Get the simplified Class 6 Maths NCERT Solutions of Ganita Prakash Chapter 6 Perimeter and Area textbook exercise questions with complete explanations.

## Ganita Prakash Class 6 Maths Chapter 6 Solutions Perimeter and Area

**NCERT Solutions for Class 6 Maths Ganita Prakash Chapter 6 Perimeter and Area**

**6.1 Perimeter Figure it Out (Page No. 132)**

Question 1.

Find the missing terms:

(a) Perimeter of a rectangle = 14 cm; breadth = 2 cm; length = ?

Solution:

We know that perimeter of a rectangle = 2(l + b)

Here, the perimeter of the rectangle = 14 cm and breadth b = 2 cm, l =?

Thus 14 = 2(l + 2)

⇒ 14 = 2l + 4

⇒ 2l = 14 – 4 = 10

⇒ l = \(\frac{10}{2}\) = 5 cm

(b) Perimeter of a square = 20 cm; side of a length = ?

Solution:

We know that the perimeter of the square = 4 × a

where a = side of the square

∴ 20 = 4 × a

⇒ a = 5 cm

(c) Perimeter of a rectangle = 12 m; length = 3 m; breadth = ?

Solution:

Perimeter of rectangle = 2(l + b)

⇒ 12 = 2(3 + b)

⇒ 12 = 6 + 2b

⇒ 12 – 6 = 2b

⇒ 2b = 6 m

⇒ b = 3 m

Question 2.

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

Solution:

Here perimeter of rectangle = 2(5 + 3)

= 2 × 8

= 16 cm

Now the wire is straightened and then bent to form a square.

∴ Perimeter of square = 16 cm

⇒ 4a = 16 cm

⇒ a = 4 cm, the required length of the side of the square.

Question 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:

Let the length of the third side of the triangle be x cm then

Perimeter of triangle = AB + BC + CA

⇒ 55 = 20 + 14 + x

⇒ 55 = 34 + x

⇒ x = 55 – 34

⇒ x = 21 cm

Question 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 meter?

Solution:

The length of the fence is the perimeter of the rectangular park.

Given that the length of the rectangular park = 150 m and breadth = 120 m

∴ Perimeter = 2(l + b)

= 2(150 + 120)

= 2(270)

= 540 m

Now cost of fencing per meter = ₹ 40

Cost of fencing the rectangular park = ₹ 40 × 540 = ₹ 21600

Question 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 oil equal length?

Solution:

(a) Given, a piece of string is 36 cm long

∴ length of each side of the square = a

perimeter = 36

⇒ 4a = 36

⇒ a = 9 cm

(b) Length of each side of the triangle = 3a (Given)

perimeter = 36

⇒ 3a = 36

⇒ a = 12 cm

(c) Length of each side of hexagon = a

perimeter = 36

⇒ 36 = 6a

⇒ a = 6 cm

Question 6.

A farmer has a rectangular field with having length of 230 m and a breadth of 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 rectangular field = 2(l + b)

Here l = 230 m, b = 160 m

∴ P = 2(230 + 160)

= 2 (390)

= 780 m

Distance covered by a farmer in one round = 780 m

∴ Total length of rope needed = 3 × 780 = 2340 m

**6.1 Perimeter Figure it Out (Page No. 133 – 134)**

Question 1.

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

Solution:

Distance covered by Akshi in 5 rounds = 5 × perimeter of PQRS

= 5 × 220

= 1100 m

Question 2.

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

Solution:

Distance covered by Toshi in 7 rounds = 7 × 180 = 1260 m

∴ 1260 m > 1100 m

Hence Toshi ran the longer distance.

Question 3.

Think and mark the positions as directed-

(a) Mark ‘A’ at the point where Akshi will be after she runs 250 m.

Solution:

Here,

1. One complete round = 220 meters.

2. Distance Akshi has run = 250 meters.

3. Extra distance beyond one round = 250 – 220 = 30 meters.

Since Akshi has already completed one full round, she will be 30 meters into her second round. So, after running 30 meters more, she will be on the length side of the track, 30 meters from the starting point. Therefore, mark ‘A’ at the point 30 meters along the length of the track from the starting point.

