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Brilliant’s Composite Mathematics Class 8 Solutions Chapter 15 Areas of Rectilinear Figures
Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Brilliant’s Composite Mathematics Class 8 Math Book, Chapter 15, Areas of Rectilinear Figures. Here students can easily find step by step solutions of all the problems for Areas of Rectilinear Figures, Exercise 15.1, 15.2 and 15.3 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily.
Areas of Rectilinear Figures Exercise 15.1 Solution
Question no – (1)
Solution :
(i) Area = 12 × 3.6
= 43.2 cm2
(ii) Area = 9.5 × 7.6
= 72.2 cm2
Question no – (2)
Solution :
Required figure :
∴ Area = 1/2 × 4 × 5
= 10 cm2
Question no – (3)
Solution :
Required figure :
(i) Area = 1/2 × 8 × 6
= 12 cm2
(ii) Length of side
= √32 + 42
= √9 + 16
= √25
= 5 cm.
Question no – (4)
Solution :
Required figure :
AO = √152 – 122
= √215 – 144 = √81 = 9
∴ AC = 18 cm
∴ Area = 1/2 × 18 × 24
= 216 cm2
Question no – (5)
Solution :
Length of side
= 220.5/17.5
= 12.6 m
Question no – (6)
Solution :
Required figure :
Let the length of the other diagonal be x cm.
∴ 1/2 × x × 24 = 84
Or, x = 84/12
= 7 cm.
∴ Side of rhombus = √122 + (3.5)2
= √144 + 12.25
= √156.25
= 12.5 cm.
Question no – (7)
Solution :
The area of the rhombus
= 1/2 × 15 × 20
= 150 cm2
The side of the rhombus
= 48/4
= 12 cm.
∴ Perpendicular distance between two sides
= 150/12
= 12.5 cm.
Question no – (8)
Solution :
The side of the rhombus
= 59/4
= 14 cm.
It’s attitude,
= 119/14
= 8.5 cm.
Question no – (9)
Solution :
Area of the trapezium
= 1/2 (10 + 6) × 5
= 40 cm2
Question no – (10)
Solution :
AB = 2 CP – 1
= 2.8 – 1
= 16 – 1
= 15
∴ Height of the trapezium
= 1/2 (15 + 8) h = 69
Or, 1/2 × 23h = 69
Or, 69 × 2/23
= 6 cm.
Question no – (11)
Solution :
Required figure :
In △ADE, AE = √102 – 62
= √100 – 36
= √64
= 8 cm.
∴ Area = 1/2 × 8 × (13 + 25)
= 4 × 38
= 152 cm2.
Question no – (13)
Solution :
Let the other side be x cm.
∴ 1/2 × 3 × (3 + x) = 12
Or, 3 + x = 8
Or, x = 8 – 3
∴ x = 5 cm.
Question no – (14)
Solution :
Let the sides be 3x, 5x cm.
∴ we have, 1/2 × 10 × (3x + 5x) = 120
Or, 5 × 8x = 120
Or, x = 120/5 × 8
= 3 cm.
∴ The sides are 3 × 5
= 15 cm.
Question no – (15)
Solution :
Area = 1/2 × 4 × (8 + 6)
= 2 × 14
= 28 m2
Question no – (16)
Solution :
Area of the triangle,
= 1/2 × 24.8 × 16.5
= 104.6 cm2.
Let the other diagonal be x cm.
∴ 1/2 × x × 22 = 104.6
Or, x = 104.6/11
= 18.6 cm.
Question no – (17)
Solution :
Let the other side be x cm.
∴ 1/2 (x + 84) × 26 = 1586
Or, x + 84 = 1586/13 = 122
∴ x = 12.2 ‘- 84 = 38 cm.
Question no – (18)
Solution :
Required figure :
Area of the quadrilateral
= 1/2 × 19.5 × (5.4 + 10.6)
= 1/2 × 19.5 × 16 = 8 × 19.5
= 156 cm2
Question no – (19)
Solution :
Required figure :
Area of quadrilateral = area of △ABC + area at △ACD
∴ √s1 (s1 – a) (s1 – b) (s2 – c) + √s2 (s2 – a) (s2 – b) (s3 – c)
= √21 (21 – 13) (21 – 14) (21 – 15) + √30 (30 – 25) (30 – 15) (30 – 20)
= √21 × 8 × 7 × 6 + √30 × 5 × 15 × 10
= √7 × 3 × 2 × 7 × 2 × 3 + √2 × 3 × 5 × 5 × 3 × 5 × 2 × 5
= (2 × 3 × 7) + (2 × 3 × 5 × 5)
= 42 × 150
= 192 cm2
Question no – (20)
Solution :
Area = 1/2 × 20 × (11 + 8.5)
= 195 cm2.
