## Selina Concise Mathematics Class 6 ICSE Solutions – Perimeter and Area of Plane Figures

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**IMPORTANT POINTS**

**1. Perimeter:** It is the length of the boundry of the given figure.

(i) Perimeter of a triangle = Sum of its three sides.

(ii) Perimeter of rectangle = 2 (length + breadth)

(iii) Perimeter of square = 4 x side.

**2. Area:** Area is the measure of surface of the plane covered by a closed plane figure. In other words, we can say that area of a closed plane figure is the measure of its interior region.

(i) Area of rectangle = length x breadth

(ii) Area of square = (side)².

**3. Units of measurement of perimeter and area :**

(i) Perimeter is measured in centimetre (cm) metre (m) or millimetre (mm).

(ii) Area is measured in square mm, square cm or square metre.

**EXERCISE 32 (A)**

**Question 1.**

**What do you understand by a plane closed figure?**

**Solution:**

Any geometrical plane figure bounded by lines (straight or curved) in a plane is called a plane closed figure.

Each of the following figures is a plane closed figure.

**Question 2.**

**The interior of a figure is called region of the figure. Is this statement true ?**

**Solution:**

Yes. The interior of the figure alongwith its boundary is called region of the figure

**Question 3.**

**Find the perimeter of each of the following closed figures :**

**Solution:**

**(i) Required perimeter**

= AB + AC + BE + EF + FH + HG + HD

= 15 + 5 + 25 + 10 + 5 + 15 + 25 = 110 cm

**(ii) Required perimeter**

= AB + AC + CD + DG + BF + EF + EH + GH

= 20 + 4 + 8 + 20 + 4 + 8 + 20 + 4 = 88 cm

**Question 4.**

**Find the perimeter of a rectangle whose:**

**(i) length = 40 cm and breadth = 35 cm**

**(ii) length = 10 m and breadth = 8 m**

**(iii) length = 8 m and breadth = 80 cm**

**(iv) length = 3.6 m and breadth = 2.4 m**

**Solution:**

**(i)** length = 40 cm and breadth = 35 cm

∴Perimeter = 2 (length + breadth)

= 2 (40 cm + 35 cm)

= 2 x 75 cm

= 150 cm =

= 1.5 m

**(ii)** length = 10 m and breadth = 8 m

∴Perimeter = 2 (length + breadth)

= 2 (10 m + 8 m)

= 2 x 18 m = 54 m

**(iii)** length = 8 m and

breadth = 80 cm

Length = 8 m

Breadth = 80 cm= m = 0.8 m

∴ Perimeter = 2 (length + breadth)

= 2 (8 m + 0.8m)

= 2 x 8.8 m = 17.6 m

**(iv)** length = 3.6 m and breadth = 2.4 m

∴ Perimeter = 2 (length + breadth)

= 2 (3.6 m + 2.4 m)

= 2 x 6 m = 12 m

**Question 5.**

**If P denotes perimeter of a rectangle, l denotes its length and b denotes its breadth, find :**

**(i) l, if P = 38cm and b = 7cm**

**(ii) b, if P = 3.2m and l = 100 cm**

**(iii) P, if l = 2 m and b = 75cm**

**Solution:**

**Question 6.**

**Find the perimeter of a square whose each side is 1.6 m.**

**Solution:**

∵ Side of the square = 1.6 m

∴ its perimeter = 4 x side

= 4 x 1.6 m

= 6.4 m

**Question 7.**

**Find the side of the square whose pe-rimeter is 5 m.**

**Solution:**

**Question 8.**

**A square field has each side 70 m whereas a rectangular field has length = 50 m and breadth = 40 m. Which of the two fields has greater perimeter and by how much?**

**Solution:**

Perimeter of the square field = 4 x side = 4 x 70m = 280m

Perimeter of rectangular field = 2 (length + breadth)

= 2 (50 m + 40 m)

= 2 x 90 m

= 180 m

∴Square field has greater perimeter by 280 m – 180 m = 100 m

**Question 9.**

**A rectangular field has length = 160m and breadth = 120 m. Find :**

**(i) the perimeter of the field.**

**(ii) the length of fence required to enclose the field.**

**(iii) the cost of fencing the field at the rate of ? 80 per metre.**

**Solution:**

Given = length = 160 m, breadth = 120m

(i) The Perimeter of the field = 2 (l + b)

