Class 6 Maths Ganita Prakash Chapter 6 Solutions (NCERT 2026–27) – Perimeter and Area

These Class 6 Maths Ganita Prakash Chapter 6 solutions cover Perimeter and Area from the new NCF-2023 textbook (Reprint 2026–27). Every Figure it Out question, Math Talk and Try This task is solved step by step — with worked perimeters of rectangles, squares and triangles, areas split into rectangles and triangles, area maze puzzles and house-plan problems — so you can master the chapter and revise it quickly.

Class: 6 Subject: Mathematics Book: Ganita Prakash Chapter: 6 Exercises: Figure it Out (p. 132, 133, 138, 144, 149) & Area Maze Session: 2026–27
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Chapter 6 Overview

Chapter 6 of Ganita Prakash, Perimeter and Area, revisits two measurements of flat shapes. The perimeter is the total distance once around the boundary of a closed figure, while the area is the amount of region the figure encloses. The chapter recalls perimeter formulas for rectangles, squares and triangles, extends them to regular polygons, then explores running-track distances, “straight and diagonal” perimeters, and split-and-rejoin puzzles. In the second half it recalls area of rectangles and squares from grid paper, derives the area of a triangle as half a rectangle, and tackles area-maze puzzles and house-plan problems where the same area can hide very different perimeters. The Class 6 Maths Ganita Prakash Chapter 6 solutions below work through every Figure it Out, Deep Dive, Let’s Explore and Math Talk question step by step.

Key Concepts & Definitions

Perimeter: the distance covered along the boundary of a closed plane figure when you go around it once. For a polygon it is the sum of the lengths of all its sides.

Area: the amount of region enclosed by a closed figure, measured in square units (sq cm, sq m, …).

Regular polygon: a closed figure with all sides equal and all angles equal — e.g. the equilateral triangle, square and regular pentagon.

Triangle from a rectangle: a diagonal of a rectangle (or square) splits it into two equal triangles, so each triangle is exactly half the rectangle’s area.

Same area, different perimeter: two figures can enclose the same area yet have different perimeters (and vice-versa); a long thin rectangle has a larger perimeter than a square of equal area.

Estimating area on a grid: count full squares as 1 sq unit, count a region as 1 sq unit when more than half a square is covered, ignore parts less than half, and count an exact half as 1 sq unit.

Important Formulas (Chapter 6)

Perimeter of a rectangle = 2 × (length + breadth).

Perimeter of a square = 4 × side.

Perimeter of a triangle = sum of its three sides; for an equilateral triangle = 3 × side.

Perimeter of a regular polygon = number of sides × length of one side.

Area of a rectangle = length × breadth (width).

Area of a square = side × side.

Area of a triangle = half the area of the rectangle that contains it.

Figure it Out — Perimeter (Page 132)

Questions are reproduced verbatim from the NCERT Ganita Prakash textbook; the worked solutions are original and verified against the answers given in the book.

1. Find the missing terms: a. Perimeter of a rectangle = 14 cm; breadth = 2 cm; length = ?. b. Perimeter of a square = 20 cm; side of a length = ?. c. Perimeter of a rectangle = 12 m; length = 3 m; breadth = ?.

SOLUTION a. Perimeter = 2 × (length + breadth), so 14 = 2 × (length + 2) → length + 2 = 7 → length = 5 cm. b. Perimeter of a square = 4 × side, so 20 = 4 × side → side = 5 cm. c. 12 = 2 × (3 + breadth) → 3 + breadth = 6 → breadth = 3 m.

2. A rectangle having sidelengths 5 cm and 3 cm is made using a piece of wire. If the wire is straightened and then bent to form a square, what will be the length of a side of the square?

SOLUTION Length of wire = perimeter of the rectangle = 2 × (5 + 3) = 2 × 8 = 16 cm. The same 16 cm forms a square, so 4 × side = 16 → side = 4 cm.

3. Find the length of the third side of a triangle having a perimeter of 55 cm and having two sides of length 20 cm and 14 cm, respectively.

SOLUTION Perimeter = sum of all three sides, so third side = 55 − (20 + 14) = 55 − 34 = 21 cm.

4. What would be the cost of fencing a rectangular park whose length is 150 m and breadth is 120 m, if the fence costs ₹40 per metre?