(b) Mark ‘B’ at the point where Akshi will be after she runs 500 m.

Solution:

Distance per round = 220 meters

Total distance Akshi runs = 500 meters.

First, we will find out how many complete rounds she runs:

Number of complete rounds = \(\frac{500}{220}\) = 2.27 (approx.)

This means Akshi completes 2 full rounds and then runs an additional distance = 500 – (2 × 220) = 60 m

Therefore, Akshi will be 60 meters along the length of the track from the starting point, we can mark point ‘B’ at this position on the track.

(c) Now, Akshi ran 1000 m. How many full rounds has she finished running around her track? Mark her position as ‘C’.

Solution:

Now, Akshi ran 1000 meters. To find out how many full rounds she completed, we divide the total distance she ran by the perimeter of the track:

Number of full rounds = \(\frac{1000}{220}\) = 4.545 rounds.

Akshi has completed 4 full rounds and is partway through her 5th round.

To find her position on the track, we calculate the remaining distance after 4 full rounds:

Remaining distance = 1000 m – (4 × 220 m)

= 1000 m – 880 m

= 120 m

Since she has run an additional 120 meters after completing 4 full rounds, her position will be 120 meters from the starting point. If we mark her starting point as ‘P’, her position after running 1000 meters can be marked as ‘C’, which is 120 meters from ‘A’ along the track.

(d) Mark ‘X’ at the point where Toshi will be after she runs 250 m.

Solution:

Here,

1. Perimeter of the track = 180 meters

2. Distance Toshi runs = 250 meters

Since 250 meters is more than one complete round (180 meters), Toshi will have completed one full round and will have 70 meters left to run (250 – 180 = 70 meters). So, Toshi will be 70 meters along the length of the track from the starting point. You can mark ‘X’ at this point on the track.

(e) Mark ‘Y’ at the point where Toshi will be after she runs 500 m.

Solution:

Given, that Toshi has run an additional 140 meters after completing 2 rounds, her position will be 140 meters from the starting point. If we mark her starting point as ‘A’, her position after running 500 meters can be marked as ‘Y’.

(f) Now, Toshi ran 1000 m. How many full rounds has she finished running around her track? Mark her position as ‘Z’.

Solution:

Here, we need to find out how many full rounds Toshi has completed by dividing the total distance she ran by the perimeter of the track:

Number of full rounds = \(\frac{1000}{180}\) = 5.56

Toshi has finished 5 full rounds.

Remaining distance = 1000 meters – (5 × 180 meters)

= 1000 meters – 900 meters

= 100 meters

Starting from the initial point, Toshi would be 100 meters into her 6th round. Since the track is 60 meters long and 30 meters wide, she would be somewhere along the length of the track.

Let’s mark this position as ‘Z’.

**6.2 Area Figure it Out (Page No. 138)**

Question 1.

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

Solution:

Given, area of rectangular garden = 300 sq.m

and length = 25 m

area of rectangular field = l × b

⇒ 300 = 25 × b

⇒ b = 12 m

Question 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:

Here, length = 500 m and breadth = 200 m

Hence the area of the rectangular plot = length × breadth

= 500 × 200

= 1,00,000 m^{2}

Now cost of tilling a rectangular plot = \(\frac{8}{100}\)

Hence the cost of tilling 1,00,000 sq. m of rectangular plot = \(\frac{8}{100}\) × 100000 = ₹ 8,000

Question 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 rectangular coconut grove = 100 × 50 = 5000 sq. m

Given each coconut tree requires 25 sq. m

then the maximum no. of trees that can be planted in this grove = \(\frac{5000}{25}\) = 200

Question 4.

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

Solution:

(a) Splitting the given figure into I, II, III, and IV rectangles as shown in the figure below, we get

Here, the area of rectangle I = length × breadth

= 4 cm × 3 cm

= 12 sq. cm

Area of rectangle II = length × breadth

= 3 cm × 2 cm

= 6 sq. cm

Area of rectangle III = length × breadth

= 4 cm × 1 cm

= 4 sq. cm

Area of rectangle IV = length × breadth

= 3 cm × 2 cm

= 6 sq. cm

The total area of the whole figure = 12 sq. cm + 6 sq. cm + 4 sq. cm + 6 sq. cm = 28 sq. cm.

