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

answer: ratio of volume of cone (with 3 times the height and equal diameter) to volume of cylinder is 2 to 1. the value of π is not used.

step-by-step explanation:

radius is diameter/2 = 4.

area of circle is π r^2.

volume of cylinder is height × area of base = hπr^2 = 3(4^2) π = 48π cubic inches.

not knowing the formula for volume of cone, i calculate it using calculus (which is for calculating things).

volume of cone is integral 0 to h area(y) dy. area(y) is π r(y)^2. r(y) runs r to zero as y runs zero to h, r(0) = r, r(h) = 0.

r(y) = r - r/h y

area(y) = π (r - r/h y)^2 = π (r^2 - 2r^2y/h + r^2y^2/h^2) = π r^2(1-2y/h+y^2/h^2)

antiderivative area(y) = π r^2 (y-y^2/h+y^3/(3h^2))

integral 0 to h = π r^2(h-h^2/h+h^3/(3h^2) - (0)) = π r^2(h - h + h/3) = π r^2(h/3) = π/3 r^2 h

volume of this cone is π/3(4^2)18 cubic inches.

= 6 16 π = 96π cubic inches.

cone has twice the volume, 96/48.

cone height is 6 times the cylinder.

ratio of volume formula: π r^2 cancels.

cone over cylinder ratio is 6h/3 to h, 2 to 1.