NCERT Solutions For Class 9 Maths Chapter 9

NCERT Solutions Class 9 Maths Areas of Parallelograms and Triangles

NCERT solutions class 9 maths chapter 9 areas of parallelograms and triangle is given so that students of the 9th standard can prepare mathematics in a better way. The NCERT solutions for class 9 maths chapter 9 areas of parallelograms and triangle is one of the most crucial section in class 9 examination. A Parallelogram is a flat shape with following properties: Opposite sides are parallel and equal. Opposite angles are equal and supplementary. Rectangles, Squares, and Rhombuses are all Parallelograms. Area of a Parallelogram is given by Area = base × height. Also, the Perimeter is twice the (side length + base). The diagonals of a parallelogram bisect each other.

Every triangle comprises of 3 vertices. The altitude is the normal from base to the opposing vertex. Every triangle has 3 bases and 3 altitudes. The 3 altitudes meet at a point known as the orthocenter. The median is a line from a vertex to the midpoint of the opposing side. The 3 medians meet at a point known as the centroid. Here, we have provided a comprehensive study material for NCERT solutions class 9 maths chapter 9 areas of parallelograms and triangle. The students can download and understand all the exercise-wise NCERT problems on areas of parallelograms and triangle from the links provided below.

NCERT Solutions For Class 9 Maths Chapter 9 Exercises

Exercise-9.1

Q.1.In the given figure, PQRS is a parallelogram, PESRandRFPS..If PQ = 16 cm, PE = 8 cm & RF = 10 cm. Calculate AD.

Solution:

qwerty

 

Area of the parallelogram PQRS=PQRS=RS×PE=16×8cm2( becausePQ=RS,PQRSisaparallelogram)=128cm2 ————(1)

Now, area of parallelogram PQRS=PS×RF=PS×10cm2 ————(2)

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

PS×10=128PS=12810PS=12.8cm

 

Q.2.If E, F, G and H are  the mid-points respectively, of the sides of a parallelogram ABCD show that Area(EFGH)=12Area(ABCD)

Solution:

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Lets join HF.

In the parallelogram, i.e ABCD, AD = BC and AD || BC (because in a parallelogram the opposite sides are equal and parallel)

 

AB=CD (opposite sides are equal)

12AD=12BCAndAH||BF

AH=BFandAH||BF(  becauseThe mid point of AD and BC are H and F)

 therefore ABFH is a parallelogram.

Since ΔHEF and parallelogram ABFH are between the same parallel lines AB and HF, and are on the same base HF.

 therefore Area (ΔHEF)= 12Area(ABFH) … (1)

Similarly, we can prove that,

Area(ΔHGF)=12Area(HDFC) … (2)

Add Equation (1) and Equation (2), we obtain

Area(ΔHEF)+Area(ΔHGF)=12Area(ABFH)+12Area(HDCF)=12[Area(ABFH)+Area(HDCF)]Area(EFGH)=12(ABCD)

 

Q.3. DC and AD are two sides on which P and Q are two points lying respectively of a parallelogram ABCD. Prove that area(APB)=area(BQC)

Solution:

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It is observed that, ∆BQC and parallelogram ABCD  are between the same parallel lines AD and BC and lie on the same base BC.

 thereforeArea(ΔBQC)=12Area(ABCD) …(1)

Similarly,we can say that ∆APB and parallelogram ABCD lie between the same lines AB and DC that are parallel and on the same base AB .

 thereforeArea(ΔAPB)=12Area(ABCD) … (2)

Equating both the equations, i.e equation(1)andequation(2),we get

Area(ΔBQC)=Area(ΔAPB)

Hence, proved.

 

Q.4. In the given figure,  intheinteriorofaparallelogramABCD,thereexistapointP.Show that

(i) area(APB)+area(PCD)=12area(ABCD)

(ii) area(APD)+area(PBC)=area(APB)+area(PCD)

[Hint: Draw a line i.e. parallel to AB, through P]

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Solution:

 

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  • A line segment EF is drawn, parallel to line segment AB and passing through point P.

In the parallelogram ABCD,

AB||EF …(1) (as ABCD is a parallelogram)

 thereforeAD||BC(oppositesidesofaparallelogram)

AE||BF …(2)

By equating , Equation (1) and Equation(2) , we get,

AB||EFandAE||BF

Therefore, quad ABFE is a parallelogram.

 ItcanbesaidthatΔAPBandparallelogramABFEarebetweenthesameparallellinesABandEFandlyingonthesamebaseAB.  thereforeArea(ΔAPB)=12Area(ABFE) …(3)

Similarly ,for ΔPCD and parallelogram EFCD,

Area(ΔPCD)=12Area(EFCD) …(4)

Add equation(3) and equation(4), we get,

Area(ΔAPB)+Area(ΔPCD)=12[Area(ABEF)+Area(EFCD)]

Area(ΔAPB)+Area(ΔPCD)=12Area(ABCD) …(5)

 

(ii)

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A line segment MN is drawn,parallel to line segment AD and passing through point P.

