+0  
 
-2
73
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avatar+-291 

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.

 Mar 30, 2020
 #1
avatar+4569 
+1

Assign coordinates to each of the points.

 

A(0,0)

 

B(6,0)

 

M(3,3)

 

C(6,-6)

 

D(0,-6)

 

 

By the Distance Formula, \(\sqrt{(6-3)^2+(-6-3)^2}=\sqrt{3^2+9^2}=\sqrt{90}=3\sqrt{10}\)

.
 Mar 30, 2020
 #2
avatar+626 
0

Another Method:

 

Draw a line segment from M to the midpoint of DC. Call the midpoint of DC \(L\)

 

By simple operations:

\(ML=9\)

\(LC=3\)

 

By pythagorean theorem.

92+32

81 + 9 = 90

 

sqrt(90) = \(3\sqrt{10}\) = MC

 Mar 30, 2020

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