Two polygons are congruent and the perimeter of the first polygon is 38 cm. if the sides of the second polygon are consecutive integers (x, x+1, x+2, x+3, what value of x makes the polygons into congruent quadrilaterals? use x as the smallest side.

In order for us to find the answer, let us analyze the given problem first.
It states that there are two polygons which are congruent. So if the polygons are congruent, therefore, both their sides, areas and perimeters are also congruent. Given the the first polygon has a perimeter of 38cm, then the second polygon has the same measurement too.
So here is how we are going to answer this:
38 = x + x +1 + x +2 + x + 3
Given that x is the smallest side.
So, 38 = 4x +6
38-6 = 4x
32 = 4x <<divide both sides by 4 and we get
x = 8
Therefore, the smallest side measures 8cm.
So the other sides measurements are:  9, 10 and 11. Add them all and you get 38cm. Hope this answer helps.

I’m going to infer you said parallel.solve for yy= -1/2x+7/4lines that are parallel have the same slope, different y intercept.so we know m is -1/2then from there solve for y intercept.