X=
area of an isosceles triangle=
area of polygon=

Area of the triangle =
We solve for half of one side
half-side^2 = 18.4^2  -17^2
half-side^2 = 338.56 -289
half-side^2 = 49.56
half-side = 7.0398863627
area of polygon = 17 * 7.0398863627 * 8 =
957.4245453298
(This matches the double-check below).


Double-Check
n = number of sides
polygon area = apothem^2 * n * tan (180/n)
polygon area = 17*17 * 8 * tan (22.5)
polygon area = 17*17 * 8 * 0.41421
polygon area = 957.65352

Source:
http://www.1728.org/polygon.htm

The measure of the angle JKL = 120°Step-by-step explanation:

Given that : A circle with minor arc JL and measure of that minor arc JL is given to be 60°

We need to find the measure of the angle JKL

Solution : By using the theorem that the sum of minor arc and the corresponding angle formed by the arc is supplementary

Also, Take circle to be centred at O

⇒ m∠JOL + m∠JKL = 180°

Now, m∠ JOL = measure of minor arc JL = 60°

⇒ m∠JKL = 180° - m∠JOL

⇒ m∠JKL = 180° - 60°

⇒ m∠JKL = 120°

Thus, The measure of the angle JKL = 120°