How do you draw 2 shapes with an area of 6 square units and a permeter of 12 units?

To draw two shapes with an area of 6 square units and a perimeter of 12 units, we need to consider the possible combinations of shapes that fulfill these criteria. Let's start by examining different options.

Option 1: Rectangle
Let's assume one shape is a rectangle. To find the dimensions, we can use the formulas for area and perimeter of a rectangle:
Area = length × width
Perimeter = 2(length + width)

Given that the area is 6, we have:
6 = length × width

Since the perimeter is 12, we have:
12 = 2(length + width)
6 = length + width

We can solve this system of equations simultaneously by substitution or elimination methods. Let's choose substitution:

From the second equation, we have:
length = 6 - width

Now substitute this value of length into the first equation:
6 = (6 - width) × width
6 = 6width - width^2
0 = width^2 - 6width + 6

Solving this quadratic equation gives us two possible values for the width. Let's use the quadratic formula:
width = (-b ± √(b^2 -4ac)) / (2a)

For this equation, a = 1, b = -6, and c = 6. Substituting these values gives us:
width = (-(-6) ± √((-6)^2 - 4(1)(6))) / (2(1))
width = (6 ± √(36 - 24)) / 2
width = (6 ± √12) / 2
width = (6 ± 2√3) / 2
width = 3 ± √3

Thus, we have two possible widths: 3 + √3 and 3 - √3. To find the corresponding lengths, we substitute these values into the equation length = 6 - width.

For width = 3 + √3:
length = 6 - (3 + √3)
length = 3 - √3

For width = 3 - √3:
length = 6 - (3 - √3)
length = 3 + √3

Therefore, we have two possible rectangles with an area of 6 square units and a perimeter of 12 units:
Rectangle 1: Length = 3 - √3, Width = 3 + √3
Rectangle 2: Length = 3 + √3, Width = 3 - √3

Option 2: Triangle
Now let's explore the possibility of one shape being a triangle. A triangle with an area of 6 square units can have various side lengths, but we need the perimeter to be 12 units.

To find a triangle with a perimeter of 12, we can consider an isosceles triangle in which two sides are equal and the third side is different. Let's assume the two equal sides have a length of x, and the remaining side has a length of y.

Since the perimeter is 12, we have:
12 = 2x + y

The area of a triangle is given by the formula:
Area = (base × height) / 2

In this case, we can consider any two sides of the triangle as base and height. So, let's consider the two equal sides (x) as base and the remaining side (y) as height:

6 = (x × y) / 2
12 = x × y

Now we have a system of equations. To solve it, we can substitute the value of y from the first equation into the second equation:

12 = 2x + y
y = 12 - 2x

Substituting this value into the second equation:
12 = x(12 - 2x)
12 = 12x - 2x^2

Rearranging the equation:
2x^2 - 12x + 12 = 0

Dividing the equation by 2 gives us a simpler form:
x^2 - 6x + 6 = 0

To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 -4ac)) / (2a)

In this case, a = 1, b = -6, and c = 6. Substituting these values gives us:
x = (-(-6) ± √((-6)^2 - 4(1)(6))) / (2(1))
x = (6 ± √(36 - 24)) / 2
x = (6 ± √12) / 2
x = (6 ± 2√3) / 2
x = 3 ± √3

Thus, we have two possible equal side lengths: 3 + √3 and 3 - √3. To find the third side (y), we substitute these values into the equation y = 12 - 2x.

For x = 3 + √3:
y = 12 - 2(3 + √3)
y = 12 - 6 - 2√3
y = 6 - 2√3

For x = 3 - √3:
y = 12 - 2(3 - √3)
y = 12 - 6 + 2√3
y = 6 + 2√3

Therefore, we have two possible triangles with an area of 6 square units and a perimeter of 12 units:
Triangle 1: Two equal sides = 3 + √3, Third side = 6 - 2√3
Triangle 2: Two equal sides = 3 - √3, Third side = 6 + 2√3

So, in conclusion, you can draw either two rectangles or two triangles to satisfy the given conditions.