If G is the circumcenter of ABC, find each measure

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:

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)

AD = 1/2(30)

AD = 15

b. FC = √(GC² - GF²)

GC = GB = 22

GF = 9

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

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)

QR = 2(25)

QR = 50

b. RZ = √(XZ² + RX²)

XZ = 9

RX = 1/2(RS) = 1/2(80) = 40

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:

link

ad= 15 fc= 9 eb= 22 i dont know what ag is im sorry, i think eg is 9

Step-by-step explanation:

sorry if i'm wrong i haven't done this in 3 years but i think they are right

the quotient is 4x^2 - 27x + 167 with a remainder of -996

step-by-step explanation:

in synthetic division, use -6 as the divisor and use the coefficients {4, -3, 5, 6}:

-6     )   4     -3     5     6

                -24   162 -1002

          4   -27   167   -996

thus, the quotient is 4x^2 - 27x + 167 with a remainder of -996.

note:   your "4x°-3x+5x+6" is ambiguous / incomplete.   i had to assume that you meant '4x^3 - 3x^2 + 5x + 6."

a

step-by-step explanation: