Question

ABCD is a parallelogram whose area is 100sq.cm.P is any point inside the parellelogram (see fig.) find the area of angleAPB+angleCPD.​

Answer 1

Answer:

Given ABCD is a parallelogram area is 100 sq. cm

P is a point inside the parallelogram

Construction : Let us drawn a line passing through P such that PQ||AB.

AB||CD _____ (1)

AB || PQ _____ (2)

from (1) & (2) CD || PQ

AB||PQ (By construction)

PA||QB (∵ PA and QB is a part of parallelogram DA and CB)

∴ ABQP is a parallelogram.

Similarly we can prove that PQCD is a parallelogram.

parallelogram ABQP and ΔAPB lie on same base and same parallels ∴ar(PAB)=

2

1

ar(ABQP)

or 2ar(ΔPAB)=ar(ABQP)

Parallelogram PQCD and ΔDPC lie on same base and same parallels ∴ar(ΔDPC)=

2

1

ar((11gmPQCD))

or 2.ar(ΔDPC)=ar(11gmPQCD)

ar (11 gm PQCD) = 100cm

2

(given)

⇒ar(11gmABPQ)+ar(11PQCD)=100cm

2

2.ar(ΔPAB)+2ar(ADPC)=100cm

2

2.ar[(ΔPAB)+(ΔDPC)]=100cm

2

ar.(ΔPAB)+ar(ΔDPC)=

2

100

cm

2

ar(ΔPAB)+ar(ΔDPC)=50cm

2

∴ar(ΔAPB)+ar(ΔCPD)=50cm

2