Area of each tile = \(\frac { 1 }{ 2 } \) x d₁ X d₂ sq. units

∴ Area of 2000 tiles = 500 x 2000 = 10,00,000 cm² = 100 m²
Cost of polishing 1 m² = Rs. 5
∴ Cost of polishing 100 m² = 5 x 100 = Rs. 500

Given perimeter of the rhombus = 180 cm
∴ One side of the rhombus = \(\frac { 180 }{ 4 } \) = 45 cm
Given area of the rhombus = 315 cm²
b x h = 315
45 x h = 315
= \(\frac { 315 }{ 45 } \)
h = 7cm
Altitude of the rhombus = 7 cm

Let, the length of the required side be ‘x’ cm.
Then, area of the trapezium = \(\cfrac { 1 }{ 2 } \times h\left( a+b \right) sq.units\)
= \(\cfrac { 1 }{ 2 } \times 16(25+x)\)
= 200 + 8x
But, the area of the trapezium = 352 sq. cm (given)
Therefore, 200 + 8x = 352
\(\Rightarrow\) 8x = 352 – 200
\(\Rightarrow\) 8x = 152
 \(\Rightarrow\) \(x=\cfrac { 152 }{ 8 } \)
x = 19
Therefore, the length of the other side is 19 cm.

Given, height (h) = 9 cm
Parallel sides are (a) = 23 cm and (b) = 12 cm
Area of the trapezium = \(\cfrac { 1 }{ 2 } \times h(a+b)\)
= \(\cfrac { 1 }{ 2 } \times 9\left( 23+12 \right) \)
= \(\cfrac { 1 }{ 2 } \times 9(35)\)
= 157.5 sq. cm
Therefore, Area of the trapezium is 157.5 sq. cm.

Given: b = 12 m, h = 8 m
Area of the parallelogram = b x h sq.units
= 12 x 8 = 96 sq.m
Therefore, Area of the parallelogram = 96 sq.m.

Given perimeter = 150 cm
Let one side of the parallelogram be 'b' cm
Then the other side = b + 25 cm
b + (b + 25) + b + (b + 25) = 150
b + b + 25 + b + b + 25 = 150
4b + 50 = 150 = 4b =100

∴ One side b = 25 cm
Other side b + 25 = 50 cm

We know that in a parallelogram opposite sides are equal.
∴ 3x = 18
\(x=\frac { 18 }{ 3 } \)   
x = 16
and     
3y - 1 = 26
3y = 26 + 1 = 27
\(y=\frac { 27 }{ 3 } \)
y = 9

Given: Area of a trapezium = 180 sq.cm
height, h = 9 cm
One of its parallel side = a
Other side is longer than first side by 6.
b = a + 5
Area of the trapezium = 180
\(\frac{1}{2} \times h(a+b) =180 \)
\(\frac{1}{2} \mid \times 9(a+a+6) =180 \)
\(2 a+6 =\frac{180 \times 2}{9} \)
2a = 40 - 6
2a = 34
\(a=\frac{34}{2}\)
One parallel side, a = 17 cm
Other side, b = a + 6 = 17 + 6
b = 23 cm
Length of parallel sides are 17 cm and 23 cm.

Given base b = 14 cm ; Height h = 9 em
Area of the rhombus = b x h sq. units
= 14 x 9
Area of rhombus = 126 sq.cm

Given : A = 735 sq.cm, b = 21 cm, h = ?
Area of the trophy = 735 sq,cm
b x h = 735
21 x h = 735
Height of trophy (h) = 35 cm

Given, the length of the diagonals of a rhombus shaped tile are 40 cm and 25 cm
The area of one tile = \(\cfrac { 1 }{ 2 } \times \left( { d }_{ 1 }\times { d }_{ 2 } \right) sq.units\)
= \(\cfrac { 1 }{ 2 } \times 40\times 25\)
= 500 sq.cm
Therefore, the area of 200 such tiles = 200 x 500
= 100000 sq. cm
= \(\cfrac { 100000 }{ 10000 } \)(1sq.m = 1000 sq.cm)
= 10 sq.m
Therefore, the cost of polishing 200 such tiles at the rate of Rs 45 per sq. cm
= 10 x 45 = Rs. 450.

Given, the length of one diagonal (d₁) = 8 cm
Let, the length of the other diagonal be d₂ cm
Area of the rhombus = 60 sq. cm (given)
\(\cfrac { 1 }{ 2 } \times ({ d }_{ 1 }\times { d }_{ 2 })=60\)
\(\cfrac { 1 }{ 2 } \times ({ 8 }\times { d }_{ 2 })=60\)
8 x d₂ = 60 x 2
\({ d }_{ 2 }=\cfrac { 120 }{ 8 } \)
= 15
Therefore, length of the other diagonal is 15 cm.