Square $ABCD$ is constructed along diameter $AB$ of a semicircle, as shown. The semicircle and square $ABCD$ are coplanar. Line segment $AB$ has a length of 6 centimeters. If point $M$ is the midpoint of arc $AB$, what is the length of segment $MC$? Express your answer in simplest radical form. [asy]
size(4cm);

dotfactor = 4;
defaultpen(linewidth(1)+fontsize(10pt));

pair A,B,C,D,M;
A = (0,1);
B = (1,1);
C = (1,0);
D = (0,0);
M = (.5,1.5);

draw(A..M..B--C--D--cycle);
draw(A--B);

dot("A",A,W);
dot("M",M,N);
dot("B",B,E);
dot("C",C,E);
dot("D",D,W);

draw(M--C,linetype("0 4"));

[/asy]
If we let $E$ be the midpoint of line segment $AB$ and $F$ be the midpoint of $CD$, then line segment $MF$ will pass through point $E$. Also, $MF$ is perpendicular to $CD$, so $\triangle MFC$ is a right triangle. Now, if we can find the lengths of $MF$ and $FC$, we can use the Pythagorean Theorem to find the length of $MC$.

[asy]
size(4cm);

dotfactor = 4;
defaultpen(linewidth(1)+fontsize(10pt));

pair A,B,C,D,E,F,M;
A = (0,1);
B = (1,1);
C = (1,0);
D = (0,0);
E = (.5,1);
F = (.5,0);
M = (.5,1.5);

draw(A..M..B--C--D--cycle);
draw(A--B);
draw(M--E--F);

dot("A",A,W);
dot("M",M,N);
dot("B",B,E);
dot("C",C,E);
dot("D",D,W);
dot("E",E,NW);
dot("F",F,NW);

draw(M--C,linetype("0 4"));
draw((.5,.1)--(.6,.1)--(.6,0));
[/asy]

Since $F$ is the midpoint of $CD$ and $CD$ has length $6$, $FC$ has length $3$. $EF$ has length $6$, because it has the same length as the side length of the square. $ME$ is the radius of the semicircle. Since the diameter of the semicircle is $6$ (the same as the side length of the square), $ME$ has length $3$. Now, $MF = ME + EF = 3 + 6 = 9$. Finally, from the Pythagorean Theorem, we have that $MC^2 = MF^2 + FC^2 = 9^2 + 3^2 = 90$, so $MC = \sqrt{90} = \boxed{3\sqrt{10}}$ cm.
3\sqrt{10}
Level 5
Prealgebra
Problem: Square $ABCD$ is constructed along diameter $AB$ of a semicircle, as shown. The semicircle and square $ABCD$ are coplanar. Line segment $AB$ has a length of 6 centimeters. If point $M$ is the midpoint of arc $AB$, what is the length of segment $MC$? Express your answer in simplest radical form. [asy]
size(4cm);

dotfactor = 4;
defaultpen(linewidth(1)+fontsize(10pt));

pair A,B,C,D,M;
A = (0,1);
B = (1,1);
C = (1,0);
D = (0,0);
M = (.5,1.5);

draw(A..M..B--C--D--cycle);
draw(A--B);

dot("A",A,W);
dot("M",M,N);
dot("B",B,E);
dot("C",C,E);
dot("D",D,W);

draw(M--C,linetype("0 4"));

[/asy]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the length of segment MC, we first determine the coordinates of points A, B, C, D, and M. Given that AB is the diameter of the semicircle and has a length of 6 cm, we can place point A at (-3, 0) and point B at (3, 0). The center of the semicircle is at the origin (0, 0), and the radius is 3 cm.

The square ABCD is constructed such that points C and D are directly below points A and B, respectively. Since the square has a side length equal to the length of AB, which is 6 cm, the coordinates of points C and D are:
C = (3, -6) and D = (-3, -6).

Point M, being the midpoint of arc AB, lies directly above the center of the semicircle. The coordinates of M can be calculated as follows:
M = (0, 3), since it is 3 cm above the x-axis (the radius of the semicircle).

Next, we calculate the length of segment MC. The coordinates of M are (0, 3) and the coordinates of C are (3, -6). We use the distance formula to find the length of MC:
Distance MC = sqrt((x2 - x1)² + (y2 - y1)²)
= sqrt((3 - 0)² + (-6 - 3)²)
= sqrt(3² + (-9)²)
= sqrt(9 + 81)
= sqrt(90)
= sqrt(9 * 10)
= 3√10.

Thus, the length of segment MC is 3√10.<|END_OF_LATENT|> The final answer is $3\sqrt{10}$.