Area

Geometry Level 3

The consecutive vertices of a quadrilateral are ( 1 , 5 ) , ( 6 , 6 ) , ( 7 , 2 ) (1, 5), (6, 6), (7, 2) , and ( 3 , 2 ) (3, 2) . Find the area in square units.

31/2 33/2 35/2 32/2

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1 solution

A systematic way to find area from given coordinates is summation of area under the straight line between two adjacent coordinates in a loop. First arrange the n n pairs of coordinates enclosing the area in a clockwise order. Let the first pair of coordinates be the ( n + 1 ) (n+1) th pair to close the loop, that is ( x 1 , y 1 ) = ( x n + 1 , y n + 1 ) (x_1, y_1) = (x_{n+1}, y_{n+1}) .

A = k = 1 4 ( y k + 1 + y k ) ( x k + 1 x k ) 2 = ( 6 + 5 ) ( 6 1 ) 2 + ( 2 + 6 ) ( 7 6 ) 2 + ( 2 + 2 ) ( 3 7 ) 2 + ( 5 + 2 ) ( 1 3 ) 2 = 55 2 + 8 2 + 16 2 + 14 2 = 33 2 \begin{aligned} A & = \sum_{k=1}^4 \frac {(y_{k+1}+y_k)(x_{k+1}-x_k)}2 \\ & = \frac {(6+5)(6-1)}2 + \frac {(2+6)(7-6)}2 + \frac {(2+2)(3-7)}2 + \frac {(5+2)(1-3)}2 \\ & = \frac {55}2 + \frac {8}2 + \frac {-16}2 + \frac {-14}2 \\ & = \boxed{\dfrac {33}2} \end{aligned}

Pick's theorem: I=number of Inner points=14 E=points on the edges=7 E/2+I-1=Area 7/2+14-1=16.5=33/2

Kris Hauchecorne - 4 years, 7 months ago

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Good solution. First time that I learned about Pick's theorem.

Chew-Seong Cheong - 4 years, 7 months ago

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