The area of an equilateral triangle is ? 243 /4 sq cm. Find the length of its side.

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

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

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%

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