In the given fig. a circle is inscribed in a AABC. If AB=三9, BC = 10, CA = 11, PQ II AC, SRII AB & UT IIBC. PQ, RS, UT are tangent to incircle, then find the perimeter of PQRSTU.I
Answers
Answer:
Step-by-step explanation:
Answer:
296/15
Step-by-step explanation:
D , E & F are ping where incircle touches triangle ABC
AD = AE = x ( tangent)
CD = AC - AD = 11 - x
BE = AB - AE = 9 - x
BC = BF + CF = BE + CD
=> 10 = 9 - x + 11 - x
=> 2x = 10
=> x = 5
AD = AE = 5
BE = 9 - 5 = 4 = BF
CD = 11 - 5 = 6 = CF
RS ║ AB
=> ΔSRC ≈ ΔABC
=> SR/AB = SC/AC = RC/BC = Perimeter ΔSRC/Perimeter ΔABC
Perimeter ΔABC = 9 + 10 + 11 = 30
Perimeter ΔSRC = CS + SR + SC = CD - SD + SR + CF - RF
= CD + CF + SR - (SD + RF)
SD + RF = SR
= 6 + 6
= 12
SR/9 = SC/11 = RC/10 = 12/30 = 2/5
=> SR = 3.6 SC = 4.4 RC = 4
Similarly
TU/BC = AT/AC = AU/AB = 10/30 = 1/3
=> TU = 10/3 , AT = 11/3 , AU = 3
Similarly
PQ/AC = PB/AB = BQ/BC = 8/30 = 4/15
PQ = 44/15 PB = 2.4 BQ = 8/3
perimeter of PQRSTU = 2 * ( PQ + RS + TU)
= 2 * (44/15 + 3.6 + 10/3) = (2/15)(44 + 54 + 50) = 296/15
or ST = 11 - CS - AT
QR = 10 - CR - BQ
PU = 9 - AU - BP
and add PQ + QR + RS + ST + TU + PU
