Quadrilateral oblique prism

What is the volume of a quadrilateral oblique prism with base edges of length a = 1m, b = 1.1m, c = 1.2m, d = 0.7m if a side edge of length h = 3.9m has a deviation from the base of 20° 35' and the edges a, b form an angle of 50.5°?

Correct answer:

V =  1.0113 m³

Step-by-step explanation:

$$\begin{gathered} a=1\text{m} \\ b=1.1\text{m} \\ c=1.2\text{m} \\ d=0.7\text{m} \\ h=3.9\text{m} \\ \phi =20+35/60=\frac{247}{12}\doteq 20.5833{}^{\circ } \\ \beta =50.5{}^{\circ } \\ {u}^2=a^2+b^2-2\cdot a\cdot b\cdot \cos \beta \\ u=\sqrt{a^2+b^2-2\cdot a\cdot b\cdot \cos \beta }=\sqrt{a^2+b^2-2\cdot a\cdot b\cdot \cos 50.5°}=\sqrt{1^2+1.1^2-2\cdot 1\cdot 1.1\cdot \cos 50.5°}=\sqrt{1^2+1.1^2-2\cdot 1\cdot 1.1\cdot 0.636078}=0.90035\text{m} \\ s=(a+b+u)/2=(1+1.1+0.9003)/2\doteq 1.5002\text{m} \\ {S}_1=\sqrt{s\cdot (s-a)\cdot (s-b)\cdot (s-u)}=\sqrt{1.5002\cdot (1.5002-1)\cdot (1.5002-1.1)\cdot (1.5002-0.9003)}\doteq 0.4244\text{m} \\ {s}_2=(c+d+u)/2=(1.2+0.7+0.9003)/2\doteq 1.4002\text{m} \\ {S}_2=\sqrt{s_2\cdot (s_2-c)\cdot (s_2-d)\cdot (s_2-u)}=\sqrt{1.4002\cdot (1.4002-1.2)\cdot (1.4002-0.7)\cdot (1.4002-0.9003)}\doteq 0.3132\text{m} \\ {h}_2=h\cdot \sin \phi =h\cdot \sin 20.583333333333°=3.9\cdot \sin 20.583333333333°=3.9\cdot 0.351569=1.37112\text{m} \\ S=S_1+S_2=0.4244+0.3132\doteq 0.7376\text{m} \\ V=S\cdot h_2=0.7376\cdot 1.3711=1.0113{\text{m}}^3 \end{gathered}$$

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