Hexagon AGHBGHCGHDGHEGHFGH is a reflected image of hexagon ABCDEF. The midpoints of the sides of hexagon ABCDEF are also shown. Drag the line of reflection, until the image coincides with the preimage. At this location, if the preimage flips about the line of reflection, it will flip onto itself. In how many different positions can you place so the image reflects onto the preimage in this manner? Describe the different positions. Be sure to pass the line of reflection through both vertices and midpoints before answering.

Hexagon AGHBGHCGHDGHEGHFGH coincides with hexagon ABCDEF when GH passes through themidpoints of opposite sides; that is, it is a perpendicular bisector of the two sides. also coincides with hexagon ABCDEF when the line of reflection joins a pair ofvertices opposite one another on the hexagon. There are three perpendicular bisectors and three pairs ofopposite vertices. In all, there are six lines of reflection that will map the hexagon back onto itself.

Hexagon AGHBGHCGHDGHEGHFGH coincides with hexagon ABCDEF when  passes through the midpoints of opposite sides; that is, it is a perpendicular bisector of the two sides. Hexagon AGHBGHCGHDGHEGHFGH also coincides with hexagon ABCDEF when the line of reflection joins a pair of vertices opposite one another on the hexagon. There are three perpendicular bisectors and three pairs of opposite vertices. In all, there are six lines of reflection that will map the hexagon back onto itself.

Step-by-step explanation:

step-by-step explanation:

use pothagarean therom

a^2+b^2=c^2 so first ac

acm is a right triangle because c= 90°

and am (the hypotenuse or c) =16+9 =25

but there is not enough info we need another angle or side length

u = -(6/7)w

step-by-step explanation:

place all values with 'u' on one side of the equation.

4u + 8w = - 3u + 2w

4u + 3u = 2w - 8w

further simplify the equation by solving the addition and subtraction.

7u = - 6w

make 'u' the independent variable by placing its co-efficient on the right-hand side.

u = - (6/7)w