**Collins Maths Solutions Class 6 Chapter 16 Perimeter and Area**

Welcome to NCTB Solution. Here with this post we are going to help 6th class students for the Solutions of Collins Maths Class 6 Mathematics, Chapter 16, Perimeter and Area. Here students can easily find Exercise wise solution for chapter 16, Perimeter and Area. Students will find proper solutions for Exercise 16.1, 16.2 and 16.3 Our teacher’s solved every problem with easily understandable methods so that every students can understand easily. Here in this post all the solutions are based on ICSE latest Syllabus.

**Perimeter and Area Exercise 16.1 Solution :**

**Question no – (1) **

**Solution : **

**(a) Perimeter will be,**

= 9125 + 125 + 125 + 125)

= 500 cm

**(b) Perimeter will be,**

= (15 + 15 + 15 + 15) cm

= 60 cm

**(c) Perimeter will be,**

= 30 + 52 + 30 + 52

= 60 + 104

= 164

**(d) Perimeter will be,**

= 4 × 10

= 40 in

**(e) Perimeter will be,**

= 2 (18 + 7)

= 2 × 25

= 50

**(f) Perimeter will be,**

= 4 × 24 in

= 96 in

**Question no – (2) **

**Solution : **

**Figure – (a)**

Perimeter = 48

= 4 × S = 48

= S = 48/4

= S = 12

**∴** S = 12

**Figure – (b)**

Perimeter = 30 cm

= 2(l + 3) = 30

= 2l + 6 = 30

= 2l = 30 – 6

= l = 24/2

**∴** l = 12

**Figure – (c)**

Perimeter = 32 cm

= 4 × S = 32

**∴** S = 8 cm

**Figure – (d)**

Perimeter = 80 cm

= 2 (b + 10) = 80

= b + 10 = 40

**∴** b = 30

**Figure – (e)**

Perimeter = 150 cm

= 2 9b + 60) = 150

= b + 60 = 75

**∴** b = 15

**Figure – (f)**

Perimeter = 104 cm

= 2 (b + 2) = 104

= b + 2 = 52

**∴** b = 50

**Question no – (3) **

**Solution : **

As per the question,

One side measures = 10 cm,

Second measures = 15 cm

Perimeter is = 45 cm

**Let,** the other side is x

= x + 10 + 15 = 45

= x = 45 – 10 – 15

= 45 – 25

= 20 cm

Therefore, Side of the triangle will be 20 cm.

**Question no – (4) **

**Solution : **

According to the question,

A rectangular field has dimensions 30 m by 40 m.

As we know that, Perimeter of rectangle = 2 (length + breadth)

**∴** Perimeter will be,

= 2 (30 + 40)

= 2 × 70

= 140 m

Therefore, 140 m fencing is needed to enclose the field.

**Question no – (5) **

**Solution : **

As per the question,

If the perimeter of a square is = 60 cm

**Let,** length of each side of a square = a cm

**∴** Perimeter = 4 a cm

= 4a = 60

= a = 60/4

= a = 15

Therefore, the length of each side of the square will be 15 cm.

**Question no – (6) **

**Solution : **

According to the given question,

The length of a rectangle is twice its breadth.

Perimeter is = 144 feet

**Let,** the breadth of rectangle x

= length is = 2x

**∴** Perimeter = 2 × (x + 2x)

**∴** 2 (x + 2x) = 144

= 2 × 3x = 144

= 6x = 144

= x = 24

**∴** The breadth = 24 feet

**∴** The length = 2 × 24 = 48 feet

Therefore, the dimension of the triangle will be breadth 24 feet and length 48 feet.

**Question no – (7) **

**Solution : **

Perimeter of table cloth,

= 4 × 25 cm

= 100 cm

**∴** The cost will be,

= 10 × 5

= 50 Rs

Therefore, the cost of the lace required to stitch will be Rs 50

**Perimeter and Area Exercise 16.3 Solution :**

**Question no – (1) **

**Solution : **

Area of the rectangle,

= (30 × 16) sq.cm

= 480 sq.cm

As we know that, Perimeter of rectangle = 2 (length + breadth)

**∴** Perimeter of the rectangle,

= 2 × (30 + 16) cm

= 2 × 46 cm

= 92 cm

Therefore, the perimeter of the rectangle will be 92 cm.

**Question no – (2) **

**Solution : **

According to the given question,

Square field of side 42 m at 15 per sq. m.

Area of field,

= (42 × 42) sq.cm

= 1764 sq.cm

**∴** Cost of cultivating the field,

= 176 × 15

= 2640 Rs

Therefore, the cost of cultivating a square field Rs 2640

**Question no – (3) **

**Solution : **

As per the given question,

A room is 15 m long and 8 m wide.

A square carpet of side 6 m is laid on its floor.

Now, Total area of floor,

= 15 × 8 sq. m

= 120 sq. m

**∴** Covered floor,

= 6 × 6 sq. m

= 36 sq. m

**∴** Area of the floor that is not carpeted,

= (120 – 36) sq. m

= 84 sq. m

Therefore, are of the floor that is not carpeted is 84 sq. m

**Question no – (4) **

**Solution : **

According to the given question,

**Let,** the length of side x

Area = x × x = x^{2}

If length is halved then length will be x/2

Area = x/2 × x/2 = x^{2}/4

**∴** x^{2} – x^{2}/4

= 3x^{2}/4

Therefore, If sides of square halved then area will reduce 3/4

**Question no – (5) **

**Solution : **

Area of the floor of room,

= 90 × 120

= 10800

Area of one tiles is,

= 9 × 4 sq.m

= 36 sq.m

**∴** Number of tiles needed to cover the floor is,

= 10800/36

= 300

Therefore, 300 tiles are needed to cover the floor.

**Question no – (6) **

**Solution : **

Let, the equal side is = x cm

**∴** Area = 1/2 × AB × BC = 128

= 1/2 × x × x = 128

= 1/2 x^{2} = 128

= x^{2} = 128 × 2

= x^{2} = 256

= x = 16

Therefore, the length of the identical side of the triangle is 16 cm.

**Next Chapter Solution : **

👉 Chapter 17 👈