# NCERT Solutions for Class 9 Maths Exercise 9.3

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NCERT solutions for Class 9 Maths Areas of Parallelograms and Triangles

## NCERT Solutions for Class 9 Mathematics Areas of Parallelograms and Triangles

###### 1. In figure, E is any point on median AD of a ABC. Show that ar (ABE) = ar (ACE).

Ans. In ABC, AD is a median.

ar (ABD) = ar (ACD) ……….(i)

[ Median divides a into two s of equal area]

Again in EBC, ED is a median

ar (EBD) = ar (ECD) ……….(ii)

Subtracting eq. (ii) from (i),

ar (ABD) – ar (EBD) = ar (ACD) – ar (ECD)

ar (ABE) = ar (ACE)

###### 2. In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) = ar (ABC).

Ans. Given: A ABC, AD is the median and E is the mid-point of median AD.

To prove: ar (BED) = ar (ABC)

Proof: In ABC, AD is the median.

[ Median divides a into two s of equal area]

ar (ABD) = ar (ABC) ……….(i)

In ABD, BE is the median.

ar (BED) = ar (BAE)

ar (BED) = ar (ABD)

ar (BED) = ar (ABC) = ar (ABC)

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

Ans. Let parallelogram be ABCD and its diagonals AC and BD intersect each other at O.

AB = DC [Opposite sides of a parallelogram]

BC = AD [Opposite sides of a parallelogram]

And AC = AC [Common]

ABC CDA [By SSS congruency]

Since, diagonals of a parallelogram bisect each other.

O is the mid-point of bisection.

Now in ADC, DO is the median.

ar (AOD) = ar (COD) ……….(i)

[Median divides a triangle into two equal areas]

Similarly, in ABC, OB is the median.

ar (AOB) = ar (BOC) ……….(ii)

And in AOB and AOD, AO is the median.

ar (AOB) = ar (AOD) ……….(iii)

From eq. (i), (ii) and (iii),

ar (AOB) = ar (AOD) = ar (BOC) = ar (COD)

Thus diagonals of parallelogram divide it into four triangles of equal area.

NCERT Solutions for Class 9 Maths Exercise 9.3

###### 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).

Ans. Draw CMAB and DNAB.

In CMO and DNO,

CMO = DNO = [By construction]

COM = DON [Vertically opposite]

OC = OD [Given]

CMO DNO [By ASA congruency]

AM = DN [By CPCT] ……(i)

Now ar (ABC) = ……….(ii)

Using eq. (i) and (iii),

From eq. (ii) and (iv),

NCERT Solutions for Class 9 Maths Exercise 9.3

###### 5. D, E and F are respectively the mid-points of the sides BC, CA and AB of a ABC. Show that:

(i) BDEF is a parallelogram.

(ii) ar (DEF) = ar (ABC)

(iii) ar (BDEF) = ar (ABC)

Ans. (i) F is the mid-point of AB and E is the mid-point of AC.

FEBC and FE = BD

[ Line joining the mid-points of two sides of a triangle is parallel to the third and half of it]

FEBD [BD is the part of BC]

And FE = BD

Also, D is the mid-point of BC.

BD = BC

And FEBC and FE = BD

###### Again E is the mid-point of AC and D is the mid-point of BC.

DEAB and DE = AB

DEAB [BF is the part of AB]

And DE = BF

Again F is the mid-point of AB.

BF = AB

But DE = AB

DE = BF

Now we have FEBD and DEBF

And FE = BD and DE = BF

Therefore, BDEF is a parallelogram.

