Class 9MathematicsAreas of Parallelograms and TrianglesTextbook - Areas of Parallelograms and Triangles
Difficult$ABCD$ is a trapezium with $AB || DC$. $A$ line parallel to $AC$ intersects $AB$ at $X$ and $BC$ at $Y$. Prove that $\operatorname{ar}(ADX) = \operatorname{ar}(ACY)$. [Hint: Join $CX$.]
(N/A) We have a trapezium $ABCD$ such that $AB || DC$.
$XY || AC$ meets $AB$ at $X$ and $BC$ at $Y$.
Let us join $CX$.
$\because AB || DC$ [Given]
$\therefore \Delta ADX$ and $\Delta ACX$ are on the same base $AX$ and between the same parallels $AB$ and $DC$.
$\therefore \operatorname{ar}(\Delta ADX) = \operatorname{ar}(\Delta ACX) \quad \dots(1)$
$\because AC || XY$
$\therefore \Delta ACX$ and $\Delta ACY$ are on the same base $AC$ and between the same parallels $AC$ and $XY$.
$\therefore \operatorname{ar}(\Delta ACX) = \operatorname{ar}(\Delta ACY) \quad \dots(2)$
From $(1)$ and $(2)$,we have
$\operatorname{ar}(\Delta ADX) = \operatorname{ar}(\Delta ACY)$
$XY || AC$ meets $AB$ at $X$ and $BC$ at $Y$.
Let us join $CX$.
$\because AB || DC$ [Given]
$\therefore \Delta ADX$ and $\Delta ACX$ are on the same base $AX$ and between the same parallels $AB$ and $DC$.
$\therefore \operatorname{ar}(\Delta ADX) = \operatorname{ar}(\Delta ACX) \quad \dots(1)$
$\because AC || XY$
$\therefore \Delta ACX$ and $\Delta ACY$ are on the same base $AC$ and between the same parallels $AC$ and $XY$.
$\therefore \operatorname{ar}(\Delta ACX) = \operatorname{ar}(\Delta ACY) \quad \dots(2)$
From $(1)$ and $(2)$,we have
$\operatorname{ar}(\Delta ADX) = \operatorname{ar}(\Delta ACY)$
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Similar Questions
In the figure,$ABC$ is a right triangle right-angled at $A$. $BCED$,$ACFG$,and $ABMN$ are squares on the sides $BC$,$CA$,and $AB$ respectively. Line segment $AX \perp DE$ meets $BC$ at $Y$. Show that: $\operatorname{ar}(BYXD) = 2 \operatorname{ar}(MBC)$
Medium
View SolutionIn the figure,$P$ is a point in the interior of a parallelogram $ABCD$. Show that:
$(i)$ $\operatorname{ar}(APB) + \operatorname{ar}(PCD) = \frac{1}{2} \operatorname{ar}(ABCD)$
$(ii)$ $\operatorname{ar}(APD) + \operatorname{ar}(PBC) = \operatorname{ar}(APB) + \operatorname{ar}(PCD)$
[Hint: Through $P$,draw a line parallel to $AB$.]
$(i)$ $\operatorname{ar}(APB) + \operatorname{ar}(PCD) = \frac{1}{2} \operatorname{ar}(ABCD)$
$(ii)$ $\operatorname{ar}(APD) + \operatorname{ar}(PBC) = \operatorname{ar}(APB) + \operatorname{ar}(PCD)$
[Hint: Through $P$,draw a line parallel to $AB$.]
Difficult
View SolutionParallelogram $ABCD$ and rectangle $ABEF$ are on the same base $AB$ and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.
Difficult
View SolutionIn the figure,$ABCD$ is a parallelogram and $BC$ is produced to a point $Q$ such that $AD = CQ$. If $AQ$ intersects $DC$ at $P$,show that $\operatorname{ar}(BPC) = \operatorname{ar}(DPQ)$. [Hint: Join $AC$.]
Medium
View SolutionIn the figure,$ABC$ is a right triangle right-angled at $A$. $BCED$,$ACFG$,and $ABMN$ are squares on the sides $BC$,$CA$,and $AB$ respectively. Line segment $AX \perp DE$ meets $BC$ at $Y$. Show that: $\operatorname{ar}(CYXE) = 2 \operatorname{ar}(FCB)$.
Medium
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