Respuesta :
Given:
The vertices of the quadrilateral ABCD are A(3,8), B(6,5), C(5,4), and D(2,7).
To find:
Whether the given quadrilateral is a rectangle, a rhombus or a square.
Solution:
Distance formula: The distance between two points is
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Using distance formula, the side lengths are:
[tex]AB=\sqrt{\left(6-3\right)^2+\left(5-8\right)^2}[/tex]
[tex]AB=\sqrt{\left(3\right)^2+\left(-3\right)^2}[/tex]
[tex]AB=\sqrt{9+9}[/tex]
[tex]AB=\sqrt{18}[/tex]
[tex]AB=3\sqrt{2}[/tex]
Similarly,
[tex]BC=\sqrt{\left(5-6\right)^2+\left(4-5\right)^2}=\sqrt{2}[/tex]
[tex]CD=\sqrt{\left(2-5\right)^2+\left(7-4\right)^2}=3\sqrt{2}[/tex]
[tex]AD=\sqrt{\left(2-3\right)^2+\left(7-8\right)^2}=\sqrt{2}[/tex]
The length of diagonals are:
[tex]AC=\sqrt{\left(5-3\right)^2+\left(4-8\right)^2}=2\sqrt{5}[/tex]
[tex]BD=\sqrt{\left(2-6\right)^2+\left(7-5\right)^2}=2\sqrt{5}[/tex]
From the above calculation, we conclude that the given quadrilateral has two pairs of congruent opposite sides and equal diagonals.
Opposite sides of a rectangle are equal and its diagonals are also equal.
Therefore, the given quadrilateral ABCD is a rectangle.
