Answer:
<edf=23(angles standing on common arc fe)
<span>Let A be the center of a circle and two angles at the adjacent center AOB and BOC. Knowing the measure of the angle AOB = 120 and the measure BOC = 150, find the measures of the angles of the ABC triangle.
</span>solution
Given the above information;
AC=AB, therefore ABC is an isosceles triangle.
therefore, BAO=ABO=(180-120)/2=30
OAC=OCA=(180-90)/2=45
OBC=BCO=(180-150)/2=15
THUS;
BAC=BAO+OAC=45+30=75
ABC=OBA+CBO=15+30=45
ACB=ACO+BCO=15+45=60
(0,0)
(1,3)
(2,4)
(3,3)
(4,0)
Vertex is (2,4)
Answer:
0
Step-by-step explanation:
Let's define 3 areas:
- S = area of semicircle with radius 6 in (diameter AB)
- T = area of quarter circle with radius 6√2 in (radius AC)
- U = area of triangle ABC (side lengths 6√2)
The white space between the "moon" and the triangle has area ...
white = T - U
Then the area of the "moon" shape is ...
moon = S -white = S -(T -U) = S -T +U
The area we're asked to find is ...
moon - triangle = (S -T +U) -U = S -T
__
The formula for the area of a circle of radius r is ...
A = πr²
So, ...
S = (1/2)π(6 in)² = 18π in²
and
T = (1/4)π(6√2 in)² = 18π in²
The difference in areas is S -T = (18π in²) -(18π in²) = 0.
There is no difference between the areas of the "moon" and the triangle.
