---
title: "NCERT Solutions for Class 6 Maths Chapter 6 Perimeter and Area"
url: https://www.swavid.com/maths/class/6/chapter/perimeter-and-area/ncert-solutions
dateModified: 2026-10-07T14:51:12+00:00
---

# NCERT Solutions for Class 6 Maths Chapter 6 Perimeter and Area

This chapter's questions cover fundamental and advanced concepts related to the perimeter and area of various closed plane figures, polygons, rectangles, squares, and triangles, along with real-life problem solving.

Free PDF (16 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-6/swavid-ncert-solutions-class-6-maths-chapter-6-perimeter-and-area-77adcc2b60.pdf

## Figure it Out

### Question 1

*3 marks · Short answer*

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 = ?.

**Part a (1 mark)**

1. Perimeter of a rectangle = $2 \times (\text{length} + \text{breadth})$
2. $14 = 2 \times (\text{length} + 2)$
3. $\text{length} + 2 = 7$, so $\text{length} = 5~\text{cm}$

Answer a: 5 cm

**Part b (1 mark)**

1. Perimeter of a square = $4 \times \text{side}$
2. $20 = 4 \times \text{side}$
3. $\text{side} = 20 \div 4 = 5~\text{cm}$

Answer b: 5 cm

**Part c (1 mark)**

1. Perimeter of a rectangle = $2 \times (\text{length} + \text{breadth})$
2. $12 = 2 \times (3 + \text{breadth})$
3. $3 + \text{breadth} = 6$, so $\text{breadth} = 3~\text{m}$

Answer c: 3 m

**Answer:** a. length = 5 cm, b. side = 5 cm, c. breadth = 3 m

> Common mistake: Forgetting to divide the perimeter by 2 before subtracting the known side in a rectangle.

### Question 2

*3 marks · Short answer*

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**

1. Perimeter of the rectangle = $2 \times (5~\text{cm} + 3~\text{cm}) = 2 \times 8~\text{cm} = 16~\text{cm}$
2. Length of the wire = Perimeter of the rectangle = $16~\text{cm}$
3. The wire is bent to form a square, so the perimeter of the square is $16~\text{cm}$
4. Length of a side of the square = $\text{Perimeter} \div 4 = 16~\text{cm} \div 4 = 4~\text{cm}$

**Answer:** 4 cm

> Common mistake: Dividing by 2 instead of 4 when finding the side of the square from its perimeter.

### Question 3

*3 marks · Short answer*

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**

1. Perimeter of a triangle = sum of the lengths of its three sides
2. Given perimeter = $55~\text{cm}$, and two sides are $20~\text{cm}$ and $14~\text{cm}$
3. Sum of the two given sides = $20~\text{cm} + 14~\text{cm} = 34~\text{cm}$
4. Length of the third side = $55~\text{cm} - 34~\text{cm} = 21~\text{cm}$

**Answer:** 21 cm

> Common mistake: Adding all given numbers to the perimeter instead of subtracting the two known sides.

### Question 4

*3 marks · Short answer*

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**

1. Length of the park = $150~\text{m}$, Breadth of the park = $120~\text{m}$
2. Perimeter of the park = $2 \times (\text{length} + \text{breadth}) = 2 \times (150~\text{m} + 120~\text{m}) = 2 \times 270~\text{m} = 540~\text{m}$
3. Cost of fencing per metre = $₹40$
4. Total cost of fencing = $540 \times 40 = ₹21,600$

**Answer:** ₹21,600

> Common mistake: Multiplying length and breadth (area) instead of finding the perimeter for fencing.

### Question 5

*3 marks · Short answer*

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?

**Part a (1 mark)**

1. A square has $4$ equal sides and its total length is $36~\text{cm}$
2. Length of each side = $36~\text{cm} \div 4 = 9~\text{cm}$

Answer a: 9 cm

**Part b (1 mark)**

1. A triangle with all sides equal (equilateral triangle) has $3$ equal sides
2. Length of each side = $36~\text{cm} \div 3 = 12~\text{cm}$

Answer b: 12 cm

**Part c (1 mark)**

1. A regular hexagon has $6$ equal sides
2. Length of each side = $36~\text{cm} \div 6 = 6~\text{cm}$

Answer c: 6 cm

**Answer:** a. 9 cm, b. 12 cm, c. 6 cm

> Common mistake: Dividing by the wrong number of sides (e.g., dividing by 4 for a hexagon).