= 2 (160 m + 120 m)

= 2 x 280

= 560 m

(ii) The length of fence required to enclose the field = The perimeter of the rectan-gular field

= 560 m

(iii) The cost of fencing the field = Length of fence x Rate of fence

= 560 m x ₹80 per metre

= ₹44, 800

**Question 10.**

**Each side of a square plot of land is 55 m. Find the cost of fencing the plot at the rate of ₹32 per metre.**

**Solution:**

∵Perimeter of square field = 4 x its side = 4 x 55 m

∴Length of required fencing = 220 m Now, the cost of fencing = its length x its rate

= 220 m x ₹32 per metre?

= ₹7040

**Question 11.**

**Each side of a square field is 70 cm. How much distance will a boy walk in order to make ?**

**(i) one complete round of this field ?**

**(ii) 8 complete rounds of this field ?**

**Solution:**

(i) Distance covered by the boy to make one complete round of the field.

Perimeter of the field : 4 x its side = 4 x 70 = 280 m

(ii) Distance covered by the boy to make 8 complete rounds of this field.

= 280 m x 8 m = 2240 m

**Question 12.**

**A school playground is rectangular in shape with length = 120 m and breadth = 90 m. Some school boys run along the boundary of the play-ground and make 15 complete rounds in 45 minutes. How much distance they run during this period.**

**Solution:**

Length of the rectangular playground = 120 mBreadth of the rectangular playground = 90 m

∴ Perimeter of the rectangular ground = 2(l + b)

= 2(120 + 90) m = 420 m

Thus, in one complete round, boys covers a distance of = 420 m

∴Distance covered in 15 complete rounds = 420 m x 15 = 6300 m

**Question 13.**

**Mohit makes 8 full rounds of a rect-angular field with length = 120 m and breadth = 75 m.**

**John makes 10 full rounds of a square field with each side 100 in. Find who covers larger distance and by how much?**

**Solution:**

**Mohit**

Length of the rectangular field = 120

Breadth of the rectangular field = 75 m

∴ Distance covered in one round (perim-eter) = 2(1 + b)

= 2(120 + 75) = 390 m Hence, distance covered in 8 rounds = 390 x 8 m = 3120 m

**John**

Side of the field = 100 m

∴Distance covered in one round = 4 x a = 4 x 100 = 400 m

Hence, Distance covered in 10 rounds = 400 x 10 m = 400 m

John a covers greater distance then Mohit by = (4000-3120) m = 880 m

**Question 14.**

**The length of a rectangle is twice of its breadth. If its perimeter is 60 cm, find its length.**

**Solution:**

Let the breadth of the field = x cm

∴ its length = 2x

and, its perimeter = 2 x (length + breadth)

= 2 x (2x + x)

= 2(3x)

= 6x cm

Perimeter = 60 cm

⇒ 60 cm = 6x cm

⇒ x = = 10 cm

∴Breadth = x = 10 cm

Length = 2x = 2 x 10 = 20 cm

**Question 15.**

**Find the perimeter of :**

**(i) an equilateral triangle of side 9.8 cm.**

**(ii) an isosceles triangle with each equal side = 13 cm and the third side = 10 cm.**

**(iii) a regular pentagon of side 8.2 cm.**

**(iv) a regular hexagon of side 6.5 cm.**

**Solution:**

**(i)** The perimeter of equilateral triangle = 3 x side

= 3 x 9.8 cm

= 29.4 cm

**(ii)** Required perimeter = 13 cm + 13 cm + 10 cm

= 36 cm

**(iii)** Perimeter of given pentagon = 5 x side = 5 x 8.2 cm

= 41 cm

**(iv)** Perimeter of given hexagon = 6 x side = 6 x 6.5 cm

= 39 cm

**Question 16.**

An equilateral triangle and d square has equal perimeter. If side of the triangle is 9.6 cm ; what is the length of the side of the square ?

**Solution:**

Perimeter of equilateral triangle = Perimeter of square Side of triangle = 9.6 cm

∴Perimeter of triangle = 3 x side

= 3 x 9.6 cm = 28.8 cm

> Perimeter of the square = 28.8 cm

4 x the side of square = 28.8 cm

⇒ The side of the square = cm

= 7.2 cm Ans.