Areas of Rectilinear Figures Exercise 15.2 Solution
Question no – (1)
Solution :
(a) a = 8, b = 11, c = 13, s = 8 + 11 + 13/2 = 32/2 = 16
∴ Area = √s (s – a) (s – b) (s – c)
= √16 (16 – 8) (16 – 11) (16 – 13)
= √16 × 8 × 5 × 3
= 8 √30 = 8 × 5.477
= 43.81 m2
(b) a = 60 cm, b = 56cm, c = 52cm,
s = 60 + 56 + 52/2 = 84 cm.
∴ Area = √84 (84 – 60) (84 – 56) (84 – 52)
= √84 × 24 × 28 × 32
= √2 × 2 × 3 × 7 × 2 × 2 × 2 × 3 × 2 × 2 × 7 × 2 × 2 × 2 × 2 × 2
= 2 × 2 × 2 × 2 × 2 × 2 × 3 × 7
= 1348 cm2.
(c) a = 17 m, b = 25 m, c = 26 m,
s = 17 + 25 + 26/2 = 34 cm.
∴ area = √s (s – a) (s – b) (s – c)
= √34 (34 – 17) (34 – 25) (34 – 26)
= √34 × 17 × 9 × 8
= √2 × 17 × 17 × 17 × 3 × 3 × 2 × 2 × 2
= 2 × 2 × 3 × 17
= 204 m2.
(d) a = 112 cm, b = 78 cm, c = 50 cm,
s = 112 + 78 + 50/2 = 120 cm.
∴ Area = √120 (120 – 112) (120 – 78) (120 – 50)
= √120 × 8 × 42 × 70
= √2 × 2 × 2 × 3 × 5 × 2 × 2 × 2 × 2 × 3 × 7 × 2 × 2 × 5 × 7
= 2 × 2 × 2 × 2 × 3 × 5 × 7
= 1680 cm2.
(e) a = 10 m, b = 12 m, c = 18 m,
s = 10 + 12 + 18/2 = 20 m.
Area = √20 (20 – 10) (20 – 12) (20 – 18)
= √20 × 10 × 8 × 2
= √2 × 2 × 5 × 2 × 5 × 2 × 2 × 2 × 2
= 8 × 5 √2
= 40 × 1414
= 56.56 m2.
Question no – (2)
Solution :
Required figure,
Let the sides e x cm.
∴ x + X + 48 = 108
Or, 2x = 108 – 48 = 60
Or, 60/2 = 30 cm.
Now, AO = √(30)2 – (24)2
= √900 – 576
= √324 = 18 cm.
∴ Area of △ABC = 1/2 × 18 × 48
= 9 × 48
= 432 cm2.
Question no – (4)
Solution :
a = 25, b = 39, c = 56, s = 25 + 39 + 56/2 = 60 cm.
∴ Area = √60 (60 – 25) (60 – 39) (60 – 56)
= √60 × 35 × 21 × 4
= √2 × 2 × 3 × 5 × 5 × 7 × 3 × 7 × 2 × 2
= 2 × 2 × 3 × 5 × 7
= 420 cm2.
Again, area at the triangle
= 1/2 × 56 × h
= 420
Or, h = 420/28 = 15 cm.
Question no – (5)
Solution :
Let the sides be 9x, 40x, 41x meters,
∴ 9x + 40x + 41x = 180
Or, The sides are 9 × 2 = 18 m,
40 × 2 = 80 m,
41 × 2 = 82 m.
Question no – (6)
Solution :
a = 30m, b= 72m, c = 78m, s = 30 + 72 + 78/2 = 90m.
