Let a = 51 m, b = 37 m, and c = 20 m
Then, s=  \(\frac { a+b+c }{ 2 } \) = \(\frac { 51+37+20 }{ 2 } =\frac { 108 }{ 2 } \) = 54 m
\(\therefore \)  Area of the triangular field = \(\sqrt { s(s-a)(s-b)(s-c) } \)
= \(\sqrt { 54(54-51)(54-37)(54-20) } \) 
=\(\sqrt { 54\times 3\times 17\times 34 } =\sqrt { 2\times 3\times 3\times 3\times 3\times 17\times 2\times 17 } \)
= \(2\times 3\times 3\times 17=306\)m²
Space occupied by one rose bed = 6 m²
Number of rose beds that can be prepared in the field = \(\frac { Area\ of\ the\ field }{ Space\ occupied\ by\ one\ rose\ bed } \)
\(=\frac { 306 }{ 6 } =51\)

For \(\Delta \)ABD
a = 13 cm, b = 14 cm, c = 15 cm

\(\therefore s=\frac { a+b+c }{ 2 } s=\frac { 13+14+15 }{ 2 } \) = 21 cm
\(\therefore \)   Area = \(=\sqrt { s(s-a)(s-b)(s-c) } \)
\(=\sqrt { 21(21-13)(21-14)(21-15) } \)
\(\sqrt { 21\times 8\times 7\times 6 } =84\) cm²
For\(\Delta \) BCD
1 = 9 cm, b = 12 cm, c = 15 cm
\( \therefore s=\frac { a+b+c }{ 2 } s=\frac { a+12+15 }{ 2 } \)=18 cm
\(\therefore \)Area \(=\sqrt { s(s-a)(s-b)(s-c) } \)
\(=\sqrt {18(18-9)(18-12)(18-15) } \)
\(\sqrt { 18\times 9\times 6\times 3 } =54\) cm²
Now, area of quadrilateral ABCD = Area of \(\Delta \)ABD + Area of \(\Delta \)BCD
= 84 cm² + 54 cm² = 138 cm²

Let each of the equal sides of the rhombus be a cm.
Then,      Perimeter = a + a + a + a = 4a m
According to the question,  4a = 200
\(\Rightarrow \)   a= \(\frac { 200 }{ 4 } \) = 50 m

d₁= 80 m
a²=\({ \left( \frac { { d }_{ 1 } }{ 2 } \right) }^{ 2 }+{ \left( \frac { { d }_{ 2 } }{ 2 } \right) }^{ 2 }\)
\(\Rightarrow \)   (50)² =( 40)²+\({ \left( \frac { { d }_{ 2 } }{ 2 } \right) }^{ 2 }\)
\(\Rightarrow \) \({ \left( \frac { { d }_{ 2 } }{ 2 } \right) }^{ 2 }\)= (50)^2-(40)² = 900= (30)²
\(\Rightarrow \)   \(\frac { { d }_{ 2 } }{ 2 } \)  = 30
\(\Rightarrow \)  d₂ = 60 m
\(\therefore \) Area of the rhombus = \(\frac { 1 }{ 2 } \) d₁d₂=\(\frac { 1 }{ 2 } \)\(\times \)80\(\times \)60 = 2400 m²

Let the depth be h m
Area of trapezium = 84 m²

\(\Rightarrow \)  Area of \(\Delta \)ABC +Area of \(\Delta \) ADC = 84 m²
\(\Rightarrow \)  \(\frac { 1 }{ 2 } \)(AB)(DE) +\(\frac { 1 }{ 2 } \) (DC)(DE) = 84 
\(\Rightarrow \) \(\frac { 1 }{ 2 } \) (12)(h) + (8)(h) = 84
\(\Rightarrow \)  6h + 4h = 84
\(\Rightarrow \)  10h = 84
\(\Rightarrow \) h =\(\frac { 84 }{ 10 } \) = 8.4
Hence, the depth of the canal is 8.4 m.

