What is the area and perimeter of the polygon with the given vertices #A) (5,0), B) (2,4), C) (-3,0), D) (-6,4)?#

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Answer 1 · 2018-07-11T13:14:04

#32# & #26#

Given that the vertices of polygon i.e. quadrilateral #ABDC# are #A(5, 0)#, #B(2, 4)#, #D(-6, 4)# & #C(-3, 0)#.

The area #\Delta_1# of #\Delta BCD# with vertices at #B(2, 4)#, #C(-3, 0)# & #D(-6, 4)# is given as

#\Delta_1=1/2|2(0-4)-3(4-4)-6(4-0)|=16#

Similarly, the area #\Delta_2# of #\Delta ABC# with vertices at #A(5, 0)#, #B(2, 4)# & #C(-3, 0)# is given as

#\Delta_2=1/2|5(4-0)+2(0-0)-3(0-4)|=16#

hence, the area of quadrilateral #ABDC# is the sum of areas of above two triangles given as

#=\Delta_1+\Delta_2#

#=16+16#

#32#

Now, the lengths of all four sides are computed by using distance formula as follows

#AB=\sqrt{(5-2)^2+(0-4)^2}=5#

#AC=\sqrt{(5+3)^2+(0-0)^2}=8#

#CD=\sqrt{(-3+6)^2+(0-4)^2}=5#

#BD=\sqrt{(2+6)^2+(4-4)^2}=8#

hence, the perimeter of the quadrilateral ABCD

#=AB+AC+CD+BD#

#=5+8+5+8#

#=26#