Question

ABCD is a rectangle in the diagram. P is a point on the side CD. Show that the area of ∆AQB is equal to the sum of the areas of ∆PQD and ∆PCB.

Answer 1

area of ∆AQB = area of ∆APB – area of ∆BQP
⇒ area of ∆ADB = area of ∆DBC
area of (∆ADQ + ∆AQB) = area of (∆PQD + ∆QBP + ∆PCB ) …….(i)
area of ∆DAP = area of ∆DBP …….(ii)
Subtract area of ∆PQD from both side of the equation (ii),
⇒ area of ∆ADQ = area of ∆QBP
⇒ area of (∆ADQ + ∆AQB) = area of (∆PQD + ∆ADQ + ∆PCB)
⇒ area of ∆AQB = area of ∆PQD + area of ∆PCB.