SOLUTION Length of fence = perimeter = 2 × (150 + 120) = 2 × 270 = 540 m. Cost = 540 × ₹40 = ₹21,600.

5. A piece of string is 36 cm long. What will be the length of each side, if it is used to form: a. A square, b. A triangle with all sides of equal length, and c. A hexagon (a six sided closed figure) with sides of equal length?

SOLUTION a. Square has 4 equal sides: 36 ÷ 4 = 9 cm each. b. Equilateral triangle has 3 equal sides: 36 ÷ 3 = 12 cm each. c. Regular hexagon has 6 equal sides: 36 ÷ 6 = 6 cm each.

6. A farmer has a rectangular field having length 230 m and breadth 160 m. He wants to fence it with 3 rounds of rope as shown. What is the total length of rope needed?

SOLUTION Perimeter of the field = 2 × (230 + 160) = 2 × 390 = 780 m. For 3 rounds: 3 × 780 = 2340 m of rope.

Figure it Out — Matha Pachchi! Running Tracks (Page 133)

Akshi runs the outer track (length 70 m, breadth 40 m); Toshi runs the inner track (length 60 m, breadth 30 m).

1. Find out the total distance Akshi has covered in 5 rounds.

SOLUTION Perimeter of the outer track = 2 × (70 + 40) = 220 m (one round). In 5 rounds: 5 × 220 = 1100 m.

2. Find out the total distance Toshi has covered in 7 rounds. Who ran a longer distance?

SOLUTION Perimeter of the inner track = 2 × (60 + 30) = 180 m (one round). In 7 rounds: 7 × 180 = 1260 m. Since 1260 m > 1100 m, Toshi ran the longer distance.

3. Think and mark the positions as directed— a. Mark ‘A’ at the point where Akshi will be after she ran 250 m. b. Mark ‘B’ at the point where Akshi will be after she ran 500 m. c. Now, Akshi ran 1000 m. How many full rounds has she finished running around her track? Mark her position as ‘C’. d. Mark ‘X’ at the point where Toshi will be after she ran 250 m. e. Mark ‘Y’ at the point where Toshi will be after she ran 500 m. f. Now, Toshi ran 1000 m. How many full rounds has she finished running around her track? Mark her position as ‘Z’.

SOLUTION Akshi’s round = 220 m. (a) 250 m = one full round (220 m) + 30 m, so A is 30 m past the start. (b) 500 m = two rounds (440 m) + 60 m, so B is 60 m past the start. (c) 1000 m ÷ 220 = 4 full rounds (880 m) with 120 m left over, so C is 120 m past the start. Toshi’s round = 180 m. (d) 250 m = one round (180 m) + 70 m, so X is 70 m past the start. (e) 500 m = two rounds (360 m) + 140 m, so Y is 140 m past the start. (f) 1000 m ÷ 180 = 5 full rounds (900 m) with 100 m left over, so Z is 100 m past the start. (The exact marks are placed on the track diagram; the distances past the starting point are computed above.)

Deep Dive, Perimeter Puzzles & Math Talk — Answered

These are the in-text exploratory and reflective tasks; the determinate ones are solved, the open ones are guided with worked examples.

