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**Frank Learning Maths Class 4 Solutions Chapter 12 Perimeter and Area**

Welcome to NCTB Solutions. Here with this post we are going to help 4th class students for the Solutions of Frank Learning Maths Class 4 Math Book, Chapter 12, Perimeter and Area. Here students can easily find step by step solutions of all the problems for Perimeter and Area, Exercise 12.1, 12.2, 12.3, 12.4, 12.5 and 12.6 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily.

**Perimeter and Area Exercise 12.1 Solution**

**Question no – (1) **

**Solution :**

**(a)** Perimeter → 7 + 7 + 7 + 7

= 28 cm

**(b)** Perimeter → 3 + 6 + 8

= 17 cm

**(c)** Perimeter rectangle → 8 + 6 + 8 + 6

= 28 cm

**(d)** Perimeter → 6 + 6 + 6

= 18 cm

**(e)** Perimeter → 4 + 4 + 4 + 4 + 4

= 20 cm

**(f)** Perimeter → 8 + 8 + 8 + 8 + 8 + 8

= 48 cm

**Perimeter and Area Exercise 12.2 Solution**

**Question no – (1) **

**Solution :**

**(a)** Perimeter → 12 + 14 + 18

= 44 cm

**(b)** Perimeter → 9 + 11 + 16

= 36 cm

**(c)** Perimeter → 10 + 14 + 15

= 39 cm

**Question no – (2)**

**Solution :**

Perimeter of equilateral triangle

= 3 × 6

= 18 cm

**Question no – (3)**

**Solution :**

Perimeter of the rectangle

= 12 + 4 + 12 + 14 cm

= 32 cm

**Question no – (4)**

**Solution :**

Perimeter of the square –

= 4 × 20

= 80 feet

**Question no – (5)**

**Solution :**

Lace of bedspread needed –

= 3 + 2 + 3 + 2

= 10

**Question no – (6)**

**Solution :**

Perimeter of square = 36

So, one side of square

= 36/4

= 9 cm

**Question no – (7)**

**Solution :**

Perimeter of pentagon = 100 cm

So, Each side length

= 100/5

= 20 cm

**Question no – (8)**

**Solution :**

Breadth = 3cm

Length = 2 × 3 = 6cm

So, perimeter = 2 (3 + 6 )

= 18 cm

**Question no – (9)**

**Solution :**

Perimeter of the park = 180m

It’s breadth = 33m

So, 2 (Length + 33) = 180

= length = 180/2 – 33

= 90 – 33

= length = 57 m

**Question no – (10)**

**Solution :**

Perimeter of side 14m

= 4 × 14

= 56 m

Perimeter of side 29m

= 4 × 29

= 116 m

Rohit runs more by

= (116 – 56)

= 60 m

**Question no – (11)**

**Solution :**

Total length of the pool

= 22 + 2 × 10

= 22 + 20

= 42

Total breadth of the pool

= 16 + 2 × 10

= 16 + 20

= 36

So, Perimeter = 2(42 + 36)

= 2 × 78

= 156 m

**Question no – (12)**

**Solution :**

Perimeter of the field

= 2(23 + 36)

= 2 × 59

= 118

**∴** Total cost will be,

= 118 × 125

= Rs 14750

**Perimeter and Area Exercise 12.6 Solution**

**Question no – (2) **

**Solution :**

**Rectangle :**

**(a)** Given, 1 = 8, b = 3

Area = 3 × 8

= 24 sq cm

**(b)** Given, 1 = 11, b = 4

Area = 4 × 11

= 44 sq cm

**(c)** 1 = 15 m, b = 12

Area = 15 × 12

= 180 sq cm

**Question no – (3) **

**Squares :**

**(a)** Area = 6 × 6

= 36 sq cm

**(b)** Area = 9 × 9

= 81 sq cm

**(c)** Area = 12 × 12

= 144 sq cm

**(d)** Area = 4 × 4

= 16 sq cm

**Question no – (4)**

**Solution :**

**(a)** Sides of square = 7m

So, area of the floor

= 7 × 7

= 49 sq m

**(b)** Length of room = 15 m

Breadth of room = 8 m

**∴** Area of the room,

= 15 × 8

= 120 sq m

**(c)** Breadth of the land = 12 m

Length of the land = 24 m

So, Area of the plot land,

= 12 × 24

= 288 sq m

**(d)** Length = 16 m, breadth = 12 m

Area = 16 × 12 = 192 sq m

**∴** Total cost

= 192 × 50

= 9600 Rs.

**(e)** Area of the square garden = 100 m2

So, length of the garden

= √100

= 10m

**(f)** Area of the rectangular field = 391 sq m

Breadth = 17 m

So, length of the field

= 391/17

= 23 m

**(g)** Area of the room = 36 sq cm

breadth = 6 cm

So, length = 36/6

= 6 cm

Since, the length and the breadth of the room become equal soil is called a square.

**Next Chapter Solution : **

👉 Chapter 13 👈