Asked by SAMK
Find the area of the quadrilateral whose vertices are at(1, 3), (-1,-6), (4, -3) & (6, 1).
Answers
Answered by
mathhelper
list the ordered pairs in a column with the points going counter-clockwise,
starting at any point. Whatever point you start with, repeat it at the end
1 3
-1 -6
4 -3
6 1
1 3
area = sum of the product of the downwards diagonals - sum of the product of the upward diagonals
= (1/2) [(-1)(-3) + 4(1) + 6(3) - (4(-6) + 6(-3) + 1(1) ) ]
= (1/2)[3 + 4 + 18 - (-24 - 18 + 1)]
= (1/2)(25 - (-41))
= 33
check my arithmetic
starting at any point. Whatever point you start with, repeat it at the end
1 3
-1 -6
4 -3
6 1
1 3
area = sum of the product of the downwards diagonals - sum of the product of the upward diagonals
= (1/2) [(-1)(-3) + 4(1) + 6(3) - (4(-6) + 6(-3) + 1(1) ) ]
= (1/2)[3 + 4 + 18 - (-24 - 18 + 1)]
= (1/2)(25 - (-41))
= 33
check my arithmetic
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