The document Ex 9.3 NCERT Solutions- Areas of Parallelograms and Triangles Class 9 Notes | EduRev is a part of the Class 9 Course Class 9 Mathematics by VP Classes.

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**Questions 1. In the figure, E is any point on median AD of a DABC. Show that ar (ABE) = ar (ACE). Solution: **We have a Î”ABC such that AD is a median. âˆ´

ar (Î”ABD) = ar (Î”ADC) ...(1)

[âˆµ A median divides the triangle into two triangles of equal areas.] Similarly, in DBEC, we have

ar (Î”BED) = ar (Î”DEC) â€¦(2)

Subtracting (2) from (1), we have

ar (Î”ABD) - ar (Î”BED) = ar (Î”ADC) - ar (Î”DEC)

â‡’ ar (Î”ABE) = ar (Î”ACE).

**Question 2. In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) = ar (ABC). Solution: **We have a Î”ABC and its median AD.

Since, a median divides the triangle into two triangles equal in area.

âˆ´ ar (Î”ABD) = (1/2) ar (Î”ABC) â€¦(1)

Let us join B and E.

Now, in Î”ABD, BE is a median. [âˆµ E is the mid-point of AD]

âˆ´ ar (Î”BED) = (1/2)ar (Î”ABD)

From (1) and (2), we have ar (Î”BED) = (1/2)[(1/2)]ar ( ABC) ]

â‡’ ar (Î”BED) = (1/4)ar (Î”ABC)

**Question 3. Show that the diagonals of a parallelogram divide it into four triangles of equal area. Solution:** We have a parallelogram ABCD such that its diagonals intersect at O.

âˆµ Diagonals of a parallelogram bisect each other.

âˆ´ AO = OC and BO = OD Let us draw CE âŠ¥ BD.

Now, ar (Î”BOC) = (1/2) BO x CE

and ar (Î”DOC) = (1/2) OD x CE

Since, BO = OD âˆ´

ar (Î”BOC) = ar (Î”DOC) â€¦(1)

Similarly, ar (Î”AOD) = ar (Î”DOC) â€¦(2)

and ar (Î”AOB) = ar (BOC) â€¦(3)

From (1), (2) and (3), we have

ar (Î”AOB) = ar (Î”BOC)

= ar (Î”BOC) = ar (Î”AOD) Thus, the diagonals of a parallelogram divide it into four triangles of equal area.

**Question 4. 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 (ABC) = ar (ABD). Solution:** We have Î”ABC and Î”ABD are on the same base AB.

âˆµ CD is bisected at O [Given]

âˆ´ AO = BO

Now, in Î”ACD, AO is a median.

âˆ´ ar (Î”OAC) = ar (Î”OAD) â€¦(1)

Again, in DBCD, BO is a median.

âˆ´ ar (Î”OBC) = ar (Î”ODC) â€¦(2)

Adding (1) and (2), we have ar (Î”OAC) + ar (Î”OBC)

= ar (Î”OAD) + ar (Î”ODC)

â‡’ ar (Î”ABC) = ar (Î”ABD)

**Question 5. D, E and F are respectively the mid-points of the sides BC, CA and AB of a Î”ABC. Show that (i) BDEF is a parallelogram. (ii) ar (DEF) = (1/4) **

**(iii) ar (BDEF) = **(1/2)**ar (ABC)**

**Solution: **We have: Î”ABC such that the mid-points of BC, CA and AB are respectively D, E and F.

(i) To prove that BDEF is a parallelogram.

In Î”ABC, E and F are the mid-points of AC and AB.

âˆ´ EF || BC

â‡’ EF || BD

Also EF =(1/2) BC

â‡’ EF = BD [âˆµ D is the mid-point of BC.]

âˆµ BDEF is a quadrilateral whose one pair of opposite sides is parallel and of equal lengths.

âˆ´ BDEF is a parallelogram.

(ii) To prove that ar (DEF) = (1/4)ar (ABC).

We have proved that BDEF is a parallelogram.

Similarly, CDEF is a parallelogram and Î”EAF is a parallelogram.

Now, parallelogram BDEF and parallelogram CDEF are on the same base EF and between the same parallels BC and EF.

