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MaRussiya [10]
3 years ago
8

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

Mathematics
1 answer:
Zigmanuir [339]3 years ago
3 0

<em>Greetings from Brasil...</em>

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

<h3>QR = 2,66</h3>

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

<h3>PS = 4,14</h3>

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

for ΔQRS = A1:

A1 = √[P.(P - QR).(P - RS).(P - QS)]

<em>where P = (QR + RS + QS)/2</em>

<em>A1 = </em>√[P.(P - QR).(P - RS).(P - QS)]

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

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

<h3>Total Area = 20,33</h3>

<em>see more:</em>

brainly.com/question/17138076

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