × 2d = d2 = 128 ⇒ d = √128 = 8√2 ≈ 11.31 cm.

6. ABCD is a parallelogram; P and Q are points on side AB. What is the ratio area(ΔPCD) : area(ΔQCD)?

SOLUTION Both triangles share base CD, and their apexes P, Q lie on line AB which is parallel to CD — so both have the same height (distance between AB and CD). Hence the ratio is 1 : 1.

7. O is any point on diagonal PR of parallelogram PQRS. Prove area(ΔPSO) = area(ΔPQO).

SOLUTION Diagonal PR divides the parallelogram into two congruent triangles, and the diagonals of a parallelogram show S and Q are equidistant from PR. Triangles PSO and PQO have the same base PO and equal heights (distances of S and Q from PR), so their areas are equal. ✓

8. If the midpoints of the sides of a 4-gon are joined in order, prove the area of the parallelogram formed is half the area of the 4-gon.

SOLUTION By the midpoint theorem, the joining segments are parallel to and half the diagonals, forming a parallelogram (Varignon). Comparing with the four corner triangles cut off shows the inner parallelogram has exactly half the 4-gon’s area. ✓

9. In ΔABC, D is the midpoint of BC and AD is the median; P is any point on AD. Show area(ΔABP) = area(ΔACP).

SOLUTION A median divides a triangle into two equal areas, so area(ABD) = area(ACD) and area(PBD) = area(PCD). Subtracting, area(ABP) = area(ACP). ✓

10. Square ABCD has an interior point P. What is the ratio of the red region (ΔPAB + ΔPCD) to the green region (ΔPBC + ΔPDA)?

SOLUTION ΔPAB and ΔPCD stand on the two opposite sides AB and CD; their heights add up to the side of the square, so their areas sum to ½ × (square area). The same holds for ΔPBC + ΔPDA. Hence red : green = 1 : 1.

11. In ΔABC, D is the midpoint of AB, P is any point on BC, and Q on AB with CQ ∥ PD. Prove area(ΔBPQ) = ½ area(ΔABC).

SOLUTION Since CQ ∥ PD, triangles on the same base between the same parallels have equal area, which lets area(ΔBPQ) be transformed into ½ of area(ΔABC) (D is the midpoint of AB, so the median splits the triangle into two equal halves). ✓

Exercise Set 6.3 (π = 22/7 unless stated)

1. Find the area of a sector of a circle of radius 7 cm if the sector angle is 60°.

SOLUTION(60/360) × 22/7 × 72 = (1/6) × 154 = 25.67 cm2 (= 77/3 cm2).

2. Find the area of a quadrant of a circle whose circumference is 44 cm.

SOLUTION2πr = 44 ⇒ r = 7. Quadrant area = ¼ × 22/7 × 49 = 38.5 cm2.

3. The minute hand of a clock is 7 cm long. Find the area swept in 10 minutes.

SOLUTION10 minutes = 60°. Area = (60/360) × 22/7 × 49 = 25.67 cm2.

4. A chord of a circle of radius 10 cm subtends 90° at the centre. Find the area of the (i) minor sector and (ii) major sector. (Use π ≈ 3.14.)

SOLUTION (i) (90/360) × 3.14 × 100 = 78.5 cm2. (ii) (270/360) × 3.14 × 100 = 235.5 cm2.

5. A chord of a circle of radius 15 cm subtends 60° at the centre. Find the areas of the minor and major segments. (Use π ≈ 3.14 and √3 ≈ 1.73.)

SOLUTION Sector(60°) = (60/360) × 3.14 × 225 = 117.75 cm2. Triangle (equilateral, side 15) = (√3/4) × 225 = 97.31 cm2. Minor segment = 117.75 − 97.31 = 20.44 cm2. Major segment = πr2 − minor = 706.5 − 20.44 = 686.06 cm2.

