Question

Quadrilateral BCDE has vertices B(-1, -1), C(6, -2), D(5, -9), and D(-2, -8). Determine the most precise classification of BCDE: a parallelogram, rectangle, rhombus, or square. Use the distance formula to justify your answer. All work must be shown for credit.

Answer 1

Given:

Quadrilateral BCDE has vertices B(-1,-1), C (6,-2),D(5,-9) and E(-2,-8).

To Find:

Determine whether it is a parallelogram,rectangle,rhombus or a square.

Solution:

Distance formula = $\sqrt{(x - x_{1} )^{2} + (y-y_{1} )^{2} }$

BC = $\sqrt{(-1-6)^{2} + (-1+2)^{2} } = \sqrt{50}$

CD= $\sqrt{(6-5)^{2} + (-2+9)^{2} } = \sqrt{50}$

DE= $\sqrt{(5+2)^{2}+(-9+8)^{2} } = \sqrt{50}$

BE = $\sqrt{(-1+2)^{2} +(-1+8)^{2} } = \sqrt{50}$

All sides of quadrilateral are equal.

Hence  it is either a square or a rhombus.

Properties of rhombus- diagonals are not equal

Diagonals of  our quadrilateral are BD and CE.

BD= $\sqrt{(-1+9)^{2}+ (-1+5)^{2} } = \sqrt{80}$

CD= $\sqrt{(6+2)^{2} + (-2+8)^{2} } =10$

Diagonals of quadrilateral are not equal.

Hence the quadrilateral BCDE is a RHOMBUS.

Answer 2

Quadrilateral BCDE is a Rhombus

Step-by-step explanation:

• It can be formulated by making the use of Distance Formula.

B(-1,-1), C (6,-2),D(5,-9) and E(-2,-8) are the vertices of the quadrilateral BCDE.

We know that,

Distance Formula = $\sqrt {(x-x_1)^2(y-y_1)^2}$

So, BC = $\sqrt {(-1-6)^2+ (-1+2)^2} = \sqrt {50}$ ""(1)

CD = $\sqrt {(6-5)^2+(-2+9)^2} = \sqrt {50}$  ""(2)

DE = $\sqrt {(5+2)^2+(-9+8)^2} = \sqrt {50}$  ""(3)

BE = $\sqrt {(-1+2)^2+(-1+8)^2} = \sqrt {50}$ ""(4)

From (1),(2),(3),(4) we get-

All Sides of the Quadrilateral are Equal.

We know that,

• Quadrilateral of equal sides could be a square or a rhombus.

• Also, Diagonals or Square are Equal and that of Rhombus are unequal.

Now, Let us find the whether the Diagonals DB and DC of quadrilateral are equal or not.

So, using Distance Formula again-

BD = $\sqrt {(-1+9)^2+(-1+5)^2} = \sqrt {80}$ """(5)  

CD = $\sqrt {(6+2)^2+(-2+8)^2} = \sqrt {100} = 10$ ""(6)

From (5) and (6) we get-

• Diagonals of Quadrilateral are unequal.

Hence, we can say that Quadrilateral BCDE is a Rhombus.