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

We know that all sides of rhombus are equal and its diagonals bisect each other.
∴ x = 5, y = 12 and z = 13.

Given, Area of the Trapezium = 828 cm²
\(\cfrac { 1 }{ 2 } \times h(a+b)=828\)
\(\cfrac { 1 }{ 2 } \times h(19.6+16.4)=828\)
\(\cfrac { 1 }{ 2 } \times h(36)=828\)
h (18) = 828
\(h=\cfrac { 828 }{ 18 } \)
h - = 46 m
Therefore, distance between the parallel sides = 46 cm

Given that the adjacent sides of parallelogram are 12 cm and 9 cm
If we choose the shorter side as base, that is b = 9 cm then distance between the shorter sides is height, that is h = 8 cm
Area of parallelogram = b x h sq.units = 9 x 8 = 72 sq. cm.
Again, if we choose longer side as base, that is b = 12 cm then distance between longer sides is height. Let it be 'h' units.

We know that, the area of the parellelogram = 72 sq.cm
b x h = 72
12 x h = 72
\(h=\cfrac { 72 }{ 12 } =6cm\)
Therefore, the distance between the longer sides = 6 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.

Given: d₁ = 6 m, d₂ = 8 m
Area of the rhombus = \(\cfrac { 1 }{ 2 } \times ({ d }_{ 1 }\times { d }_{ 2 })sq.units\)
= \(\cfrac { 1 }{ 2 } \times (6\times 8)\)
= \(\cfrac { 48 }{ 2 } \)
= 24 sq.m
Hence, area of the rhombus is 24 sq.m.

Base = 17 cm, height = 8 cm
Area of the rhombus = b x h sq. units
= 17 x 8 = 136
Therefore, area of the rhombus = 136 sq. cm