The one line of symmetry is vertical, so we could fold the hexagon in half in such a way that the vertices A and B would meet at the same point, and the same goes for the pairs C,F and D,E. Because of this symmetry, we know angle AFE is congruent to BCD, and angle FED is congruent ot CDE.
Let <em>x</em> be the measure of angle CDE.
In any convex polygon with <em>n</em> sides, the interior angles sum to (<em>n</em> - 2)*180º in measure. ABCDEF is a hexagon, so <em>n</em> = 6.
We have 2 angles of measure 123º, 2 of measure <em>x</em>, and 2 of measure 2<em>x</em>. So
2(123º + <em>x</em> + 2<em>x</em> ) = (6 - 2)*180º
246º + 2<em>x</em> + 4<em>x</em> = 720º
6<em>x</em> = 474º
<em>x</em> = 79º
Angle AFE is congruent to angle BCD, which is twice the measure of CDE, so angle AFE has measure 2*79º = 158º.
Answer:
a) Circle 1: 3.14275
Circle 2: 3.1425
Circle 3: 3.143
b)
c) C= d •
Step-by-step explanation:
a) you just divide the circumference over the diameter
Hope this helps!
It is center point I think
Answer:
the graph is in the attachment.
the coordinates of the centroid : (2/3,2/3)
Step-by-step explanation:
- y=0 represents x-axis ( you can easily mark it on the graph)
- now draw x=1 line.( It is a line parallel to y axis and passing through the point (1,0) )
- y=2x is a line which passes through origin and has a slope "2"
by using these sketch the region.
I have uploaded the region bounded in the attachment. You may refer it. The region shaded with grey is the required region.
it can be easily identified that the formed region is a triangle
- the coordinates of three vertices of the triangle are
(1,2) , (0,0) , (1,0)
( See the graph. the three intersection points of the lines are the three vertices of the triangle)
- for general FORMULA, let the coordinates of three vertices of a triangle PQR be P(a,b) , Q(c,d) , R(e,f)
- then the coordinates of the centroid( let say , G) of the triangle is given by
G =
- therefore , the exact coordinates of the centroid =
this point is marked as G in the graph uploaded.