The vertices of a quadrilateral are A(-2,3), B(5,4), C(4, -3) and D(-3, -4).

Question

Prove that ABCD is a rhombus. Find the area of ABCD.

Please show working out, thanks.

Answer 1

1. A rhombus is a parallelogram with adjacent sides equal.
2. A rhombus is a quadrilateral with four equal sides.

To prove that a figure is a rhombus, use either 1 or 2 above.
In this case, proving that the four sides are equal is relatively easy even by visual inspection.
The distance of each side is obtained by the difference of coordinates of consecutive sides squared. However, the differences are either (7,1), or (1,7), or variants with sign. The length of each side is therefore √50, hence all sides are equal.

The area of the figure can be found using the property of the rhombus, namely the product of the lengths of diagonals divided by two. That means
Area = mAC*mBD/2.

There is also a general formula for any figure given the coordinates of the vertices. Post and request it if you are interested.

Answer 2

@MathMate would love that, thanks.

Answer 3

Area = 24?

Answer 4

sorry, 24 is not correct.
one diagonal is 6√2, and the other is 8√2.
The product of the two diagonals/2=?

Answer 5

For the given figure of
A(-2,3), B(5,4), C(4, -3) and D(-3, -4).
we write it out as, repeating A at the end:

A(-2,3),
B(5,4),
C(4,-3)
D(-3,-4).
A(-2,3),

Find the four products:
Ax*By-Bx*Ay=(-2*4)-(5*3)=-23
Bx*Cy-Cx*By=(5*-3)-(4*4)=-31
Cx*Dy-Dx*Cy=(4*-4)-(-3*-3)=-25
Dx*Ay-Ax*Dy=(-3*3)-(-2*-4)=-17
Sum the products = -96

Area = |sum of products|/2=48<-b>

If the figure has more (or less) vertices, just list them and repeat the first vertice at the bottom (circular list).