Question 4 (i)
In the given figure, P is a point in the interior of a parallelogram ABCD. Show that
$(i) a r (A P B) + a r (P C D) = \frac{1}{2} a r (A B C D)$

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Solution

(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)
$\Rightarrow$ 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 $Δ A P B$ and parallelogram ABFE are lying on the same base AB
and between the same parallel lines AB and EF.
$∴ A r e a (Δ A P B) = \frac{1}{2} A r e a (A B F E) . . . (3)$
Similarly, for ΔPCD and parallelogram EFCD,
$A r e a (Δ P C D) = \frac{1}{2} A r e a (E F C D) . . . (4)$
Adding equations (3) and (4), we obtain,
$A r e a (Δ A P B) + A r e a (Δ P C D) - \frac{1}{2} [A r e a (A B F E) + A r e a (E F C D)]$
$A r e a (Δ A P B) + A r e a (Δ P C D) = \frac{1}{2} A r e a (A B C D) . . . . (5)$

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Similar questions

(i) a r (A P B) + a r (P C D) = \frac{1}{2} a r (A B C D)

P

A B C D

a r (△ A P B) + a r (△ P C D) = \frac{1}{2} a r (A B C D)

(i) a r (A P B) + a r (P C D) = \frac{1}{2} a r (A B C D)

(ii) a r (A P D) + a r (P B C) = a r (A P B) + a r (P C D)

In Fig, P is a point in the interior of a parallelogram ABCD. Show that
(i) ar(APB)+ar(PCD)=½ ar(ABCD)
(ii) ar(APD)+ar(PBC)=ar(APB)+ar(PCD)

In the given figure, P is a point in the interior of a parallelogram ABCD. Show that

(i) ar (APB) + ar (PCD) = ar (ABCD)

(ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD)

[Hint: Through. P, draw a line parallel to AB]

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