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Mathsight Class 7 Solutions Chapter 15 Perimeter and Area
Welcome to NCTB Solutions. Here with this post we are going to help 7th class students to solve MathSight Class 7 Mathematics Book, Chapter 15, Perimeter and Area. Here students can easily get all the exercise questions solution for Chapter 15, Perimeter and Area Exercise 15.1, 15.2 and 15.3
Perimeter and Area Exercise 15.1 Solution :
Question no – (1)
Solution :
Area of triangle,
= 1/2 × b × 2h
bh = b × h
(a) b × h
Question no – (2)
Solution :
Area = 50 × 40
= 2000 m2
Question no – (3)
Solution :
Area = 1/2 × b × h
= 1/2 × 5 × 8
= 60 cm2.
Question no – (4)
Solution :
Let, OX = H
We know that △XOY and △XOZ –
YO/XY = XO/XZ = XY/YZ
∴ XO/XZ = XY/YZ
Or, H = XZ × XY/YZ = 12 × 9/15
= 36/5 = 7.2
∴ OX = 7.2 cm.
Question no – (5)
Solution :
Let CF = X
∴ base = 2x
Height = 3x
According to question,
1/2 × 2x × 3x = 108
Or, 3x2 = 108
Or, x2 = 108/3 = 36
Or, x2 = 62
Or, x = 6
Question no – (6)
Solution :
Area of △ACB = 1/2 × base × height
= 1/2 × 25 × 8 = 100 cm2
Area of △BCD = 1/2 × base × height
= 1/2 × 16 × 5 = 40 cm2
∴ Area of quadrilateral ABCD = △ACB + △BCD
= 100 + 40
= 140 cm2
Question no – (8)
Solution :
Figure,
Area of ABCD = 30 × 20 = 600 cm2
∴ OQ, PS, R is midpoint
AQ = QB = DS = SC
= 30/2
= 15 cm.
AP = PD = BR = RC
= 20/2
= 10 cm.
Area of △APQ = 1/2 × base × height
= 1/2 × 10 × 15 = 75 cm2
∴ Area △APQ = Area of △BQR
= Area of △CRS = Area of △PDS
= 75 cm2
Area of △APS × △BQR × △CRS × △PDS
= 75 × 4
= 300 cm2
∴ Area of PQRS
= 600 – 300
= 300 cm2
Question no – (9)
Solution :
Required,
Total area of ABCD = (16 + 10) × (18 × 8)
= 26 × 26 = 676 cm2
Area of △AFE = 1/2 × 8 × 16 = 64 cm2
Area of △EGB = 1/2 × 10 × 18 = 90 cm2
∴ △AFE + △EGB = 64 + 90 = 154 cm2
∴ shaded region = 678 – 154 = 522 cm2
Question no – (10)
Solution :
Length of court yard = 19.5 m
Breath of court yard = 17 m
Area of court yard
= 19.5 × 17
= 331.5 m2
Side of square slab = 0.5 m
∴ Area of square slab
= 0.5 × 0.5
= 0.25 m2
∴ Number of square slab,
= 331.5/0.25
= 1326
Question no – (11)
Solution :
Aman’s rooms length = 9 m
Aman’s rooms Breath = 7 m
∴ Aman’s rooms Area
= 9 × 7
= 63 m2
Riddhi’s rooms side length = 8 m
∴ Riddhi’s rooms Area
= 8 × 8
= 64 m2
64 > 63
Riddhi’s study room is greater than Aman.
Perimeter and Area Exercise 15.2 Solution :
Question no – (1)
Solution :
Area of ABCD
= 100 × 80
= 8000 m2
Path wide = 10 m.
∴ Length of PQRS,
= 100 – (10 × 2)
= 100 – 20
Breath of PQRS,
= 80 – (10 × 2)
= 80 – 20
= 40 m
∴ Area of PQRS,
= 80 × 60
= 4800 m2
∴ Area of path,
= 8000 – 4800
= 3200 m2
Question no – (2)
Solution :
Area of ABCD
= 200 × 100
= 20000 m2
Path wide = 20 m
Length of PQRS,
= 200 – (20 × 2)
= 200 – 40
= 160 m.
Breadth of PQRS,
= 100 – (20 × 2)
= 100 – 40
= 60 m.
∴ Area of PQRS,
= 160 × 60
= 9600 m2
Question no – (3)
Solution :
Path wide = 5 m
Area of lawn (PQRS) = 400 m2
Side of lawn (PQRS)
= √400 = 20 m
Side of lawn (ABCD),
= 20 + (2 × 5)
= 20 + 10
= 30 m
∴ Area of (ABCD),
= 30 × 30
= 900 m2
∴ Area of path,
= 900 – 400
= 500 m2
Question no – (4)
Solution :
Area of room (ABCD)
= 120 × 15
= 300 m2
Margin wide = 2 m
∴ Length of carpet area (PQRS)
= 20 – (2 × 2)
= 20 – 4
= 16 m.
Breadth of carpet area (PQRS) ,
= 15 – (2 × 2)
= 15 – 4
= 11 m.
∴ Area of carpet area (PQRS),
= 16 × 11
= 176 m2.
∴ Cost of carpeting,
= 176 × 800
= 140,800
Question no – (5)
Solution :
Border width,
= 50 cm.
= 0.50 m
∴ Length of PQ,
= 20 + (0.50 × 2)
= 20 + 1
= 21 m.
Length of DD,
= 15 + (0.50 × 2)
= 15 + 1
= 16 m.