**Question 17.**

**A rectangle with length = 18 cm and breadth = 12 cm has same perimeter as that of a regular pentagon. Find the side of the pentagon.**

**Solution:**

Length of rectangle = 18 cm

Breadth of rectangle = 12 cm

∴ Perimeter of rectangle = 2 x (l + b)

= 2 x (18+12)

= 2 x 30 = 60 cm

∵Perimeter, of rectangle = Perimeter of pentagon

60 cm = 5 x side

side = cm = 12 cm

∴Side of the pentagon = 12 cm Ans.

**Question 18.**

**A regular pentagon of each side 12 cm has same perimeter as that of a regular hexagon. Find the length of each side of the hexagon.**

**Solution:**

Perimeter of regular pentagon = 5 x length of the side

= 5 x 12 cm = 60 cm

Clearly, perimeter of the given pentagon = 60 cm

⇒ 6 x side of hexagon = 60 cm 60

⇒ side of hexagon = cm = 10 cm

**Question 19.**

**Each side of a square is 45 cm and a rectangle has length 50 cm. If the perimeters of both (square and rectangle) are same, find the breadth of the rectangle.**

**Solution:**

**Question 20.**

**A wire is bent in the form of an equilateral triangle of each side 20 cm. If the same wire is bent in the form of a square, find the side of the square.**

**Solution:**

∵Each side of the given equilateral triangle = 20 cm

∴Perimeter of the triangle = 3 x side = 3 x 20 cm = 60 cm ,

∴ Perimeter of the square = Perimeter of equilateral triangle

⇒ 4 x side of square = 60 cm

⇒ The side of the square =

=15 cm

**EXERCISE 32 (B)**

**Question 1.**

**Find the area of a rectangle whose :**

**(i) length = 15 cm breadth = 6.4 cm**

**(ii) Length = 8.5 m breadth = 5 m**

**(iii) Length = 3.6 m breadth = 90 cm**

**(iv) Length = 24 cm breadth =180 mm**

**Solution:**

**Question 2.**

**Find the area of a square, whose each side is :**

**(i) 7.2 cm**

**(ii) 4.5 m**

**(iii) 4.1 cm**

**Solution:**

**(i)** 7.2 cm

Area of the square = (side)² = (7.2 cm)² = 7.2 cm x 7.2 cm = 51.84 cm²

**(ii)** 4.5 m

Area of the square = (side)² = (4.5 m)² = 4.5 m x 4.5 m = 20.25 m²

**(iii)** 4.1 cm

Area of the square = (side)² = (4.1 cm)² = 4.1 cm x 4.1 cm = 16.81 cm²

**Question 3.**

If A denotes area of a rectangle, l represents its length and b represents its breadth, find :

(i) l, if A = 48 cm² and b = 6 cm

(ii) b, if A = 88 m² and l = 8m

**Solution:**

**Question 4.**

**Each side of a square is 3.6 cm; find its**

**(i) perimeter**

**(ii) area.**

**Solution:**

(i) Perimeter = 4 x side

= 4 x 3.6 cm = 14.4 cm

(ii) Area = (side)²

= (3.6 cm)²

= 12.96 cm²

**Question 5.**

**The perimeter of a square is 60 m, find :**

**(i) its each side its area**

**(ii) its new area obtained on increasing**

**(iii) each of its sides by 2 m.**

**Solution:**

Perimeter of a square = 60 m

(i) Perimeter of a square = 4 x side

60 m = 4 x side

= side 4

∴side = 15 m

(ii) Area of square = (side)² = (15 m)²

= 15 m x 15 m

= 225 m²

(iii) Increased each side = 2 m

Side of square = 15 m

New length of side = (2m + 15m)