(ii) BDEF is a parallelogram.

ar (BDF) = ar (DEF) ……….(i)

[diagonals of parallelogram divides it in two triangles of equal area]

DCEF is also parallelogram.

ar (DEF) = ar (DEC) ……….(ii)

Also, AEDF is also parallelogram.

ar (AFE) = ar (DEF) ……….(iii)

From eq. (i), (ii) and (iii),

ar (DEF) = ar (BDF) = ar (DEC) = ar (AFE) ……….(iv)

Now, ar (ABC) = ar (DEF) + ar (BDF) + ar (DEC) + ar (AFE) ……….(v)

ar (ABC) = ar (DEF) + ar (DEF) + ar (DEF) + ar (DEF)

[Using (iv) & (v)]

ar (ABC) = 4 ar (DEF)

ar (DEF) = ar (ABC)

(iii) ar (gm BDEF) = ar (BDF) + ar (DEF) = ar (DEF) + ar (DEF) [Using (iv)]

ar (gm BDEF) = 2 ar (DEF)

ar (gm BDEF) = ar (ABC)

ar (gm BDEF) = ar (ABC)

NCERT Solutions for Class 9 Maths Exercise 9.3

###### 6. In figure, diagonals AC and BD of 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) DACB or ABCD is a parallelogram.

Ans. (i) Draw BM AC and DN AC.

In DON and BOM,

OD = OB [Given]

DNO = BMO = [By construction]

DON = BOM [Vertically opposite]

DON BOM [By RHS congruency]

DN = BM [By CPCT]

Also ar (DON) = ar (BOM) ……….(i)

Again, In DCN and ABM,

###### CD = AB [Given]

DNC = BMA = [By construction]

DN = BM [Prove above]

DCN BAM [By RHS congruency]

ar (DCN) = ar (BAM) ……….(ii)

ar (DON) + ar (DCN) = ar (BOM) + ar (BAM)

ar (DOC) = ar (AOB)

(ii) Since ar (DOC) = ar (AOB)

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

ar (DCB) = ar (ACB)

(iii) Since ar (DCB) = ar (ACB)

Therefore, these two triangles in addition to be on the same base CB lie between two same parallels CB and DA.

DACB

Now AB = CD and DACB

Therefore, ABCD is a parallelogram.

NCERT Solutions for Class 9 Maths Exercise 9.3

###### 7. D and E are points on sides AB and AC respectively of ABC such that ar (DBC) = ar (EBC). Prove that DEBC.

Ans. Given: ar (DBC) = ar (EBC)

Since two triangles of equal area have common base BC.

Therefore, DEBC [ Two triangles having same base (or equal bases) and equal areas lie between the same parallel]

NCERT Solutions for Class 9 Maths Exercise 9.3

###### 8. XY is a line parallel to side BC of triangle ABC. If BEAC and CFAB meet XY at E and F respectively, show that ar (ABE) = ar (ACF).

Ans. ABE and parallelogram BCYE lie on the same base BE and between the same parallels BE and AC.

ar (ABE) = ar (gm BCYE) ……….(i)

Also ACF and gm BCFX lie on the same base CF and between same parallel BX and CF.

ar (ACF) = ar (gm BCFX) ……….(ii)

But gm BCYE and gm BCFX lie on the same base BC and between the same parallels BC and EF.

ar (gm BCYE) = ar (gm BCFX) ……….(iii)

From eq. (i), (ii) and (iii), we get,

ar (ABE) = ar (ACF)

###### 9. The side AB of 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 (ABCD) = ar (PBQR).

Ans. Given: ABCD is a parallelogram, CPAQ and PBQR is a parallelogram.

To prove: ar (ABCD) = ar (PBQR)

Construction: Join AC and QP.

Proof: Since AQCP

ar (AQC) = ar (AQP)

[Triangles on the same base and between the same parallels are equal in area]

Subtracting ar (ABQ) from both sides, we get

ar (AQC – ar (ABQ) = ar (AQP) – ar (ABQ)

ar (ABC) = ar (QBP) ……….(i)

Now ar (ABC) = ar (gm ABCD)

[Diagonal divides a parallelogram in two parts of equal area]

And ar (PQB) = ar (gm PBQR)

From eq. (i), (ii) and (iii), we get

ar (gm ABCD) = ar (gm PBQR)

NCERT Solutions for Class 9 Maths Exercise 9.3

###### 10. Diagonals AC and BD of a trapezium ABCD with ABDC intersect each other at O. Prove that ar(AOD) = ar (BOC).

Ans. ABD and ABC lie on the same base AB and between the same parallels AB and DC.

