Q. 94.3( 18 Votes )

In the figure, if ar ∆RAS = ar ∆RBS and [ar (∆QRB) = ar(∆PAS) then show that both the quadrilaterals PQSR and RSBA are trapeziums.


Answer :

Extend lines R and S to points J and K as shown



Given that area(∆RAS) = area(∆RBS) …(i)


Common base is RS


Let height of ∆RAS be h1 and ∆RBS be h2 as shown


area(∆RAS) = × RS × h1


area(∆RBS) = × RS × h2


by given × RS × h1 = × RS × h2


h1 = h2


As the distance between two lines is constant everywhere then lines are parallel


RS || AB …(*)


Therefore, ABSR is a trapezium


Given area(∆QRB) = area(∆PAS) …(ii)


area(∆QRB) = area(∆RBS) + area(∆QRS) …(iii)


area(∆PAS) = area(∆RAS) + area(∆RPS) …(iv)


subtract (iii) from (iv)


area(∆QRB) - area(∆PAS) = area(∆RBS) + area(∆QRS) - area(∆RAS) - area(∆RPS)


using (i) and (ii)


0 = area(∆QRS) - area(∆RPS)


area(∆QRS) = area(∆RPS)


Common base for ∆QRS and ∆RPS is RS


Let height of ∆RPS be h3 and ∆RQS be h4 as shown


area(∆RPS) = × RS × h3


area(∆RQS) = × RS × h4


by given × RS × h3 = × RS × h4


h3 = h4


As the distance between two lines is constant everywhere then lines are parallel


RS || PQ


PQSR is a trapezium


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