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

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 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

b + b + 6 = 24
 \(\Rightarrow \)   b = 9 cm
\(\therefore \) Area \(=\frac { 9 }{ 4 } \sqrt { 4{ b }^{ 2 }-{ a }^{ 2 } } =\frac { 6 }{ 4 } \sqrt { 4{ (9) }^{ 2 }-{ (6) }^{ 2 } } \)
 \(=\frac { 3 }{ 2 } \sqrt { 288 } =\frac { 3 }{ 2 } .12\sqrt { 2 } \)
\( 18\sqrt { 2 } \)  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.

\(s=\frac { a+b+c }{ 2 } =\frac { 13+14+15 }{ 2 } =21\)cm
\(\therefore \) \(\quad =\sqrt { s(s-a)(s-b)(s-c) } \)
\(=\sqrt { 21(21-13)(21-14)(21-15) } =\sqrt { 21\times 8\times 7\times 6 } =84\)cm²

\(\because \)  6²+8²=10²
\(\because \) The triangle is right angled with hypotenuse 10 cm
\(\therefore \)  Area \(=\frac { 6\times 8 }{ 2 } \)= 24 cm²
\(\therefore \)  Cost of painting =24 \(\times \) 0.09 = Rs. 2.16