PR = √((5+2)^2 + (3+3)^2)
= √85
You do SQ and let me know what you got
(( 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
= √85
You do SQ and let me know what you got
i will do it
:)
QS = �ã((-4+5)^2 + (3+4)^2)
=�ã50
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)
Let's start with the diagonal connecting points P and R. The distance formula is used to find the distance between two points in a Cartesian coordinate system. The formula is:
distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
For the diagonal PR, we have the coordinates P(3,5) and R(-3,-2). Substituting these values into the distance formula, we get:
distance PR = sqrt((-3 - 3)^2 + (-2 - 5)^2)
= sqrt((-6)^2 + (-7)^2)
= sqrt(36 + 49)
= sqrt(85)
≈ 9.2 (rounded to the nearest tenth)
Now let's find the length of the other diagonal, QS. We have the coordinates Q(-4,3) and S(5,-4). Substituting these values into the distance formula, we get:
distance QS = sqrt((5 - (-4))^2 + (-4 - 3)^2)
= sqrt((5 + 4)^2 + (-4 - 3)^2)
= sqrt((9)^2 + (-7)^2)
= sqrt(81 + 49)
= sqrt(130)
≈ 11.4 (rounded to the nearest tenth)
Therefore, the lengths of the diagonals PR and QS are approximately 9.2 and 11.4 units, respectively.