A room 5.44 m x 3.75 m is to be paved with square tiles. the least number of tiles required to cover the floor is?

Area of park = 100 x 100 = 10000 m²
Area of circular lawn = Area of park - area of park excluding circular lawn
= 10000 - 8614
= 1386

Now again area of circular lawn = (22/7) x r² = 1386 m²
? r² = (1386 x 7) / 22
= 63 x 7
= 3 x 3 x 7 x 7

? r = 21 m

? 22/7 x r² = 462
? r² = (462 x 7) /22 = 147
? r = 7?3 cm

? Height of the triangle = 3r = 21?3 cm
Now, ? a² = a²/4 + (3r)²
? 3a²/4 = (21?3)²
? a² = (1323 x 4)/3
? a = 21x 2 = 42 cm

? Perimeter = 3a = 3 x 42
=126 cm

Angle swept in 30 min= 180°
Area swept = [(22/7) x 7 x 7] x [180°/360°] cm²
= 77 cm²

Let circumference = 100 cm .
Then, ? 2?r = 100
? r = 100/2?
=50/?

? New circumference
= 105 cm

Then, 2?R = 105
? R = 105 / (2?)
&rArr Original area = [ ? x (50/?) x (50/?) ]
= 2500/? cm²

? New Area = [? x (105/2?) x (105/2?)]
= 11025 / (4?) cm²

? Increase in area = [11025/(4?)] - 2500/? cm²
= 1025 / 4? cm²

Required increase percent [1025/(4?)] x 2500/? x 100 = 41/4%
= 10.25%

? 1/2 x 2 ? x h

= ?3/4 x (2 ?3)²

? h = 3 cm.

Ratio of similar triangle
= Ratio of the square of corresponding sides
= (3x)² / (4x)² = 9x² / 16x²
= 9/16 = 9 : 16

Let there be n sides of the polygon. Then it has n vertices.
The total number of straight lines obtained by joining n vertices by talking 2 at a time is ⁿC₂
These ⁿC₂ lines also include n sides of polygon.
Therefore, the number of diagonals formed is ⁿC₂ - n.
Thus, ⁿC₂ - n = 44
? [n(n - 1)/2] - n = 44
? ( n² - 3n) / 2 = 44
? n² - 3n = 88
? n² - 3n - 88 = 0
?(n - 11) (n + 8) = 0
? n = 11

Let the radius of circle is 'r' and a side of a square is 'a',
then given condition
2?r = 4a
? a = ?r/2

? Area of square = (?r/2)² = ?² /4r² = 9.86r²/4 = 2.46r²
and area of circle = ?r² = 3.14;r²
and let the side of equilateral triangle is x.
Then, given condition,
3x = 2?r
? x = 2?r/3

? Area of equilateral triangle = ?3/4 x ²
= ?3/4 x 4?²r²/9
= ?²/3?3r²
= 1.89r²

Hence, Area of circle > Area of square > Area of equilateral triangle.

Area of equilateral triangle = ?3a²/4 = x ......(i)

And perimeter = 3a = y
? a = y/3 ....(ii)

Now, Putting the value of a from Eq. (ii) in Eq. (i). we get
?3 (y/3)²/4 = x
? x = ?3 x y²/36
? x = y²/3?3x = y²/12?3
12?3 x = y²
On squaring both sides, we get
y⁴ = 432x²

We know that, the radius of a circle inscribed in a equilateral triangle = a/[2?3]
Where, a be the length of the side of an equilateral triangle.

Given that, area of a circle inscribed in an equilateral tringle = 154 cm²
? ?(a/2?3)² = 154
? (a/2?3)² = 154 x (7/22) = (7)a²
? a = 42?3 cm

Perimeter of an equilateral triangle = 3a
= 3(14?3)
= 42?3 cm