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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Answer: 11.6
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Explanation:
See the attached image for a visual reference
The distance from point D to E is 21 units. Point C is the midpoint, so CD is 10.5 units long (21/2 = 10.5)
We have a right triangle ACD. The legs are
AC = 5
CD = 10.5
The hypotenuse is
AD = x
Because AD is another radius of the same circle
Use the pythagorean theorem to find x
a^2 + b^2 = c^2
5^2 + 10.5^2 = x^2
25 + 110.25 = x^2
135.25 = x^2
x^2 = 135.25
x = sqrt(135.25)
x = 11.629703349613
which rounds to 11.6 when rounding to the nearest tenth (one decimal place)
--------------------------------------
Explanation:
See the attached image for a visual reference
The distance from point D to E is 21 units. Point C is the midpoint, so CD is 10.5 units long (21/2 = 10.5)
We have a right triangle ACD. The legs are
AC = 5
CD = 10.5
The hypotenuse is
AD = x
Because AD is another radius of the same circle
Use the pythagorean theorem to find x
a^2 + b^2 = c^2
5^2 + 10.5^2 = x^2
25 + 110.25 = x^2
135.25 = x^2
x^2 = 135.25
x = sqrt(135.25)
x = 11.629703349613
which rounds to 11.6 when rounding to the nearest tenth (one decimal place)