ar (ABD) = ar (ABC)

Subtracting ar (AOB) from both sides,

ar (ABD) – ar (AOB)

= ar (ABC) – ar (AOB)

ar (AOD) = ar (BOC)

11. In 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)

Ans. (i) Given that BFAC

ACB and ACF lie on the same base AC and between the same parallels AC and BF.

ar (ACB) = ar (ACF) ……….(i)

(ii) Now ar (ABCDE) = ar (trap. AEDC) + ar (ABC) ……….(ii)

ar (ABCDE) = ar (trap. AEDC) + ar (ACF) = ar (quad. AEDF) [Using (i)]

ar (AEDF) = ar (ABCDE)

NCERT Solutions for Class 9 Maths Exercise 9.3

###### 12. A villager Itwaari has a plot of land of the shape of quadrilateral. The Gram Panchyat of two villages decided to take over some portion of his plot from one of the corners to construct a health centre. Itwaari agrees to the above personal 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.

Ans. Let Itwari has land in shape of quadrilateral PQRS.

Draw a line through 5 parallel to PR, which meets QR produced at M.

Let diagonals PM and RS of new formed quadrilateral intersect each other at point N.

We have PRSM [By construction]

ar (PRS) = ar (PMR)

[Triangles on the same base and same parallel are equal in area]

Subtracting ar (PNR) from both sides,

ar (PRS) – ar (PNR) = ar (PMR) – ar (PNR)

ar (PSN) = ar (MNR)

It implies that Itwari will give corner triangular shaped plot PSN to the Grampanchayat for health centre and will take equal amount of land (denoted by MNR) adjoining his plot so as to form a triangular plot PQM.

NCERT Solutions for Class 9 Maths Exercise 9.3

###### 13. ABCD is a trapezium with ABDC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).

Ans. Join CX, ADX and ACX lie on the same base XA and between the same parallels XA and DC.

ar (ADX) = ar (ACX) ……….(i)

Also ACX and ACY lie on the same base

AC and between same parallels CY and XA.

ar (ACX) = ar (ACY) ……….(ii)

From (i) and (ii),

NCERT Solutions for Class 9 Maths Exercise 9.3

###### 14. In figure, APBQCR. Prove that ar (AQC) = ar (PBR).

Ans. ABQ and BPQ lie on the same base BQ and between same parallels AP and BQ.

ar (ABQ) = ar (BPQ) ……….(i)

BQC and BQR lie on the same base BQ and between same parallels BQ and CR.

ar (BQC) = ar (BQR) ……….(ii)

Adding eq (i) and (ii), ar (ABQ) + ar (BQC) = ar (BPQ) + ar (BQR)

ar (AQC) = ar (PBR)

NCERT Solutions for Class 9 Maths Exercise 9.3

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.

Ans. Given that ar (AOD) = ar (BOC)

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

ar (ABD) = ar (ABC)

Since if two triangles equal in area, lie on the same base then, they lie between same parallels. We have ABD and ABC lie on common base AB and are equal in area.

They lie in same parallels AB and DC.

ABDC

Now in quadrilateral ABCD, we have ABDC

Therefore, ABCD is trapezium. [ In trapezium one pair of opposite sides is parallel]

NCERT Solutions for Class 9 Maths Exercise 9.3

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

Ans. Given that DRC and DPC lie on the same base DC and ar (DPC) = ar (DRC) …..(i)

DCRP

[If two triangles equal in area, lie on the same base then, they lie between same parallels]

Therefore, DCPR is trapezium. [ In trapezium one pair of opposite sides is parallel]

Also ar (BDP) = ar (ARC) ……….(ii)

Subtracting eq. (i) from (ii),

ar (BDP) – ar (DPC) = ar (ARC) – ar (DRC)

Therefore, ABDC [If two triangles equal in area, lie on the same base then, they lie between same parallels]

Therefore, ABCD is trapezium.

## NCERT Solutions for Class 9 Maths Exercise 9.3

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