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

Perimeter = AB + BC + CD + AD
∴ [AB = DC & AD = BC]
= 7 cm + 5 cm + 7 cm + 5 cm = 24 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

base = 4 times of the height
b = 4h
Area of parallelogram = 576
b x h = 576
4h x h = 576
4h² = 576
h² = I44 cm
h = 12 cm
Ans : Height of the parallelogram is 12 cm.

Given : One of the diagonals of a rhombus is thrice the other
\(\Rightarrow \ \mathrm{d}_{2}=3 \mathrm{~d}_{1}\)       ............(1)
Sum of the length of the diagonals = 24 cm
d₁+ d₂ = 24cm
3 d₁ + 3d₂ = 24
4d₁ = 24
\(d_{1}=\frac{24}{4}=6 \mathrm{~cm}\)
d₁ = 6 cm
d₂  = 3d₁
= 3 x  6
d₂ = 18 cm
Area of the rhombus \(=\frac{1}{2} \mathrm{~d}_{1} \times \mathrm{d}_{2} \text { sq.units } \)
\(=\frac{1}{2} \times 6 \times 18 \)
= 54 sq.cm

Consider PM as height (h) and QR as base (b)
Then Area of the parallelogram = 900
b x h = 900
b x 20 = 900
b = 45 cm
QR = 45 cm
In diagram (ii),
Consider PN as height (h) and SR as its base (b)
Area of the parallelogram = 900
b x h = 900
b x 36 = 900
SR = 25 cm
QR = 45 cm, SR = 25 cm

Length of one diagonal \(=\frac{1}{2}\) length of other diagonal
\(\mathrm{d}_{2}=\frac{1}{2} \mathrm{~d}_{1}\)
Area of the rhombus = 576
\(\frac{1}{2} \times\left(d_{1} \times d_{2}\right)=576 \)
\(\frac{1}{2} \times d_{1} \times \frac{1}{2} d_{1}=576 \)
d₁ x d₁ = 576 x 2 x 2
d₁ x d₁ = 24 x 24 x 2 x 2
d₁ = 24 x 2
d₁ = 48 cm
\(\mathrm{d}_{2}=\frac{1}{2} \mathrm{~d}_{1}=\frac{1}{2} \times 48\)
d₂ = 24 cm
The length of the diagonals are 24 cm and 48 cm.

Given : Area of a parallelogram = Area of a square
b x h = a x a
b x 18 = 48 x 48
\(b=\frac{48 \times 48}{18}\)
Base (b) = 128 m

Given : Area of parallel shaped agricultural field
= 68.75 sq.hm
Distance between the parallel sides (h) = 6.25 hm
Area of agriculture field = 68.75
b x h = 68.75
b x 6.25 = 68.75
\(\mathrm{b}=\frac{68.75 \times 100}{6.25 \times 100}=\frac{6875}{625}\)
Base (b) = 11 hm

Given : Parallel sides of window,
a = 105 cm, b = 50 cm
Distance between parallel sides,
h = 60 cm
Area of trapezium shaped window
\(=\frac{1}{2} \times h(a+b) \text { sq.units } \)
\(=\frac{1}{2} \times 6 p(105+50) \)
\(=\frac{1}{2} \times 60 \times 155=4650 \text { sq.cm } \)
Cost of the glass to cover the window per 100 sq.cm = Rs. 15
Cost of the glass to cover the window for 4650 sq.cm \(=\frac{15}{100} \times 4650= Rs. \ 697.50\)

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: A = 1586 sq.cm, h = 26 cm, a = 84 cm, b = ?
Area of trapezium = 1586
\(\frac{1}{2} \times \mathrm{h}(\mathrm{a}+\mathrm{b}) =1586 \)
\(\frac{1}{2} \times 26(84+\mathrm{b}) =1586 \)
\(84+\mathrm{b} =\frac{1586 \times 2}{26} \)
b = 122 - 84
b = 38 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
\(\mathrm{d}_{2}=\frac{100 \times 2}{8} \Rightarrow \mathrm{d}_{2}=25 \mathrm{~cm}\)

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

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