Proving that ABCD is a Rhombus and Finding its Area
To begin, we will use the distance formula to find the lengths of each side.
Side AB = sqrt((11-7)^2 + (2-4)^2) = sqrt(16 + 4) = sqrt(20)
Side BC = sqrt((7-11)^2 + (0-2)^2) = sqrt(16 + 4) = sqrt(20)
Side CD = sqrt((3-7)^2 + (2-0)^2) = sqrt(16 + 4) = sqrt(20)
Side DA = sqrt((3-7)^2 + (2-4)^2) = sqrt(16 + 4) = sqrt(20)
As we can see, all four sides have the same length, which proves that ABCD is a rhombus.
Next, we can use the slope formula to find the slopes of each side.
Slope AB = (2-4)/(11-7) = -0.5
Slope BC = (0-2)/(7-11) = -0.5
Slope CD = (2-0)/(3-7) = -0.5
Slope DA = (2-4)/(3-7) = -0.5
Since all four sides have the same slope, we can conclude that all four angles are congruent and ABCD is a rhombus.
Now, to find the area of the rhombus, we can use the formula A = (1/2) * d1 *d2, where d1 and d2 are the lengths of the diagonals.
Diagonal AC = sqrt((7-7)^2 + (4-0)^2) = sqrt(16) = 4
Diagonal BD = sqrt((11-3)^2 + (2-2)^2) = sqrt(64) = 8
Therefore, the area of ABCD = (1/2) * 4 * 8 = 16 square units.
Hence, we have proven that ABCD is a rhombus and have found its area to be 16 square units.
Disclaimer: Always double check your work and make sure to cite any outside sources used in your proof.