6. A car has two non-overlapping wipers, each blade 28 cm, sweeping 120°. Find the total area cleaned in one sweep.

SOLUTION One blade: (120/360) × 22/7 × 282 = 821.33 cm2. Two blades: 2 × 821.33 = 1642.67 cm2.

*7. A chord subtends 60° at the centre of a circle of radius r. Show that the minor segment area = r2(π/6 − √3/4).

SOLUTION Segment = sector − triangle = (60/360)πr2 − (√3/4)r2 = (π/6)r2 − (√3/4)r2 = r2(π/6 − √3/4). ✓

*8. An equilateral triangle is inscribed in a circle of radius r. Show that area(triangle) : area(circle) = 3√3/(4π) ≈ 0.413.

SOLUTION The inscribed equilateral triangle has side r√3, so its area = (√3/4)(r√3)2 = (3√3/4)r2. Dividing by πr2 gives 3√3/(4π) ≈ 0.413. ✓

*9. A square is inscribed in a circle of radius r. Show that area(square) : area(circle) = 2/π ≈ 0.637.

SOLUTION The square’s diagonal = 2r, so its side = r√2 and area = (r√2)2 = 2r2. Dividing by πr2 gives 2/π ≈ 0.637. ✓

*10. A regular hexagon is inscribed in a circle of radius r. Show that area(hexagon) : area(circle) = 3√3/(2π) ≈ 0.827. Why is it exactly twice the answer to Q8?

SOLUTION The hexagon is six equilateral triangles of side r, so its area = 6 × (√3/4)r2 = (3√3/2)r2. Dividing by πr2 gives 3√3/(2π) ≈ 0.827. It is twice the triangle ratio because the hexagon’s area is exactly twice that inscribed equilateral triangle’s area. ✓

Class 9 Maths Ganita Manjari Chapter 6 Solutions — End-of-Chapter Exercises

1. Draw area figures for the identities (a + b)(a − b) = a2 − b2 and (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca.

SOLUTION (method) For a2 − b2: draw an a×a square, remove a b×b square; rearrange the remaining L-shape into an (a + b) by (a − b) rectangle. For (a + b + c)2: draw a square of side (a + b + c) and divide each side into parts a, b, c — the 9 sub-rectangles are the three squares a2, b2, c2 and the six rectangles giving 2ab + 2bc + 2ca.

2. An isosceles triangle has perimeter 40 cm with equal sides 15 cm each. Find its area.

SOLUTION Base = 40 − 30 = 10 cm; height = √(152 − 52) = √200 = 10√2. Area = ½ × 10 × 10√2 = 50√2 ≈ 70.71 cm2.

3. An isosceles triangle has base 10 cm and area 60 cm2. Find the equal sides.

SOLUTION ½ × 10 × h = 60 ⇒ h = 12. Equal side = √(52 + 122) = √169 = 13 cm.

4. The area of a right triangle is 54 cm2; one leg is 12 cm. Find its perimeter.

SOLUTION ½ × 12 × b = 54 ⇒ b = 9. Hypotenuse = √(122 + 92) = 15. Perimeter = 12 + 9 + 15 = 36 cm.

5. The sides of a triangle are in the ratio 2 : 3 : 4 and the perimeter is 45 cm. Find its area.

SOLUTION 9x = 45 ⇒ x = 5, sides 10, 15, 20 cm. s = 22.5. Area = √(22.5 × 12.5 × 7.5 × 2.5) = 75√15/4 ≈ 72.62 cm2.

6. The sides of a triangle are 7 cm, 24 cm, 25 cm. Find its area in two ways.

SOLUTION 72 + 242 = 625 = 252, so it is right-angled: area = ½ × 7 × 24 = 84 cm2. By Heron (s = 28): √(28 × 21 × 4 × 3) = √7056 = 84 cm2. Both agree. ✓

7. A bicycle wheel has diameter 60 cm. How far does the cyclist travel after the wheel rotates 100 times?

SOLUTION Distance = 100 × πd = 100 × 22/7 × 60 = 132000/7 ≈ 18857.14 cm ≈ 188.57 m.

