The sides of a triangle area in the ratio of 1/3 : 1/4 : 1/5 and its perimeter is 94 cm.
Find the length of the smallest side of the triangle .

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

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

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

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.

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

Diagonal of square = ?2a [a = side]
4?2 = ?2 a
a = 4 cm
Now, area of square = a² = (4²) = 16
Side of a square whose area is 2 x 16.
a₁² = 32
? a₁ = ?32 ?a₁4?2

Now, diagonal of new square = ?2a
= ?2x 4 ?2
= 8 cm

Let the diagonals of the squares be 3x and 2x.
? Ratio of their areas = [(1/2) (3x²)] / [(1/2) (2x²)] = 9/4

Required increment = 2a + [a² / 100] %
= 2 x 25 + [(25²)/100)] %
= 50 + (625/100)%
= 56.25%