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Areas of Parallelograms and Triangles

Short Answer Type

31. In the figure, ABCD is a parallelogram. P is a point on AB produced and DN ⊥ AB. If AB = 8 cm and DN = 3 cm. Find the area of ΔCPD.

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32. ABCD is a parallelogram in which BC is produced to E such that CE = BC (see figure). AE intersects CD at F. Prove that ar(ΔBFC)

equals 1 fourth space ar left parenthesis ABCD right parenthesis.

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33. In the figure, AB || CD. E and F are the points on the sides CD and AB respectively. If ar(ΔABE) = ar(ΔDCF), then prove that AB = DC.

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34. In figure, E is any point on median AD of a ΔABC. Show that ar(ΔABE) = ar(ΔACE).

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35. In a triangle ABC, E is the midpoint of median AD. Show that

a r left parenthesis increment B E D right parenthesis space equals 1 fourth a r left parenthesis increment A B C right parenthesis

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36. Show that the diagonals of a parallelogram divide it into four triangles of equal area.

Given: ABCD is a parallelogram whose diagonals AC and BD intersecting at O divide it into four trianlges ΔOAB, ΔOBC, ΔOCD and ΔODA.

To Prove: ar(ΔOAB) = ar(ΔOBC)
= ar(ΔOCD) = ar(ΔODA). Construction: Draw BE ⊥ AC.
Proof: ∵ ABCD is a parallelogram ∴ OA = OC and OB = OD
| ∵ Diagonals of a parallelogram bisect each other
Now,

ar left parenthesis increment OAB right parenthesis equals space fraction numerator Base space straight x space Corresponding space altitude over denominator 2 end fraction space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator left parenthesis OA right parenthesis left parenthesis BE right parenthesis over denominator 2 end fraction and space space space ar left parenthesis increment OBC right parenthesis equals space fraction numerator Base space straight x space Corresponding space altitude over denominator 2 end fraction space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator left parenthesis OC right parenthesis left parenthesis BE right parenthesis over denominator 2 end fraction

But OA = OC
∴ ar(ΔOAB) = ar(ΔOBC)    ...(1)
Similarly,
ar(ΔOBC) = ar(ΔOCD)    ...(2)
and, ar(ΔOCD) = ar(ΔODA)    ...(3)
From (1), (2) and (3), ar(ΔOAB) = ar(ΔOBC)
= ar(ΔOCD) = ar(ΔODA).

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37. In figure, ABC and ABD are two triangles on the same base AB. If line-segment CD is bisected by AB at O. Show that ar(AABC) = ar(ΔABD).

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Long Answer Type

38.

D, E and F are respectively the midpoints of the sides BC, CA and AB of a ΔABC. Show that:

(i) BDEF is a parallelogram

$left parenthesis ii right parenthesis space ar left parenthesis increment DEF right parenthesis equals 1 fourth ar left parenthesis increment ABC right parenthesis left parenthesis iii right parenthesis space ar left parenthesis square space space BDEF right parenthesis equals 1 half ar left parenthesis increment ABC right parenthesis$