Question

triangles.)
ar (PAB) = ar (BQP)
(2)
So,
1
(3)
Therefore,
ar (PAB) =
ar (ABQP) [From (2)]
2
ar (PAB) =
ar (ABCD)
[From (1) and (3)
This gives
EXERCISE 9.2
In Fig. 9.15, ABCD is a parallelogram, AE I DC
and CF I AD. If AB = 16 cm, AE = 8 cm and
CF= 10 cm, find AD.
If E,F,G and H are respectively the mid-points of
the sides of a parallelogram ABCD, show that
2.
D
Fig. 9.15
1
ar (EFGH)=
2
1
ar (ABCD)
3. Pand Q are any two points lying on the sides DC and AD respectively of a parallelogram
ABCD. Show that ar (APB) = ar (BQC).
4. In Fig. 9.16, P is a point in the interior of a
parallelogram ABCD. Show that​

Answer 1

abcd is Bo's 16 cm ab 8 can