Therefore, the total area of Figure (a) is 28 sq. cm.

(b) Similarly, by splitting figure (b) into I, II, and III rectangles as shown in the figure below, we get

Area of the rectangle I = length × breadth

= 3 cm × 1 cm

= 3 sq. cm

Area of rectangle II = length × breadth

= 3 cm × 1 cm

= 3 sq. cm

Area of rectangle III = length × breadth

= 3 cm × 1 cm

= 3 sq. cm

The total area of the figure = 3 sq. cm + 3 sq. cm + 3 sq. cm = 9 sq. cm.

Therefore, the total area of Figure (b) is 9 sq. cm.

**6.2 Area Figure it Out (Page No. 139)**

Cut out the tangram pieces given at the end of your textbook.

Question 1.

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

Solution:

There are two pieces (A and B) that have the same area.

Question 2.

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

Solution:

Shape D is two times bigger than shape C. Clearly from the figure, the area of shapes C and E is equal to the area of shape D.

Question 3.

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

Solution:

Since the medium triangle and the square are each made up of two small tangram triangles, they each have an area 2x that of the small triangle. Hence both have the same area.

Question 4.

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

Solution:

Since the medium triangle and the rhomboid are each made up of two small tangram triangles, they each have an area 2x that of the small triangle. Hence both have the same area.

Question 5.

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

[Hint: In the tangram pieces, by placing the shapes over each other, we can find out that Shapes A and B have the same area, and Shapes C and E have the same area. You would have also figured out that Shape D can be exactly covered using Shapes C and E, which means Shape D has twice the area of Shape C or Shape E, etc.]

Solution:

Shape A has twice the area of shape G.

Question 6.

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

Answer:

Let’s say the area of C = x

Area of D = Area of 2C = 2x

Area of E = Area of C = x

Area of F = Area of 2C = 2x

Area of G = Area of 2C = 2x

Area of A = Area of 2F = 2 × 2x = 4x

Area of B = Area of A = 4x

Hence total area of big shape = Area of A + B + C + D + E + F + G

= 4x + 4x + x + 2x + x + 2x + 2x

= 16x

= 16C

That means the area of a big square is 16 times the area of shape C.

Question 7.

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:

The tangram rectangle with all 7 pieces is a tangram square with 5 pieces extended with two big triangles. All seven tans fit together to form a rectangle. Hence area of this rectangle in terms of Shape C is 16 small triangles.

Question 8.

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:

The perimeter of the square is equal to the square formed from these 7 pieces because these are the arrangements of pieces.

**6.3 Area of A Triangle Figure it Out (Page 144)**

Question 1.

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

Solution:

(a)

∴ Total area of the figure = 20 + 4 = 24 sq. units

(b)

∴ Total area of the figure = 25 + 4 = 29 sq. units

(c)

∴ Total area of the figure = 36 + 1 + 8 = 45 sq. units

(d)

∴ Total area of the figure = 13 + 3 = 16 sq. units

(e)

∴ Total area of the figure = 5 + 2 + 4 = 11 sq. units

**6.3 Area of A Triangle Figure it Out (Page 149)**

Question 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:

Here, Area of rectangle 1 = 5 × 10 = 50 sq m

Area of rectangle 2 = 2 × 7 = 14 sq m

The Sum of the areas of these 2 rectangles = 50 + 14 = 64 sq m

Now, the total area of the rectangle = 64

Let’s say the sides of the rectangle are Length = x and Width = y

Area of rectangle = x × y

Hence x × y = 64

xy = 64

Let’s say x = 1, then y = \(\frac{64}{1}\) = 64

if x = 2, then y = \(\frac{64}{2}\) = 32

Hence the dimensions of the rectangle are (1 × 64), (2 × 32)

Question 2.

