Since D,E and F are the midpoints of AB, BC and CA of triangle respectively
Coordinate of D=(20+2,2−1+1)=(1,0)
Coordinate of E=(22+0,21+3)=(1,2)
Coordinate of F=(20+0,2−1+3)=(0,1)
∴ar(△DEF) = 21
[1(2−1)+1(1−0)+0(0−2)]=1sq.unitandar(△ABC)=21
[0(1−3)+2(3+1)+0(−1−1)]=4sq.unitar(△ABC) ar (△DEF)
=41⇒ar (△DEF) :ar (△ABC)= 1 : 4
I hope my answer helps you....✌
Step-by-step explanation:
h
[0(1−3)+2(3+1)+0(−1−1)]=4sq.unitar(△ABC) ar
(△DEF)
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Answers & Comments
Since D,E and F are the midpoints of AB, BC and CA of triangle respectively
Coordinate of D=(20+2,2−1+1)=(1,0)
Coordinate of E=(22+0,21+3)=(1,2)
Coordinate of F=(20+0,2−1+3)=(0,1)
∴ar(△DEF) = 21
[1(2−1)+1(1−0)+0(0−2)]=1sq.unitandar(△ABC)=21
[0(1−3)+2(3+1)+0(−1−1)]=4sq.unitar(△ABC) ar (△DEF)
=41⇒ar (△DEF) :ar (△ABC)= 1 : 4
I hope my answer helps you....✌
Step-by-step explanation:
h
Since D,E and F are the midpoints of AB, BC and CA of triangle respectively
Coordinate of D=(20+2,2−1+1)=(1,0)
Coordinate of E=(22+0,21+3)=(1,2)
Coordinate of F=(20+0,2−1+3)=(0,1)
∴ar(△DEF) = 21
[1(2−1)+1(1−0)+0(0−2)]=1sq.unitandar(△ABC)=21
[0(1−3)+2(3+1)+0(−1−1)]=4sq.unitar(△ABC) ar
(△DEF)
=41⇒ar (△DEF) :ar (△ABC)= 1 : 4