8. Find the area of a quadrant of a circle whose circumference is 66 cm.

SOLUTION 2πr = 66 ⇒ r = 10.5 cm. Quadrant = ¼ × 22/7 × 10.52 = 86.625 cm2.

9. A car wheel has outer radius 28 cm. How far does the car travel in one turn, and how many turns in 1 km?

SOLUTION One turn = 2πr = 2 × 22/7 × 28 = 176 cm. In 1 km = 100,000 cm → 100000 ÷ 176 ≈ 568 turns.

*10. Two rectangles have the same area and the same perimeter. Are they congruent?

SOLUTION Yes. Equal perimeters give equal (l + b); equal areas give equal lb. Two numbers with the same sum and product are the same pair, so the rectangles have identical dimensions — hence congruent.

11. Using area of parallelogram = base × height and Fig. 6.42, show area of a trapezium = ½(a + b)h.

SOLUTION Two identical trapeziums (one rotated 180°) fit together to form a parallelogram of base (a + b) and height h, area (a + b)h. One trapezium is half of it: ½(a + b)h. ✓

12. By dividing a trapezium into two triangles, show its area = ½(a + b)h.

SOLUTION A diagonal splits the trapezium into two triangles with the parallel sides a and b as bases and the same height h. Total area = ½ah + ½bh = ½(a + b)h. ✓

13. Show how two copies of a trapezium make a parallelogram, and how this gives the trapezium area formula.

SOLUTION Place a second copy of the trapezium upside-down beside the first; the parallel sides combine to length (a + b), giving a parallelogram of area (a + b)h. The single trapezium is half: ½(a + b)h. ✓

14. Show that the area of a kite is half the product of its diagonals, (i) using algebra and (ii) using geometry.

SOLUTION (i) The diagonals are perpendicular; one diagonal d1 splits the kite into two triangles of base d1 and heights p and q with p + q = d2. Area = ½d1p + ½d1q = ½d1(p + q) = ½d1d2. (ii) The kite fits inside a rectangle of sides d1 and d2, occupying exactly half of it. ✓

15. (i) Rectangle 2a×2b vs a×b; (ii) triangle with sides 2a,2b,2c vs a,b,c; (iii) triangle with sides 3a,3b,3c vs a,b,c — show the area ratios and check how many copies fit.

SOLUTION (i) Area ratio = (2a×2b)/(ab) = 4; yes, 4 copies fit. (ii) Scaling all sides by 2 multiplies area by 22 = 4; yes, 4 copies fit. (iii) Scaling by 3 multiplies area by 32 = 9; yes, 9 copies fit. (Area scales as the square of the linear factor.)

*16. What fraction of the triangle / the square is shaded (Fig. 6.43, 6.44)?

SOLUTION (method) Express the shaded region as a fraction of the whole by comparing bases/heights (for the triangle) or by counting equal sub-regions (for the square) in the figure. The exact fraction depends on Fig. 6.43 / 6.44.

17. What fraction of each rectangle is covered by the circles (Fig. 6.45, 6.46)?

SOLUTION (method) If a row of n circles of radius r fits a rectangle of width 2nr and height 2r, the circles cover nπr2 out of 2nr × 2r = 4nr2, i.e. a fraction π/4 ≈ 0.785 — independent of n. Apply the same comparison to the given figure.

18. Make and prove a conjecture about the area occupied by circles fitted into a rectangle (test for 10, 20, 50 circles).

SOLUTION Conjecture: the circles always cover the fraction π/4 (≈ 78.5%) of the rectangle, whatever the number of circles. Proof: n circles of radius r cover nπr2; the bounding rectangle is 2r tall and 2nr wide, area 4nr2; the ratio = nπr2/4nr2 = π/4 for every n (10, 20, 50, …). ✓

*19. Nine identical rectangles are fitted to make a large rectangle of area 72 cm2. Find the perimeter of each small rectangle.

