Since E is the mid point of DC
DE = 12 m ; EC = 12 m
ADEB is a trapezium of parallel side
a = 12 m, b = 24 m and h = 18 m
Area of trapezium = \(\frac{1}{2} \times \mathrm{h}(\mathrm{a}+\mathrm{b}) \text { sq.units }\)
\(=\frac{1}{2} \times 18(12+24)\)
= 9 x 36
Cultivated Area = 324 m²

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

d₁ = 4 cm, d₂ = 5 cm
Area of sweet (rhombus) \(=\frac{1}{2} \times\left(d_{1} \times d_{2}\right) \text { sq.units } \)
\(=\frac{1}{2} \times(4 \times 5)\)
= 10 sq.cm
Area of aluminium foil = Area of 400 sweets
= 400 x 10 sq.cm
= 4000 sq. cm
Cost of aluminum foil per 100 sq.cm = Rs. 7
Cost of aluminum foil for 4000 sq.cm
\(=\frac{7}{100} \times 4000\)
Cost of aluminum foil used for 400 sweets
= Rs. 280

(i) Given: d₁ = 8 cm, d₂ = 16 cm
Area of the rhombus
\(=\frac{1}{2} \times\left(\mathrm{d}_{1} \times \mathrm{d}_{2}\right) \text { sq.units } \)
\(=\frac{1}{2} \times(8 \times 16)=\frac{128}{2} \)
Area of the rhombus = 64 sq.cm
(ii) b = 15 cm, h = 11 cm
Area of the rhombus = b x h sq.units
= 15 x 11
Area of the rhombus = 165 sq.cm