Areas of Parallelograms and Triangles| NCERT Solutions| School Help by Gunjan

NCERT Solutions of Chapter: Areas of Parallelograms and Triangles. NCERT Solutions along with worksheets and notes for Class 9.

Exercise 9.1

(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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(i)

(ii)

(iii)

(iv)

(v)

(vi)

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(3)P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar(APB) = ar(BQC).

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(i)

(ii)

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(6) 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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(3) Show that the diagonals of a parallelogram divide it into four triangles of equal area.

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(4) In figure below, ABC and ABD are two triangles on the same base AB. If line – segment CD is bisected by AB at O, show that ar(ABC) = ar (ABD).

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Draw perpendiculars DE and BF from points D and B to AC respectively.

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(9) 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 (Figure below). Show that ar(ABCD) = ar (PBQR).

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

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(12)A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to rake over some portion of his plot from one of the corners 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 from a triangular plot. Explain how this proposal will be implemented.

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(15) Diagonals AC and BD of a quadrilateral ABCD intersects at O in such a way that ar(AOD) = ar(BOC). Prove that ABCD is a trapezium.

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(16) In figure below, ar(DRC) = ar(DPC) and ar(BDP) = ar(ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

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Exercise 9.4

(1) Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.

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(2) In figure below, D and E are two points on BC such that BD = DE = EC. Show that ar(ABD) = ar(ADE) = ar(AEC).

Can you now answer the question that you have left in the ‘Introduction’ of this chapter, whether the field of the Budhia has been actually divided into three parts of equal area?

[Remark: Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into n equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide ABC into n triangles of equal areas.]

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(3)In figure below, ABCD and DCFE and ABFE are parallelograms. Show that ar(ADE) = ar(BCF).

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(4) In figure below, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersects DC at P, show that ar(BPC) = ar(DPQ).

[Hint: Join AC]