Question

P, Q, and R are points on the sides BC, CA, and AB of Δ ABC respectively such that
AQ ≡ AR and BR ≡ BP . If ∠BCA =68o
, find the measure, in degrees of ∠QRP .

Answer 1

Answer:BP=AP, AQ=QC AND BR=CR.

=> P , Q & R are mid points of AB , AC & BC

=> PQ ║ BC   , PR ║ AC  & QR ║ AB

PR ║ AC => PR ║ AQ

QR ║ AB =>  QR ║ AP

also  AP/AB  = AQ/AC = PQ/BC

=> PQ = BC/2    

Simialrly QR = AB/2 = AP

& PR = AC/2 = AQ

PR ║ AQ & PR =  AQ    and   QR ║ AP  & QR = AP

=> PARQ is a parallelogram

Area of ΔAPQ = Area of ΔBPR = Area of ΔCRQ = Area of ΔABC / 4

=> 4 = Area of ΔABC / 4

=> Area of ΔABC  = 16 cm²

=> Area of Δ PQR = Area of ΔABC - 3 Area of ΔABC / 4

=> Area of Δ PQR = 16 - 12 = 4 cm²

Area of PAQR = Area of ΔAPQ  + Area of Δ PQR

= 4 + 4

= 8 cm²

Step-by-step explanation: