What is the area of the polygon with the vertices #G(4,1), H(4,-2), J(-1, -2)#?

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Answer 1 · 2016-11-13T12:22:08

Area of polygon is #7.5#

As the polygon has three corners #G(4,1)#, #H(4,-2)# and #J(-1,-2)#, it is a triangle, whose sides are

#sqrt((4-4)^2+(1-(-2))^2)=sqrt(0+9)=sqrt9=3#

#sqrt((4-(-1))^2+(1-(-2))^2)=sqrt(25+9)=sqrt34=5.831#

#sqrt((4-(-1))^2+(-2-(-2))^2)=sqrt(25+0)=sqrt25=5#

As area of a triangle #Delta# is given by

#Delta=sqrt(s(s-a)(s-b)(s-c))#, where #s# is semiperimeter

i.e. #1/2(3+5.831+5)=1/2xx13.831=6.9155#

and area of triangle is

#Delta=sqrt(6.9155(6.9155-3)(6.9155-5.831)(6.9tt155-5))#

= #sqrt(6.9155xx3.9155xx1.0845xx1.9155)#

= #sqrt56.25=7.5#