In the parallelogram ABCD,

MN||AD …(6) (as ABCD is a parallelogram)

 thereforeAB||DC(oppositesidesofaparallelogram)

AM||DN …(7)

By equating , Equation (6) and Equation(7) , we get,

MN||ADandAM||DN

Therefore , quad AMND AMNDisaparallelogram.

It can be said that ∆APD and parallelogram AMND are  between  thesameparallellinesADandMN and lying on the same base AD.

 thereforeArea(ΔAPD)=12Area(AMND) …(8)

Similarly ,for ΔPCB and parallelogram MNCB,

Area(ΔPCB)=12Area(MNCB) …(9)

Add equation(8) and equation(9), we get,

Area(ΔAPD)+Area(ΔPCB)=12[Area(AMND)+Area(MNCB)]

Area(ΔAPD)+Area(ΔPCB)=12Area(ABCD) …(10)

 

Now compare equation(5) with equation(10), we get

Area(ΔAPD)+Area(ΔPBC)=Area(ΔAPB)+Area(ΔPCD)

 

Q.5. In the figure given below  X is a point on the side BR and ABRS and PQRS are parallelograms. Prove that

(i) area (PQRS) = area (ABRS)

(ii)  area(ΔPXS)=12area(PQRS)

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Solution:

(i)It can be said that  parallelogramPQRS and the parallelogram ABRS lie inbetweenthesameparallellinesSRandPB and also, on the same base SR.

 thereforeArea(PQRS)=Area(ABRS) …(1)

(ii)Consider ΔAXS and parallelogram ABRS.

As these lie on the same base and are between the same parallel lines AS and BR,

 therefore12Area(ΔAXS)=Area(ABRS) … (2)

By equating , equation (1) and equation(2) ,we get

Area(ΔAXS)=12Area(PQRS)

 

Q.6. A farmer had a field and that was in  a parallelogram shape PQRS.He took a point A on RS and he joined it to points P and Q. In how many parts the field 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 he do it ?

Solution:

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From the figure , it can be observed that point A divides the field into three parts.

The parts which are triangular in shape are – ΔPSA,ΔPAQ,andΔQRA

AreaofΔPSA+AreaofΔPAQ+AreaofΔQRA=AreaofparallelogramPQRS …(i)

We know that if a parallelogram and a triangle are between the same parallels and on the same base , then the area of the triangle becomes half the area of the parallelogram.

 thereforeArea(ΔPAQ)=12Area(PQRS) … (ii)

By equation , equation (i) and (ii) , we get

Area(ΔPSA)+Area(ΔQRA)=12Area(PQRS) … (iii)

Clearly , it can be said that the farmer should sow wheat in triangular part PAQ and he should sow pulses in other two triangular parts PSA and QRA or wheat in triangular parts PSA and QRA and in triangular part PAQ he should sow pulses.

 

Exercise – 9.2

Q.1.In the given figure, E is any point on median AD of a ΔABC. Prove that

Area(ABE)=Area(ACE) 

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Solution:

Given: AD is the median of ΔABC.

To Prove: Area(ΔABC)=Area(ΔACE)

Proof: In ΔABC,

AD is a median.

 thereforeArea(ΔABD)=Area(ΔACD) ….(i)

In ΔEBC,

ED is a median.

 thereforeArea(ΔEBD)=Area(ΔECD) ….(ii)

Subtracting equation (ii) from equation (i) , we get

Area(ΔABD)Area(ΔEBD)=Area(ΔACD)Area(ΔECD)Area(ΔABE)=Area(ΔACE)

 

Q.2.In a ΔABC ,E is the mid-point of median AD. Prove that

 Area(ΔBED)=14Area(ΔABC).

Solution:

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Given: E is the mid-point of median AD in ΔABC.

To Prove: Area(ΔBED)=Area(ΔABC)

Proof: In ΔABC,

AD is a median.

 thereforeArea(ΔABD)=Area(ΔABC) ….(i) ( because median of a triangle divides it into two triangles of equal area.)

In ΔABD,

BE is a median.

 thereforeArea(ΔBED)=Area(ΔBEA) =  12Area(ΔABD)

 Area(ΔBED)=12Area(ΔABD)=1212Area(ΔABC)(From(i)))=14Area(ΔABC)  Area(ΔBED)=14Area(ΔABC)

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

Solution:

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We know that diagonals of parallelogram bisect each other.

 thereforeO is the mid-point of AC and BD.

