Let the side of the square plot be a ft.a² = 289 => a = 17Length of the fence = Perimeter of the plot = 4a = 68 ft.Cost of building the fence = 68 * 58 = Rs. 3944.

Let the length and the breadth of the rectangle be l cm and b cm respectively. Let the side of the square be a cm. a2 = 4096 = 2¹² a = (2¹²)^1/2 = 2⁶ = 64 L = 2a and b = a - 24 b : l = a - 24 : 2a = 40 : 128 = 5 : 16

The circumference of the circle is equal to the permeter of the rectangle. Let l = 6x and b = 5x
2(6x + 5x) = 2 * 22/7 * 3.5  =>  x = 1 Therefore l = 6 cm and b = 5 cm
Area of the rectangle = 6 * 5 = 30 cm²

Let the side of the square be a cm. Let the length and the breadth of the rectangle be l cm and b cm respectively. 4a = 2(l + b) 2a = l + b l . b = 480 We cannot find ( l + b) only with the help of l . b.
Therefore a cannot be found . Area of the square cannot be found.

Area of the four walls = 2h(l + b)Since there are doors and windows, area of the walls = 2 * 12 (15 + 25) - (6 * 3) - 3(4 * 3) = 906 sq.ft.Total cost = 906 * 5 = Rs. 4530

The volume of the cone = (1/3)Ï€r²h Only radius (r) and height (h) are varying.Hence, (1/3)Ï€ may be ignored. V₁/V₂ = r₁²h₁/r₂²h₂  => 1/10  =  (1)²h₁/(2)²h₂ => h₁/h₂ = 2/5  i.e. h₁ : h₂ = 2 : 5

Volume of the wire (in Cylindrical shape) is equal to the volume of the sphere.  Ï€(16)² * h = (4/3)Ï€ (12)³ => h = 9 cm

In one resolution, the distance covered by the wheel is its own circumference. Distance covered in 500 resolutions. = 500 * 2 * 22/7 * 22.4
= 70400 cm = 704 m

Ratio of the sides = ³√729 : ³√1331 = 9 : 11 Ratio of surface areas = 9² : 11² = 81 : 121

Consider the rhombus ABCD. Let the diagonals intersect at E. Since diagonals bisect at right angles in a rhombus. BE² + AE² = AB² 25² = 15² + AE²
AE = √(625 - 225) = √400 = 20, AC = 20 + 20 = 40 cm. Area of a rhombus = 1/2 * d₁d₂ = 1/2 * 40 * 30 = 600 sq.cm.