Question

i have a question in math
(( distance between two points ))

A quardilateral has vertices P(3,5), Q(-4,3), R(-3,-2), and S(5,-4). find the lengths of the diagonals,to the nearest tenth .

please in detales .. thank u

Answer 1

PR = √((5+2)^2 + (3+3)^2)
= √85

You do SQ and let me know what you got

Answer 2

it is easy thank u
i will do it
:)

QS = ã((-4+5)^2 + (3+4)^2)
=ã50

Answer 3

hi
i think theere is something wrong in the answer

PR = ã((5+2)^2 + (3+3)^2)
= ã85

because the rule is
ã((x2-x1)^2 + (y2-y1)^2)

Answer 4

3892