ABCD is a parallelogram and P is any point on the side CD. Prove that area of AAPD + area of ABCP = area of AABP
Answers
Answer:
Step-by-step explanation:
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