Let the parallel sides be 3a and 5a.
So Area of trapezium = 1/2 x sum of parallel side x perpendicular distance between them.
? 1/2 (3a +5a) x 12 = 384
? 8a = 64
? a =8

? Smaller side = 3x = 3 x 8 = 24 cm.

Area of trapezium = 1/2 x sum of parallel sides x perpendicular distance between them
= 1/2 (1 + 2) x 6 m²
= 9 m²

AB = 60 m, BC = 40 m and AC = 80 m
? s = (60 + 40 + 80 ) / 2 m = 90 m
(s-a) = 90 - 60 = 30 m,
(s-b) = 90 - 40 = 50 m and
(s-c) = 90 - 80 = 10 m

? Area of ? ABC =
?s(s-a)(s-b)(s-c)
= ?90 x 30 x 50 x 10 m²
= 300?15 m²

? Area of parallelogram ABCD = 2 x area of ? ABC
= 600?15 m²

AB = 6 cm ; AD = 4 cm and ? BAD = 30°
Area of parallelogram ABCD = AB x AD x sin 30°
= 6 x 4 x sin 30° = 12 cm²

Area of parallelogram = Side of parallelogram x distance from the opposite side
= 14 x 16 cm²
= 224 cm²

Let length of the longer diagonal = d cm
Then, length of other diagonal = 0.8 x d cm
Area of rhombus = (1/2) x d x 0.8 x d = 2/5 d²
= 2/5 d²

Area of square of the length of the longer diagonal = d²

So the area of the rhombus is 2/5 times the square of the length of the longer diagonal.

Area of rhombus = (d₁ x d₂) / 2
? (d x 2d ) / 2 = 144
? d² = 144
? d = 12
? Length of diagonal = 12 cm, 24 cm

We know that in any triangle
"the sum of two sides is always greater than its third side" and
"the difference of two sides is always less than its third side".
(i) 2 + 3 is not greater than 5
(ii) |5 - 2| not less than 3

In a triangle
Sum of two sides is always greater than 3rd side
i.e., x < 25 + 15 = 40 .....(i)

Difference of two sides is always less than 3rd side
i.e., 25 - 15 = 10 < x ...(ii)

From Eqs. (i) and (ii) , we get
10 < x < 40

According to the question,
?3a²/4 = 3?3 [ side = a, and area = ?3a²/4]
? a²/4 = 3
? a² = 3 x 4
? a = 2?3

? Required perimeter = 3a
=3 x 2?3
= 6?3 cm