âˆ´ ar (BDEF) = ar (CDEF)

â‡’(1/2) ar (BDEF) = (1/2)ar (DEF)

â‡’ ar (Î”BDF) = ar (Î”CDF) â€¦(1)

[âˆµ Diagonal of a parallelogram divides it into two triangles of equal area]

Similarly, ar (Î”CDE) = ar (Î”DEF) â€¦(2)

ar (Î”AEF) = ar (Î”DEF) â€¦(3)

From (1), (2) and (3), we have

ar (Î”AEF) = ar (Î”FBD)

= ar (Î”DEF) = ar (Î”CDE)

Thus, ar (Î”ABC) = ar (Î”AEF) + ar (Î”FBD) + ar (Î”DEF) + ar (Î”CDE)

= 4 ar (Î”DEF)

â‡’ ar (Î”DEF) =(1/4) ar (Î”ABC)

(iii) To prove that ar (BDEF) = (1/2)ar (ABC).

We have ar (BDEF)

= ar (Î”BDF) + ar (Î”DEF)

= ar (Î”DEF) + ar (Î”DEF) [âˆµ ar (Î”DEF) = ar (Î”BDF)]

= 2 ar (Î”DEF)

= 2 [ (1/4)ar ( ABC)]

= 2 x (1/4)ar (Î”ABC)

= (1/2)ar (Î”ABC)

Thus, ar (BDEF) =(1/2)ar (Î”ABC)

**Question 6. In the figure, diagonals AC and BD of a quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that**

**(i) ar (DOC) = ar (AOB) (ii) ar (DCB) = ar (ACB) (iii) DA || CB or ABCD is a parallelogram. Hint: From D and B, draw perpendiculars to AC. Solution:** We have a quadrilateral ABCD whose diagonals AC and BD intersect at O. We also have that OB = OD

AB = CD

Let us draw DE âŠ¥ AC and BF âŠ¥ AC

(i) To prove that ar (Î”DOC) = ar (Î”AOB)

In Î”DEO and Î”BFO, we have

DO = BO [Given]

â€“DOE = â€“BOF

[Vertically opposite angles]

â€“DEO = â€“BFO [Each = 90Âº]

âˆ´ Î”DEO â‰Œ Î”BFO [ASA criteria]

â‡’ DE = BF [c.p.c.t]

and ar (Î”DEO) = ar (Î”BFO) â€¦(1)

[âˆµ Congruent triangle are equal in areas]

Now, in Î”DEC and Î”BFA, we have

â€“DEC = â€“BFA [Each = 90Âº]

DE = BF [Proved]

DC = BA [Given]

âˆ´ Î”DEC â‰Œ Î”BFA [RHS criteria]

â‡’ ar (Î”DEC) = ar (Î”BFA) â€¦(2)

Adding (1) and (2), we have ar (Î”DEO) + ar (Î”DEC) = ar (Î”BFO) + ar (Î”BFA)

â‡’ ar (Î”DOC) = ar (Î”AOB)

(ii) To prove that ar (DCB) = ar (ACB)

âˆµ ar (Î”DOC) = ar (Î”AOB) [Proved]

âˆ´ Adding ar (Î”BOC), on both sides, we have

ar (Î”DOC) + ar (Î”BOC) = ar (Î”AOB) + ar (Î”BOC)

â‡’ ar (Î”DCB) = ar (DACB)

(iii) To prove that DA || CB

âˆµ Î”DCB and Î”ACB are on the same base CB and having equal areas.

âˆ´ They lie between the same parallels CB and DA.

â‡’ CB || DA (or ABCD is a parallelogram)

**Question 7. D and E are points on sides AB and AC respectively of Î”ABC such that ar (DBC) = ar (EBC). Prove that DE || BC.**

**Solution: **We have Î”ABC and points D and E are such that ar (DBC) = ar (EBC) Since, Î”DBC and Î”EBC are on the same base DE and having same area.

âˆ´ They lie between the same parallels DE and BC. â‡’ DE || BC.

