In Fig. 9.16. P is a point in the interior of a
parallelogram ABCD. Show that
B
© ar (APB) + ar (PCD) a (ABCD)
D
C
War (APD)+ ar (PBC)=ar (APB) + ar (PCD)
(Hint: Through P. draw a line parallel to AB.
Fig. 9.16
Answers
Step-by-step explanation:
(i) Let us draw a line segment EF, passing through point P and
parallel to line segment AB.
In
parallelogram ABCD,
AB
|| EF (By construction) ... (1)
ABCD
is a parallelogram.
∴
AD || BC (Opposite sides of a parallelogram)
⇒
AE || BF ... (2)
From
equations (1) and (2), we obtain
AB
|| EF and AE || BF
Therefore,
quadrilateral ABFE is a parallelogram.
It
can be observed that ΔAPB
and parallelogram ABFE are lying on the same base AB and between the
same parallel lines AB and EF.
∴
Area (ΔAPB) =
Area
(ABFE) ... (3)
Similarly,
for ΔPCD and parallelogram
EFCD,
Area
(ΔPCD) =
Area
(EFCD) ... (4)
Adding
equations (3) and (4), we obtain
(ii)
Let
us draw a line segment MN, passing through point P and parallel to
line segment AD.
In
parallelogram ABCD,
MN
|| AD (By construction) ... (6)
ABCD
is a parallelogram.
∴
AB || DC (Opposite sides of a parallelogram)
⇒
AM || DN ... (7)
From
equations (6) and (7), we obtain
MN
|| AD and AM || DN
Therefore,
quadrilateral AMND is a parallelogram.
It
can be observed that ΔAPD
and parallelogram AMND are lying on the same base AD and between the
same parallel lines AD and MN.
∴
Area (ΔAPD) =
Area
(AMND) ... (8)
Similarly,
for ΔPCB and parallelogram
MNCB,
Area
(ΔPCB) =
Area
(MNCB) ... (9)
Adding
equations (8) and (9), we obtain
On
comparing equations (5) and (10), we obtain
Area
(ΔAPD) + Area (ΔPBC)
= Area (ΔAPB) + Area (ΔPCD)