Deep Dive — Common finishing line (Page 134) Two square tracks: inner of side 100 m, outer of side 150 m, with a common finishing line at the centre of one side. For a 350 m race, where should the two runners start so both finish together at the flag? Mark the inner-track start as ‘A’ and the outer-track start as ‘B’. Answer. Work backwards 350 m from the flag along each track. On the inner track: 100 + 100 + 100 + 50 = 350 m, so start A is reached after tracing three full sides and half of the fourth. On the outer track: 125 + 150 + 75 = 350 m, so start B is half-way along one side, then a full side, then 75 m of the next. Both runners then cover exactly 350 m to the common flag.
Estimate and Verify — The 9-unit triangle (Page 134) Akshi says the perimeter of this triangle shape is 9 units. Toshi says it can’t be 9 units and the perimeter will be more than 9 units. What do you think? Answer. Toshi is right — the perimeter is more than 9 units. The triangle is drawn on a grid using both straight (horizontal/vertical) and diagonal segments. A diagonal of a unit square is always longer than a side of the square, so counting a diagonal as 1 straight unit undercounts. The true perimeter is therefore greater than 9 units.
Straight & diagonal units (Page 135) Write the perimeters of the figures below in terms of straight and diagonal units (s = straight unit, d = diagonal unit). Answer. The four figures have perimeters 8s + 2d, 4s + 6d, 12s + 6d and 18s + 6d — counting the red (straight) and blue (diagonal) segments separately, since the two unit lengths are different.
Regular polygons (Page 135) Find objects around you with regular shapes and find their perimeters. Generalise your understanding for the perimeter of other regular polygons. Answer. Because every side of a regular polygon is the same length, the perimeter is simply repeated addition: Perimeter of a regular polygon = number of sides × length of one side. For example, a regular pentagon of side 6 cm has perimeter 5 × 6 = 30 cm.
Split and rejoin (Page 136) A 6 cm × 4 cm rectangular chit is cut into two equal pieces, then joined in different ways. Arrangement a. has a perimeter of 28 cm. Find the perimeter of the other arrangements (b, c, d), then arrange the two pieces to form a figure with a perimeter of 22 cm. Answer. The original rectangle has perimeter 2 × (6 + 4) = 20 cm. Cutting it into two 6 cm × 2 cm pieces creates two new 6 cm cut-edges (total 12 cm of new edge). When the pieces are rejoined, the perimeter depends on how much of those cut-edges is left exposed: joining along a short 2 cm edge exposes more boundary (larger perimeter, e.g. 28 cm), while joining along a long 6 cm edge hides more (smaller perimeter). To get exactly 22 cm, rejoin the two 6 cm × 2 cm pieces side-by-side along a full 2 cm edge to re-form the original 6 cm × 4 cm rectangle — its perimeter is 20 cm — or stagger them so the overlap along the 6 cm edges leaves a boundary of 22 cm. (Hands-on arrangement task; the reasoning about exposed cut-edges is the key idea.)
Let’s Explore! — Rectangles of area 24 sq units / Math Talk (Page 141) Make every whole-number rectangle of area 24 square units. a. Which has the greatest perimeter? b. Which has the least perimeter? c. Repeat for area 32 sq cm. Given an area, can you predict the rectangle with the greatest and least perimeter? Answer. Rectangles of area 24: 1×24, 2×12, 3×8, 4×6, with perimeters 50, 28, 22 and 20 units. (a) greatest perimeter is the long thin 1 × 24 (perimeter 50). (b) least perimeter is the nearest-to-square 4 × 6 (perimeter 20). (c) For area 32: 1×32, 2×16, 4×8, with perimeters 66, 36, 24; greatest is 1 × 32, least is 4 × 8. Prediction: for a fixed area, the longest, thinnest rectangle has the greatest perimeter and the rectangle closest to a square has the least perimeter.
Area of a triangle — inference (Page 142–143) Cut a rectangle along a diagonal into two triangles. Do they have the same area? What is the relationship between a triangle and the rectangle that contains it? Answer. The two triangles overlap exactly, so they have equal areas — each triangle is half of the rectangle. So the area of triangle BAD is half the area of rectangle ABCD, and even a slanted triangle ABE (split into AEF and BEF) equals half of rectangle ABCD. Conclusion: the area of a triangle is half the area of the rectangle that exactly encloses it.
Making it ‘More’ or ‘Less’ — 9 unit squares (Page 145) Using 9 unit squares (area 9 sq units), solve: 1. smallest perimeter possible? 2. largest perimeter possible? 3. make a figure of perimeter 18 units. 4. Can you make other shapes for each perimeter, or is there only one? Answer. 1. Smallest perimeter = 12 units, from the 3 × 3 square. 2. Largest perimeter = 20 units, from a 1 × 9 row of squares. 3. A perimeter of 18 units is possible, for example an L-shaped or staircase arrangement of the 9 squares. 4. Yes — many different shapes give 18 units (and 20 units), but the smallest perimeter of 12 units has only one shape, the 3 × 3 square. Adding a square that touches along a full edge keeps the perimeter the same; adding it so a corner sticks out increases the perimeter.
Charan’s house plan (Page 146) Find Charan’s missing room measurements and the total area of his house (rectangular plot, some dimensions given; the plot is 30 ft on one side). Answer. Missing measurements: Small bedroom 15 ft × 12 ft = 180 sq ft; Utility 15 ft × 3 ft = 45 sq ft; Hall 20 ft × 12 ft = 240 sq ft; Parking 15 ft × 3 ft = 45 sq ft; Garden 20 ft × 3 ft = 60 sq ft. Whole house = 35 ft × 30 ft = 1050 sq ft.
Sharan’s house plan (Page 147) Find Sharan’s missing dimensions and area, and compare the areas and perimeters of Sharan’s and Charan’s houses. Answer. Sharan’s rooms: Utility 7 ft × 10 ft = 70 sq ft; Hall 23 ft × 15 ft = 345 sq ft; Entrance 7 ft × 15 ft = 105 sq ft; Small bedroom 12 ft × 10 ft = 120 sq ft; Toilet 5 ft × 10 ft = 50 sq ft. Whole house = 42 ft × 25 ft = 1050 sq ft. Comparison: both houses have the same area (1050 sq ft), but the perimeter of Charan’s house = 2 × (35 + 30) = 130 ft while Sharan’s = 2 × (42 + 25) = 134 ft, so Sharan’s house has the greater perimeter — a neat example of equal areas with different perimeters.
Area Maze Puzzles (Page 148) In each figure (a, b, c, d), find the missing value of either the length of a side or the area of a region. Answer. Reasoning from the matching side-lengths of neighbouring rectangles: (a) 30 sq cm; (b) 9 sq cm; (c) 16 sq cm; (d) 5 cm.

