Question

Consider the figure below. Abraham cut regular pentagon ABCDE into quadrilateral BEGC. Which of the following statements is true about quadrilateral BEGC? (picture included)

Answer 1

Answer:

Correct: 2, 3, 4, 6

Step-by-step explanation:

ABCDE is a regular pentagon. The sum of the measures of all interior angles in the pentagon is

$(n-2)\cdot 180^{\circ}=(5-2)\cdot 180^{\circ}=540^{\circ},$

then the measure of each interior angles in the regular pentegon is

$\dfrac{540^{\circ}}{5}=108^{\circ}$

Triangle ABe is isosceles triangle, so the adjacent to the bas BE angle has the measure

$m\angle ABE=\dfrac{180^{\circ}-108^{\circ}}{2}=36^{\circ}$

Therefore,

$m\angle CBE=m\angle ABC-m\angle ABE=108^{\circ}-36^{\circ}=72^{\circ}$

(option 2 is correct)

Since $m\angle BCG=108^{\circ},$ then

$m\angle EGC=108^{\circ }+4x$

The sum of the measures of all interior angles in quadrilateral BCGE is

$(n-2)\cdot 180^{\circ}=(4-2)\cdot 180^{\circ}=360^{\circ},$

(option 6 is correct), then

$m\angle CBE+m\angle BEG+m\angle EGC+m\angle BCG=360^{\circ}\\ \\72^{\circ}+19x+3^{\circ}+108^{\circ}+4x+108^{\circ}=360^{\circ}\\ \\23x=69^{\circ}\\ \\x=3$

(option 1 is false)

$m\angle EGC=108^{\circ}+4\cdot 3^{\circ}=120^{\circ}$

(option 3 is correct)

$m\angle GCB+m\angle EGC=108^{\circ}+72^{\circ}=180^{\circ}$

(option 4 is correct)

$m\angle BEC+m\angle EGC=(19\cdot 3+3+108+4\cdot 3)^{\circ}=180^{\circ}$

(option 5 is false)