The diagonals of a squares is 4?2 cm. The diagonal of another square whose area is double that of the first square is

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

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

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

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%

Ratio of area of 2 squares = (ratio of perimeter of 2 squares)²
= (24/12)²
= 4

Area of the field =1215/135 = 9 hec
= 90000 m² [1 hec =10000 m²]
? Side of the field = ?90000 = 300 m
Perimeter of the field = 4 x 300 = 1200 m

Now, cost of putting a fence around field = (1200 x 75)/100 = ? 900

Area of rhombus =(d₁ x d₂)/2
=(1 x 1.5)/2 m²
= 0.75 m²