Question

MNOP is a parallelogram. X is any point on MO. RQ parallel MN and BS parallel PM. Show that area of PSXR equal to area of BNQX

Answer 1

Assuming MNOP as ABCD and O as X, RQ as PQ and BS as ML Here is the answer,

 
To prove :Area (Parallelogram DLOP) = Area (Parallelogram BMOQ)

 
Proof : Given ABCD is a ||gm and AC is one of its diagonal.

∴ Area (∆ABC) = Area (∆ADC)           ... (1)         (∵ the diagonal of a parallelogram divides it into two equal triangles)

 Now,

LM is a parallel to AD                (∵ AD = BC and AD || BC as ABCD is a ||gm  )

∴ LM || BC ⇒ LO || CQ

 
Similarly PQ || DC ⇒ OQ || LC

∴ CLOQ is also as parallelogram, as its opposites sides are parallel to each other.

 

In ||gm CLOQ, CO is the diagonal

∴ Area (∆COQ) = Area (∆CLO)       ... (2)             (∵ the diagonal of a parallelogram divides it into true equal triangles)

 

Again, AD || LM ⇒ AP || OM

and AB || PQ ⇒ AM || PO

∴ APOM is also as ||gm, as its opposite sides are parallel to each other.

∴ In ||gm AOPM, AO is the diagonal so, Area (∆AOP) = Area (∆AOM)         ... (3)         (∵ the diagonal of a parallelogram divides it into two equal triangles)

 

Now using (1)

Area (∆ABC) = Area (∆ADC)

⇒ Area (∆COQ) + Area (AOM) + Area (BMOQ) = Area (∆CLO) + Area (AOP) + Area (DLOP)

 

Using (2) and (3), we get

Area (BMOQ) = Area (DLOP)

 

Now, BMOQ is an ||gm,

as PQ || AB ⇒ OQ || BM and BC || AD ⇒ OM || BQ

 

Similarly DLOP is also a ||gm

as AD || LM ⇒ DP || LO and PQ || DC ⇒ PO || DL

 

∴ Area (Parallelogram BMOQ) = Area (Parallelogram DLOP)