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

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

Please show working out, thanks.

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.

@MathMate would love that, thanks.

Area = 24?

sorry, 24 is not correct.

one diagonal is 6√2, and the other is 8√2.
The product of the two diagonals/2=?

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).

To prove that ABCD is a rhombus, we need to show that all four sides of the quadrilateral are equal. Here's how we can do that:

1. Calculate the length of each side:
- Side AB: Using the distance formula, d = √((x2 - x1)^2 + (y2 - y1)^2)
- dAB = √((5 - (-2))^2 + (4 - 3)^2) = √(7^2 + 1^2) = √(49 + 1) = √50 = 5√2

- Side BC:
- dBC = √((4 - 5)^2 + (-3 - 4)^2) = √((-1)^2 + (-7)^2) = √(1 + 49) = √50 = 5√2

- Side CD:
- dCD = √((-3 - 4)^2 + (-4 - (-3))^2) = √((-7)^2 + (-1)^2) = √(49 + 1) = √50 = 5√2

- Side DA:
- dDA = √((-3 - (-2))^2 + (-4 - 3)^2) = √((-1)^2 + (-7)^2) = √(1 + 49) = √50 = 5√2

2. Compare the lengths:
Since dAB = dBC = dCD = dDA = 5√2, we can conclude that all four sides of ABCD are equal. Hence, ABCD is a rhombus.

To find the area of ABCD, we can use the coordinates of the vertices to calculate the area of the quadrilateral using the Shoelace Formula or the coordinate geometry method.

Using the Shoelace Formula:
- Let's label the vertices as A(-2,3), B(5,4), C(4,-3), and D(-3,-4).
- The formula for the area of a polygon using the Shoelace Formula is:
- Area = (1/2) * |(x1y2 + x2y3 + x3y4 + x4y1) - (x2y1 + x3y2 + x4y3 + x1y4)|
- Plugging in the coordinates, we have:
- Area = (1/2) * |((-2 * 4) + (5 * (-3)) + (4 * (-4)) + ((-3) * 3)) - ((5 * 3) + (4 * 4) + ((-3) * (-4)) + ((-2) * (-3)))|
- Simplifying, we get:
- Area = (1/2) * |(-8 - 15 - 16 - 9) - (15 + 16 + 12 + 6)|
- Area = (1/2) * |-48 - 49|
- Area = (1/2) * |-97|
- Area = (1/2) * 97
- Area = 97/2 = 48.5 square units

Therefore, the area of the quadrilateral ABCD is 48.5 square units.