arcs and chords. a chord is a segment with endpoints on a circle. if a chord is not a diameter, then...
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SECTION 10.3Arcs and Chords
A chord is a segment with endpoints on a circle. If a chord is not a diameter, then its endpoints divide the circle into a major and minor arc.
Example 1:
a) A circular piece of jade is hung from a chain by two wires around the stone. and 90 . Find .JM KL mKL mJM
and are congurent chords, so the
corresponding arcs and are congruent.
90
JM KL
JM KL
mKL mJM
Example 1:
b) has congruent chords and . If 85 . Find .W RS TV mRS mTV
and are congurent chords, so the
corresponding arcs and are congruent.
85
RS TV
RS TV
mRS mTV
Example 2: a) In the figure and . Find .A B WX YZ WX
and are congruent arcs in congruent circles, so the
corresponding chords and are congruent.
WX YZ
WX YZ
WX = YZ Definition of congruent segments
7x – 2 = 5x + 6 Substitution
7x = 5x + 8 Add 2 to each side.
2x = 8 Subtract 5x from each side.
x = 4 Divide each side by 2.
So, WX = 7x – 2 = 7(4) – 2 or 26.
Example 2: b) In the figure and . Find .G H RT LM LM
and are congruent arcs in congruent circles, so the
corresponding chords and are congruent.
RT LM
RT LM
RT = LM Definition of congruent segments
3x – 5 = 2x + 1 Substitution
3x = 2x + 6 Add 5 to each side.
x = 6 Subtract 2x from each side.
So, LM = 2x + 1 = 2(6) + 1 or 13.
Example 3:
a) In , 150 . Find .G mDEF mDE
Radius is perpendicular to chord . So by Theorem 10.3,
bisects . Therefore . By substitution,
150 or 75 .
2
EG DF
EG DEF mDE mEF
mDE
Example 3:
b) In , 60 . Find .Z mWUX mUX
Radius is perpendicular to chord . So by Theorem 10.3,
bisects . Therefore . By substitution,
160 or 80 .
2
UZ WX
UZ WUX mWU mUX
mUX
Example 4: a) In the ceramic stepping stone below, diameter AB is 18 inches long and chord EF is 8 inches long. Find CD.
Draw radius CE.
This forms right ΔCDE.
Find CE and DE.Since AB = 18 inches, CB = 9 inches. All radii of a circle arecongruent, so CE = 9 inches.
Since diameter AB is perpendicular to EF, AB bisects chord EF by Theorem 10.3. So, DE = 0.5(8) or 4 inches.
Use the Pythagorean Theorem to find CD.
CD2 + DE2 = CE2 Pythagorean TheoremCD2 + 42 = 92 SubstitutionCD2 + 16 = 81 Simplify.CD2 = 65 Subtract 16 from each side.
Take the positive square root.65CD
Example 4: b) In the circle below, diameter QS is 14 inches long and chord RT is 10 inches long. Find VU.
Draw radius VR.
This forms right ΔVRU.
Find VR and UR.Since QS = 14 inches, VS = 7 inches. All radii of a circle arecongruent, so VR = 7 inches.
Since diameter QS is perpendicular to RT, QS bisects chord RT by Theorem 10.3. So, UR = 0.5(10) or 5 inches.
Use the Pythagorean Theorem to find VU.
VU2 + UR2 = VR2 Pythagorean TheoremVU2 + 52 = 72 SubstitutionVU2 + 25 = 49 Simplify.VU2 = 24 Subtract 25 from each side.
Take the positive square root.24VU
Example 5:
a) In , . Find .P EF GH PQ
Since chords EF and GH are congruent, they are equidistant from
P. So, PQ = PR.
PQ = PR
4x – 3 = 2x + 3 Substitution
x = 3 Simplify.
So, PQ = 4(3) – 3 or 9
Example 5:
b) In , 29. Find .R MN PO RS
RS = 15