Instruction

1

For square or rhombus task is to find

**side**of the perimeter is solved very simply. It is known that these two shapes with 4**sides**and all of them are equal, so the perimeter p of a square and a rhombus is 4a, where a is the side of the square or rhombus. Then the length of**side**equal to one-fourth of the perimeter: a = p/4.2

Easy to solve this problem for an equilateral triangle. He's got three equal length

**sides**, so the perimeter p of an equilateral triangle is equal to 3a. Then the side of an equilateral triangle a = p/3.3

For the rest of the figures need additional information. For example, you may find

p = 2(a+b)

s = ab.Express from the first equation: a = p/2 - b. Substitute into the second equation and find b s = pb/2 - b2. The discriminant of this equation D = p2/4 - 4s. Then b = (p/2±D^1/2)/2. Discard the root that is less than zero, and substitute in the expression for

**the sides**of a rectangle, knowing its perimeter and area. Suppose that the length of the two opposite sides of the rectangle equal a, and the length of other two sides is b. Then the perimeter p of a rectangle is 2(a+b), and the area s is equal to ab. Get a system of equations with two unknowns:p = 2(a+b)

s = ab.Express from the first equation: a = p/2 - b. Substitute into the second equation and find b s = pb/2 - b2. The discriminant of this equation D = p2/4 - 4s. Then b = (p/2±D^1/2)/2. Discard the root that is less than zero, and substitute in the expression for

**side**a.