1. If V is the circumcenter of ∆PQR, PR = 46, TV = 15, and VR = 25, find each measure. 2. If L is the incenter of ∆EFG, JL = 16, EH = 22, and LG = 34, find each measure. 3. If Z is the centroid of ∆RST, RZ = 42, ST = 74, TW = 51, ZY = 23 and find each measure.

1 If V is the circumcenter of PQR PR 46 TV 15 and VR 25 find each measure 2 If L is the incenter of EFG JL 16 EH 22 and LG 34 find each measure 3 If Z is the ce class=

Respuesta :

Answer:

1. a) SR = 23

b) QV = 25

c) QT = 20

d) PQ = 40

e) VS = 4·√6

2. a) LH = 16

b) EL = 2·√185

c) JG = 30

d) EK = 22

e) KG = 30

3. a) XT = 37

b) TZ = 34

c) ZW = 17

d) XZ = 21

e) SY = 69

Step-by-step explanation:

The circumcenter ΔPQR is the center of the circle that circumscribes ΔPQR

The length of the radius of the circle ≡ VR = VP = QV = 25

a) Given that VR ≅ VP - Radius of circumcircle

VS ≅ VS Reflective property

∠VPS ≅ ∠VRS - Base angles of an isosceles triangle

Right triangle VPS ≅ Right triangle VRS -Hypotenuse and one Leg HL congruency

Therefore, SR ≅ PS -Corresponding parts of congruent triangles are congruent CPCTC

SR + PS = PR = 46

SR + PS = SR + SR = 2·SR = 46

∴ SR = 46/2 = 23

b) QV = VR = 25 = Radius of circumcircle of ΔPQR -Given V = center and Q = vertices of the triangle circumscribed by the circle referred to in the question

c) QT = √(QV² - TV²) = √(25² - 15²) = √400 = 20

d) TV ≅ TV - Reflexive property of congruency

ΔTQV ≅ ΔTVP - Hypotenuse and one Leg (HL) congruency

QT ≅ TP -Corresponding parts of congruent triangles are congruent CPCTC

PQ = QT + TP Given

∴ PQ = QT + QT since QT = TP

PQ = 2·QT = 2 × 20 = 40

e) VS = √(VR² - SR²) = √(25² - 23²) = √96 = 4·√6

2.  The incenter is the center of the incircle of ΔEFG

  a) LH = LK = JL = 16 -Radius of incircle of ΔEFG

b) EL = Hypotenuse of right triangle LHE = √(LH² + EH²) = √(16² + 22²) = √740 = 2·√185

c) JG = Leg length of right triangle JGL = √(LG² - JL²) = √(34² - 16²) = √900 = 30

d) EK = Leg length of right triangle LKE = √(EL² - LK²) = √(740 - 256) = 22

e) KG = Leg length of right triangle LKG = √(LG² - LK²) = √(34²- 16²) = √900 = 30

3. Point Z id the centroid of ΔRST

a) XT = XS - point X on ST bisected by median line RX

ST = XT + XS = XT + XT = 2.XT = 74

XT = 74/2 = 37

b) TZ = 2/3×TW  - Length from a vertex to the centroid on a median line is equal to two third the length of the median line

TZ = 2/3×51 = 34

c) TZ + ZW = TW

∴ ZW = TW - TZ = 51 - 34 = 17

d) RZ = 42 = 2/3×RX - Length from a vertex to the centroid on a median line is equal to two third the length of the median line

∴ RX = 3/2×42 = 63

RZ + XZ = RX - Given

XZ = RX - RZ = 63 - 42 = 21

e) SZ = 2/3×SY - Length from a vertex to the centroid on a median line is equal to two third the length of the median line

SZ + ZY = SY

∴ ZY = SY - SZ = SY - 2/3×SY = 1/3×SY = 23

Which gives;

SY = 3 × 23 = 69.