### Question 6

*3 marks · Short answer*

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**

1. Length of the rectangular field = $230~\text{m}$, Breadth = $160~\text{m}$
2. Perimeter of the field = $2 \times (230~\text{m} + 160~\text{m}) = 2 \times 390~\text{m} = 780~\text{m}$
3. The farmer wants to fence it with $3$ rounds of rope
4. Total length of rope needed = $3 \times 780~\text{m} = 2340~\text{m}$

**Answer:** 2340 m

> Common mistake: Calculating the perimeter for only 1 round and forgetting to multiply by 3.

## Figure it Out

### Question 1

*3 marks · Short answer*

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

**Solution**

1. Perimeter of Akshi's outer track = $2 \times (70 + 40) \text{ m} = 220 \text{ m}$
2. Total distance covered by Akshi in 5 rounds = $5 \times 220 \text{ m} = 1100 \text{ m}$
3. Hence, the total distance Akshi covered is $1100 \text{ m}$.

**Answer:** $1100 \text{ m}$

> Common mistake: Multiplying by the inner track dimensions instead of the outer track dimensions.

### Question 2

*3 marks · Short answer*

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

**Solution**

1. Perimeter of the inner track = $2 \times (60 + 30) \text{ m} = 180 \text{ m}$.
2. Total distance Toshi has covered in 7 rounds = $7 \times 180 \text{ m} = 1260 \text{ m}$.
3. Comparing the distances, Akshi covered $1100 \text{ m}$ and Toshi covered $1260 \text{ m}$, so Toshi ran a longer distance.

**Answer:** Toshi ran a longer distance of 1260 m.

> Common mistake: Using the dimensions of the outer track instead of the inner track for Toshi.

### Question 3

*5 marks · Long answer*

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'.

**Part a**

1. Akshi's one round is $220 \text{ m}$ and she starts at the given starting point.
2. After $250 \text{ m}$, she completes 1 full round ($220 \text{ m}$) and travels an extra $30 \text{ m}$ along the track.
3. Mark 'A' at $30 \text{ m}$ past the starting point.

Answer a: Mark 'A' at $30 \text{ m}$ past the starting point along the outer track.

**Part b**

1. Akshi's one round is $220 \text{ m}$.
2. After $500 \text{ m}$, she completes 2 full rounds ($440 \text{ m}$) and travels an extra $60 \text{ m}$ ($500 - 440$).
3. Mark 'B' at $60 \text{ m}$ past the starting point.

Answer b: Mark 'B' at $60 \text{ m}$ past the starting point along the outer track.

**Part c**

1. Akshi runs $1000 \text{ m}$ around a $220 \text{ m}$ track.
2. Dividing $1000$ by $220$ gives 4 full rounds ($880 \text{ m}$) with a remainder of $120 \text{ m}$.
3. Mark position 'C' at $120 \text{ m}$ past the starting point after completing 4 full rounds.

Answer c: 4 full rounds completed; mark 'C' at $120 \text{ m}$ past the starting point.

**Part d**

1. Toshi's one round is $180 \text{ m}$ and she starts at the inner track starting point.
2. After $250 \text{ m}$, she completes 1 full round ($180 \text{ m}$) and travels an extra $70 \text{ m}$ ($250 - 180$).
3. Mark 'X' at $70 \text{ m}$ past the starting point.

Answer d: Mark 'X' at $70 \text{ m}$ past the starting point along the inner track.

**Part e**

1. Toshi's one round is $180 \text{ m}$.
2. After $500 \text{ m}$, she completes 2 full rounds ($360 \text{ m}$) and travels an extra $140 \text{ m}$ ($500 - 360$).
3. Mark 'Y' at $140 \text{ m}$ past the starting point.

Answer e: Mark 'Y' at $140 \text{ m}$ past the starting point along the inner track.

**Part f**

1. Toshi runs $1000 \text{ m}$ around a $180 \text{ m}$ track.
2. Dividing $1000$ by $180$ gives 5 full rounds ($900 \text{ m}$) with a remainder of $100 \text{ m}$.
3. Mark position 'Z' at $100 \text{ m}$ past the starting point after completing 5 full rounds.

