Question

Column II gives the area of triangles whose vertices are given in column I, match them correctly.

Answer 1

$Weshallapplyar\Delta formulawhichisar\Delta ABC=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)].when\Delta ABChasverticesA(x_1,y_1),B(x_2,y_2)\&C(x_3,y_3)A)Letthe\Delta ABChaveverticesA(x_1,y_1)=(2,3),B(x_2,y_2)=(-1,0)\&C(x_3,y_3)=(2,-4).\therefore ar\Delta ABC=\frac{1}{2}[2(0+4)+(-1)(-4-3)+2(3-0)]sq.units=10.5sq.units.⟹ar\Delta ABC=10.5sq.unitswhichmatcheswith4.B)Letthe\Delta ABChaveverticesA(x_1,y_1)=(-5,-1),B(x_2,y_2)=(3,-5)\&C(x_3,y_3)=(5,2).\therefore ar\Delta ABC=\frac{1}{2}[-5(-5-2)+3(2+1)+5(-1+5)]sq.units=32sq.units.⟹ar\Delta ABC=32sq.unitswhichmatcheswith3.C)Letthe\Delta ABChaveverticesA(x_1,y_1)=(1,-1),B(x_2,y_2)=(-4,6)\&C(x_3,y_3)=(-3,-5).\therefore ar\Delta ABC=\frac{1}{2}[1(6+5)+(-4)(-5+1)+(-3\left)(-1-6)\right]sq.units=24sq.units.⟹ar\Delta ABC=24sq.unitswhichmatcheswith2.D)Letthe\Delta ABChaveverticesA(x_1,y_1)=(0,0),B(x_2,y_2)=(8,0)\&C(x_3,y_3)=(0,10).\therefore ar\Delta ABC=\frac{1}{2}[0(0-0)+8(10-0)+0(0-0)]sq.units=40sq.units.⟹ar\Delta ABC=40sq.unitswhichmatcheswith1.Ans-A⟶4B⟶3C⟶2D⟶1$