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Answer:
m∠EGF = 65° and m∠CGF = 115°
Step-by-step explanation:
Given;
∠EFG = 50°
EF = FG
Solution,
In ΔEFG m∠EFG = 50° and EF = FG.
Since triangle is an isosceles triangle hence their base angles are always equal.
∴
Let the measure of ∠EGF be x.
∴
Now by angle Sum property which states "The sum of all the angles of a triangle is 180°."
m∠EFG + m∠FEG + ∠EGF = 180
Hence
m∠EGF = 65°
Also 'The sum of angles that are formed on a straight line is equal to 180°."
m∠EGF + m∠CGF = 180°
65° + m∠CGF = 180°
m∠CGF = 180° - 65° = 115°
Hence m∠EGF = 65° m∠CGF = 115°
The <em>second order</em> polynomial that involves the variable <em>x</em> (border inside the rectangle) and associated to the <em>unshaded</em> area is x² - 62 · x + 232 = 0.
<h3>How to derive an expression for the area of an unshaded region of a rectangle</h3>
The area of a rectangle (<em>A</em>), in square inches, is equal to the product of its width (<em>w</em>), in inches, and its height (<em>h</em>), in inches. According to the figure, we have two <em>proportional</em> rectangles and we need to derive an expression that describes the value of the <em>unshaded</em> area.
If we know that <em>A =</em> 648 in², <em>w =</em> 22 - x and <em>h =</em> 40 - x, then the expression is derived below:
<em>A = w · h</em>
(22 - x) · (40 - x) = 648
40 · (22 - x) - x · (22 - x) = 648
880 - 40 · x - 22 · x + x² = 648
x² - 62 · x + 232 = 0
The <em>second order</em> polynomial that involves the variable <em>x</em> (border inside the rectangle) and associated to the <em>unshaded</em> area is x² - 62 · x + 232 = 0.
To learn more on polynomials, we kindly invite to check this verified question: brainly.com/question/11536910
