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Areas of Parallelograms and Triangles

Short Answer Type

11. Parallelograms on the same base and between same parallels are equal in area. Prove this.

Given: Two parallelograms ABCD and EFCD, on the same base DC and between the same parallels AF and DC.
To Prove: ar(ABCD) = ar(EFCD)

Proof: In ΔADE and ΔBCF,
∠DAE = ∠CBF    ...(1)
Corresponding angles
(∵ AD || BC and a transversal AF intersects them)
∠AED = ∠BFC    ...(2)
Corresponding angles (∵ ED || FC and a transversal AF intersects them)
∴ ∠ADE = ∠BCF    ...(3)
| Angle sum property of a triangle Also AD = BC    ...(4)
Opposite sides of the parallelogram ABCD
∴ ΔADE ≅ ΔBCF | By ASA congruence rule, using (1), (3) and (4) ∴ ar(ΔADE) = ar(ΔBCF)
| ∵ Congruent figures have equal areas ⇒ ar(ΔADE) + ar(EDCB)
= ar(ΔBCF) + ar(EDCB)
| Adding ar(EDCB) to both sides ⇒ ar(ABCD) = ar(EFCD)
So, parallelograms ABCD and EFCD are equal in area.

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Long Answer Type

12. In figure, AD is median of triangle ABC, E is the mid-point of AD and F is the mid-point of AE.
Prove that

ar open parentheses ABF close parentheses equals 1 over 8 ar left parenthesis ABC right parenthesis.

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Short Answer Type

13. ABCD is a quadrilateral and BD is one of its diagonals as shown in figure. Show that ABCD is a parallelogram and find its area.

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14. In the given figure, AB | | DC. Show that ar(BDE) = ar(ACED).

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15.

In the given figure, ABED is a parallelogram in which DE = EC. Show that area (ABF) = area (BEC)

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16. Areas of triangles on the same bases and between the same parallels are equal in. Prove it.

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17. Prove that the area of a trapezium is equal to half of the product of its height and sum of parallel sides.

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18.

In the following figure, ABCD is a parallelogram and EFCD is a rectangle. Also, AL ⊥ DC. Prove that
(i) ar(ABCD) = ar(EFCD)
(ii) ar(ABCD) = DC x AL.

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19. If a triangle and a parallelogram are on the same base and between the same parallels, then prove that the area of the triangle is equal to half the area of the parallelogram.

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20. In the figure, PQRS is a parallelogram with PQ = 12 cm, altitudes corresponding to PQ and SP are respectively 8 cm and 10 cm. Find SP.

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