the perimter of a pentagon is outside of the the pentegon. count what is in the outside
All you have to do is... Add all the sides together! Simples
Many shapes could have a perimeter of 16. A square or rhombus with side = 4, or a rectangle 6 x 2 would have a perimeter of 16. A regular octagon with side length = 2. An isosceles triangle with sides 7, 7 and 2. There are an infinite number of possibilities.
The area is given by xy and the perimter is given by 2x + 2y, where x is the side length and y is the height. So we have two equations: A = xy, P = 2x + 2y. Because the square and the rectangle have the same perimeter, P will be a constant, say 40. We can simplify this to the equivalent 20 = x + y by dividing both sides by 2. We can then rearrange it to y = 20 - x or vice versa for x and y. By substituting this into the first equation, we get A = x(20 - x) = 20x - x2. By using differentiation, we can find the x value for which A is maximum: A' = 20 - 2x Set A' to 0 (we know this will be a maximum because the x2 coefficient is negative): 0 = 20 - 2x 2x = 20 x = 10. Now we back-substitute this value for x into the equation for y: y = 20 - x = 20 - 10 = 10. Thus, the maximum area occurs when x = 10 and y = 10 i.e. when the shape is a square. This works for any starting value of the perimeter, and means that a square will always have more area than any rectangle with an equal perimeter.
etyghviugui
40 cm
It is the sum of its 4 equal sides.
measure the distance around the outside
add all of the sides together.
the forumal is to add (sum) all the sides of the polygon.
Perimeter = 4*1.25 = 5m
the outside of the shape
18
All you have to do is... Add all the sides together! Simples
creates rough edges
It is: 4*12 = 48 inches