Question

A triangle is formed by joining the coordinates B(0,0), C(8,0) and A(4,8). DE is joined where D is the mid-point of AB, and E is the mid-point of AC. DC and BE are joined to intersect at F.
Then, Area of $\Delta$ DCB = $\frac{n}{2}$ $\times$ Area of $\Delta$ EBC, where n is an integer. The value of n is __.

Answer 1

D is the mid-point of AB and E is the mid-point of AC in $\Delta$ ABC. From the mid-point theorem, the line joining the mid points of two sides is parallel to the third side and is equal to half of it.

Thus, DE $\parallel$ BC.

We know that the triangles between the same base and same parallels have equal area.
Thus, Area of $\Delta$ DCB = Area of $\Delta$ BEC.
$\Rightarrow$ Area of $\Delta$DCB = $\frac{2}{2}$ $\times$Area of $\Delta$ EBC.

Hence value of n = 2.