# Mathsight Class 6 Solutions Chapter 16

## Mathsight Class 6 Solutions Chapter 16 Perimeter and Area

Welcome to NCTB Solutions. Here with this post we are going to help 6th class students to solve MathSight Class 6 Mathematics Book, Chapter 16, Perimeter and Area. Here students can easily get all the exercise questions solution for Chapter 16, Perimeter and Area Exercise 16.1 and 16.2

Perimeter and Area Exercise 16.1 Solution :

Question no – (1)

Solution :

Required figures, (a) ∴ Perimeters,

= 12 + 2 + 5 + 4 + 5 + 2 + 12 + 2 + 5 + 4 +5

= 58 cm .

(b) ∴ Perimeters,

= 2 + 15 + 2 + 13 + 13 + 2

= 47 cm .

(c) ∴ Perimeters,

= 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4

= 32 cm .

(d) ∴ Perimeters,

= 10 + 8 + 2 + 6 + 2 + 2 + 2 + 6 + 2 + 8

= 48 cm .

Question no – (2)

Solution :

3rd side of triangl

e

= 85 – (21 + 32)

= 85 – 53

= 32 cm .

Question no – (3)

Solution :

(a) Side = 84/4

= 21 m .

(b) side = 124/4

= 31 m .

(c) side = 10.2/4

= 2.55 cm

(d) side = 1.04/4

= 0.01 m .

Question no – (4)

Solution :

Length of room

= 3 m 55 cm

= 355cm

= 3m 25 cm

= 325 cm .

∴ Perimeter of room,

= (355 + 325)

= 380 cm .

Question no – (5)

Solution :

(a) Length of rope using plot,

= 36/4

= 9 cm .

(b) Length of rope wed equilateral triangle,

= 36/3

= 12 cm .

Question no – (6)

Solution :

Length of Penfago x

= 20.5/5

= 4.1 cm

Question no – (7)

Solution :

According to question

2 (82 + B) = 246

Or, 82 + B = 246/2 = 123

Or, B = 123 – 82

= 41 cm .

Question no – (8)

Solution :

Perimeter of sq park

= 30 × 4

= 120 m .

Perimeter of rect park

= 2 (35 + 20)

= 2 × 55

= 110 m

∴ 120 > 110 m .

∴ Sonam cover more distance.

Question no – (9)

Solution :

Perimeter of plot

= 2 (0.4 + 0.3)

= 2 × 0.7

= 1.4 km.

∴ Wire needed

= 1.4 × 3

= 4.2 km.

Question no – (10)

Solution :

Perimeter of square plot

= 145 × 4

= 580 m .

∴ Tencing cost

= 580 × 22

= 12,760 Rs

Perimeter and Area Exercise 16.2 Solution :

Question no – (1)

Solution :

Required figure : (a) Area of ABCD = 8 × Area of 1 sq

= 8 sq unit

Area of EGF

= 2 + 0 + 1/2 × 4

= 2 + 2

= 4 unit sq .

∴ Total Area of ABCGFED

= 8 + 4

= 12 unit sq .

(b) Area of ABCD

= 5 × 1

= 5 sq unit .

Area of JDEF

= 2 × 1

= 2 sq unit .

Area of EFGH

= 12 × 1

= 12 sq unit .

∴ Area of ABCIJFGHED

= 5 + 2 + 12

= 19

(c) Area of ABKI

= 3 × 1

= 3 sq unit .

Area of HKGL

= 6 × 1

= 6 sq unit .

Area of CDOL

= 4 × 1

= 4 sq unit .

Area of EFOQ

= 1 × 1

= 1 sq unit

Total rea

= 3 + 6 + 4 + 1 + 1

= 15 sq unit .

Question no – (2)

Solution :

According to question

2 (34 + B) = 120

Or, 32 + B = 120/2 = 60

Or, B = 60 – 32 = 28 m

Area of recf

= 34 × 28

= 952 m2

Question no – (3)

Solution :

Let, width = x = 15 m

Length = 3x = 3 × 15 = 45 ,

∴ Area = 15 × 45 = 675 m2

Cost of flooring

= 675  × 45

= 30,375 Rs

Question no – (4)

Solution :

Area of floor

= 288 × 72

= 20736 cm2

Area of 1 Tile

= 12 × 6

= 72 cm2

Now, number of tile,

= 20736/72

= 288

Question no – (5)

Solution :

Required figures : (a) Area of ABCD

= 9 × 5

= 45 cm2

Area of MNOP

= 4 × 5

= 20 cm2

Area of ABMNOPC

= 45 + 20

= 65 cm2

(b) Area of ABIJ

= 7 × 4

= 28 cm2

Area of CIHD

= 4 × (7 – 3)

= 4 × 4

= 16 cm2

Area of EFGH

= 3 × 7

= 21 cm2

∴ Area of ABCDEFGHIJ

= 28 + 16 + 21

= 55 cm2

Question no – (6)

Solution :

Total area of room

= (10)2

= 100 m2

Carpeted area

= (8 × 8)

= 64 m2

∴ are without carate

= 100 – 64

= 36 m2

∴ Cost of carpeted area

= 64 × 120

= 7,680 Rs

Question no – (7)

Solution :

Required figures : (a) Area of ABCD

= 4 × 5

= 20 cm2

Area of DEFG

= 3 × 1

= 3 cm2

Area of BKJI

= 3 × 2

= 6 cm2

∴ Area of ABKJICGFED

= 20 + 3 + 6

= 29 cm2

(b) Area of ABCD

= 12 × 2

= 24 cm2

Area of CDEF

= 2 × (12 – 6)

= 2 × 6

= 12 cm2

Area of EFGH

= 2 × 2

= 4 cm2

∴ Area of ABCFGHED

= 24 + 12 + 4

= 40 cm2

(c) Area of AHGI

= 5 × 2

= 10 cm2

Area of EDJF

= 5 × 2

= 10 cm2

Area of BCIJ

= (10-5) × 12

= 5 × 12

= 60 cm2

Area of AIBCJDEFG

= 10 + 10 + 60

= 80 cm2

Question no – (9)

Solution :

Required figure : Area of Total park

= 6 × 40

= 2400 m2

Path width = 5 m .

∴ Length of EFGH

= 60 – 5

= 55 m

= 40 – 5

= 35 m

Area of EFGH

= 55 × 35

= 1925 m2

Floral bed width = 3 m

Length of MNOP

= 55 – 3

= 52 m

= 35 – 3

= 52 m

Area of MNOP

= 52 × 32

= 1664 m2

Area of only constructed path

= (2400 – 1925)

= 475 m2

Area of only flower bed

= (1925 – 1625)

= 216 m2

cost of constructing path

= 475 × 7

= 3325 Rs

Cost of planting flower.

= 216 × 4

= 1044 Rs

Question no – (10)

Solution :

Area of small s.q area,

= (2)2

= 4 cm2

Area of rectangle,

= 20 × 10 = 200 cm2

= 200 – 4

= 196 cm2

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Updated: May 25, 2023 — 4:05 pm