Question

6. Quadrilateral ABCD has vertices at A(0,6), B(4, -1), C(-4,0) and D(-8,7). Prove that:

(a) ABCD is a rhombus using the distance formula

Answer 1

Step-by-step explanation:

AB = sqrt((4-0)^2 + (6-(-1))^2) = sqrt(16+49)=sqrt(65)

CD=sqrt((-8-(-4))^2 + (0-7)^2) = sqrt(49+16)=sqrt(65)

Therefore, sides AB and CD are congruent because they have the same measure.

BC = sqrt((4-(-4)^2 + (-1-0)^2) = sqrt(64+1)=sqrt(65)

AD=sqrt((0-(-8))^2 + (7-6)^2) = sqrt(64+1)=sqrt(65)

Therefore, sides BC and AD were congruent because they have the same measure.

This means ABCD is a parallelogram because there are two pairs of opposite congruent sides.

Now, notice that AB=BC, meaning sides AB and BC are congruent.

This means that parallelogram ABCD has one pair of consecutive congruent sides, and is thus a rhombus.