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

Solution:

Width of rectangular garden = \(\frac{\text { area }}{\text { length of garden }}\)

= \(\frac{1000}{50}\)

= 20 m.

Question 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:

We have a floor with dimensions 4 m width and 5 m length.

A square carpet of side 3 m.

Area of the floor = length × breath

Area of the floor = 5 × 4 = 20 m^{2}

Area of the square carpet = 3 × 3 = 9 m^{2}

Now, we will subtract the square carpet area from the floor’s area to get the area of the floor that is not carpeted.

Hence, the area of the floor that is not carpeted = 20 – 9 = 11 m^{2}

Thus, the area of the floor that is not carpeted is 11 m^{2}.

Question 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:

Here, Length of garden = 15 m

Width of garden = 12 m

So, the area of the garden = 15 × 12 sq m = 180 sq m

Now, the length of the flower bed = 2 m

Width of flower bed = 1 m

Area of the flower bed = 2 × 1 sq m = 2 sq m

Since, the area of four flower beds = 2 × 4 sq m = 8 sq m

Now the area is available for laying down a lawn = (180 – 8) sq m = 172 sq m

Question 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 has an area of 18 sq. units.

∴ Possible sides are 18 × 1, 2 × 9, 6 × 3

Also, shape B has an area of 20 sq. units.

∴ Possible sides are 20 × 1, 4 × 5, 10 × 2

Given shape A has a longer perimeter than shape B, hence two such shapes satisfying the given conditions are:

Here Perimeter of shape A = 9 + 2 + 9 + 2 = 22 units

Here Perimeter of shape B = 5 + 4 + 5 + 4 = 18 units.

Question 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:

Perimeter of the rectangular border = 2 × [length + width]

= 2 × [1 + 1.5]

= 2 × 2.5

= 5 cm

Question 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 given rectangle = 12 × 8 = 96 units^{2}

and area of new rectangle = \(\frac{1}{2}\) × 96 = 48 sq. units

∴ Possible sides of new rectangle are 12 × 4, 16 × 3, 8 × 6, 1 × 48

∴ Hence dimensions of the new rectangle fill in the rectangle of 12 × 8 units^{2} = 8 units × 6 units.

Question 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 are always 1\(\frac{1}{2}\) 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:

Now in the above square piece

side of square = 1 unit

area of square = 1 × 1 = 1 sq. unit.

and perimeter of square = 1 + 1 + 1 + 1 = 4 units.

Now after folding the above square piece in half becomes 2 rectangles

Perimeter of rectangle R_{1} = 1 + \(\frac{1}{2}\) + 1 + \(\frac{1}{2}\) = 3 units.

Area of rectangle R_{1} = \(\frac{1}{2}\) × 1 = \(\frac{1}{2}\) sq. unit.

Perimeter of rectangle R_{2} = 1 + \(\frac{1}{2}\) + 1 + \(\frac{1}{2}\) = 3 units.

Area of rectangle R_{2} = \(\frac{1}{2}\) × 1 = \(\frac{1}{2}\) sq. unit.

(a) Now, area of rectangle R_{1} = area of rectangle R_{2} = \(\frac{1}{2}\) < 1.

Hence, option (a) is not true.

(b) Here perimeter of square = 4 units

and perimeters of both the rectangles = 3 + 3 = 6 units.

which is greater than 4 units.

Hence option (b) is not true.

(c) Here perimeters of both the rectangles = 6 units

and perimeter of square = 4 units × 1\(\frac{1}{2}\) = 4 × \(\frac{3}{2}\) = 6 units.

The perimeters of both the rectangles added together are 1\(\frac{1}{2}\) times the perimeter of the square.

Hence, option (c) is true.

(d) Here, the area of the square = 4 units

and areas of both the rectangles = \(\frac{1}{2}\) + \(\frac{1}{2}\) = 1 unit.

The area of the square is four times the area of both rectangles.

Hence, option (d) is not true.

Intext Questions

**Matha Pachchi! (Page No. 133)**

Akshi and Toshi start running along the rectangular tracks as shown in the figure. Akshi runs along the outer track and completes 5 rounds. Toshi runs along the inner track and completes 7 rounds. Now, they are wondering who ran more. Find out who ran the longer distance.

