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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 cm^{2}

**(ii)** Area = 9.5 × 7.6

= 72.2 cm^{2}

**Question no – (2)**

**Solution :**

**Required figure :**

**∴** Area = 1/2 × 4 × 5

= 10 cm^{2}

**Question no – (3)**

**Solution :**

**Required figure :**

**(i)** Area = 1/2 × 8 × 6

= 12 cm^{2}

**(ii)** Length of side

= √32 + 42

= √9 + 16

= √25

= 5 cm.

**Question no – (4)**

**Solution :**

**Required figure :**

AO = √15^{2} – 12^{2}

= √215 – 144 = √81 = 9

**∴** AC = 18 cm

**∴** Area = 1/2 × 18 × 24

= 216 cm^{2}

**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}

= √14^{4} + 12.25

= √156.25

= 12.5 cm.

**Question no – (7)**

**Solution :**

The area of the rhombus

= 1/2 × 15 × 20

= 150 cm^{2}

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 cm^{2}

**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 = √10^{2} – 6^{2}

= √100 – 36

= √64

= 8 cm.

**∴** Area = 1/2 × 8 × (13 + 25)

= 4 × 38

= 152 cm^{2}.

**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 m^{2}

**Question no – (16)**

**Solution :**

Area of the triangle,

= 1/2 × 24.8 × 16.5

= 104.6 cm^{2}.

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 cm^{2}

**Question no – (19)**

**Solution : **

**Required figure :**

Area of quadrilateral = area of △ABC + area at △ACD

∴ √s1 (s_{1} – a) (s_{1} – b) (s_{2} – c) + √s_{2} (s_{2} – a) (s_{2} – b) (s_{3} – 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 cm^{2}

**Question no – (20)**

**Solution :**

Area = 1/2 × 20 × (11 + 8.5)

= 195 cm^{2}.

**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 m^{2}

**(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 cm^{2}.

**(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 m^{2}.

**(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 cm^{2}.

**(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 m^{2}.

**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 cm^{2}.

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 m^{2}.

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 m^{2}.

**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 cm^{2}.

b, c, d are not done in khata :

**Question no – (10)**

**Solution :**

Area at the triangle = √3/4 × (2a)^{2}

Or, ah = √3/4 × 4a^{2}

Or, ah = √3 a^{2}

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 × x^{2} = 81√3

Or, x^{2} = 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

= √S_{1} (s_{1} – a) (s_{1} – b_{1}) (s_{1} – c_{1}) + √S_{2} (s_{2 }– a_{2}) (s_{2} – b_{2}) (s_{2} – c_{2}) + √S_{3} (s_{3} – a_{3}) (s_{3} – b_{3}) (s_{3} – c_{3})

= √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 m^{2}.

**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 m^{2}

**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 m^{2}

**Previous Chapter Solution : **