The bases

The bases of the isosceles trapezoid ABCD have lengths of 10 cm and 6 cm. Its arms form an angle α = 50˚ with a longer base. Calculate the circumference and content of the ABCD trapezoid.

Result

o =  22.223 cm
S =  19.068 cm2

Solution:

a=10 cm c=6 cm α=50   x=(ac)/2=(106)/2=2 cm  cosα=x:r  r=x/cos(α)=2/cos(50)=3.11145  o=a+c+2 r=10+6+2 3.111422.222922.223 cma=10 \ \text{cm} \ \\ c=6 \ \text{cm} \ \\ α=50 \ ^\circ \ \\ \ \\ x=(a-c)/2=(10-6)/2=2 \ \text{cm} \ \\ \ \\ \cos α=x:r \ \\ \ \\ r=x / \cos ( α )=2 / \cos ( 50^\circ )=3.11145 \ \\ \ \\ o=a+c+2 \cdot \ r=10+6+2 \cdot \ 3.1114 \doteq 22.2229 \doteq 22.223 \ \text{cm}
tanα=h:x  h=x tan(α)=2 tan(50)=2.38351  S=a+c2 h=10+62 2.383519.068119.068 cm2\tan α=h:x \ \\ \ \\ h=x \cdot \ \tan( α )=2 \cdot \ \tan( 50^\circ )=2.38351 \ \\ \ \\ S=\dfrac{ a+c }{ 2 } \cdot \ h=\dfrac{ 10+6 }{ 2 } \cdot \ 2.3835 \doteq 19.0681 \doteq 19.068 \ \text{cm}^2



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