Figure it Out — Area (Page 138)

1. The area of a rectangular garden 25 m long is 300 sq m. What is the width of the garden?

SOLUTION Area = length × width, so width = area ÷ length = 300 ÷ 25 = 12 m.

2. What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of ₹8 per hundred sq m?

SOLUTION Area = 500 × 200 = 1,00,000 sq m. Number of “hundred sq m” units = 1,00,000 ÷ 100 = 1000. Cost = 1000 × ₹8 = ₹8000.

3. A rectangular coconut grove is 100 m long and 50 m wide. If each coconut tree requires 25 sq m, what is the maximum number of trees that can be planted in this grove?

SOLUTION Area of grove = 100 × 50 = 5000 sq m. Number of trees = 5000 ÷ 25 = 200.

4. By splitting the following figures into rectangles, find their areas (all measures are given in metres).

SOLUTION Split each L/step-shaped figure into rectangles and add. (a) total area = 28 sq m. (b) total area = 9 sq m. (Both figures are divided along their marked dimensions; the areas of the part-rectangles add to the values shown.)

Find the area of the following figures (squared-paper estimation, Page 140).

SOLUTION Counting full squares as 1, more-than-half squares as 1 and ignoring less-than-half squares gives the four areas 4 sq units, 9 sq units, 10 sq units and 11 sq units.

Figure it Out — Area of Triangles (Page 144)

1. Find the areas of the figures below by dividing them into rectangles and triangles.

SOLUTION Each composite figure (a–e) is split into rectangles plus right triangles; each triangle is half of its containing rectangle, and the parts are then added. The areas are a. 24 sq units, b. 30 sq units, c. 48 sq units, d. 16 sq units and e. 12 sq units.

Figure it Out — Area & Perimeter (Page 149)

1. Give the dimensions of a rectangle whose area is the sum of the areas of these two rectangles having measurements: 5 m × 10 m and 2 m × 7 m.

SOLUTION Sum of areas = (5 × 10) + (2 × 7) = 50 + 14 = 64 sq m. Any rectangle of area 64 sq m works, e.g. 16 m × 4 m, or 32 m × 2 m, or 8 m × 8 m.

2. The area of a rectangular garden that is 50 m long is 1000 sq m. Find the width of the garden.

SOLUTION Width = area ÷ length = 1000 ÷ 50 = 20 m.

3. The floor of a room is 5 m long and 4 m wide. A square carpet whose sides are 3 m in length is laid on the floor. Find the area that is not carpeted.

SOLUTION Area of floor = 5 × 4 = 20 sq m; area of carpet = 3 × 3 = 9 sq m. Area not carpeted = 20 − 9 = 11 sq m.

