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°
You might be interested in
Answer:
1st option
Step-by-step explanation:
let y = f(x) and rearrange making x the subject
y = 2x + 1 ( subtract 1 from both sides )
y - 1 = 2x ( divide both sides by 2 )
= x
Change y back into terms of x with x being the inverse h(x)
h(x) = =
x -