Drawing this square and then drawing in the four radii from the center of the cirble to each of the vertices of the square results in the construction of four triangular areas whose hypotenuse is 3 sqrt(2). Draw this to verify this statement. Note that the height of each such triangular area is (3 sqrt(2))/2.
So now we have the base and height of one of the triangular sections.
The area of a triangle is A = (1/2) (base) (height). Subst. the values discussed above, A = (1/2) (3 sqrt(2) ) (3/2) sqrt(2). Show that this boils down to A = 9/2.
You could also use the fact that the area of a square is (length of one side)^2, and then take (1/4) of this area to obtain the area of ONE triangular section. Doing the problem this way, we get (1/4) (3 sqrt(2) )^2. Thus,
A = (1/4) (9 * 2) = (9/2). Same answer as before.
Step-by-step explanation:
hope this helps you
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Answer: The distace between midpoints of AP and QB is 
.
Step-by-step explanation: Points P and Q are between points A and B and the segment AB measures a, then:
AP + PQ + QB = a
According to the question, AP = 2 PQ = 2QB, so:
PQ = 
QB =
Substituing:
AP + 2*() = a
2AP = a
AP = 
Since the distance is between midpoints of AP and QB:
2QB = AP
QB =
QB = 
QB = 
MIdpoint is the point that divides the segment in half, so:
<u>Midpoint of AP</u>:
<u>Midpoint of QB</u>:
The distance is:
d = 
d =
Answer:
2) 10/9 = 1.111111111
3) 1/2 = 0.5
4) 2
5) 3/7 = 0.428571429
6) 4/5 = 0.8
7) 1
8) 42/11 = 3.818181818
9) 5/3 = 1.666666667
10) 7/4 = 1.75
Step-by-step explanation:
Hope this helps
B. 4 1/4 <_ 2K because the point of the arrow is going to the less amount and when you ad the line under it, that makes it equal to or less than. In other words no more than.
