## NCERT Solutions Class 6 Math Ganita Prakash Chapter 6 Perimeter and Area

Get NCERT Class 6 Math Ganita Prakash 6th chapter ‘Perimeter and Area’ solutions. This new NCERT book solutions will help students.

**Figure it out:**

**1) Find the missing terms: **

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

**Solution:**

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

2 × length = Perimeter – 2 × breadth

2 × length = (14 – 2×2)

Length = 10/2 = 5

Length of perimeter = 5 cm

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

**Solution:**

Perimeter of a square = 4 × length of a side

Length of a side = Perimeter/4

Length of a side = 20/4 = 5

Side of a length = 5 cm

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

**Solution:**

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

Breadth = (Perimeter – 2×Length)/2

Breadth = (12 – 3×2)/2

Breadth = 3

Breadth of perimeter = 3 m

**2.) A rectangle having side lengths 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 and breadths of rectangle = 5 cm and 3 cm

The rectangle is made using a piece of wire, so find the total length of wire we need to calculate the length of perimeter.

So, Perimeter of rectangle = 2× (5+3) cm = 16 cm

Now, the wire is used to made a square.

Hence, Length of side = Perimeter/4

Length of side = 16/4 = 4 cm

The length of a side of the square is 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 of the triangle = 55 cm

Lengths of its two sides = 20 cm and 14 cm

Let, the length of its another side = X

Perimeter of a triangle = sum of the lengths of its three sides

55 = 20 + 14 + X

55 = 34 + X

X = 55 – 34

X = 21

So, the length of the third side of a triangle is 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 rectangular park = 150 m

Breadth of rectangular park = 120 m

For fencing a rectangular park, we find the perimeter of the park.

So, the perimeter of the rectangular park = 2 × (length + breadth)

Perimeter = 2 × (150+120) m

Perimeter = 540 m

If the fence costs is ₹ 40 per metre, then cost of fencing 540 m rectangular park = ₹ 540 × 40 = ₹ 21600

**5) A piece of string is 36 cm long. What will be the length of each side, if it is used to form:**

**Solution:**

The length of a piece of string = 36 cm

When we create a shape using this string, it is treated as the perimeter of the shape.

**a) A square, **

**Ans:**

Perimeter of square = 36 cm

So, the length of each sides = Perimeter/4 = 36/4 cm = 9 cm

**b) A triangle with all sides of equal length, and **

**Ans:**

Perimeter ofa triangle with all sides of equal length = 36

So, length of each sides = Perimeter/3 = 36/3 cm = 12 cm

(A triangle has 3 sides)

**c) A hexagon (a six sided closed figure) with sides of equal length?**

**Ans:**

Perimeter of a hexagon with sides of equal length = 36 cm

So, the length of each sides = 36/6 cm = 6 cm

**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:**

Length of rectangular field = 230 m

Breadth of rectangular field = 160 m

For fencing the field, we find the perimeter of the field.

So, Perimeter of the rectangular field = 2 × (length + breadth)

Perimeter = 2 × (230+160) cm

Perimeter = 780 cm

Now, 780 cm is the length of one round of rope.

Hence, the total length of 3 rounds of rope = 3×780 cm = 2340 cm

**Figure it out:**

**Each track is a rectangle. Akshi’s track has length 70 m and breadth 40 m. Running one complete round on this track would cover 220 m, i.e., 2 × (70 + 40) m = 220 m. This is the distance covered by Akshi in one round.**

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

**Solution:**

The distance covered by Akshi in one round = 220 m

So, the total distance covered by Akshiin 5 rounds = 220 × 5 m = 1100 m

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

**Solution:**

The distance covered by Toshi in one round = 2 × (60+30) m = 180 m

The distance covered by Toshi in 7 rounds = 7 × 180 m = 1260 m

**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’.**

**Ans: **Akshi ran 1000 m

She finished = 1000/220 = 4.54 ≈ 4 full rounds around her trrack.

**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’.**

**Ans: **Toshi ran 1000 m

She finished = 1000/180 = 5.55 ≈ 5 full rounds around her track.

**Figure it out:**

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

**Solution:**

Length of the rectangular garden = 25 m

Area of the rectangular garden = 300 sq m

We know,

Area of the rectangular garden = length × width

So, width = Area/length

Width = 300/25 m = 12 m

The width of the garden is 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:**

Length of the rectangular plot of land = 500 m

Width of the rectangular plot of land = 200 m

For calculate the cost of tiling a rectangular plot of land, we find the area of land.

So, area of the rectangular plot of land = 500 × 300 sq m = 150000 sq m

Now, tiling cost of 1000 sq m = ₹ 8

Tiling cost of 1 sq m = ₹ 8/1000

Tiling cost of 150000 sq m = ₹ 8/1000 × 150000 = ₹ 1200

The cost of tiling a rectangular plot of land is ₹ 1200.

**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:**

Length of the rectangular coconut grove = 100 m

Width of the rectangular coconut grove = 50 m

Area of the rectangular coconut grove = (100×50) sq m = 5000 sq m

25 sq m is required for 1 tree

5000 sq m is required for = 5000/25 trees = 200 trees

So, the maximum number of trees that can be planted in this grove is 200.

