The formula for the perimeter of a rhombus is fairly simple. The perimeter of an object is the sum of its sides. Since you specify a rhombus with four sides of 22, the sum of those sides would be 88.
Perimeter = 4*22 = 88 cm
The perimeter of a 22 ft square is 88 ft.
The perimeter will be 88cm. The square root of 484 is 22 - the length of one side of the square, multiplied by 4 is 88 !
4
88 + 22 = 110
Perimeter = 4*22 = 88 cm
88/4=22
The perimeter of a 22 ft square is 88 ft.
The perimeter will be 88cm. The square root of 484 is 22 - the length of one side of the square, multiplied by 4 is 88 !
A square of side 22 has an area of 484. Rectangle 23 x 21 has an area of 483...
you cant really answer that if you don't know what type of quadrilateral it is. if it is a square then each side is 22 inches.
Which quadrilateral? What is x?
A ziggurat could be oval, rectangular, square, or any other shape. If happened to have a square base, its perimeter would be four times its length (4 x 22 meters = 88 it meters).
4
88 + 22 = 110
22
To find the slant height of a square pyramid, we can use the formula for the lateral area, which is given by ( \text{Lateral Area} = \frac{1}{2} \times \text{Perimeter of base} \times \text{Slant height} ). The perimeter of the base for a square pyramid with a side length of 22 feet is ( 4 \times 22 = 88 ) feet. Setting the lateral area to 836 square feet gives us the equation: ( 836 = \frac{1}{2} \times 88 \times \text{slant height} ). Solving for the slant height yields ( \text{slant height} = \frac{836 \times 2}{88} = 19 ) feet.