Applying the properties of circumcenter, incenter, and centroid of a triangle, the missing measures are:
1. a. AD = 15
b. FC = 20.1
c. EB = 18
d. AG = 22
e. EG = 12.6
2. a. QR = 50
b. RZ = 41
c. XS = 40
d. ZS = 41
e. WZ = 32.5
3. a. m∠CML = 20°
b. m∠MNP = 60°
c. m∠NPC = 40°
d. JC = 4
e. MC = 11.7
4. a. VT = 24
b. YW = 10
c. SX = 27.2
d. YX = 10
e. SV = 27.2
5. a. GF = 13
b. FA = 39
c. FC = 22
d. GB = 24
e. DB = 36
6. a. LQ = 48
b. QN = 24
c. QP = 31
d. JQ = 62
e. QK = 52
Recall:
The circumcenter is the point where the three perpendicular bisectors of the sides of a triangle meet.The incenter of a triangle is the point where the three angle bisectors of a triangle's vertices meet.The centroid of a triangle is the point where the three medians that connect the vertices to the midpoint of the opposite sides meet.
1. G is given as the circumcenter of ΔABC, to find each measure, we will apply the properties of the circumcenter of a triangle.
AG = GC = GB (equidistance)
AE = EB, AD = DC, and FC = FB based on the circumcenter property
a. AD = 1/2(AC)
Substitute
AD = 1/2(30)
AD = 15
b. FC = √(GC² - GF²)
GC = GB = 22
GF = 9
Substitute
FC = √(GC² - GF²)
FC = √(22² - 9²)
FC = 20.1
c. EB = AE = 18
EB = 18
d. AG = GB = 22
AG = 22
e. EG = √(GB² - EB²)
GB = 22
EB = 18
Substitute
EG = √(22² - 18²)
EG = 12.6
2. Z is given as the circumcenter of ΔQRS, to find each measure, we will apply the properties of the circumcenter of a triangle.
RZ = ZS = QZ (equidistance)
RX = XS, QY = YS, and RW = WQ based on the circumcenter property.
a. QR = 2(WQ)
Substitute
QR = 2(25)
QR = 50
b. RZ = √(XZ² + RX²)
XZ = 9
RX = 1/2(RS) = 1/2(80) = 40
Substitute
RZ = √(9² + 40²)
RZ = 41
c. XS = RS = 40
XS = 40
d. ZS = RZ = 41
ZS = 41
e. WZ = √(RZ² - RW²)
RZ = 41
RW = WQ = 25
Substitute
WZ = √(41² - 25²)
WZ = 32.5
3. C is given as the incenter of ΔMNP, to find each measure, we will apply the properties of the incenter of a triangle.
a. m∠CML = m∠CMJ = 20°
m∠CML = 20°
b. m∠MNP = 2(m∠KNC)
Substitute
m∠MNP = 2(30°)
m∠MNP = 60°
c. m∠NPC = 1/2(m∠NPM)
m∠NPC = 1/2(180 - (40+60))
m∠NPC = 40°
d. JC = KC = 4
JC = 4
e. MC = √(CL² + ML²)
CL = KC = 4
ML = 11
Substitute
MC = √(4² + 11²)
MC = 11.7
4. Y is given as the incenter of ΔSTU, to find each measure, we will apply the properties of the incenter of a triangle.
a. VT = WT = 24
VT = 24
b. YW = √(26² - 24²)
YW = 10
c. SX = √(29² - 10²)
SX = 27.2
d. YX = YW = 1-
YX = 10
e. SV = SX = 27.2
SV = 27.2
5. Given: AG = 26, BC = 44, DG = 12
AG = 1/3(FA)
26 = 2/3(FA)
(26)(3) = 2(FA)
2FA = 78
FA = 39
DG = 1/3(DB)
12 = 1/3(DB)
DB = 36
a. GF = 1/3(FA)
GF = 1/3(39)
GF = 13
b. FA = GF + AG
FA = 13 + 26
FA = 39
c. FC = 1/2(BC)
FC = 1/2(44)
FC = 22
d. GB = 2/3(DB)
GB = 2/3(36)
GB = 24
e. DB = 36
6. Given: LN = 72, JP = 93, MK = 78
a. LQ = 2/3(LN)
LQ = 2/3(72)
LQ = 48
b. QN = LN - LQ
QN = 74 - 48
QN = 24
c. QP = 1/3(JP)
QP = 1/3(93)
QP = 31
d. JQ = JP - QP
JQ = 93 - 31
JQ = 62
e. QK = 2/3(MK)
QK = 2/3(78)
QK = 52
Learn more about circumcenter, incenter, and centroid on:
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