Answer:
Given: A parallelogram EFGH in which diagonals intersect at point J.
To Find: A segment congruent to EJ.
Solution: In parallelogram EFGH
Point of intersection of Diagonals EG and FH are point J.
As we know diagonals of parallelogram bisect each other.
So, EJ=JG and FJ=JH
∵ Solution is EJ≅ JG
now, bear in mind that all these ones are lines, and to graph a line all you need is two points, so let's pick a couple of random values for say "x" and let's see what we get for "y" and that's our x,y point.
3)
check the red line in the picture below.
4)
check the blue line in the picture below.
Answer:
C 10 in
Step-by-step explanation:
side of P is = 6
side of Q is = 8
for missing side use Pythagorean theorem
=
r^{2} = 36 + 64
r^{2} = 100
r = 10 in