Problem 1 :
Find x.
Problem 2 :
Find x.
Problem 3 :
Find x.
Problem 4 :
If MP = 5x - 34, PN = 2x - 4, find MP.
Problem 5 :
If QM = 6x - 11 and MR = 2x + 9, find MN.
Problem 6 :
In circle Z, if RS = 18, and m∠arcTY = 42°, find the following measures.
(i) TU, (ii) TU, (iii) WS, (iv) m∠arc YU, (v) m∠arc RS
Problem 7 :
Use the circle below to find :
(i) VW and (ii) m∠arc YW
Problem 8 :
In the circle below HK = 30 and PM = 8. Find the following measures :
(i) PH and (ii) m∠arc GJ
1. Answer :
In the diagram above, the two chords ST and RS are congruent.
Then,
ST = RS
7x + 24 = 115
Subtract 24 from each side.
7x = 91
Divide each side by 7.
x = 13
2. Answer :
In the diagram above, the two chords VW and XY are congruent.
Then,
WV = XY
9x - 34 = 4x + 1
Subtract 4x from each side.
5x - 34 = 1
Add 34 to each side.
5x = 35
Divide each side by 5.
x = 7
3. Answer :
In the diagram above,
m∠arc YZ = (6x - 20)°
And also, the two chords YZ and XY are congruent.
Then,
m∠arc XY = m∠arc YZ
m∠arc XY = (6x - 20)°
In the circle above,
m∠arc XY + m∠arc YZ + m∠arc ZX = 360°
(6x - 20)° + (6x - 20)° + 76° = 360°
6x - 20 + 6x - 20 + 76 = 360
Combine the like terms.
12x + 36 = 360
Subtract 36 from each side.
12x = 324
Divide each side by 12.
x = 27
4. Answer :
In the diagram above,
m∠arc JK = m∠arc KL
Then, the two chords JK and KL are congruent.
Because the two chords JK and KL are congruent, they are equidistant from the center.
Then,
MP = PN
5x - 34 = 2x - 4
Subtract 2x from each side.
3x - 34 = -4
Add 34 to each side.
3x = 30
Divide each side by 3.
x = 10
Find MP :
MP = 5x - 34
Substitute x = 10.
MP = 5(10) - 34
MP = 50 - 34
MP = 16
5. Answer :
In the diagram above, the two chords LM and MN are equidistant from the center.
Then, the two chords LM and MN are congruent.
The radii JP and KP are perpendicular to the chords LM and MN respectively.
Then, they bisect the chords.
Because the chords LM and MN are congruent and they are bisected by the radii,
QM = MR
6x - 11 = 2x + 9
Subtract 2x from each side.
4x - 11 = 9
Add 11 to each side.
4x = 20
Divide each side by 4.
x = 5
Find MN :
MN = 2(MR)
MN = 2(2x + 9)
MN = 4x + 18
Substitute x = 5.
MN = 4(5) + 18
MN = 20 + 18
MN = 38
6. Answer :
(i) TU :
Because the two chords RS and TU are equidistant from the center, they are congruent.
TU = RS
TU = 18
(ii) TV :
Because the radius YZ is perpendicular to the chord TU, the radius YZ bisects the chord TU.
Then,
TV = TU/2
TV = 18/2
TV = 9
(iii) WS :
Because the radius XZ is perpendicular to the chord RS, the radius XZ bisects the chord RS.
Then,
WS = RS/2
WS = 18/2
WS = 9
(iv) m∠arc YU :
m∠arc YU = m∠arc TY
m∠arc YU = 42°
(iv) m∠arc RS :
m∠arc TU = m∠arc TY + m∠arc YU
m∠arc TU = 42° + 42°
m∠arc TU = 84°
Because the two chords RS and TU are equidistant from the center, they are congruent.
Then,
m∠arc RS = m∠arc TU
m∠arc RS = 84°
7. Answer :
(i) VW :
Using the Pythagorean theorem in right triangle WXZ,
WZ2 + XZ2 = WX2
WZ2 + 52 = 132
WZ2 + 25 = 169
Subtract 25 from each side.
WZ2 = 144
WZ2 = 122
WZ = 12
Because the radius XY is perpendicular to the chord VW, the radius XY bisects the chord VW.
Then,
VW = 2(WZ)
VW = 2(12)
VW = 2
(ii) m∠arc YW :
In right triangle WXZ,
cos X = 5/13
m∠ X = cos-1(5/13)
m∠ X = 67.4°
Then,
m∠arc YW = 67.4°
8. Answer :
(i) PH :
In the circle above,
PN = PM
PN = 8
Because the radius PJ is perpendicular to the chord HK, the radius PJ bisects the chord HK.
Then,
HN = HK/2
HN = 30/2
HN = 15
Using the Pythagorean theorem in right triangle HNP,
HN2 + PN2 = PH2
152 + 82 = PH2
225 + 64 = PH2
289 = PH2
172 = PH2
17 = PH
(ii) m∠arc GJ :
In right triangle HNP,
tan P = 15/8
m∠ P = tan-1(15/8)
m∠ X = 61.9°
Then,
m∠arc GJ = 2 ⋅ m∠ X
m∠arc GJ = 2 ⋅ 61.9°
m∠arc GJ = 123.8°
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