Diagonals of a rhombus area 1 m and 1.5 m in lengths. The area of the rhombus is

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

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

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

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

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

Perimeter of rhombus
= 2 ? d₁ ² + d₂ ²
= 2 ?(4.8)² + (1.4)²
= 2 ?23.04 + 1.96
= 2 ?25
= 2 x 5 = 10 cm

Let First angle = 3k
Second angle = 4k
Third angle = 5k and
Fourth angle = 8k

We know 3k + 4k + 5k + 8k = 360°
? 20k = 360°
? k = 18 °

Measure of smallest angle = 3k
=3 x 18 °
= 54 °

Area of square room = (10)² = 100 sq m
= 100 x (100)² sq cm
= 100 x 100 x 100 sq cm

Now, area of tiles = (50)² = 50 x 50 sq cm
? Number of tiles needed = (Area of square room) / (Area of tile)
= (100 x 100 x 100) / (50 x 50)
= 400

Hence, 400 tiles will be needed.

Let the third side be x.
According to the question,
x + 20 = 26
? x = 26 - 20 = 6 cm
? Each equal side = 20/2 = 10 cm

Let length of rectangle = 5k
and breadth of rectangle = 3k

According to the quecation,
5k - 3k = 8
? 2k =8
? k = 4
? Lenght = 5k = 5 x 4 = 20 m
Breadth = 3k = 3 x 4 = 12 m

? Required area = Lenght x Breadth
= 20 x 12 = 240 sq m