+0

pythagorean theorem

-2
73
3
+-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
+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, $$\sqrt{(6-3)^2+(-6-3)^2}=\sqrt{3^2+9^2}=\sqrt{90}=3\sqrt{10}$$

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Mar 30, 2020
#2
+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