Answer f: 5 full rounds completed; mark 'Z' at $100 \text{ m}$ past the starting point.

**Answer:** Positions marked based on track perimeters.

> Common mistake: Dividing the total distance incorrectly by the perimeter or miscalculating the remainder distance along the rectangular sides.

## Deep Dive

### Question 1

*3 marks · Short answer*

If the total race is of 350 m, then we have to find out where the starting positions of the two runners should be on these two tracks so that they both have a common finishing line after they run for 350 m. Mark the starting points of the runner on the inner track as 'A' and the runner on the outer track as 'B'.

**Solution**

1. The inner square track has a side of $100\text{ m}$, so the distance from the starting point to the common finishing line for the inner runner is $100\text{ m} + 100\text{ m} + 100\text{ m} + 50\text{ m} = 350\text{ m}$.
2. Thus, the starting point 'A' for the inner runner is placed at a distance of $50\text{ m}$ before the corner along the side, such that the total distance to the finish line is $350\text{ m}$.
3. The outer square track has a side of $150\text{ m}$, so the starting point 'B' for the outer runner is placed such that the distance along the track to the common finishing line equals $125\text{ m} + 150\text{ m} + 75\text{ m} = 350\text{ m}$.

**Answer:** The starting points 'A' and 'B' are placed at distances such that the path along each track to the common finishing line measures exactly $350\text{ m}$.

> Common mistake: Counting the full perimeter of the track instead of measuring the distance backwards from the finishing line.

## Figure it Out

### Question 1

*3 marks · Short answer*

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

**Solution**

1. Given length of the rectangular garden = $25\text{ m}$ and area = $300\text{ sq m}$
2. Area of a rectangle = $\text{length} \times \text{width}$
3. Width = $\frac{\text{Area}}{\text{length}} = \frac{300}{25} = 12\text{ m}$

**Answer:** $12\text{ m}$

> Common mistake: Multiplying length and area instead of dividing the area by the length to find the width.

### Question 2

*3 marks · Short answer*

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**

1. Length of the rectangular plot = $500\text{ m}$ and width = $200\text{ m}$
2. Area of the plot = $\text{length} \times \text{width} = 500\text{ m} \times 200\text{ m} = 100,000\text{ sq m}$
3. Cost of tiling per hundred sq m = $₹8$
4. Total cost = $\frac{100,000}{100} \times 8 = 1000 \times 8 = ₹8,000$

**Answer:** $₹8,000$

> Common mistake: Forgetting to divide the total area by 100 before multiplying by the rate given per hundred square metres.

### Question 3

*3 marks · Short answer*

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**

1. Length of the rectangular coconut grove = $100\text{ m}$ and width = $50\text{ m}$
2. Area of the grove = $\text{length} \times \text{width} = 100\text{ m} \times 50\text{ m} = 5,000\text{ sq m}$
3. Area required for each coconut tree = $25\text{ sq m}$
4. Maximum number of trees = $\frac{5,000}{25} = 200$

**Answer:** $200\text{ trees}$

> Common mistake: Adding the length and width instead of multiplying to find the total area.

### Question 4

*3 marks · Short answer*

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

**Part a (1.5 marks)**

1. Split the figure into three horizontal rectangles: top rectangle of dimensions $3\text{ m} \times 2\text{ m}$, middle rectangle of $5\text{ m} \times 2\text{ m}$, and bottom rectangle of $3\text{ m} \times 2\text{ m}$ (or using the dimensions shown in the textbook).
2. Area of top rectangle = $3\text{ m} \times 2\text{ m} = 6\text{ sq m}$.
3. Area of middle rectangle = $4\text{ m} \times 4\text{ m}$ or as per grid = $16\text{ sq m}$, and bottom part $3\text{ m} \times 2\text{ m} = 6\text{ sq m}$, giving total area = $6 + 16 + 6 = 28\text{ sq m}$.

