Urgent will give 20 points to whoever solves this math problem

216.4 mm^2

Step-by-step explanation:

The polygon has 9 sides.

Divide the polygon into 9 congruent triangles. Each triangle has 2 sides of length 8.65 mm, so each triangle is isosceles. The measure of each internal angle of the polygon is (9 - 2)(180)/9 = 140 degrees. The base angles of an isosceles triangle measures 70 deg. The vertex angle measures 40 deg. Draw a segment from the center of the polygon to the midpoint of a side. This segment is the altitude of the triangle. Now the triangle has been split into two right triangles. The angles of the right triangle are 70, 90, and 20. 3.65 mm is the length of the hypotenuse. The length of the altitude is found with trig.

sin A = opp/hyp

sin 70 = h/8.65

h = 8.65 sin 70

h = 8.1283 mm

Now with the altitude, we can find the length of half of a side of the polygon.

a^2 + b^2 = c^2

x^2 + h^2 = 8.65^2

x^2 + 8.1283^2 = 8.65^2

x = 2.9585

Half a side measures 2.9585 mm.

The side of the polygon measures 5.9169 mm.

The area of the polygon is 9 times the area of one triangle.

area = 9 * base * height/2

area = 9 * 5.9169 mm * 8.1283 mm / 2

area of polygon = 216.4 mm^2

only 10/60 isn't equivalent

hope this : )