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

**Solution:**

By splitting the following figures into rectangles, we find

Area of the rectangle A = 3 × 3 m = 9 m

Area of the rectangle B = 2 × (4-3) m = 2 × 1 m = 2 m

Area of the rectangle C = (4+1) × (2+1) m = 5×3 m = 15 m

Area of the rectangle D = {(4+1) – 3} ×1 = 2 × 1 m = 2 m

By splitting the following figures into rectangles, we find

Area of the rectangle (I) = 5 ×(3-2) m = 5×1 m = 5 m

Area of the rectangle (II) = (2 × 1) m = 2 m

Area of the rectangle (III) =(2 × 1) m = 2 m

**Figure it out:**

**Cut out the tangram pieces given at the following figure.**

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

**Ans:**

There are two pieces have the same area in this figure.

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

**Ans: **

2 times bigger is Shape D as compared to Shape C.

The relationship between Shapes C, D and E,

∴ C + D = E

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

**Ans: **

Shape D and F has the same area.

The shapes are divided into 2 triangles of equal areas. Now, we see that the area of each triangle is same as other. So, that we can say the area of that two figure is same.

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

**Ans:**

Shape F and G has the same area.

The shapes are divided into 2 triangles of equal areas. Now, we see that the area of each triangle is same as other. So, that we can say the area of that two figure is same.

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

**Ans:**

The area of Shape A is 2 times of the area of shape G.

**(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, 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)**

**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? **

**Ans:**

The area of the big square is 16 times of the area of shape C.

**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.**

**Ans:**

The area of the rectangle is 16 times of the area of shape C.

**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.**

**Ans:**

The perimeter of triangle is greater than the perimeter of the square.

**6.3 Area of a Triangle**

**Figure it out:**

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

**Solution:**

Let, the 1 square = 1 square unit

**a)**

Length of rectangle ABCD = 5 units

Width of rectangle ABCD = 4 units

So, area of rectangle ABCD = 5 × 4 square units = 20 sq units

Now, Height oftriangle AXD = 1 unit

Base of triangle AXD = 4 units

So, the area of triangle AXD = ½ × 1 × 4 sq units = 2sq units

As same as the area of triangle BCY = 2 sq units

Ans:

Hence, the area of the figure XBYD = (20+2+2) sq units = 24 sq units

**b)**

Area of rectangle ABCD = 5 × 4 sq units = 20 sq units

Area of triangle BCX = ½ × 2 × 4 sq units = 4 sq units

Area of triangle ADY = ½ × 4 × 4 sq units = 8 sq units

So, the area of the figure ABXY = (20+4+8) sq units = 32sq units

**c)**

Area of the rectangle ABCD = 6 × 10 sq units = 60 sq units

Area of triangle BCO = ½ × 10 × 3 sq units = 15 sq units

Area of triangle PDO = ½ × 2 ×3 sq units = 3 sq units

So, the area of figure ABOP = Area of the rectangle ABCD – (Area of triangle BCO + Area of triangle PDO)

∴Area of figure ABOP = 60 – (15+3) sq units = 42 sq units

Now, Area of triangle AMN = ½ × 2 × 3 sq units = 3sq units

∴Area of triangle AMB = 2 × 3 sq units = 6sq units

Ans:

Hence, area of the figure AMBOP = (42 + 6) sq units = 48 sq units

**d)**

Area of rectangle ABCD = 5 × 4 sq units = 20 sq units

Area of triangle AXO = ½ × 2 × 1 sq units = 1 sq units

Area of triangle DOY = ½ × 2 × 3 sq units = 3 sq units

Hence, the area of the figure ABCDO = (20+1+3) sq units = 24 sq units

**d)**

Area of rectangle MNOP = 6 × 4 sq units = 24 sq units

Area of triangle MBA = ½ × 2 × 2 sq units = 2 sq units

Area of triangle ADP = ½ × 2 × 2 sq units = 2 sq units

Area of triangle BNC = ½ × 4 × 3 sq units = 6 sq units

Area of triangle DOC = ½ × 4 × 1 sq units = 2 sq units

So, the area of figure ABCD = Area of rectangle MNOP – Area of other 4 triangle

∴Area of figure ABCD = 24 – (2+2+6+2) sq units = 12 sq units

**Figure it out:**

**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:**

The sum of the areas of these two rectangles = (5 × 10 + 2 × 7) sq m = 64 sq m

Now, the rectangle has the area of 64 sq m

So, the dimensions of rectangle = 16 m × 4 m

(64 = 2 × 2 × 2 × 2 × 2 × 2)

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

**Solution:**

Length of a rectangular garden = 50 m

Area of the rectangular garden = 1000 sq m

Now,

∴ Area of the rectangular garden = length × width

∴ 1000 = 50 × width

∴ Width = 1000/50 = 20

So, the width of the rectangular garden = 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:**

Length of a room = 5 m

Width of the room = 4 m

So, area of the room = 5 × 4 sq m = 20 sq m

Now, length of sides of a square carpet = 3 m

Area of the carpet = 3 × 3 sq m = 9 sq m

The area that is not carpeted is = (20-9) sq m = 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:**

Length of garden = 15 m

Width of garden = 12 m

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

Now, length of 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

**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:**

**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:**

So, the perimeter of the rectangle border = 2 × (length × width)

∴Perimeter = 2 × (1 + 1.5) cm = 2 × 2.5 cm = 5 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 rectangle = 12 × 8 sq units = 96sq units

The area of new rectangle is half of this rectangle, so the area of new rectangle = ½ × 96 sq m = 48 sq m

(48 = 2 × 2 × 2 × 2 × 3)

Dimensions of new rectangle = 8 units × 6 units

**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 1+(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:**

The statement:-c) The perimeters of both the rectangles added together is always 1+(1/2) times the perimeter of the square.( True )

Also See: NCERT class 6 Ganita Prakash all chapter-wise solutions