Isosceles

Isosceles trapezium ABCD ABC = 12 angle ABC = 40 ° b=6. Calculate the circumference and area.

Result

o =  26.807
S =  28.554

Solution:

a=12 b=6 B=40 ∘  x=b⋅ cos⁡(B∘→ rad)=b⋅ cos⁡(B∘⋅ π180 )=6⋅ cos⁡(40∘⋅ 3.1415926180 )=4.59627 c=a−2⋅ x=12−2⋅ 4.5963≐2.8075  o=a+2⋅ b+c=12+2⋅ 6+2.8075≐26.8075≐26.807a=12 \ \\ b=6 \ \\ B=40 \ ^\circ \ \\ \ \\ x=b \cdot \ \cos ( B ^\circ \rightarrow\ \text{rad})=b \cdot \ \cos ( B ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ )=6 \cdot \ \cos ( 40 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ )=4.59627 \ \\ c=a - 2 \cdot \ x=12 - 2 \cdot \ 4.5963 \doteq 2.8075 \ \\ \ \\ o=a + 2 \cdot \ b + c=12 + 2 \cdot \ 6 + 2.8075 \doteq 26.8075 \doteq 26.807
h=b⋅ sin⁡(B∘→ rad)=b⋅ sin⁡(B∘⋅ π180 )=6⋅ sin⁡(40∘⋅ 3.1415926180 )=3.85673 S=a+c2⋅ h=12+2.80752⋅ 3.8567≐28.5542≐28.554h=b \cdot \ \sin( B ^\circ \rightarrow\ \text{rad})=b \cdot \ \sin( B ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ )=6 \cdot \ \sin( 40 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ )=3.85673 \ \\ S=\dfrac{ a+c }{ 2 } \cdot \ h=\dfrac{ 12+2.8075 }{ 2 } \cdot \ 3.8567 \doteq 28.5542 \doteq 28.554

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