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# In the given figure, $PQRS$ and $ABRS$ are parallelogram and $X$ is any point on $BR$. Show that,$(i)$area$(PQRS)$$= area(ABRS),(ii) area(AXS)$$ = $$\frac{1}{2}area(PQRS). Last updated date: 19th Jul 2024 Total views: 452.4k Views today: 12.52k Answer Verified 452.4k+ views Hint: Both the parallelograms are lying on the same base and within the same parallel lines. (i)Consider parallelograms PQRS and ABRS. They are lying on the same base RS and between the same parallel lines RS and PB. And according to the property of parallelogram, we know that parallelograms lying on the same base and between the same parallel lines have equal area. Therefore: \Rightarrow area(PQRS)$$ =$area$(ABRS)$.
$(ii)$Now, consider $\Delta AXS$ and parallelogram $ABRS$. Both are lying on the same base $AS$ and between the same parallel lines $AS$ and $BR$. And we know that the area of the triangle having the base as a parallelogram and its third vertex lying on the opposite parallel side of the parallelogram is half the area of the same parallelogram.
$\therefore$ area$(AXS)$$=$$\frac{1}{2}$ area$(ABRS)$
And we have already proved that, area$(PQRS)$$= area(ABRS). Using this, we’ll get: \Rightarrow area(AXS)$$ =$$\frac{1}{2}$area$(PQRS)$

Note: Area of a parallelogram is base times its height. So, if two parallelograms are having the same base and lying between the same parallel lines, their heights will also be the same and consequently their areas will be the same. Triangle on the other hand is having an area as half of base times its height. Therefore, a triangle having such property as mentioned above will have its area as half of the area of the parallelogram.