Solution:

Here, perimeter of rectangular track PQRS = 2 × (l + b)

= 2 × (70 + 40)

= 2 × 110

= 220 m

and perimeter of rectangular track ABCD = 2 × (60 + 30)

= 2 × 90

= 180 m

**Deep Dive: (Page No. 134)**

In races, usually, there is a common finish line for all the runners. Here are two square running tracks with an inner track of 100 m on each side and an outer track of 150 m on each side. The common finishing line for both runners is shown by the flags in the figure which are in the center of one of the sides of the tracks. If the total race is 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:

Inner Track (100 m per side)

Perimeter Calculation: The perimeter of the inner track is (4 times 100 = 400) meters.

Distance to Run: The runner on the inner track needs to run 350 meters.

Starting Position (A): Since the perimeter is 400 meters, the runner will start 50 meters before the common finish line (400 – 350 = 50 meters).

Outer Track (150 m per side)

Perimeter Calculation: The perimeter of the outer track is (4 times 150 = 600) meters.

Distance to Run: The runner on the outer track also needs to run 350 meters.

Starting Position (B): Since the perimeter is 600 meters, the runner will start 250 meters before the common finish line (600 – 350 = 250 meters).

**Perimeter of a Regular Polygon (Page No. 135)**

Find various objects from your surroundings that have regular shapes and find their perimeters. Also, generalize your understanding of the perimeter of other regular polygons.

Solution:

Some common objects with regular shapes and calculating their perimeters:

Here are a few examples:

1. Square Table:

Shape: Square

Side Length = 1 meter

Perimeter = 4 × 1 = 4 meters

2. Equilateral Triangle Clock:

Shape: Equilateral Triangle

Side Length = 30 cm

Perimeter = 3 × 30 = 90 cm

3. Hexagonal Tile:

Shape: Regular Hexagon

Side Length = 10 cm

Perimeter = 6 × 10 = 60 cm

In general, the Perimeter of a Regular Polygon = (Number of sides) × (Side length of a polygon) units.

**Split and Rejoin (Page No. 136)**

A rectangular paper chit of dimension 6 cm × 4 cm is cut as shown into two equal pieces. These two pieces are joined in different ways.

For example, the arrangement a. has a perimeter of 28 cm.

Find out the length of the boundary (i.e., the perimeter) of each of the other arrangements below.

Solution:

(b)

∴ Length of boundary = AB + BC + CD + DE + EF + FG + GA

= 6 + 2 + 6 + 2 + 4 + 6 + 2

= 28 cm

(c)

Total length of boundary = AB + BC + CD + DE +EF + FG + GH + HA

= 2 + 6 + 2 + 2 + 6 + 2 + 6 + 2

= 28 cm

(d)

Total length of boundary = AB + BC + CD + DE + EF + FG + GH + HA

= 6 + 2 + 3 + 2 + 6 + 2 + 3 + 2

= 26 cm

Arrange the two pieces to form a figure with a perimeter of 22 cm.

Solution:

Arranging the two pieces in such a way that they form a new shape with the desired perimeter:

Total length of boundary = AB + BC + CD + DE + EF + FG + GH + HA

= 2 + 1 + 2 + 6 + 2 + 1 + 2 + 6

= 22 cm

**Find the area of the following figures. (Page No. 140)**

Solution:

(i)

∴ Total area of the figure = 3 + 1 = 4 sq. units

(ii)

∴ Total area of the figure = 6 + 3 = 9 sq. units

(iii)

∴ Total area of the figure = 7 + 3 = 10 sq. units

(iv)

∴ Total area of the figure = 8 + 3 = 11 sq. unit

**Let’s Explore! (Page No. 141)**

On a squared grid paper (1 square = 1 square unit), make as many rectangles as you can whose lengths and widths are a whole number of units such that the area of the rectangle is 12 square units.

(a) Which rectangle has the greatest perimeter?

(b) Which rectangle has the least perimeter?