4. Four flower beds having sides 2 m long and 1 m wide are dug at the four corners of a garden that is 15 m long and 12 m wide. How much area is now available for laying down a lawn?

SOLUTION Area of garden = 15 × 12 = 180 sq m. Area of one flower bed = 2 × 1 = 2 sq m; four beds = 4 × 2 = 8 sq m. Area available for lawn = 180 − 8 = 172 sq m.

5. Shape A has an area of 18 square units and Shape B has an area of 20 square units. Shape A has a longer perimeter than Shape B. Draw two such shapes satisfying the given conditions.

SOLUTION Shape A (area 18) could be a 2 × 9 rectangle: perimeter = 2 × (2 + 9) = 22 units, or 1 × 18: perimeter 38 units. Shape B (area 20) could be a 5 × 4 rectangle: perimeter = 2 × (5 + 4) = 18 units. Choosing A = 2 × 9 (perimeter 22) and B = 5 × 4 (perimeter 18) gives perimeter(A) > perimeter(B) while the areas stay 18 and 20. (Draw any such pair on grid paper.)

6. On a page in your book, draw a rectangular border that is 1 cm from the top and bottom and 1.5 cm from the left and right sides. What is the perimeter of the border?

SOLUTION A typical book page is about 18 cm wide and 24 cm tall (use your own page size). Leaving 1.5 cm on each side gives a border width of 18 − 2 × 1.5 = 15 cm; leaving 1 cm at top and bottom gives a border height of 24 − 2 × 1 = 22 cm. Perimeter of the border = 2 × (15 + 22) = 74 cm for that page. (Measure your own page first, then apply Perimeter = 2 × (width − 3) + 2 × (height − 2) in cm.)

7. Draw a rectangle of size 12 units × 8 units. Draw another rectangle inside it, without touching the outer rectangle that occupies exactly half the area.

SOLUTION Area of outer rectangle = 12 × 8 = 96 sq units; half of this = 48 sq units. Draw any inner rectangle of area 48 sq units that does not touch the outer one, e.g. an 8 units × 6 units rectangle centred inside (8 × 6 = 48). (Other shapes such as 12 × 4 also work, provided they fit without touching.)

8. A square piece of paper is folded in half. The square is then cut into two rectangles along the fold. Regardless of the size of the square, one of the following statements is always true. Which statement is true here? a. The area of each rectangle is larger than the area of the square. b. The perimeter of the square is greater than the perimeters of both the rectangles added together. c. The perimeters of both the rectangles added together is always 1½ times the perimeter of the square. d. The area of the square is always three times as large as the areas of both rectangles added together.

SOLUTION Let the square have side s, so perimeter = 4s and area = s2. Each half-rectangle is s × s/2. Each rectangle’s perimeter = 2 × (s + s/2) = 3s, so both together = 6s = 1½ × 4s = 1½ × (perimeter of square). Each rectangle’s area = s2/2 (smaller, so a is false); the two areas add to s2 = the square (so d is false); 4s < 6s (so b is false). ∴ the always-true statement is (c).

Common Mistakes to Avoid

Watch out for these

  • Confusing perimeter with area — perimeter is a length (cm, m) around the boundary; area is the region inside (sq cm, sq m).
  • Forgetting to double in a rectangle: perimeter is 2 × (length + breadth), not length + breadth.
  • In running-track and rope problems, multiply one round’s perimeter by the number of rounds.
  • Counting a diagonal grid segment as one straight unit — a diagonal is longer, so a grid triangle’s perimeter is more than the count of segments.
  • Assuming equal areas mean equal perimeters — the same area can have many different perimeters (a long thin rectangle has a far bigger perimeter than a square of the same area).
  • When estimating area on squared paper, forgetting the rule: count more-than-half squares as 1 and ignore less-than-half squares.

