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**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

Breadth of room

= 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 : **

Let, Breadth = B

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 : **

Let, Breadth = B

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 m^{2}

**Question no – (3)**

**Solution : **

Let, width = x = 15 m

Length = 3x = 3 × 15 = 45 ,

∴ Area = 15 × 45 = 675 m^{2 }

**∴** Cost of flooring

= 675 × 45

= 30,375 Rs

**Question no – (4)**

**Solution : **

Area of floor

= 288 × 72

= 20736 cm^{2}

Area of 1 Tile

= 12 × 6

= 72 cm^{2}

Now, number of tile,

= 20736/72

= 288

**Question no – (5)**

**Solution : **

**Required figures : **

**(a)** Area of ABCD

= 9 × 5

= 45 cm^{2}

Area of MNOP

= 4 × 5

= 20 cm^{2 }

**∴** Area of ABMNOPC

= 45 + 20

= 65 cm^{2}

**(b)** Area of ABIJ

= 7 × 4

= 28 cm^{2}

Area of CIHD

= 4 × (7 – 3)

= 4 × 4

= 16 cm^{2}

Area of EFGH

= 3 × 7

= 21 cm^{2}

∴ Area of ABCDEFGHIJ

= 28 + 16 + 21

= 55 cm^{2}

**Question no – (6)**

**Solution : **

Total area of room

= (10)^{2}

= 100 m^{2}

Carpeted area

= (8 × 8)

= 64 m^{2}

∴ are without carate

= 100 – 64

= 36 m^{2}

∴ Cost of carpeted area

= 64 × 120

= 7,680 Rs

**Question no – (7)**

**Solution : **

**Required figures : **

**(a)** Area of ABCD

= 4 × 5

= 20 cm^{2}

Area of DEFG

= 3 × 1

= 3 cm^{2}

Area of BKJI

= 3 × 2

= 6 cm^{2}

∴ Area of ABKJICGFED

= 20 + 3 + 6

= 29 cm^{2}

**(b)** Area of ABCD

= 12 × 2

= 24 cm^{2 }

Area of CDEF

= 2 × (12 – 6)

= 2 × 6

= 12 cm^{2}

Area of EFGH

= 2 × 2

= 4 cm^{2}

∴ Area of ABCFGHED

= 24 + 12 + 4

= 40 cm^{2}

**(c)** Area of AHGI

= 5 × 2

= 10 cm^{2}

Area of EDJF

= 5 × 2

= 10 cm^{2}

Area of BCIJ

= (10-5) × 12

= 5 × 12

= 60 cm^{2 }

**∴** Area of AIBCJDEFG

= 10 + 10 + 60

= 80 cm^{2}

**Question no – (9)**

**Solution : **

**Required figure : **

Area of Total park

= 6 × 40

= 2400 m^{2}

Path width = 5 m .

∴ Length of EFGH

= 60 – 5

= 55 m

Breadth of EFGH

= 40 – 5

= 35 m

**∴** Area of EFGH

= 55 × 35

= 1925 m^{2}

Floral bed width = 3 m

**∴** Length of MNOP

= 55 – 3

= 52 m

Breadth of MNOP

= 35 – 3

= 52 m

**∴** Area of MNOP

= 52 × 32

= 1664 m^{2}

**∴** Area of only constructed path

= (2400 – 1925)

= 475 m^{2}

**∴ **Area of only flower bed

= (1925 – 1625)

= 216 m^{2}

**∴** 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 cm^{2 }

**∴** Area of rectangle,

= 20 × 10 = 200 cm^{2}

= 200 – 4

= 196 cm^{2}

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