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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 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
Breadth of EFGH
= 40 – 5
= 35 m
∴ Area of EFGH
= 55 × 35
= 1925 m2
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 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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