Answer a: $28\text{ sq m}$

**Part b (1.5 marks)**

1. Split the U-shaped figure into three simple rectangles: left vertical rectangle, right vertical rectangle, and top horizontal connecting bar.
2. Calculate the area of each part using the given dimensions: left vertical rectangle = $3\text{ m} \times 1\text{ m} = 3\text{ sq m}$, right vertical rectangle = $3\text{ m} \times 1\text{ m} = 3\text{ sq m}$, and top bar = $3\text{ m} \times 1\text{ m} = 3\text{ sq m}$.
3. Add the individual areas to find the total area: $3\text{ sq m} + 3\text{ sq m} + 3\text{ sq m} = 9\text{ sq m}$.

Answer b: $9\text{ sq m}$

**Answer:** (a) $28\text{ sq m}$, (b) $9\text{ sq m}$

> Common mistake: Forgetting to correctly determine the missing side lengths of the split rectangles before calculating their individual areas.

## Figure it Out

### Question 1

*3 marks · Short answer*

Explore and figure out how many pieces have the same area.

**Solution**

1. Observe the tangram pieces in the figure in the textbook (Fig. 6.11).
2. Shapes A and B are congruent large triangles and have the same area.
3. Shapes C and E are congruent small triangles and also have the same area.
4. Thus, two pairs of pieces have the same area (A and B; C and E).

**Answer:** Two pairs of pieces have the same area: Shapes A and B, and Shapes C and E.

> Common mistake: Confusing different shapes that have the same area with congruent shapes.

### Question 2

*3 marks · Short answer*

How many times bigger is Shape D as compared to Shape C? What is the relationship between Shapes C, D and E?

**Solution**

1. Observe the tangram pieces in the figure in the textbook (Fig. 6.11).
2. Shape D can be exactly covered using Shapes C and E.
3. Since Shapes C and E have the same area, Shape D has twice the area of Shape C.
4. The relationship is that Shape D is 2 times bigger than Shape C, and Shape D is equal to the sum of the areas of Shapes C and E.

**Answer:** Shape D is 2 times bigger than Shape C; Shape D is equal to the combined area of Shapes C and E.

> Common mistake: Assuming shape D is four times bigger because of linear dimensions.

### Question 3

*3 marks · Short answer*

Which shape has more area: Shape D or F? Give reasons for your answer.

**Solution**

1. Shape D is a square formed by combining two smaller triangles (Shapes C and E), so its area is equal to the sum of the areas of Shape C and Shape E.
2. Shape F is a parallelogram whose area can be shown to be equal to the area of Shape C plus Shape E using standard tangram area relations.
3. Therefore, both Shape D and Shape F have the same area, as each is composed of two basic triangle units of the tangram.

**Answer:** Both Shape D and Shape F have the same area because each is made up of two smaller triangle units (equivalent to two of Shape C).

> Common mistake: Assuming visual shape differences mean different areas without checking their composition from smaller tangram pieces.

### Question 4

*3 marks · Short answer*

Which shape has more area: Shape F or G? Give reasons for your answer.

**Solution**

1. Observe Shapes F and G in the tangram puzzle in the figure in the textbook (Fig. 6.11).
2. Shape G is a medium triangle formed by combining two small triangles (Shapes C and E).
3. Shape F is a parallelogram which is also equal in area to Shape D, which is formed by Shapes C and E.
4. Therefore, both Shape F and Shape G have the same area.

**Answer:** Both Shape F and Shape G have the same area because each is equivalent in size to two small triangle units.

> Common mistake: Assuming triangles always have smaller areas than parallelograms in a tangram.

### Question 5

*3 marks · Short answer*

What is the area of Shape A as compared to Shape G? Is it twice as big? Four times as big?

**Solution**

1. Observe Shapes A and G in the tangram puzzle in the figure in the textbook (Fig. 6.11).
2. Shape G is a medium triangle and Shape A is a large triangle.
3. A large triangle (Shape A) is made up of two medium triangles (like Shape G).
4. Therefore, the area of Shape A is twice as big as the area of Shape G.

**Answer:** Shape A is twice as big as Shape G.

> Common mistake: Saying Shape A is four times as big by confusing length scale with area scale.

### Question 6

*3 marks · Short answer*

Can you now figure out the area of the big square formed with all seven pieces in terms of the area of Shape C?

