If Delta_1 is the area of the triangle formed | vedclass

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(C) Let the vertices of $\Delta ABC$ be $A(x_1, y_1)$,$B(x_2, y_2)$,and $C(x_3, y_3)$,and let $\Delta$ be the area of $\Delta ABC$.$\Delta_1$ is the a

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$4 : 3$

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(C) Let the vertices of $\Delta ABC$ be $A(x_1, y_1)$,$B(x_2, y_2)$,and $C(x_3, y_3)$,and let $\Delta$ be the area of $\Delta ABC$.$\Delta_1$ is the area of the triangle formed by the centroid $G$ and two vertices,say $B$ and $C$. It is a known property that the area of the triangle formed by the centroid and any two vertices is $\frac{1}{3}$ of the area of the original triangle.$\Rightarrow \Delta_1 = \frac{1}{3} \Delta$.$\Delta_2$ is the area of the triangle formed by the mid-points of the sides of $\Delta ABC$. The area of the triangle formed by joining the mid-points of the sides is $\frac{1}{4}$ of the area of the original triangle.$\Rightarrow \Delta_2 = \frac{1}{4} \Delta$.Therefore,$\Delta_1 : \Delta_2 = \frac{\Delta}{3} : \frac{\Delta}{4} = \frac{1}{3} : \frac{1}{4} = 4 : 3$.

If $\Delta_1$ is the area of the triangle formed by the centroid and two vertices of a triangle,and $\Delta_2$ is the area of the triangle formed by the mid-points of the sides of the same triangle,then $\Delta_1 : \Delta_2 =$

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