Find the area of equilateral triangle with radius length of 8. PLEASE answer in simplified radical form

16√3 units²

Step-by-step explanation:

I believe you meant, "with side length of 8."

Such a triangle has three equal angles, all 60 degrees.  Using this fact we can find the height of the triangle when the bottom side is horizontal:

sin Ф = (opposite side) / (hypotenuse), which here is

sin Ф = (height of triangle) / (8 units), from which we learn that

(height of triangle) = (8 units)(sin 60 degrees) = 8(√3)/2 = 4√3

Now we have all the information necessary to find the area of this triangle A = (1/2)(base)(height)  =  (1/2)(8 units)(4√3 units)  = 16√3 units²

b. t=-25h+3

step-by-step explanation:

x = central angle apc = measure of minor arc ac

the two angles x and 30 add to 180, so x+30 = 180 and x = 180-30 = 150. the measure of minor arc ac is 150 degrees.

what we need to do from here is find the circumference and then multiply it by x/360 since there are 360 degrees in a circle. in this case, x = 150, so x/360 = 150/360 = 15/36 = 5/12. the shortest distance along the curve from a to c is 5/12 of the distance around the entire circle (aka circumference)

c = circumference

c = pi*d

c = pi*24

c = 24pi

now multiply this by 5/12 to get

distance along minor arc ac = (5/12)*(circumference)

distance along minor arc ac = (5/12)*(24pi)

distance along minor arc ac = 10pi

the formula you can use is

l = (x/360)*pi*d

with

l = length of minor arc

x = central angle

d = diameter

which essentially is the steps we took though a bit differently.