**Solution**

1. Recall that the big square is made of all 7 tangram pieces.
2. Shape C is one small triangle, and two small triangles make up Shape D or Shape F or Shape G.
3. Shape A and Shape B are each equal to 4 small triangles, and Shape E is equal to 1 small triangle.
4. Adding them up, the big square contains 16 small triangles (Shape C units).

**Answer:** The area of the big square is equal to 16 times the area of Shape C.

> Common mistake: Counting the number of pieces (7) instead of the area in terms of Shape C units.

### Question 7

*3 marks · Short answer*

Arrange these 7 pieces to form a rectangle. What will be the area of this rectangle in terms of the area of Shape C now? Give reasons for your answer.

**Solution**

1. The big square is formed by combining all 7 tangram pieces.
2. Since area is conserved when shapes are rearranged, the area of the rectangle formed using all 7 pieces is equal to the area of the big square.
3. In terms of Shape C, the total area of the 7 pieces (the big square or the rectangle) is equal to 16 times the area of Shape C.

**Answer:** The area of the rectangle is $16 \times \text{Area of Shape C}$ because the area remains conserved when rearranging the same pieces.

> Common mistake: Confusing area with perimeter and assuming changing the shape changes its total area.

### Question 8

*3 marks · Short answer*

Are the perimeters of the square and the rectangle formed from these 7 pieces different or the same? Give an explanation for your answer.

**Solution**

1. The perimeters of the square and the rectangle formed from the same 7 pieces are different.
2. Perimeter depends on the outer boundary length and how the sides of the individual pieces are joined together.
3. When the pieces are rearranged from a square to a rectangle, different lengths of the pieces form the outer boundary, resulting in a different total perimeter.

**Answer:** The perimeters are different because perimeter measures the outer boundary, which changes when the same pieces are arranged into different shapes.

> Common mistake: Assuming that shapes with the same area must also have the same perimeter.

## Figure it Out

### Question 1

*3 marks · Short answer*

What is the smallest perimeter possible?

**Solution**

1. Using 9 unit squares, the shape that is most compact and closest to a square is a $3 \times 3$ square.
2. The perimeter of a $3 \times 3$ square is $4 \times 3 = 12$ units.
3. Therefore, the smallest possible perimeter is 12 units.

**Answer:** 12 units

> Common mistake: Confusing area with perimeter.

### Question 2

*3 marks · Short answer*

What is the largest perimeter possible?

**Solution**

1. Using 9 unit squares, the shape with the longest boundary is a single straight line of 9 squares in a row.
2. The length of this rectangle is 9 units and the breadth is 1 unit.
3. The perimeter is $2 \times (9 + 1) = 20$ units.

**Answer:** 20 units

> Common mistake: Calculating perimeter by just adding lengths of squares without considering shared sides.

### Question 3

*3 marks · Short answer*

Make a figure with a perimeter of 18 units.

**Solution**

1. We need to arrange 9 unit squares such that the total length of the outer boundary is 18 units.
2. A rectangle of length 4 units and breadth 2.25 units is not possible since dimensions must be whole numbers, so we try other configurations like a T-shape or L-shape.
3. A figure made by attaching squares to form a boundary of 18 units can be drawn on grid paper, such as a rectangle of length 7 and breadth 2 which gives $2 \times (7 + 2) = 18$ units.

**Answer:** A rectangle of length 7 units and breadth 2 units (or any valid connected figure with perimeter 18 units)

> Common mistake: Drawing a figure that is not a single connected shape.

### Question 4

*3 marks · Short answer*

Can you make other shaped figures for each of the above three perimeters, or is there only one shape with that perimeter? What is your reasoning?

**Solution**

1. Yes, we can make other shaped figures for the perimeters of 20 units and 18 units.
2. For the perimeter of 12 units, only a $3 \times 3$ square is possible because any other arrangement of 9 squares will have a longer boundary.
3. For 18 and 20 units, different arrangements of the 9 unit squares give the same perimeter.

**Answer:** Yes, except for the smallest perimeter of 12 units, multiple shapes can have the same perimeter.

> Common mistake: Assuming that a given perimeter is always tied to a single unique shape.

## Figure it Out

### Question 1

*3 marks · Short answer*

Give the dimensions of a rectangle whose area is the sum of the areas of these two rectangles having measurements: $5\text{ m} \times 10\text{ m}$ and $2\text{ m} \times 7\text{ m}$.

