SOLUTION: Show that (-1,-1), (9,4), (20,6), and (10,1) are the vertices of a rhombus, and then find the area of the rhombus.

Algebra ->  Surface-area -> SOLUTION: Show that (-1,-1), (9,4), (20,6), and (10,1) are the vertices of a rhombus, and then find the area of the rhombus.      Log On


   

Question 266118: Show that (-1,-1), (9,4), (20,6), and (10,1) are the vertices of a rhombus, and then find the area of the rhombus.
Found 2 solutions by jim_thompson5910, drk:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Hint: find the distances between each vertex (excluding the vertices that are opposite one another). If those distances are equal, then the figure is a rhombus.

Answer by drk(1908) About Me  (Show Source):
You can put this solution on YOUR website!
we need the distance formula as
d+=+sqrt%28%28y2-y1%29%5E2+%2B+%28x2-x1%29%5E2%29
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distance between: (-1,-1), (9,4) is
d+=+sqrt%28%284-%28-1%29%29%5E2+%2B+%289-%28-1%29%29%5E2%29
or
d+=+sqrt%28%285%29%5E2+%2B+%2810%29%5E2%29
and then
d+=+sqrt%28125%29
-----
distance between: (9,4), (20,6) is
d+=+sqrt%28%286-%284%29%29%5E2+%2B+%289-%2820%29%29%5E2%29
or
d+=+sqrt%28%282%29%5E2+%2B+%28-11%29%5E2%29
and then
d+=+sqrt%28125%29
-----
distance between: (20,6), and (10,1) is
d+=+sqrt%28%281-%286%29%29%5E2+%2B+%2810-%2820%29%29%5E2%29
or
d+=+sqrt%28%28-5%29%5E2+%2B+%28-10%29%5E2%29
and then
d+=+sqrt%28125%29
-----
distance between: (-1,-1), and (10,1) is
d+=+sqrt%28%281-%28-1%29%29%5E2+%2B+%2810-%28-1%29%29%5E2%29
or
d+=+sqrt%28%282%29%5E2+%2B+%28-11%29%5E2%29
and then
d+=+sqrt%28125%29
-----
So we have shown we have a square with all sides equal. SInce the angles are all not 90 degrees, we have a rhombus.
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area = diagonal 1 x diagonal 2 / 2
diagonal 1 = sqrt(21^2 + 7^2) = sqrt(490)
diagonal 2 = sqrt(1^2 + 3^2) = sqrt(10)
area = sqrt(490) * sqrt(10)/2 = 35