**Question 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 (ABE) = ar (ACF).**

**Solution: **We have a DABC such that XY || BC, BE || AC and CF || AB.

Since, XY || BC and BE || CY

âˆ´ BCYE is a parallelogram.

Now, the parallelogram BCYE and Î”ABE are on the same base BE and between the same parallels BE and AC.

âˆ´

ar (Î”ABE) = (1/2)ar (BCYE) â€¦(1)

Again, CF || AB [Given]

XY || BC [Given]

â‡’ CF || AB and XF || BC

âˆ´ BCFX is a parallelogram.

Now, Î”ACF and parallelogram BCFX are on the same base CF and between the same parallels AB and FC.

âˆ´ ar (Î”ACF) = (1/2)ar (BCFX) â€¦(2)

Also parallelogram BCFX and parallelogram BCYE are on the same base BC and between the same parallels.

âˆ´ ar (BCFX) = ar (BCYE) â€¦(3)

From (1), (2) and (3) we get

ar (Î”ABE) = ar(Î”ACF)

**Question 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 (see figure). Show that ar (ABCD) = ar (PBQR).**

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

Solution: We have a parallelogram ABCD. AB is produced to P. CB is produced to Q and PBQR is completed.

Let us join AC and PQ.

âˆµ ABCD is a parallelogram [Given] and AC is its diagonal.

ar (Î”ABC) = (1/2)ar (parallelogram ABCD) â€¦(1)

âˆµ BQRP is a parallelogram [Given] and QP is its diagonal.

âˆ´ ar (Î”BPQ) = (1/2)ar (parallelogram BQRP) â€¦(2)

Since Î”ACQ and Î”APQ are on the same base AQ and between the same parallels AQ and CP.

âˆ´ ar (Î”ACQ) = ar (Î”APQ) [âˆµ Triangles on the same base and between the same parallels are equal.]

â‡’ ar (Î”ACQ) - ar (Î”ABQ) = ar (DAPQ) - ar (DABQ) [Subtracting ar (Î”ABQ) from both sides]

â‡’ ar (Î”ABC) = ar (Î”BPQ) â€¦(3)

From (1), (2) and (3),

we get (1/2)ar (parallelogram ABCD) = (1/2)ar (parallelogram BQRP)

â‡’ 2 [ (1/2)ar (parallelogram ABCD)] = 2 [(1/2)ar (parallelogram BQRP)]

â‡’ ar (parallelogram ABCD) = ar (parallelogram BQRP)

**Question 10. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC). Solution: **We have a trapezium ABCD having AB || CD and its diagonals AD and BC are joined.

Since, triangles on the same base and between the same parallels have equal areas.

âˆµ Î”ABD and Î”ABC are on the same base AB and between the same parallels AB and DC.

âˆ´ ar (Î”ABD) = ar (Î”ABC)

Subtracting ar (Î”AOB) from both sides, we get

[ar (Î”ABD) â€“ ar (Î”AOB)] = [ar (Î”ABC) â€“ ar (Î”AOB)]

â‡’ ar (Î”AOD) = ar (Î”BOC)

**Question 11. In the figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that (i) ar (ACB) = ar (ACF) (ii) ar (AEDF) = ar (ABCDE)**

**Solution:** We have a pentagon ABCDE in which BF || AC and DC is produced to F.

(i) To prove ar (Î”ACB) = ar (Î”ACF)

Since, the triangles between the same parallels and on the same base are equal in area.

âˆµ Î”ACB and Î”ACF are on the same base AC and between AC and BF.

âˆ´ ar (Î”ACB) = ar (Î”ACF)

(ii) Since, ar (Î”ACB) = ar (Î”ACF) [Proved] Adding ar (AEDC) to both sides, we get

â‡’ [ar (Î”ACB) + ar (AEDC)] = [ar (Î”ACF) + ar (AEDC)]

â‡’ ar (AEDF) = ar (ABCDE)

**Question 12. 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 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 form a triangular plot. Explain how this proposal will be implemented.**

**Solution:** We have a plot in the form of a quadrilateral ABCD.

Let us draw DF || AC and join A and F.

Now, Î”DAF and Î”DCF are on the same base DF and between the same parallels AC and DF.