∴ Area = √90 (90 – 30) (90 – 72) (90 – 78)
= √90 × 60 × 18 × 12
= √2 × 3 × 3 × 5 × 2 × 2 × 3 × 5 × 2 × 3 × 3 × 2 × 2 × 3
= 2 × 2 × 2 × 3 × 3 × 5 = 1080 m2.
Now, total expense = 1080 × 10 = 10800 paise
= 10800/100
= 108 rupees.
Question no – (7)
Solution :
Let the sides be 13x, 14x, 15x m.
∴ 13x + 14x + 15x = 84
Or, 42x = 84
Or, x = 84/42 = 2m.
∴ Sides are 13 × 2 = 26m, 14 × 2 = 28m, 15 × 2 = 30m.
∴ Area = √42 (42 – 26) (42 – 28) (42 – 30)
= √42 × 16 × 14 × 12
= √2 × 3 × 7 × 2 × 2 × 2 × 2 × 2 × 7 × 2 × 2 × 3
= 2 × 2 × 2 × 2 × 3 × 7
= 380 m2.
Question no – (8)
Solution :
a = 120m, b = 150m, c = 200m,
s = 120 + 150 + 200/2 = 235m.
∴ Area = √235 (235 – 120) (235 – 150) (235 – 200)
= √235 × 115 × 85 × 35
= √5 × 47 × 5 × 23 × 5 × 17 × 5 × 7
= 5 × 5 × √47 × 23 × 17 × 7
= 25 × √47 × 23 × 17 × 7 m2.
∴ Expense = 25 × √47 × 23 × 17 × 7 × 25/100 rupees
= 422416.25 rupees.
Question no – (9)
Solution :
(a) side = 5.6 cm.
∴ area = √3/4 × 5.6 × 5.6
= 13.56 cm2.
b, c, d are not done in khata :
Question no – (10)
Solution :
Area at the triangle = √3/4 × (2a)2
Or, ah = √3/4 × 4a2
Or, ah = √3 a2
Or, h = √3 a
Question no – (11)
Solution :
Altitude = √3/2 × 32 = 16 √3 cm.
Question no – (12)
Solution :
Let the length at each side be x cm.
∴ √3/4 × x2 = 81√3
Or, x2 = 81 × 4
Or, x = √81 × 4
= 9 × 2
= 18 cm.
Question no – (13)
Solution :
Let the length of each side be x cm.
∴ √3/4 x2 = 36 √3
Or, x2 = 36 × 4
Or, x = √36 × 4
Or, x = 6 × 2 = 12 cm
∴ Perimeter,
= 12 × 3
= 36 cm.
Areas of Rectilinear Figures Exercise 15.3 Solution
Question no – (1)
Solution :
Area of the polygon ABCDE = area of △ADE + area of △BCD
= √S1 (s1 – a) (s1 – b1) (s1 – c1) + √S2 (s2 – a2) (s2 – b2) (s2 – c2) + √S3 (s3 – a3) (s3 – b3) (s3 – c3)
= √30 (30 – 25) (30 – 20) (30 – 15) + √24.5 (24.5 – 15) (24.5 – 20) (24.5 – 14) + √16 (16 – 14) (16 – 8) (16 – 10)
= √30 × 5 × 10 × 15 + √14.5 × 9.5 × 4.5 × 10.5 + √16 × 2 × 8 × 6
= 150 + ___ + 16 √6
Question no – (2)
Solution :
Area of ABCDE = area of △AEF + △BCD + rectangle ABCDE
∴ (1/2 × 250 × 40) + (1/2 × 150 × 200) + (250 × 150)
= 5000 + 15000 + 32500
= 52500 m2.
Question no – (3)
Solution :
Total area = (1/2 × 18 × 20 + (1/2 × 52 × 45) + (1/2 × 48 × 27) + 1/2 × 34 × 15) + (1/2 × 26 × 37 + (1/2 × 22 × 58 )
= 180 + 1134 + 648 + 255 + 518 + 638
= 3409 m2
Question no – (5)
Solution :
Required photo :
Area = area of △AGM + area of _ FGMO + _ DEFO + area of △ABN + _ BCPN
∴ (1/2 × 50 × 400) + (1/2 × 110 × 200) + (1/2 × 130 × 200) + (1/2 × 50 × 570) + (1/2 × 90 × 180)
= 10000 + 11000 + 13000 + 14250 + 8100
= 56350 m2
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