1. To solve this we are going to use the formula for the area of a sector of a circle: [tex]A= \frac{1}{2} r^2 \alpha [/tex]
where
[tex]A[/tex] is the area of the sector
[tex]r[/tex] is the radius of the circle
[tex] \alpha [/tex] is the angle in radians
We know from our problem that the radius of the circle is 5 cm and the angle of the sector is [tex] \frac{ \pi }{4} [/tex], so [tex]r=[/tex] and [tex] \alpha = \frac{ \pi }{4} [/tex]. Lets replace those values in our formula:
[tex]A= \frac{1}{2} r^2 \alpha [/tex]
[tex]A= \frac{1}{2} (5)^2( \frac{ \pi }{4} )[/tex]
[tex]A= \frac{25 \pi }{8} [/tex]
[tex]A=9.8[/tex]
We can conclude that the area of sector GHJ in terms of pi is [tex] \frac{25 \pi }{8}[/tex] [tex]cm^2[/tex], and as a decimal rounded to the nearest tenth is 9.8 [tex]cm^2[/tex].
2. To construct the circle that circumscribes triangle DEF, we are going to draw the perpendicular bisectors of triangle DEF, and then we are going to draw the circle with radius at the interception point of the bisectors. Remember that the perpendicular bisector are the lines that passes trough the midpoint of the segment and are perpendicular to the segment.