= 17m

∴New Area of square = (17m)² = 17m x 17m = 289 m²

**Question 6.**

**Each side of a square is 7 m. If its each side be increased by 3 m, what will be the increase in its area.**

**Solution:**

Each side of square = 7 m

∴Area of square = (side)²= (7 m)²

= 7m x 7m =49m²

∵ Side increased by 3 m

∴Total length of side will be = 3 m + 7 m = 10m

∴Area of square = (10 m)²= 10m x 10 m = 100 m²

∴Increase in area = 100 m² – 49 m² = 51 m²

**Question 7.**

**The perimeter of a square field is numerically equal to its area. Find each side of the square.**

**Solution:**

**Question 8.**

**A rectangular piece of paper has area = 24 cm² and length = 5 cm. Find its perimeter.**

**Solution:**

**Question 9.**

**Find the perimeter of a rectangle whose area = 2600 m² and breadth = 50 m.**

**Solution:**

**Question 10.**

**What will happen to the area of a rectangle, if its length and breadth both are trebled?**

**Solution:**

Let the original length of the rectangle = l and its original breadth = b

∴ its original area = length x breadth i.e A = l – b i. e.

Since,

Increased length -=3l

and, increased breadth = 3b

∴ New area = 3l x 3b = 9 x l x b [∵A = l x b]

⇒ Area of the new rectangle = 9 times than area of original rectangle

**Question 11.**

**Length of a rectangle is 30 m and its breadth is 20 m. Find the increase in its area if its length is increased by 10 m and its breadth is doubled.**

**Solution:**

Length of a rectangle (l) = 30 m,

Breadth of the rectangle (b) = 20 m

Area of rectangle = l x b

= 30 x 20 = 600 m2

Since, the length its increased by 10 m and breadth is doubled

∴New length (l) = (30 + 10) m = 40 m

and new breadth = (20 x 2) m = 40 m

∴New area = l x b = 40 x 40 m2 = 1600 m2

Hence, the increase in the area = (1600 – 600) m2

= 1000 m2

**Question 12.**

**The side of a square field is 16 m. What will be increase in its area, if:**

**(i) each of its sides is increased by 4 m**

**(ii) each of its sides is doubled.**

**Solution:**

**Question 13.**

**Each rectangular tile is 40 cm long and 30 cm wide. How many tiles will be required to cover the floor of a room with length = 4.8 m and breadth = 2.4 m.**

**Solution:**

**Question 14.**

**Each side of a square tile is 60 cm. How many tiles will be required to cover the floor of a hall with length = 50 m and breadth = 36 m.**

**Solution:**

**Question 15.**

**The perimeter of a square plot = 360 m. Find :**

**(i) its area.**

**(ii) cost of fencing its boundary at the rate of ₹ 40 per metre.**

**(iii) cost of levelling the plot at ₹60 per square metre.**

**Solution:**

Given, perimeter of square plot = 360 m

∵ Perimeter of the square = 4 x its side

∴4 x side of square = 360 m

⇒ side of the square = = 90 m

(i) The area of the square field = (side)²

= (90 m)²

= 90 m x 90 m

= 8100 m²

Cost of fencing at ₹ 40 per metre

= 8100 m2 x ₹ 40 per metre

= ₹ 324000

Cost of levelling at₹ 60 per m²

= 8100 m² x ₹60 per m²

= ₹ 486000

**Question 16.**

**The perimeter of a rectangular field is 500 m and its length = 150 m. Find:**

**(i) its breadth,**

**(ii) its area.**

**(iii) cost of ploughing the field at the rate of ₹1.20 per square metre.**

**Solution:**

**(i)** Perimeter of a rectangle = 2 x (length + breadth)

⇒500 m = 2x(i50m + breadth)

⇒250 m – 150 m = breadth

∴breadth = 100 m

**(ii)** Area of rectangular field = length x breadth

= 150 m x 100 m = 15000 m²

**(iii)** Cost of ploughing the field at the rate of

= ₹1.20 per square m²= area of the field x rate of ploughing = 15000 m² x ₹1.20 per square metre = ₹15000 x 1.20 = ₹18000

**Question 17.**

**The cost of flooring a hall of ₹64 per square metre is ₹2,048. If the breadth of the hall is 5m, find :**

**(i) its length.**

**(ii) its perimeter.**

**(iii) cost of fixing a border of very small width along its boundary at the rate of ₹60 per square metre.**

**Solution:**

**Question 18.**

**The length of a rectangle is three times its breadth. If the area of the rectangle is 1875 sq. cm, find its perimeter.**

**Solution:**

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