(c) If you take a rectangle of area 32 sq cm, what will your answers be? Given any area, is it possible to predict the shape of the rectangle with the greatest perimeter as well as the least perimeter? Give examples and reasons for your answer.

Solution:

Perimeter of (a) = 2(1 + 24) = 2 × 25 = 50 units

Perimeter of (b) = 2(2 + 12) = 2 × 14 = 28 units

Perimeter of (c) = 2(4 + 6) = 2 × 10 = 20 units

Perimeter of (d) = 2(3 + 8) = 2 × 11 = 22 units

(a) Clearly rectangle (a) has the greatest perimeter.

(b) Obviously rectangle (c) has the least perimeter.

(c) Yes, it is possible to predict the shape of a rectangle with the greatest and least perimeter for a given area. Here’s how:

Greatest Perimeter: For a given area, the rectangle with the greatest perimeter will have one side as small as possible. This essentially means that the rectangle becomes very elongated.

For example, if the area is 24 square units, a rectangle with dimensions 1 unit by 24 units will have the greatest perimeter.

Example: Area = 24 square units

Dimensions = 1 unit by 24 units

Perimeter = 2(1 + 24) = 50 units

Least Perimeter: The rectangle with the least perimeter for a given area will be as close to a square as possible. This is because a square has the smallest perimeter for a given area among all rectangles.

Example: Area = 24 square units

Dimensions = 4 units by 6 units (since 4 × 6 = 24)

Perimeter = 2(4 + 6) = 20 units

Reasoning

Greatest Perimeter: When one side is minimized, the other side must be maximized to maintain the same area. This increases the sum of the sides, thus increasing the perimeter.

Least Perimeter: A square or a shape close to a square minimizes the sum of the sides for a given area, thus minimizing the perimeter.

Check! whether the two triangles overlap each other exactly. Do they have the same area? **(Page No. 142)**

Solution:

Two triangles overlap each other exactly, which means they are congruent. Congruent triangles have the same shape and size, which implies that they also have the same area.

Can you draw any inferences from this exercise? Please write it here. **(Page No. 142)**

Solution:

Congruence: The triangles are congruent, meaning all corresponding sides and angles are equal.

Area: Since the triangles are congruent, their areas are identical.

Use your understanding from previous grades to calculate the area of any closed figure using grid paper and- **(Page No. 143)**

1. Find the area of the blue triangle BAD.

2. Find the area of the red triangle ABE.

Solution:

1. Area of blue triangle BAD

∴ Total area of BAD = 6 + 1 + 3 = 10 sq. units

2. Area of red triangle ABE

∴ Total area of ABE = 5 + 1 + 4 = 10 sq. units

Area of rectangle ABCD = Number of fully tilled squares

= 20 × 1

= 20 sq. units

**Making it ‘More’ or ‘Less’ (Page No. 145)**

**Using 9 unit squares, solve the following.**

Question 1.

What is the smallest perimeter possible?

Solution:

The smallest perimeter is achieved by forming a 3 × 3 square:

Perimeter = 3 + 3 + 3 + 3 = 12 units.

Question 2.

What is the largest perimeter possible?

Solution:

The largest perimeter is achieved by arranging the squares in a straight line:

Perimeter = 1 + 9 + 1 + 9 = 20 units.

Question 3.

Make a figure with a perimeter of 18 units.

Solution:

One possible figure is an L-shaped arrangement:

Arrange 6 squares in a vertical line and 3 squares in a horizontal line at the bottom.

Perimeter = 6 + 3 + 1 + 1 + 1 + 1 + 1 + 1 + 3 = 18 units.

Question 4.

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:

Smallest Perimeter (12 units): Only the 3 × 3 square achieves this.

Largest Perimeter (20 units): Only the straight line achieves this or any pattern made by folding two straight lines.

Perimeter of 18 units: Multiple shapes can achieve this. For example, a T-shaped figure or other L-shaped configurations.

The reasoning is based on the arrangement of the unit squares and the number of exposed edges. The more compact the shape, the smaller the perimeter; the more elongated, the larger the perimeter. For intermediate perimeters like 18 units, various configurations can be created by adjusting the arrangement of the squares.