Class 9MathematicsAreas of Parallelograms and TrianglesTextbook - Areas of Parallelograms and Triangles
MediumIn the figure,$ABCDE$ is a pentagon. $A$ line through $B$ parallel to $AC$ meets $DC$ produced at $F$. Show that:
$(i)$ $ar(ACB) = ar(ACF)$
$(ii)$ $ar(AEDF) = ar(ABCDE)$
$(i)$ $ar(ACB) = ar(ACF)$
$(ii)$ $ar(AEDF) = ar(ABCDE)$
(N/A) We have a pentagon $ABCDE$ in which $BF \parallel AC$ and $DC$ is produced to $F$.
$(i)$ To prove $ar(\Delta ACB) = ar(\Delta ACF)$:
Since triangles on the same base and between the same parallels are equal in area.
$\because \Delta ACB$ and $\Delta ACF$ are on the same base $AC$ and between the same parallels $AC$ and $BF$.
$\therefore ar(\Delta ACB) = ar(\Delta ACF)$.
$(ii)$ Since $ar(\Delta ACB) = ar(\Delta ACF)$:
Adding $ar(AEDC)$ to both sides,we get:
$ar(\Delta ACB) + ar(AEDC) = ar(\Delta ACF) + ar(AEDC)$
$\Rightarrow ar(ABCDE) = ar(AEDF)$.
$(i)$ To prove $ar(\Delta ACB) = ar(\Delta ACF)$:
Since triangles on the same base and between the same parallels are equal in area.
$\because \Delta ACB$ and $\Delta ACF$ are on the same base $AC$ and between the same parallels $AC$ and $BF$.
$\therefore ar(\Delta ACB) = ar(\Delta ACF)$.
$(ii)$ Since $ar(\Delta ACB) = ar(\Delta ACF)$:
Adding $ar(AEDC)$ to both sides,we get:
$ar(\Delta ACB) + ar(AEDC) = ar(\Delta ACF) + ar(AEDC)$
$\Rightarrow ar(ABCDE) = ar(AEDF)$.
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