Well, a distance-preserving transformation is called a rigid motion, and the name suggests that it <em>moves the points of the plane around in a rigid fashion.</em>
A transformation is distance-preserving if the distance between the images of any two points and the distance between the two original points are equal.
If that's confusing, I get it; basically if you transform something, the points from the transformation are image points. Take the distance from a pair of image points, and take the distance from a pair of original points, and they should be the same for a <em>rigid </em>motion.
I keep emphasizing this b/c not all transformations preserve distance; a dilation can grow or shrink things. But if you didn't go over dilations, don't say nothin XD
Here is the answer but basically the work you just plug in 6 for A, 12 for B, and 5 for C..into the quadratic formula and solve from there
For #4, we know V = lwh
We have the values
5184 = 2(18)h
Solve for h
5184 = 36h
144 = h.
for # 3:
we know A = pi r^2
We have
A = 3.14 * 6^2
A = 3.14 * 36
A = 113.04
for #2:
We know C = 2pi *r
C = 2(3.14)(14)
C = 28(3.14)
C = 87.92
for #1:
C = 2pi * r
C = 2(3.14)(26.3/2)
C = 2(3.14)(13.15)
C = 26.3(3.14)
C = 82.582
