The lengths of a square's sides if its area is 5 is: 2.236
Wiki User
∙ 14y agosquares have equal sides. Each side will be 5 meters because 20 divided by 4 = 5. Take length x width = area. 5x5=25. The area is 25 m2
A squares sides are equal in length. The answer is 4*5"=20".
1 square . . . 4 sides2 squares . . . 8 sides..5 squares . . . 20 sides
how many squares with sides that are 6 inches long are needed to cover a squae with a side length of 30 inches without overlapping
Since all sides of the square are the same length, if the length is 5, the width is five. Therefore area= length x width = 5 x 5 = 25. The answer is 25 square feet.
squares have equal sides. Each side will be 5 meters because 20 divided by 4 = 5. Take length x width = area. 5x5=25. The area is 25 m2
A squares sides are equal in length. The answer is 4*5"=20".
how many squares with sides that are 6 inches long are needed to cover a squae with a side length of 30 inches without overlapping
1 square . . . 4 sides2 squares . . . 8 sides..5 squares . . . 20 sides
Imagine the rectangle divided into squares corresponding to length and width... Eg a 6" x 5" rectangle would have 5 rows of six one-inch squares, total 30, which would make its area 30 squinches
Since all sides of the square are the same length, if the length is 5, the width is five. Therefore area= length x width = 5 x 5 = 25. The answer is 25 square feet.
The area of a rectangle = length x width = 5 x 8 = 40 square inches.
Assuming a regular hexagon (all sides equal) with sides of length L: Area = (6.75).5 * L2 =~ 2.5981 * L2
To find the area of joined squares, add up the area of each individual square. For instance, I have a square with a side length of 5 attached to another square with a side length of 2 A = 52 + 22 = 25 + 4 = 29 units2
With a rectangular prism with these dimensions, you'd have 2 sides with an area of 15*5, two sides with an area of 5*4 and two sides with an area of 4*15. The surface area is 15*5+15*5+5*4+5*4+4*15+4*15 = 310.
The area depends on the length of each sides as well as the angles between them.
No, because the biggest length (hypotenuse) has to be equal to the square root of the sum of the squares of the other two sides, which it is not