Consider the figure below. Abraham cut regular pentagon ABCDE into quadrilateral BEGC. Which of the following statements is true
about quadrilateral BEGC? (picture included)
2 answers:
Let 'n' be the number of side of a regular polygon each angle are equal and sum of each angle equal to (n-2)×180. ∴For a regular pentagon have five equal side, n=5
∴ sum of each angle = (n-2)×180=3×180 = 540. and ecah angle = 540÷5=108°. Therefore ∠BCG = 108° and ∠ABC =108°
Here BE is one of the diagonal of regular pentagon ABCDE. BE divide ∠ABC in 1:2 ratio
Therefore ∠ABE : ∠CBE = 36: 72
∴ ∠GCB + ∠CBE = 108° +72°=180°
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Step-by-step explanation:
We have
Let's factor the denomiator first,
the denomaitor is a perfect square so we get
Now, we must think of two fractions that
We use a perfect square term for one fraction, then a linear one for the next, because if we set both of the denomiator to the same factor, we would get a inconsistent system.
So right now, we have
so that means that a is
So our equation is