Question

P and Q are respectively points lying on the side DC and AD respectively of a parallelogram ABCD. Show that ar(APB) = ar(BCQ)

Answer 1

Hello Mate!

In ||gm ABCD, its given that P and Q are points on side CD and AD respectively.

Since ∆APB and ||gm ABCD lies on same base i.e AB and same parallels i.e AB and DC so,

ar(∆APB) = ½ ar(||gm ABCD) __(i)

Since ∆CQB and ||gm ABCD lies on same base i.e BC and same parallels i.e AD and BC so,

ar(∆CQB) = ½ ar(||gm ABCD) ___(ii)

Hence from equation (i) and (ii) we get

ar(∆APB) = ar(∆CQB)

Hence Proved.

Have great future ahead!

Answer 2

here is your answer OK ☺☺☺☺☺☺


$here \: is \: your \: solution$
given: ABCD is a parallelogram . P and Q are any points on the sides DC and AD respectively.

TPT: area(ΔAPB) = area(ΔBQC)

proof;

if a triangle and a parallelogram are on the same base and between same set of parallel lines. the area of the triangle is half the area of the parallelogram.

ΔAPB and parallelogram ABCD are on the same base AB and between same parallels AB and CD.

therefore area(ΔAPB)= 1/2 area(ABCD)..............(1)

similarly

triangle BQC and parallelogram ABCD are on the same base BC and between same parallels AD and BC.

area(ΔBQC)=1/2 area(ABCD).......................(2)

from (1) and (2): area(ΔAPB) = area(ΔBQC)

which is the required result.

hope this helps you.