the vertices of a triangle ABC are A(5,3),B(-9,-3) and C(-3,-5),if P,Q,R are the midpoints of BC,AC and AB.Prove that area of triangle ABC=4×area of triangle PQR
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then area of the ∆ABC=5 -9 -3 5
3 -3 -5 3
=1/2|[-15+45+-9]-[-27+9-25]|=1/2|[21]-[-43]|=1/2(64)=32sq.units
and again coordinates of P=(-12/2,-8/2)=(-6,-4).and the coordinates of the point Q=(5-3/2,3-5/2)=(1,-1) also the coordinates of point R=(-9+5/2,0/2)=(-2,0). so the area of ∆PQR=-6 1 -2 -6
-4 -1 0 -4
=1/2[6+0+8]-[-4+2]=1/2(14+2)=8sq.units hence the area of ∆ABC=4×[ar(∆PQR)].
3 -3 -5 3
=1/2|[-15+45+-9]-[-27+9-25]|=1/2|[21]-[-43]|=1/2(64)=32sq.units
and again coordinates of P=(-12/2,-8/2)=(-6,-4).and the coordinates of the point Q=(5-3/2,3-5/2)=(1,-1) also the coordinates of point R=(-9+5/2,0/2)=(-2,0). so the area of ∆PQR=-6 1 -2 -6
-4 -1 0 -4
=1/2[6+0+8]-[-4+2]=1/2(14+2)=8sq.units hence the area of ∆ABC=4×[ar(∆PQR)].
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