**Solution**

1. Given the measurements of the first rectangle as $5\text{ m} \times 10\text{ m}$ and the second as $2\text{ m} \times 7\text{ m}$.
2. Area of the first rectangle = $5\text{ m} \times 10\text{ m} = 50\text{ sq m}$.
3. Area of the second rectangle = $2\text{ m} \times 7\text{ m} = 14\text{ sq m}$.
4. Total area required = $50\text{ sq m} + 14\text{ sq m} = 64\text{ sq m}$.
5. Possible dimensions of a rectangle with area $64\text{ sq m}$ are length $= 16\text{ m}$ and breadth $= 4\text{ m}$ (or $32\text{ m} \times 2\text{ m}$, or $8\text{ m} \times 8\text{ m}$).
6. Hence, the dimensions can be $16\text{ m} \times 4\text{ m}$.

**Answer:** $16\text{ m} \times 4\text{ m}$ (or $32\text{ m} \times 2\text{ m}$ or $8\text{ m} \times 8\text{ m}$)

> Common mistake: Adding lengths and breadths instead of calculating areas.

### Question 2

*3 marks · Short answer*

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

**Solution**

1. Area of the rectangular garden $= 1000\text{ sq m}$.
2. Length of the garden $= 50\text{ m}$.
3. Area of a rectangle $= \text{length} \times \text{width}$.
4. Width $= \text{Area} \div \text{length} = 1000\text{ sq m} \div 50\text{ m} = 20\text{ m}$.

**Answer:** 20 m

> Common mistake: Dividing length by area instead of area by length.

### Question 3

*3 marks · Short answer*

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**

1. Length of the floor $= 5\text{ m}$ and width of the floor $= 4\text{ m}$.
2. Area of the floor $= 5\text{ m} \times 4\text{ m} = 20\text{ sq m}$.
3. Side of the square carpet $= 3\text{ m}$.
4. Area of the carpet $= 3\text{ m} \times 3\text{ m} = 9\text{ sq m}$.
5. Area of the floor not carpeted $= 20\text{ sq m} - 9\text{ sq m} = 11\text{ sq m}$.

**Answer:** 11 sq m

> Common mistake: Adding the areas instead of subtracting.

### Question 4

*3 marks · Short answer*

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**

1. Length of the garden $= 15\text{ m}$ and width of the garden $= 12\text{ m}$.
2. Area of the whole garden $= 15\text{ m} \times 12\text{ m} = 180\text{ sq m}$.
3. Dimensions of each flower bed $= 2\text{ m} \times 1\text{ m}$, so area of one flower bed $= 2\text{ sq m}$.
4. Total area of four flower beds $= 4 \times 2\text{ sq m} = 8\text{ sq m}$.
5. Available area for lawn $= 180\text{ sq m} - 8\text{ sq m} = 172\text{ sq m}$.

**Answer:** 172 sq m

> Common mistake: Multiplying the area of one flower bed by two instead of four.

### Question 5

*3 marks · Short answer*

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**

1. Shape A has an area of $18\text{ square units}$ and Shape B has an area of $20\text{ square units}$.
2. Possible dimensions for Shape A (rectangle) can be $6\text{ units} \times 3\text{ units}$, giving a perimeter of $2 \times (6 + 3) = 18\text{ units}$.
3. Alternatively, dimensions of Shape A can be $2\text{ units} \times 9\text{ units}$, giving a perimeter of $2 \times (2 + 9) = 22\text{ units}$.
4. Possible dimensions for Shape B (rectangle) can be $5\text{ units} \times 4\text{ units}$, giving a perimeter of $2 \times (5 + 4) = 18\text{ units}$.
5. Since the perimeter of Shape A must be greater than Shape B, we can choose Shape A of dimensions $2\text{ units} \times 9\text{ units}$ (perimeter $= 22\text{ units}$) and Shape B of dimensions $5\text{ units} \times 4\text{ units}$ (perimeter $= 18\text{ units}$).
6. Thus, the required dimensions satisfy both area and perimeter conditions.

**Answer:** Shape A: $2\text{ units} \times 9\text{ units}$ and Shape B: $5\text{ units} \times 4\text{ units}$

> Common mistake: Confusing area with perimeter.

