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Answer :
We begin by noting that the arc [tex]\( \widehat{AB} \)[/tex] covers [tex]\(\frac{1}{4}\)[/tex] of the entire circumference. Since a full circle measures [tex]\(360^\circ\)[/tex], the central angle [tex]\( \theta \)[/tex] for the sector [tex]\( AOB \)[/tex] is
[tex]$$
\theta = \frac{1}{4} \times 360^\circ = 90^\circ.
$$[/tex]
The area [tex]\( A \)[/tex] of a sector with central angle [tex]\( \theta \)[/tex] in a circle of radius [tex]\( r \)[/tex] is given by
[tex]$$
A = \frac{\theta}{360^\circ} \times \pi r^2.
$$[/tex]
Given that [tex]\( r = 5 \)[/tex] and using [tex]\( \pi = 3.14 \)[/tex], we calculate the area of the full circle first:
[tex]$$
\text{Area of circle} = \pi r^2 = 3.14 \times 5^2 = 3.14 \times 25 = 78.5 \text{ square units}.
$$[/tex]
Now, since the sector represents [tex]\(\frac{1}{4}\)[/tex] of the circle, its area is
[tex]$$
\text{Area of sector } AOB = \frac{90^\circ}{360^\circ} \times 78.5 = \frac{1}{4} \times 78.5 = 19.625 \text{ square units}.
$$[/tex]
Rounding to the closest answer, we have
[tex]$$
19.625 \approx 19.6 \text{ square units}.
$$[/tex]
Thus, the area of sector [tex]\( AOB \)[/tex] is approximately [tex]\(\boxed{19.6}\)[/tex] square units.
[tex]$$
\theta = \frac{1}{4} \times 360^\circ = 90^\circ.
$$[/tex]
The area [tex]\( A \)[/tex] of a sector with central angle [tex]\( \theta \)[/tex] in a circle of radius [tex]\( r \)[/tex] is given by
[tex]$$
A = \frac{\theta}{360^\circ} \times \pi r^2.
$$[/tex]
Given that [tex]\( r = 5 \)[/tex] and using [tex]\( \pi = 3.14 \)[/tex], we calculate the area of the full circle first:
[tex]$$
\text{Area of circle} = \pi r^2 = 3.14 \times 5^2 = 3.14 \times 25 = 78.5 \text{ square units}.
$$[/tex]
Now, since the sector represents [tex]\(\frac{1}{4}\)[/tex] of the circle, its area is
[tex]$$
\text{Area of sector } AOB = \frac{90^\circ}{360^\circ} \times 78.5 = \frac{1}{4} \times 78.5 = 19.625 \text{ square units}.
$$[/tex]
Rounding to the closest answer, we have
[tex]$$
19.625 \approx 19.6 \text{ square units}.
$$[/tex]
Thus, the area of sector [tex]\( AOB \)[/tex] is approximately [tex]\(\boxed{19.6}\)[/tex] square units.
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