Line segments ab and cd intersect each other at point o. prove that the sum of the distances between any point on the plane to points a, b, c, and d is not lesser than oa+ob+oc+od.

Solution: Two line segments AB and CD of length m and n intersect each other at point O.

Considering it is a two dimensional plane and O being the origin, and AB being X axis and CD being Y axis.

Coordinates of A=(m/2,0), Coordinate of B=(-m/2,0)

Coordinate of C=(0,n/2), Coordinate of D=(0,-n/2)

Let P(m/2, n/2) be any point in that plane.

OA+OB+OC+OD=m/2+m/2+n/2+n/2=m+n----------------(1)

PA+PB+PC+PD= n/2++m/2+

As, is always greater than m and   is always greater than n .

So,PA+PB+PC+PD will always be>m+n.      ----------(2)

Combining (1) and (2), we arrived at the result

PA+PB+PC+PD>OA+OB+OC+OD

d 25 pi/6

step-by-step explanation:

the diameter is 10 inches

the radius is 1/2 of the diameter or 5 inches

the area of the circle is

a = pi r^2

  = pi 5^2

  = 25 pi

the slice is 60 degrees of the entire circle.

60 degrees out of 360   60/360 = 1/6

the area of the slice is 1/6 of the circle

a slice = 1/6 a circle

a slice = 1/6 * 25 pi

could you provide figure.