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

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 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.

(i) From the Fig.2.4
Base of a parallelogram (b) = 15 cm, Height of a parallelogram (h) = 4 cm
Area of a parallelogram = b x h sq.units. Therefore, Area = 15 × 4 = 60 sq. cm.
Thus, area of the parallelogram is 60 sq. cm.
Perimeter of the parallelogram = sum of the length of the four sides.
= (15 + 5 +15 + 5) = 40 cm.
(ii) From the Fig.2.5
Base of a parallelogram (b) = 9 m, Height of a parallelogram (h) = 16 cm.
Area of a parallelogram = b × h sq.units Therefore, Area = 9 × 16 = 144 sq. cm
Thus, area of the parallelogram is 144 sq. cm.
Perimeter of the parallelogram = sum of the length of the four sides.
= (18 + 9 +18 + 9) = 54 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 : Area = 100 sq.cm, d₁ = 8 cm, d₂ = ?
Area of rhombus = 100 sq.cm
\(\frac{1}{2} \times\left(d_{1} \times d_{2}\right)=100 \)
\(\frac{1}{2} \times\left(8 \times d_{2}\right)=100 \)
8 x d₂ = 100 x 2
\(\mathrm{d}_{2}=\frac{100 \times 2}{8} \Rightarrow \mathrm{d}_{2}=25 \mathrm{~cm}\)

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

h = 14 m,b = 14 + 8 = 22 m
Area of ground (parallelogram) = b x h sq.units
= 22 x 14 = 308 m²
Cost of levelling the ground = Rs. 15  per sq.m
Cost of levelling the ground = 308 x 15 = Rs. 4620

h = 12 m,b = 18 m
Area of a parallelogram (fabric) = b x h = 18 x 12
= 216 m² (or) sq.m
She cuts the fabric into 4 equal parallelograms
Area of a new parallelogram = 54 m²

(i) b = 18 cm, h = 5 cm
Area of parallelogram = b x h sq.units
= 18 x 5
= 120 sq.cm
(ii) b = 8 m, Area = 56 sq.m, h = ?
Area of a parallelogram = 56
b x h = 56
8 x h = 56
h = 7m
(iii)  h = 17 mm, Area = 221 sq.mm
Area of a parallelogram = 221 sq.mm
b x h = 221
b x 17 = 221
b = 13 mm

(i) Area of a parallelogram
= b x h sq.units
= 11 x 3 = 33 cm²
Perimeter of the parallelogram
= sum of the length of the four sides.
= 11+ 4 + 11 + 4 = 30 cm.
(ii) Area of a parallelogram
= b x h sq.units
= 7 x 10 = 70 cm²
Perimeter of the parallelogram
= sum of the length of the four sides.
= 13 + 7 + 13 + 7 = 40 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.

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