∴ Area of PQRS,
= 21 × 16
= 336 m2
∴ cost of printing of border
= 336 × 200
= 67,200
Question no – (6)
Solution :
Area of ABCD,
= 100 × 7000 m2
Area of QRST
= 70 × 5
= 350 m2
Area of MNOP
= 100 × 2
= 200 m2
Area of WXYZ
= 5 × 2
= 10
∴ Area of not covered by cross road,
= 7000 – [350 + (200 – 10)]
= 7000 – [350 + 190]
= 7000 – 540
= 6460 m2
Question no – (8)
Solution :
Area of ABCD,
= 30 × 20
= 600 cm2
Frame wide = 2 cm
Length of PQ,
= 30 + (2 × 2)
= 30 + 4
= 34 cm
Breadth of PQ,
= 20 + (2 × 2)
= 24 cm.
∴ Area of PQRS,
= 34 × 24
= 816 cm2
Cost of training,
= 816 × 20
= 16,320
Question no – (9)
Solution :
Area of ABCD,
= 18.5 × 14
= 259 m2
Length of PQ,
= 18.5 + (4.5 × 2)
= 27.5 m
Breadth of PQ
= 14 + (4.5 × 2)
= 14 + 9
= 23 m
Area of PQRS
= 27.5 × 23
= 632.5 m2
(i) Area of verandah,
= 632.5 – 259
= 373.5 m2
(ii) cost of paring tile in verandah,
= 373.5 × 20
= 7,470
Perimeter and Area Exercise 15.3 Solution :
Question no – (1)
Solution :
(a) 2 π r = 2 × 22/7 × 42/2
= 132 cm.
(b) 2 π r = 2 × 22/7 × 9.1/2
= 28.6 cm
(c) 2 π r = 2 × 22/7 × 105/2
= 330 cm
(d) 2 π r = 2 × 22/7 × 14.7/2
= 46.2 cm
Question no – (2)
Solution :
(a) r = c/2 π
= 44 × 7/2 × 22
= 7 cm.
(b) r = c/2 π = 15.4 × 7/2 × 22
= 2.45 cm
(c) r = c/2 π
= 242 × 7/2 × 22
= 38.5 cm.
(d) r = c/2 π
= 30.8 × 7/2 × 22
= 4.9 cm.
Question no – (3)
Solution :
Let, radi = r
∴ r = c/2 π = 88 × 7/2 × 22
= 14 cm.
∴ Area of circle = π r2
= 22/7 × 14 × 14
= 161 cm2
Question no – (4)
Solution :
C = 2 π r
= 2 × 22/7 × 35
= 220 cm.
∴ Distance covered in 100 revolution,
= 220 × 100
= 22000 cm.
Question no – (5)
Solution :
Inner radi = r = c/2 π
= 264 × 7/2 × 22
= 42 cm.
Outer radi = R = c/2 π
= 352 × 7/2 × 22
= 56 cm.
∴ Width of racing track,
= 56 – 42
= 14 cm.
Question no – (6)
Solution :
Perimeter of wire,
= 22 × 4
= 88 cm.
∴ Circumference of circle = 88 cm.
∴ radi = r = c/2 π
= 88 × 7/2 × 22
= 14 cm.
∴ Area = π r2
= 22/7 × 14 × 14
= 616 cm2
Question no – (7)
Solution :
Perimeter of square
= 28 × 4
= 112 cm.
Area of square
= 28 × 28
= 784 cm2
Side f square = dia of circle = 28 cm.
∴ radi of circle
= 28/2
= 14 cm
∴ Area of circle = π r2
= 22/7 × 14 × 14
= 616 cm2
∴ reaming area
= 784 – 616
= 168 cm2
Question no – (8)
Solution :
Radi of total lawn = R
= 28/2
= 14 cm
∴ Area of total lawn = π × 14 × 14
= 21/7 × 14 × 14
= 616 cm2
Radi of cow graining lawn = 7 cm.
∴ Area of cow graining lawn = π × (7)2
= 22/7 × 7 × 7
= 154 cm2
∴ Area of cow not graining,
= 616 – 154
= 462 cm2
Question no – (9)
Solution :
Perimeter square = 44
Side of square = 44/4 = 11 cm
Area of square = (11)2 = 121 cm2
Perimeter square = circumference = 44 cm.
∴ radi = r = 44 × 7/2 × 22
= 7 cm.
∴ Area of circle,
= π (7)2 = 22/7 × 7
= 154 cm2
∴ Circle has more area,
= 154 – 121
= 33 m2
Question no – (10)
Solution :
(a) radi of semi circle
= 4 × 2
= 8 cm
Area of semi circle = π R2
= 22/7 × 8 × 8
= 201.14 cm2
Area of small circle = π r2
= 22/7 × 4 × 4
= 50.28 cm2
∴ Shaded area
= 201.14 – 50 – 28
= 150.86 cm2
(b) Area of triangle
= 1/2 × 15 × 20
= 150 cm2
Area of rectangle
= 25 × 40
= 1000 cm2
∴ shaded Area
= 1000 – 150 = 850
(c) ∴ Area square
= 20 × 20 = 400 cm2
R = 20/2 = 10 cm.
Area of circle = π r2
= 22/7 × 10 × 10
= 314.28 cm2
∴ Area of no of 1 half circle
= 314.28/2
= 157.14 cm2
∴ Area of no of 4 half circle
= 157.14 × 4
= 628.56 cm2
Shaded Area,
= 628.56 + 400
= 1028.56 cm2
Question no – (11)
Solution :
Long a minute had = 15 cm
1 hr = 60 min
∴ Distance covered,
= 60 × 15
= 900 cm
Question no – (12)
Solution :
Perimeter = 2 π r
= 2 × 22/7 × 18
= 113.14 cm
Perimeter of semi circular silence = p/2 + 2r
= 113.14/2 + (2 × 18)
= 56.57 + 36
= 92.57
Previous Chapter Solution :