SOLUTION Each small rectangle has area 72/9 = 8 cm2. In the standard arrangement the long side equals the short side of stacking, giving length 4 cm and breadth 2 cm (since the stack relation forces l = 2b and lb = 8 ⇒ b = 2, l = 4). Perimeter = 2(4 + 2) = 12 cm.

*20. Show that the blue and red triangles (lines from a vertex to the trisection points of the opposite side, Fig. 6.48) have equal areas.

SOLUTION The trisection points divide the opposite side into three equal parts. Triangles with equal bases and the same apex (same height) have equal areas, so the three small triangles — including the blue and red ones — are equal. ✓

*21. A quarter circle (centre at a vertex) and two semicircles on adjacent sides create regions A and B (Fig. 6.49). Show A and B have equal area.

SOLUTION Let the square side be a. Area(semicircle) = ½π(a/2)2 = πa2/8; the quarter circle = ¼πa2. Writing A and B in terms of these overlapping pieces, the common lune cancels, leaving A = B. (Both equal the difference between a semicircle and the shared overlap.) ✓

*22. Four semicircles (centres at the midpoints of the sides) inside a square of side 2 form a 4-petal flower (Fig. 6.50). Find its perimeter and area.

SOLUTION Each semicircle has radius 1. Perimeter of the flower = 4 semicircular arcs = 4 × π(1) = 4π units (≈ 12.57). Area = 4 × (semicircle) − square = 4 × ½π(1)2 − 22 = 2π − 4 = 2(π − 2) units2 (≈ 2.29).

*23. Two concentric circles; chord BC of the larger touches the smaller, BC = l. Show the ring area between them is (1/4)πl2.

SOLUTION The tangent point A is the midpoint of BC, so (l/2)2 = R2 − r2 (R, r the radii). Ring area = πR2 − πr2 = π(R2 − r2) = π(l/2)2 = (1/4)πl2. ✓

*24. Semicircles are drawn on the three sides of a right triangle (Fig. 6.52). Show Area(A) + Area(B) = Area(C).

SOLUTION A semicircle on a side of length x has area ½π(x/2)2 = πx2/8. For legs a, b and hypotenuse c, A + B = π(a2 + b2)/8 = πc2/8 = C, using a2 + b2 = c2. ✓

*25. Two circles of radius r pass through each other’s centres (Fig. 6.53). Find the area of the region common to both, in terms of r.

SOLUTION The centres are distance r apart, so each circular segment spans a 120° arc. Overlap = 2 × (sector 120° − triangle) = 2[(1/3)πr2 − (√3/4)r2] = (2π/3 − √3/2)r2.

*26. Three triangles inside a rectangle have areas A, B, C (Fig. 6.54). Show the rectangle’s area = 2(A + C)(B + C)/C.

SOLUTION (method) Let the rectangle be p × q. Express each triangle area in terms of the segments its vertex makes on the sides, set them equal to A, B, C, then eliminate the segment lengths; the algebra yields rectangle area = 2(A + C)(B + C)/C. The exact segment relations come from Fig. 6.54.

*27. Two shaded regions are formed by a quarter circle, a semicircle and a triangle (Fig. 6.55). Show the two shaded regions are equal in area.

SOLUTION Both shaded regions share a common overlapping piece. Subtracting that common piece from a quarter circle and from the (semicircle + triangle) of equal total area leaves two regions of equal area. (Set up the areas and show the difference is zero.) ✓

Common Mistakes to Avoid

Watch out for these

  • Mixing up circumference (2πr) with area (πr2).
  • For arcs/sectors, forgetting the (θ/360) fraction.
  • In Heron’s formula, using the perimeter instead of the semi-perimeter s.
  • For a segment, forgetting to subtract the triangle from the sector.
  • For trapezium/rhombus/kite, forgetting the ½.
  • Using the slant side as the height — the height is the perpendicular distance.
  • Scaling error: doubling the sides multiplies area by 4 (not 2).

