find the area of quadrilateral ABCD, the co-ordinate of whose vertices are A(5,-2), B(-3,-1), C(2,1) and D(6,0)
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consider it as two triangles,
triangle 1:
=1/2{-3(1-0)+2(0+1)+6(-1-1)}
=1/2{-3+2-12}
=1/2{-13}
area1=-13/2
triangle 2:
=1/2{-3(-2+0)+5(0+1)+6(-1+2)}
=1/2{6+5+6}
=1/2{17}
area2=17/2
area of quadrilateral= A1+A2
=(-13=17)/2
=4/2=2sq.unit
hope it helpsss
triangle 1:
=1/2{-3(1-0)+2(0+1)+6(-1-1)}
=1/2{-3+2-12}
=1/2{-13}
area1=-13/2
triangle 2:
=1/2{-3(-2+0)+5(0+1)+6(-1+2)}
=1/2{6+5+6}
=1/2{17}
area2=17/2
area of quadrilateral= A1+A2
=(-13=17)/2
=4/2=2sq.unit
hope it helpsss
rk78685848p65407:
No, the answer is 15sq. Unit
let me check it again
consider it as two triangles,
triangle 1:
=1/2{-3(1-0)+2(0+1)+6(-1-1)}
=1/2{-3+2-12}
=1/2{-13}
area1=-13/2sq.u (as area can't be nagative A1=13/2sq.u)
triangle 2:
=1/2{-3(-2+0)+5(0+1)+6(-1+2)}
=1/2{6+5+6}
=1/2{17}
area2=17/2
area of quadrilateral= A1+A2
=(13+17)/2
=30/2=15sq.unit
hope it helpsss
triangle 1:
=1/2{-3(1-0)+2(0+1)+6(-1-1)}
=1/2{-3+2-12}
=1/2{-13}
area1=-13/2sq.u (as area can't be nagative A1=13/2sq.u)
triangle 2:
=1/2{-3(-2+0)+5(0+1)+6(-1+2)}
=1/2{6+5+6}
=1/2{17}
area2=17/2
area of quadrilateral= A1+A2
=(13+17)/2
=30/2=15sq.unit
hope it helpsss
i'm sry for the wrong answer
Ok, thanks
pls mark my answer as the brainliest
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