If XY=32, XZ=28, JQ=12, and the radius of the circumscribed circle of (triangle)XYZ is 20, find QK.
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Answered by
1
Answer:
In a circle, the line joining the center of the circle to the mid-point of a chord is perpendicular to the chord.
Given that
D
,
E
,
F
are the midpoint of chords
A
B
,
A
C
,
and
B
C
, and
O
D
,
O
E
,
and
O
F
are perpendicular to chords
A
B
,
A
C
,
and
B
C
, respectively,
⇒
point
O
is the center of the circumscribed circle.
In other words, the three perpendicular bisectors of
A
B
,
A
C
,
and
B
C
meet at point O,
⇒
point
O
is the center of the circumscribed circle.
⇒
O
A
,
O
B
=
the radius of the circle
⇒
O
A
=
O
B
Given that
O
A
=
5
x
−
8
,
and
O
B
=
3
x
,
⇒
5
x
−
8
=
3
x
⇒
x
=
4
⇒
O
A
=
O
B
=
3
x
=
3
⋅
4
=
12
Hence, the radius of the circumcircle
=
12
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Answered by
3
Answer:
14.3
Step-by-step explanation:
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