For one black colour sheet  a= 4 cm, b = 6 cm
\(\therefore \) Area \(=\frac { a }{ 4 } \sqrt { 4{ b }^{ 2 }-{ a }^{ 2 } } \)     | Sheet is an isosceles triangle
 \(=\frac { 4 }{ 4 } \sqrt { 4{ (6) }^{ 2 }-4^{ 2 } } \)  \(=\sqrt { 128 } =8\sqrt { 2 } \) cm²
\(\therefore \) Area of 2 black colour sheets = 2\(\times 8\sqrt { 2 } \) = \( 16\sqrt { 2 } \) cm²
Similarly, area of 2 white colour sheets =  \( 16\sqrt { 2 } \) cm²
\(\therefore \) Total area of black and white colour sheets =  \( 16\sqrt { 2 } \) + \( 16\sqrt { 2 } \) =  \( 32\sqrt { 2 } \) cm²

a:b:c = 5:12:13
5+12+13 = 30
a+b+c = 300 cm
\(\therefore a=\frac { 5 }{ 30 } \times 300=50\) cm
 \(b=\frac { 12 }{ 30 } \times 300=120\) cm
\(c=\frac { 13 }{ 30 } \times 300=130\)
\(\therefore s=\frac { a+b+c }{ 2 } =\frac { 50+120+130 }{ 2 } =150\) cm
\(\therefore \)  Area of the triangular field \(=\sqrt { s(s-a)(s-b)(s-c) } \)
\(=\sqrt { 150(150-50)(150-120)(150-130) } \)
\(\\ =\sqrt { 150\times 100\times 30\times 20 } \)
= 3000 cm²           ----(1)
Let the length of the perpendicular from the opposite vertex to the side whose length is 130 cm be h cm.Then, 
Area of the triangular field  \(=\frac { 130\times h }{ 2 } \) = 65h cm²   ...(2)
From (1) and (2)
65h = 3000
\(\Rightarrow h=\frac { 3000 }{ 65 } =\frac { 600 }{ 13 } \) cm = 46.15 cm

Let the sides of the triangle be 13k, 14k, and 15k (in cm).Then,
Perimeter =  a + b + c = 13K + 14k + 15k = 42k cm    
According to the question, 42k = 84
\(\Rightarrow k=\frac { 84 }{ 42 } =2\)
\(\therefore \)  Sides are 26 cm, 28 cm, and 30 cm.
\(\therefore s=\frac { a+b+c }{ 2 } =\frac { 26+28+30 }{ 2 } \)
= 42 cm  
\(\therefore \)   Area \(=\sqrt { s(s-a)(s-b)(s-c) } \)
\(=\sqrt { 42(42-26)(42-28)(42-30) } \)
\(\\ =\sqrt { (42)(16)(14)(12) } \)
= 336 cm²

Let the third side be x cm. Then, 12 + 12 + x = 30
24 + x = 30
x = 6 cm
So, a = 12 cm, b = 12 cms, c = 6 cm
\(\therefore s=\frac { a+b+c }{ 2 } =\frac { 12+12+6 }{ 2 } \)
= 15 cm
\(\therefore \)   Area \(=\sqrt { s(s-a)(s-b)(s-c) } \)
\(=\sqrt { 15(15-12)(15-12)(15-6) } \)
 \(=9\sqrt { 15 } \) cm²

Let the side of the equilateral triangle be a cm.
Then it's area = \(\frac { \sqrt { 3 } }{ 4 } \) a² cm²
 \(\frac { \sqrt { 3 } }{ 4 } \) a² = \(81\sqrt { 3 } \)   \(\Rightarrow \) a²= 81\(\times \)4
\(\Rightarrow \)  a = \(\sqrt { 81\times 4 } \)
\(\Rightarrow \)  a = 9\(\times \)2 = 18 cm
\(\therefore \) Perimeter of the equilateral triangle = 31 = 3 \(\times \)18 = 54 cm

Area of \(\Delta \)ABC = \(=\frac { AB\times BC }{ 2 } =\frac { 24\times 10 }{ 2 } \) = 120 m².