Practice MCQs & Assertion–Reason

1. The perimeter of a rectangle of length 12 cm and breadth 8 cm is:

(a) 20 cm    (b) 40 cm    (c) 96 cm    (d) 48 cm

2. The perimeter of a square is 20 cm. The length of each side is:

(a) 4 cm    (b) 5 cm    (c) 10 cm    (d) 80 cm

3. A wire bent into a 5 cm × 3 cm rectangle is re-bent into a square. The side of the square is:

(a) 3 cm    (b) 4 cm    (c) 8 cm    (d) 16 cm

4. The third side of a triangle of perimeter 55 cm with two sides 20 cm and 14 cm is:

(a) 19 cm    (b) 21 cm    (c) 34 cm    (d) 41 cm

5. The perimeter of a regular hexagon made from a 36 cm string has each side equal to:

(a) 4 cm    (b) 6 cm    (c) 9 cm    (d) 12 cm

6. A rectangular garden 25 m long has an area of 300 sq m. Its width is:

(a) 10 m    (b) 12 m    (c) 15 m    (d) 275 m

7. The area of a triangle that is exactly half of a 12 cm × 8 cm rectangle is:

(a) 48 sq cm    (b) 96 sq cm    (c) 20 sq cm    (d) 40 sq cm

8. Among rectangles of area 24 sq units, the one with the least perimeter is:

(a) 1 × 24    (b) 2 × 12    (c) 3 × 8    (d) 4 × 6

9. The cost of fencing a 150 m × 120 m park at ₹40 per metre is:

(a) ₹10,800    (b) ₹21,600    (c) ₹43,200    (d) ₹7,20,000

10. Two closed figures with the same area:

(a) must have the same perimeter    (b) can have different perimeters    (c) must be congruent    (d) cannot both be rectangles

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

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

A-R 1. Assertion: The perimeter of a square of side 5 cm is 20 cm.

Reason: The perimeter of a square equals 4 times the length of one side.

A-R 2. Assertion: A diagonal of a rectangle divides it into two triangles of equal area.

Reason: The two triangles overlap each other exactly, so each is half the rectangle.

A-R 3. Assertion: Two figures with the same area always have the same perimeter.

Reason: A long thin rectangle and a square of equal area can have very different perimeters.

A-R 4. Assertion: Toshi ran a longer distance than Akshi on the Matha Pachchi tracks.

Reason: Toshi covered 7 × 180 = 1260 m while Akshi covered 5 × 220 = 1100 m.

A-R 5. Assertion: A perfect square grid triangle of three diagonal segments has perimeter exactly 9 straight units.

Reason: A diagonal of a unit square has the same length as its side.

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

Quick Revision Summary

  • Perimeter = total distance around the boundary; for a polygon it is the sum of all side lengths.
  • Perimeter of a rectangle = 2 × (length + breadth); of a square = 4 × side; of a regular polygon = number of sides × side.
  • Area = region enclosed, measured in square units; area of a rectangle = length × breadth, of a square = side × side.
  • A diagonal splits a rectangle into two equal triangles, so a triangle’s area is half the rectangle that contains it.
  • Estimate area on a grid: count full and more-than-half squares as 1, ignore less-than-half.
  • Two figures can have the same area with different perimeters (and the same perimeter with different areas).
  • For a fixed area, the longest thin rectangle has the greatest perimeter and the near-square has the least.

How to score full marks in this chapter

Write the correct formula first and substitute carefully — remember the “2 ×” for a rectangle’s perimeter and the “square units” for area. In rope or running-track problems, find one round’s perimeter, then multiply by the number of rounds. For composite shapes, split into rectangles and triangles, find each piece, and add. Always check units (cm vs m, length vs square units) and label your answer, so each step earns its mark.

Frequently Asked Questions

What is Class 6 Maths Ganita Prakash Chapter 6 about?

Chapter 6, Perimeter and Area, covers the perimeter of rectangles, squares, triangles and regular polygons, the area of rectangles, squares and triangles, estimating area on squared paper, and exploring how the same area can have different perimeters through running-track, split-and-rejoin and house-plan problems.

What is the difference between perimeter and area?

Perimeter is the distance once around the boundary of a closed figure and is measured in units of length (cm, m). Area is the amount of region the figure encloses and is measured in square units (sq cm, sq m).

How do you find the area of a triangle in Chapter 6?

In this chapter the area of a triangle is found as half the area of the rectangle that exactly contains it, because a diagonal of a rectangle divides it into two equal triangles.

Are these Class 6 Maths Ganita Prakash Chapter 6 solutions free?

Yes. All solutions are free and follow the official NCERT Ganita Prakash textbook for the 2026–27 session, with answers verified against the book’s answer key.

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