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