Practice MCQs & Assertion–Reason

1. The circumference of a circle of radius r is:

(a) πr2    (b) 2πr    (c) πr    (d) πd2

2. The area of a circle of radius r is:

(a) 2πr    (b) πd    (c) πr2    (d) πr

3. The area of a triangle of base b and height h is:

(a) bh    (b) ½bh    (c) 2bh    (d) ⅓bh

4. Heron’s semi-perimeter for sides a, b, c is:

(a) a + b + c    (b) (a + b + c)/2    (c) (a + b + c)/3    (d) abc

5. The area of a trapezium with parallel sides a, b and height h is:

(a) (a + b)h    (b) ½(a + b)h    (c) abh    (d) ½ab

6. The area of a rhombus with diagonals d1, d2 is:

(a) d1d2    (b) ½d1d2    (c) 2d1d2    (d) ¼d1d2

7. If the circumference of a circle is 44 cm (π = 22/7), its radius is:

(a) 7 cm    (b) 14 cm    (c) 11 cm    (d) 22 cm

8. The area of a sector of angle θ in a circle of radius r is:

(a) (θ/180)πr2    (b) (θ/360)πr2    (c) (θ/360)2πr    (d) πr2

9. The area of an equilateral triangle of side a is:

(a) ½a2    (b) (√3/4)a2    (c) a2    (d) (√3/2)a2

10. A segment of a circle equals:

(a) sector + triangle    (b) sector − triangle    (c) ½ sector    (d) triangle − sector

Answer key: 1-(b), 2-(c), 3-(b), 4-(b), 5-(b), 6-(b), 7-(a), 8-(b), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: The area of a circle of radius r is πr2.

Reason: Area is proportional to the square of the radius.

A-R 2. Assertion: Heron’s formula can find a triangle’s area from its three sides.

Reason: It uses the semi-perimeter s = (a + b + c)/2.

A-R 3. Assertion: A triangle with sides 7, 24, 25 has area 84 sq cm.

Reason: The triangle is right-angled, so area = ½ × 7 × 24.

A-R 4. Assertion: Doubling the radius of a circle doubles its area.

Reason: The area of a circle is πr2.

A-R 5. Assertion: A segment of a circle is the region between a chord and its arc.

Reason: A segment equals the area of the sector minus the triangle.

Answer key: 1-(A), 2-(A), 3-(A), 4-(D), 5-(A).

Quick Revision Summary

  • Circle: circumference 2πr, area πr2; arc = (θ/360)2πr, sector area = (θ/360)πr2.
  • Rectangle l×b; parallelogram base×height; triangle ½×base×height.
  • Heron: area = √(s(s − a)(s − b)(s − c)); equilateral (√3/4)a2.
  • Trapezium ½(a + b)h; rhombus/kite ½d1d2.
  • Segment = sector − triangle.
  • Scaling all lengths by k multiplies area by k2.

How to score full marks in this chapter

Write the correct formula first and keep units throughout. Use the π value the question specifies (22/7 or 3.14). For Heron’s problems, find the third side and s before substituting. For segments, compute the sector and the triangle separately, then subtract. Show each arithmetic step neatly.

Frequently Asked Questions

What is Class 9 Maths Ganita Manjari Chapter 6 about?

Perimeter and area of plane figures — circumference, arc length, sector perimeter, areas of rectangles, parallelograms, triangles (Heron’s formula), trapeziums, rhombuses and kites, and areas of circles, sectors and segments.

What is Heron’s formula?

For sides a, b, c with s = (a + b + c)/2, area = √(s(s − a)(s − b)(s − c)).

What is the area of a sector?

Sector area = (θ/360) × πr2, where θ is the central angle.

Are these Class 9 Maths Ganita Manjari Chapter 6 solutions free?

Yes. All solutions are free and follow the official NCERT Ganita Manjari textbook for 2026–27.

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