### Question 6

*3 marks · Short answer*

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**

1. Let the book's page have length $L$ and breadth $B$.
2. The border is $1\text{ cm}$ from the top and bottom, so its length is $L - 1 - 1 = L - 2$.
3. The border is $1.5\text{ cm}$ from the left and right, so its breadth is $B - 1.5 - 1.5 = B - 3$.
4. Using standard textbook dimensions for a notebook page (approx $28\text{ cm} \times 21\text{ cm}$), the border dimensions are length $= 28 - 2 = 26\text{ cm}$ and breadth $= 21 - 3 = 18\text{ cm}$.
5. Perimeter of the border $= 2 \times (26 + 18) = 2 \times 44 = 88\text{ cm}$.

**Answer:** 88 cm (depending on page size)

> Common mistake: Subtracting the margins only once instead of from both sides.

### Question 7

*3 marks · Short answer*

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

**Solution**

1. Area of outer rectangle = length $\times$ breadth = $12 \times 8 = 96 \text{ sq. units}$.
2. Area of inner rectangle = exactly half of the outer rectangle = $96 \div 2 = 48 \text{ sq. units}$.
3. Draw an outer rectangle of dimensions 12 units $\times$ 8 units and an inner rectangle inside it such that it does not touch the boundary and has an area of 48 sq. units.

**Answer:** Area of outer rectangle = 96 sq. units, Area of inner rectangle = 48 sq. units

> Common mistake: Making the inner rectangle touch the outer boundary.

### Question 8

*1 mark · MCQ*

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?

- The area of each rectangle is larger than the area of the square.
- The perimeter of the square is greater than the perimeters of both the rectangles added together.
- The perimeters of both the rectangles added together is always $1\frac{1}{2}$ times the perimeter of the square.
- The area of the square is always three times as large as the areas of both rectangles added together.

**Solution**

1. Let the side of the square be $s$. Its perimeter is $4s$.
2. When cut in half, it forms two identical rectangles of dimensions $s$ and $\frac{s}{2}$.
3. The sum of their perimeters is $2(s + \frac{s}{2}) + 2(s + \frac{s}{2}) = 6s$, which is $1\frac{1}{2}$ times the perimeter of the square ($4s$).

**Answer:** (c) The perimeters of both the rectangles added together is always $1\frac{1}{2}$ times the perimeter of the square.

> Common mistake: Confusing area with perimeter when folding and cutting paper.

## Frequently asked questions

### How many exercises and total questions are in Class 6 Maths Chapter 6 Perimeter and Area?

This chapter in the new NCERT book for the 2026-27 session contains multiple sections of 'Figure it Out' and 'Deep Dive' exercises. In total, there are 34 questions spread across various topics like rectangle areas, perimeters, and tangram shapes. You can find step-by-step solutions for all these questions in SwaVid's free PDF available on this page only.

### What main topics and concepts do the questions cover in this chapter?

The questions cover essential topics such as perimeter formulas for squares and rectangles, finding the cost of fencing, and calculating the area of rectangles. Other concepts include splitting composite figures into rectangles, working with unit squares, and comparing areas and perimeters using tangram pieces.

### Which question types are considered the most challenging, and how should students approach them?

Questions involving composite figures, track positions, and tangram area relationships are generally considered the trickiest. To approach them, you should carefully split complex shapes into simpler rectangles or standard units before applying the relevant perimeter or area formulas.

### How should students write their answers to score full marks in exams?

To score full marks, always start by writing down the given values and the relevant formula, such as $\text{Perimeter} = 2 \times (\text{Length} + \text{Breadth})$ or $\text{Area} = \text{Length} \times \text{Breadth}$. Show every step of your calculation clearly and remember to include the correct units like $\text{cm}$ or $\text{m}^2$ in your final answer. You can refer to SwaVid's free PDF on this page only to learn the exact presentation format.

### Is a free PDF for these NCERT solutions available for download?

Yes, SwaVid provides a complete and reliable free PDF for this chapter aligned with the new NCERT book for the 2026-27 session. You can easily access and download these detailed step-by-step solutions right here on this page only to help with your Class 6 Maths revision.

## Related pages

- [Class 6 Maths chapters](https://www.swavid.com/maths/class/6)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student