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

Question

Philoid · Accepted 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