Answer:
(A)
Step-by-step explanation:
It is given that A quadrilateral has vertices A(3, 5), B(2, 0), C(7, 0), and D(8, 5).
Therefore, using the distance formula,
AB=[tex]\sqrt{(0-5)^2+(2-3)^2}=\sqrt{25+1}=\sqrt{26}[/tex],
BC=[tex]\sqrt{(0-0)^2+(7-2)^2}=\sqrt{25}=5[/tex],
CD=[tex]\sqrt{(5-0)^2+(8-7)^2}=\sqrt{25+1}=\sqrt{26}[/tex] and
DA=[tex]\sqrt{(5-5)^2+(8-3)^2}=\sqrt{25}=5[/tex]
Now, it can be seen that opposite sides of the quadrilateral are equal but not all the sides are equal, thus the given quadrilateral is not rhombus.
We have to check whether the quadrilateral have perpendicular or non perpendicular adjacent sides.
For this, we take
[tex]AB{\cdot}BC=(-1){\cdot}5+(-5){\cdot}0=-5{\neq}0[/tex]
and [tex]AB{\cdot}DA=-1{\cdot}(-5)+(-5)0=5{\neq}0[/tex]
The dot products are not equal to zero, then angles A and B are not right. This means that option C is false and option A is correct.
