Calculate:
a) QR
b) PS
c) The area of quadrilateral PQRS

Greetings from Brasil...

For QR: we need to use the Sine Law in Any Triangle....

QS/SEN 72 = QR/SEN 25

6/SEN 72 = QR/SEN 25

6,3 = QR/SEN 25

For PS: we need to use the Cosine Law in Any Triangle....

PS² = PQ² + QS² - 2.PQ.QS.COS Q

PS² = 7,4² + 6² - 2.(7,4).6.COS 34

PS² = 90,76 - 73,61

PS = √17,14

For area we use Heron's Formula 2x.....

for ΔQRS = A1:

A1 = √[P.(P - QR).(P - RS).(P - QS)]

where P = (QR + RS + QS)/2

A1 = √[P.(P - QR).(P - RS).(P - QS)]

     RS = 6,26 (using RS/SEN 97 = QS/SEN 72)

P = (2,66 + 6,26 + 6)/2

P = 14,92/2 ⇒ P = 7,46

A1 = √[7,46.(7,46 - 2,66).(7,46 - 6,26).(7,46 - 6)]

A1 = 7,92

for ΔPQS = A2:

A2 = √[P.(P - PQ).(P - PS).(P - QS)]

P = (7,4 + 4,14 + 6)/2 = 8,77

A2 = √[8,77.(8,77 - 7,4).(8,77 - 4,14).(8,77 - 6)]

A2 = 12,41

Total Area = A1 + A2

Total Area = 7,92 + 12,41

see more:

link

{3 + i, 3 - i, 3, -1}

step-by-step explanation: