In circle L, points E and F lie on the circle such that E, F and L are not collinear. If LE, LF and EF are drawn then which of the following must be true about LEF?
(1) It is a right triangle.
(2) it is and isosceles triangle.
(3) it is an equilateral triangle.
(4) it is an obtuse triangle.

option (2) it is and isosceles triangle.

Step-by-step explanation:

In circle L, points E and F lie on the circle such that E, F and L are not collinear

If LE, LF and EF are drawn

So, LE = EF = the radius of the circle L

So, LEF is an isosceles triangle.

The answer is option (2) it is and isosceles triangle.

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.

51840000 cm 2

step-by-step explanation: