The diagnonals of a parallelogram bisect each other ,
so
=》 BD = 3x × 2
=》BD = 6x
and there's given that
=》

=》

=》

=》

so, if x = 4 , then the given quadrilateral is a parallelogram.
I believe its the first one because you can divide the H and pi which would be V/h×pi then you square root r2 and v/h×pi so r= square root of V/H×pi
Answer:
P2 affirms P1 and the conclusion is in the same direction.
P1--->P2--->C
This argument is valid.
Step-by-step explanation: using the syllogism rules.
Premises 1 (P1) = Some foreign emissaries are persons without diplomatic immunity,
Premises 2 (P2) = so some persons invulnerable to arrest and prosecution are foreign emissaries
Conclusion (C) = because no persons with diplomatic immunity are persons vulnerable to arrest and prosecution.
From the argument:
P1 uses "some", that means it's not "all" foreign emissaries person that does not have diplomatic immunity. This means that some other foreign emissaries have diplomatic immunity
P2 uses "some", that means it's affirms to that part of P1 which states that some foreign emissaries have diplomatic immunity.
The conclusion is valid because the part of P2 which states that some foreign emissaries are vulnerable to arrest, which affirms with P1 which states that Some foreign emissaries are persons without diplomatic immunity. That means no persons with diplomatic immunity are persons vulnerable to arrest and prosecution. This conclusion literally means that if you don't have diplomatic immunity, you are vulnerable to arrest and prosecution.
Therefore;
P2 affirms P1 and the conclusion is in the same direction.
P1--->P2--->C
This argument is valid.
Given
centre of circle O(3,2)=O(x0,y0)
point on circle P(6,-2)
Standard equation of circle:
(x-x0)^2+(y-y0)^2=r^2
r=radius of circle
= (distance OP)
= sqrt((6-3)^2+(-2-2)^2)
=sqrt(3^2+(-4)^2)
=sqrt(25)
=>
r^2=(sqrt(25))^2=25
Equation of circle
(x-x0)^2+(y-y0)^2=r^2
(x-3)^2+(y-2)^2=25 ............... standard equation of circle
The area of the entire sector is 
The area of the triangle OAB is
.
So, the area of the segment is 