The lengths of three line segments (in cm) are given in each of the four cases. Which one of the following cases is not suitable to be the three sides of a triangle?

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

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

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

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

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

Given ratio = 1/3 : 1/4 : 1/5 = 20 : 15 : 12
Let length of the sides be 20k, 15k and 12k.

Then, according to the question,
20k + 15k + 12k = 94
? 47k = 94
? k = 94/47 = 2

Smallest side = 12k = 12 x 2 = 24 cm

Given that, perimeter = 90 cm
? 3a = 90 cm       [a = side]
? a = 90/3 = 30 cm
? Area = ?3a² / 4 = ?3/4 x (30)²
= 900 x ?3/4 = 225?3 sq cm

According to the question,
area = ?3a²/4
= ?243 /4
? a² = ?81 x 3/?3
? a = ?9
= 3 cm