For triangle   a = 15 cm, b = 14 cm, c = 13 cm
\( \therefore \ =\frac { a+b+c }{ 2 } s=\frac { 15+14+13 }{ 2 } =21\)c m
\(\therefore \)  Area = \(\sqrt { s(s-a)(s-b)(s-c) } \)
 \(=\sqrt { 21(21-15)(21-14)(12-13) } \)
\(=\sqrt { 21(6)(7)(8) } =84\) cm²
Let the height of the parallelogram be h cm.
Then, area of the parallelogram = Base \(\times \) Height = 15 \(\times \) h = 15 cm²
According to the question, Area of the parallelogram = Area of the triangle
\(\Rightarrow \)  15h = 84
\(\Rightarrow \)  \(\frac { 84 }{ 15 } \) = 5.6 cm
Hence, the height of the parallelogram is 5.6 cm.

For \(\Delta \) ABC
Area of right triangle  ABC = \(\frac { 1 }{ 2 } \times \) Base\(\times \) Height = \(\frac { 1 }{ 2 } \times \) 9 \(\times \)40 = 180 m²
 
For \(\Delta \) ACD
a = 28cm, b = 41 m, c = 15 m
\( \therefore s=\frac { a+b+c }{ 2 } s=\frac { 28+41+15 }{ 2 } =\frac { 84 }{ 2 } \) = 42 m
\(\therefore \)   Area of \(\Delta \)ACD = \(\sqrt { s(s-a)(s-b)(s-c) } \)
= \(\sqrt { 42(42-28)(42-41)(42-15) } \)
=\(\sqrt { 42(14)(1)(27) } =\sqrt { (14\times 3)(14)(1)(9\times 3) } \)
=\(14\times 3\times 3=126\) m²
\(\therefore \)   Area of the quadrilateral ABCD = Area of \(\Delta \) ABC +Area of \(\Delta \) ACD
= 180 m² + 126 m² = 306 m²

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

For \(\Delta \) ABC
a = 34 cm, b = 42 cm, c = 20 cm

\(\therefore s=\frac { a+b+c }{ 2 } =\frac { 34+42+20 }{ 2 } =48\) cm
\(\therefore \)  Area of  \(\Delta \) ABC = \(\sqrt { s(s-a)(s-b)(s-c) } \)
\(=\sqrt { 48(48-34)(48-42)(48-20) } \)
\(=\sqrt { 48(14)(6)(28) } =\quad 336\) cm²
\(\therefore \) Area of parallelogram ABCD = 2 area of triangle ABC
= 2\(\times \) 336 cm² = 672 cm²

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

For triangular park
\(\because \)   6² + 8² = 10²
\(\therefore \)  Angular between sides of length 6 m and 8 m = \(90°\)
 \(\therefore \)  Area of the triangular park = \(\frac { 6\times 8 }{ 2 } \) = 24 m²
Radius of circular area (r) = \(\frac { 2 }{ 2 } \) m = 1m
\(\therefore \) Circular area = \(\pi \)r² = \(\pi \)(1)²=\(\pi \)=3.14 m²
\(\therefore \) Area used for growing roses = Area of the triangular part - Circular area 
= 24 - 3.14 = 20.86 m²

In right triangle PSQ,  PQ\(\frac { 1 }{ 2 } \)² = PS² + QS²     |By Pythagoras Theorem
= (12)² + (16)²
= 144 + 256 =400
\(\Rightarrow \)  PQ = \(\sqrt { 400 } \) = 20 cm
Now, for  \(\Delta \)PQR
a = 20cm, b = 48cm, c = 52cm
\(\therefore s=\frac { a+b+c }{ 2 } =\frac { 20+48+52 }{ 2 } =60\) cm
\(\therefore \)  Area of  \(\Delta \)PQR \(=\sqrt { s(s-a)(s-b)(s-c) } \)
\(=\sqrt { 60(60-20)(60-48)(60-52) }\)
\( \\ =\sqrt { (60)(40)(12)(8) } \)
 \(=\sqrt { \left( 6\times 10 \right) \left( 4\times 10 \right) \left( 6\times 2 \right) \left( 8 \right) } \)
\(=6\times 10\times 8=480\) cm²
Area of \(\Delta \) PSQ = \(\frac { 1 }{ 2 } \)\(\times \)Base\(\times \)Altitude
=\(\frac { 1 }{ 2 } \)\(\times \)16\(\times \)12=96 cm²
\(\therefore \) Area of the shaded portion =Area of \(\Delta \)PQR - Area of \(\Delta \)PSQ
= 480 - 96 = 384 cm²

Let the sides forming the right angle be a cm and b cm. Then,
      
a + b + 10 = 24
\(\Rightarrow \)  a + b = 14               ....(1)
Also,  a² + b² = (10)²     |By Pythagoras Theorem
\(\Rightarrow \)   a² + b² = 100      ...(2)
We know that   (a + b)² = a² + b² + 2ab
\(\Rightarrow \)  (14)² = 100+2ab
\(\Rightarrow \) 2ab = 96
\(\Rightarrow \)  ab=48                   ...(3)
Also,    (a - b)² = a² + b²- 2ab
= 100 - 2\(\times \)48
= 100 - 96 = 4       | if a>b  
\(\Rightarrow \)  a-  b = 2
Solving (1) and (4), we get   a=8cm, b=6cm
\(\therefore \)   Area = \(\frac { 1 }{ 2 } \)ab = \(\frac { 1 }{ 2 } \).8.6 = 24 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

Area =  \(=\frac { a }{ 4 } \sqrt { 4{ b }^{ 2 }-{ a }^{ 2 } } \)
 \(\Rightarrow 60=\frac { 24 }{ 4 } \sqrt { 4{ b }^{ 2 }-{ (24) }^{ 2 } } \)
\(\Rightarrow 10=\sqrt { 4{ b }^{ 2 }-576 } \)
\(\Rightarrow 100=4{ b }^{ 2 }-576\)      | Squaring
\(\Rightarrow 4{ b }^{ 2 }=676\)
\(\\ \Rightarrow { b }^{ 2 }=\frac { 676 }{ 4 } =169\)
\(\Rightarrow b=\sqrt { 169 } \) = 13 cm
\(\therefore\) Perimeter = a + b + b
= 24 + 13 + 13 = 50 cm

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.