A box with 3 metal balls weighs 1840 kg.
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to get the equation of any straight line, we simply need two points off of it, let's use the points in the picture below.
<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>
