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Areas of Parallelograms and Triangles

Long Answer Type

41. 8. XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively, show that ar(ΔABE) = ar(ΔACF).

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42. 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. Show that ar(||gm ABCD) = ar(||gm PBQR).

[Hint. Join AC and PQ. Now compare ar(ACQ) and ar(APQ)].

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Short Answer Type

43. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar(ΔAOD) = ar(ΔBOC).

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44.

In figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that
(i) ar(ΔACB) = ar(ΔACF)
(ii) ar(□AEDF) = ar(ABCDE).

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45. A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the comers to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.

Let ABCD be the plot of land of the shape of a quadrilateral. Let the portion ADE be taken over by the Gram Panchayat of the village from one comer D to construct a Health Centre.
Join AC. Draw a line through D parallel to AC to meet BC produced in P. Then Itwaari must be given the land ECP adjoining his plot so as to form a triangular plot ABP as then
ar(ΔADE) = ar(ΔPEC).

Let ABCD be the plot of land of the shape of a quadrilateral. Let the

Proof: ∵ ΔDAP and ΔDCP are on the same base DP and between the same parallels DP and AC.
∴ ar(ΔDAP) = ar(ΔDCP)
Two triangles on the same base (or equal bases) and between the same parallels are equal in area
⇒ ar(ΔDAP) – ar(ΔDEP)
= ar(ΔDCP) – ar(ΔDEP)
| Subtracting the same areas from both sides
⇒ ar(ΔADE) = ar(ΔPCE)
⇒ ar(ΔDAE) + ar(□ABCE)
= ar(ΔPCE) + ar(□ABCE)
| Adding the same areas to both sides
⇒ ar(□ABCD) = ar(ΔABP).

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46. ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar(ΔADX) = ar(ΔACY).

[Hint. Join CX.]

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47. In figure, AP || BQ || CR. Prove that ar(ΔAQC) = ar(ΔPBR).

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48. Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar(ΔAOD) = ar(ΔBOC). Prove that ABCD is a trapezium.

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49. Infigure, ar(ΔDRC) = ar(ΔDPC) and ar(ΔBDP) = ar(ΔARC). Show that both the quadrilateral ABCD and DCPR are trapeziums.

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50.

Show that the area of a rhombus is half the product of the lengths of its diagonals.

Prove that the area of a rhombus is equal to half the rectangle contained by its diagonals.

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