Q. 84.1( 17 Votes )

PQRS and ABRS are parallelograms and X is any point on the side BR. Show that

(i) ar(PQRS) = ar(ABRS)

(ii) ar(∆AXS) = ar(PQRS)


Answer :


Constructions:


Extend the common base SR to C


Drop perpendicular from point B on the extended line mark intersection point as D


BD will be the height of both the parallelograms PQRS and ABRS with common base SR


Drop perpendicular on AS from point X thus XF will be the height for ΔAXS and also height for parallelogram ABRS if we consider AS as the base


i) consider parallelogram ABSR


base = SR


height = BD


area of parallelogram = base × height


area(ABSR) = SR × BD …(i)


consider parallelogram PQRS


base = SR


height = BD


area of parallelogram = base × height


area(PQRS) = SR × BD …(ii)


from (i) and (ii)


area(ABSR) = area(PQRS) …(*)


ii) Consider parallelogram ABRS


Let base = AS


Then Height = XF


Area of parallelogram = base × height


Area of parallelogram ABRS = AS × XF …(i)


For ΔAXS


Base = AS


Height = XF


Area of ΔAXS = × AS × XF


Using (i)


Area of ΔAXS = × Area of parallelogram ABRS


Using equation (*) from first part of question


Area of ΔAXS = × area of parallelogram PQRS


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