P is a point in the interior of a
parallelogram ABCD. Show that
(1) ar(APB)-ar(PCD) = ar (ABCD)
2
(i) ar (APD)-ar (PBC)=ar (APB)+ar (PCD)
(Hint: Through P. draw a line parallel to AB.]
Answers
Answered by
1
Answer:
Assume two lines EF and GH such that EF∥AB and GH∥AD.
In ∥gmAEFB;ar(APB)=
2
1
ar(AEFB)⟶(1)
In ∥g,EFCD;ar(DPC)=
2
1
ar(EFCD)⟶(2)
From (1) and (2) (adding )
⇒ar(APB)+ar(PDC)=
2
1
[ar(AEFB)+ar(EFCD)]
⇒ar(APB)+ar(PDC)=
2
1
ar(ABCD)−
In ∥gmAGHD;ar(APD)=
2
1
ar(AGHD)⟶(3)
In ∥gmGBCH;ar(PBC)=
2
1
ar(GBCH)⟶(4)
Adding (3) and (4)
⇒ar(APD)+ar(PBC)=
2
1
[ar(AGHD)+ar(GBCH)]
⇒ar(APD)+ar(PBC)=
2
1
ar(ABCD)
From previous part -
⇒ar(APD)+ar(PBC)=ar(APB)+ar(PCD).
Hence, solved.
Similar questions