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Areas of Parallelograms and Triangles

Short Answer Type

1. Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and the two parallels.

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2. In figure, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.

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Long Answer Type

If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar(EFGH) = $1 half$ ar(ABCD).

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Short Answer Type

4. P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar(ΔAPB) = ar{ΔBQC).

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

(1) ar(

ar left parenthesis increment APB right parenthesis plus ar left parenthesis increment PCD right parenthesis equals 1 half ar left parenthesis space parallel to space gm space ABCD right parenthesis

(ii) ar(ΔAPD) + ar(ΔPBC) = ar(ΔAPB) + ar(ΔPCD). [CBSE 2012 (March)]

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

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Long Answer Type

In figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that:

(i) ar(|| gm PQRS) = ar(|| gm ABRS)

(ii) $ar left parenthesis increment AXS right parenthesis equals 1 half ar left parenthesis space parallel to space gm space PQRS right parenthesis.$

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Short Answer Type

7. A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the fields is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?

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8. The diagonals of a parallelogram ABCD intersect at a point O. Through O, a line is drawn to intersect AD at P and BC at Q. Show that PQ divides the parallelogram into two parts of equal area.

Given: The diagonals of a parallelogram ABCD intersect at a point O. Through O, a line is drawn to intersect AD at P and BC at Q.

To Prove: ar(□PDCQ) = ar(□PQBA). Proof: ∵ AC is a diagonal of || gm ABCD ∴ ar(ΔABC) = ar(ΔACD)

equals space 1 half space ar left parenthesis space parallel to space gm space ABCD right parenthesis space space space space space space space space space... left parenthesis 1 right parenthesis

In ΔAOP and ΔCOQ,
AO = CO
| ∵ Diagonals of a parallelogram bisect each other
∠AOP = ∠COQ
| Vertically opposite angles ∠OAP = ∠OCQ
| Alternate interior angles ∴ ΔAOP = ΔCOQ
| By ASA Congruence Rule ∴ ar(ΔAOP) = ar(ΔCOQ)
| ∵ Congruent figures have equal areas ⇒ ar(ΔAOP) + ar(□OPDC)
= ar(ΔCOQ) + ar(□OPDC)
⇒ ar(ΔACD) = ar(□PDCQ)
⇒ 1/2 ar(|| gm ABCD) = ar(□PDCQ)
| From (1) ⇒ ar(□PQBA) = ar(□PDCQ)
⇒ ar(□PDCQ) = ar(□PQBA).

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9. Prove that of all parallelograms of which the sides are given, the parallelogram which is a rectangle, has the greatest area.

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10. In the figure, diagonals AC and BD of a trapezium ABCD with AB || CD intersect each other at O. Show that ar(Δ AOD) = ar(Δ BOC).

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