Question

А
ar (ABE)=ar (ACF)
9.
The side AB of a parallelogram ABCD is produced
to any point P. A line through A and parallel to CP
meets CB produced at Q and then parallelogram
PBQR is completed (see Fig. 11.26). Show that
ar (ABCD)=ar (PBQR).
[Hint: Join AC and PQ. Now compare ar (ACQ)
B
and ar (APQ).]
Q
R
Fig. 11.26
0. Diagonals AC and RD of a tranazium Ann​

Answer 1

$\huge \underline {\boxed{Answer}}$

Construction: Draw diagonals AC and PQ.

Proof: AQ II CP Triangles ACQ and APQ are on the same base AQ and between the same parallels AQ and PC.

∴arACQ)=ar(APQ)

⇒ar(ACB)+ar(ABQ)=ar(ABQ)+ar(PBQ)

⇒ar(ACB)=ar(PBQ)----(i)

$Also,\\ ar(ACB)= \frac{1}{2} ar(ABCD)$----(ii)

Diagonals of parallelogram divide the parallelogram into two equal parts of equal area.

$ar(ACB)= \frac{1}{2} ar(PBQR)$---(iii )

Reason same as above(diagonals of parallelogram divide the parallelogram into two equal parts of equal area.)

From i, ii, iii, we get

$\frac{1}{2} ar(ABCD)= \frac{1}{2} ar(PBQR)$

⇒ ar(ABCD)=ar(PBQR)