âˆ´ ar (Î”DAF) = ar (Î”DCF) Subtracting ar (Î”DEF) from both sides, we get

[ar (Î”DAF) - ar (Î”DEF)] = [ar (Î”DCF) - ar (Î”DEF)]

â‡’ ar (Î”ADE) = ar (Î”CEF)

The portion Î”ADE can be taken over by the Gram Panchayat by adding the land (Î”CEF) to his (Itwaari) land so as to form a triangular plot,

i.e. Î”ABF.

Let us prove that ar (Î”ABF) = ar (quadrilateral ABCD), we have

ar (Î”CEF) = ar (Î”ADE) [Proved] Adding ar (ABCE) to both sides, we get

[ar (Î”CEF) + ar (ABCE)] = [ar (Î”ADE) + ar (ABCE)]

â‡’ ar (Î”ABF) = ar (quadrilateral ABCD)

**Question 13. ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY). Hint: Join CX. Solution: **We have a trapezium ABCD such that AB || DC.

XY || AC meets AB at X and BC at Y.

Let us join CX.

âˆµ AB || DC [Given]

âˆ´ Î”ADX and Î”ACX are on the same base AX and between the same parallels AB and DC.

âˆ´ ar (Î”ADX) = ar (Î”ACX) â€¦(1)

âˆµ AC || XY [Given]

âˆ´ Î”ACX and Î”ACY are on the same base AC and between the same parallels AC and XY.

âˆ´ ar (Î”ACX) = ar (Î”ACY) â€¦(2)

From (1) and (2), we have

ar (Î”ADX) = ar (Î”ACY).

**Question 14. In the figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR). Solution: **We have AP || BQ || CR

Since, BQ || CR [Given]

âˆ´ Î”BCQ and Î”BQR are on the same base BQ and between the same parallels BQ and CR.

â‡’ ar (Î”BCQ) = ar (Î”BQR) â€¦(1)

Again, AP || BQ [Given]

âˆµ Î”ABQ and Î”PBQ are on the same base BQ and between the same parallels.

âˆ´ ar (Î”ABQ) = ar (Î”PBQ) â€¦(2)

Adding (1) and (2), we have

[ar (Î”BCQ) + ar (Î”ABQ)] = [ar (Î”BQR) + ar (Î”PBQ)]

â‡’ ar (Î”AQC) = ar (Î”PBR)

**Question 15. Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a trapezium. Solution:** We have a quadrilateral ABCD and its diagonals AC and BD intersect at O such that ar (Î”AOD) = ar (Î”BOC).

Now, ar (Î”AOD) = ar (Î”BOC) [Given]

Adding ar (Î”AOB) to both sides, we have

[ar (AOD) + ar (AOB)] = [ar (BOC) + ar (AOB)]

â‡’ ar (Î”ABD) = ar (Î”ABC)

But they are on the same base AB.

Since, the triangles on the same base and having equal area lie between the same parallels.

âˆ´ AB || DC

Now, ABCD is a quadrilateral having a pair of opposite sides parallel.

âˆ´ ABCD is a trapezium.

**Question 16. In the figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums. Solution:** We have ar (Î”DRC) = ar (Î”DPC)

[Given] But they are on the same base DC.

âˆ´ Î”DRC and Î”DPC must lie between the same parallels.

â‡’ DC || RP

â‡’ One pair of opposite sides of quadrilateral DCPR is parallel.

âˆ´ Quadrilateral DCPR is a trapezium.

Again, we have ar Î”(BDP) = ar Î”(ARC) [Given] .â€¦(1)

Also ar (Î”DPC) = ar Î”(DRC) [Given] â€¦.(2)

Subtracting (2) from (1),

we get [ar (Î”BDP) - ar (Î”DPC)] = [ar (Î”ARC) - ar (Î”DRC)]

â‡’ ar (Î”BDC) = ar (Î”ADC) But they are on the same base DC.

âˆ´ Î”BDC and Î”ADC must lie between the same parallels.

â‡’ AB || DC i.e. A pair of quadrilateral ABCD is parallel

âˆ´ Quadrilateral ABCD is a trapezium.

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