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
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.