# 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)  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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