Question

in a parallelogram PQRS, L and M are the mid-points of QR and RS respectively. Prove that : ar( ΔPLM) = 3/8 ar( IIgm PQRS).

Answer 1

see figure, in a parallelogram PQRS, L and M are the mid-points of QR and RS respectively.
join SQ and draw LN || PQ,
in ∆SQR,
L and M are the mid points of QR and RS respectively.
from midpoint theorem,
so, LM || SQ and LM = 1/2 SQ
also it can be considered as ar(LMR)/ar(SRQ) = LM²/SQ² = 1/4
so, ar(LMR) = 1/4 × ar(SRQ) ......(1)

we also know, a diagonal divides parallelogram into two equal areas.
so, ar(SRQ) = 1/2 × ar(||gm PQRS)

or, ar(LMR) = 1/4 × 1/2 × ar(||gm PQRS)

or, ar(LMR) = 1/8 × ar (||gm PQRS) ......(ii)

also ar(PQLN) = ar(SRLN) = 1/2 ar(||gm PQRS) [Parallelograms on equal base (LQ = LR) and between the same parallels are equal in area]

area of ∆PQL = 1/2 × area of PQLN = 1/4 × area of PQRS

so, area of ∆PQL = 1/4 × area of PQRS
similarly, area of ∆PMS = 1/4 × area of PQRS.

now, area of ∆PLN = area of PQRS - [ area of ∆PQL + area of ∆LMR + area of ∆PMS]

= area of PQRS - [ 1/4 × area of PQRS + 1/8 × area of PQRS + 1/4 × area of PQRS ]

= area of PQRS - 5/8 × area of PQRS
= 3/8 × area of PQRS

hence, proved/