Answer:
4.6 in² (nearest tenth)
Step-by-step explanation:
To find the <u>area of the unshaded region</u>, subtract the <u>area of ΔAGB</u> from the <u>area of sector AGB</u>.
The measure of an arc is equal to its corresponding central angle measure. Therefore, the <u>central angle of sector</u> AGB is 90°.
As the two sides of ΔAGB adjacent the central angle are the radii of the circle they are therefore equal in length ⇒ ∠GAB = ∠GBA.
Therefore, ΔAGB is an isosceles triangle.
<u>Area of triangle (using the Sine Rule):</u>

(where a and b are the side lengths and C is the included angle)
Given:
- a = b = radius = 4 in
- C = 90°

<u>Area of a sector of a circle</u>


Substituting the given angle and radius:

<u>Area of the unshaded region:</u>