BO is the median in ∆ABC. Therefore, it will divide it into two triangles of equal areas.

 therefore  Area(ΔAOB)=Area(ΔBOC) …(i)

In  ΔBCD,CO is the median.

 therefore  Area(ΔBOC)=Area(ΔCOD) …(ii)

Similarly ,  Area(ΔCOD)=Area(ΔAOD) …(iii)

By Equating equation(i)  , (ii)  inline,and (iii)  , we get

 Area(ΔAOB)=Area(ΔBOC)=Area(ΔCOD)=Area(ΔAOD)

Therefore , it is evident that the diagonals of a parallelogram divide it into four triangles of equal area.

 

Q.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  Area(ΔABC)=Area(ΔABD).

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Solution:

Given:  ΔABCandΔDBC are on the same base AB. Line segment CD is bisected  by AB at O.

To Prove:  Area(ΔABC)=Area(ΔABD)

Proof: Line-segment CD is bisected by AB at O.

 therefore AO is the median of ∆ACD.

 Area(ΔACO)=Area(ΔADO)

BO is the median of  ΔBCD.

 thereforeArea(ΔBCO)=Area(ΔBDO) …(ii)

Adding (i) and (ii), we get

 Area(ΔACO)+Area(ΔBCO)=Area(ΔADO)+Area(ΔBDO)Area(ΔABC)=Area(ΔABD)

 

Q.5. D, E and F are respectively the mid-points of the sides BC, CA and AB of a ΔABC.
Prove that
(i) BDEF is a parallelogram

(ii)  Area(ΔDEF)=14Area(ΔABC)

(iii)  Area(BDEF)=12Area(ΔABC)

Solution:

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Given: D, E and F are respectively the mid-points of the sides BC, CA and AB of a ΔABC.

To Prove: (i) BDEF is a parallelogram

(ii)  Area(ΔDEF)=14Area(ΔABC)

(iii)  Area(BDEF)=12Area(ΔABC)

Proof:  In ΔABC,

F is the mid-point of side AB and E is the mid-point of side AC.
 thereforeEF || BC | In a triangle, the line segment joining the mid-points of any two sides is parallel to the 3rd side.

EF||BD ……(1)

Similarly, EF||BD ……(2)

From (1) and (2) , we can say that ,

BDEF is a parallelogram.

(ii)As in (i), we can prove that

AFCE and AFDE are parallelograms.
FD is diagonal of the parallelogram BDEF.

 therefore Area(ΔFBD)=Area(ΔDEF) ….(3)

Similarly , Area(ΔDEF)=Area(ΔFAE) …..(4)

Area(ΔDEF)=Area(ΔDCE) …..(5)

From equation(3),(4) and (5), we have

Area(ΔFBD)=Area(ΔDEF)=Area(ΔFAE)=Area(ΔDCE) …..(6)

 therefore ΔABC is divided into four non-overlapping triangles ΔFBD,ΔDEF,ΔFAEandΔDCE

 therefore Area(ΔABC)=Area(ΔFBD)+Area(ΔDEF)+Area(ΔFAE)+Area(ΔDCE)

=4Area(ΔDEF)          (From equation(6))

Area(ΔDEF)=14Area(ΔABC) ……(7)

 

(iii) Area(BDEF)=Area(ΔFBD)+Area(ΔDEF) (From equation(3))

==Area(ΔDEF)+Area(ΔDEF)=2Area(ΔDEF)=2.14Area(ΔABC)(Fromequation(7))=12Area(ΔABC)

 

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

(i) Area(ΔDOC)=Area(ΔAOB)

(ii) Area(ΔDCB)=Area(ΔACB)

(iii) DA||CB or ABCD is a parallelogram.

[Hint: From D and B,draw perpendiculars to AC]

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Solution:

Given: Diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD.

To Prove: If AB = CD, then

(i) Area(ΔDOC)=Area(ΔAOB)

(ii) Area(ΔDCB)=Area(ΔACB)

(iii) DA||CB or ABCD is a parallelogram.

Construction: Draw DEACandBFAC.

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Proof: In ΔDONandΔBOM,

DNO=BMO(By Construction)

DON=BOM (vertically opposite angles)

OD = OB (Given)

By AAS congruence rule,

ΔDONΔBOM

DN=BM ….(1)

We know that congruent triangles have equal areas.

Area(ΔDON)=Area(ΔBOM)  …..(2)

In ΔDNCandΔBMA,

DNC=BMA (By Construction)

CD=AB (Given)

DN=BM [using equation(1)]

 ΔDNCΔBMA (RHS congruence rule)

 Area(ΔDNC)=Area(ΔBMA) …..(3)

Add equation(2) and (3)  , we get

 Area(ΔDON)+Area(ΔDNC)=Area(ΔBOM)+Area(ΔBMA)  thereforeArea(ΔDOC)=Area(ΔAOB)

 

(ii)We got ,

 Area(ΔDOC)=Area(ΔAOB)

Now, add  Area(ΔOCB) on both the sides.

 Area(ΔDOC)+Area(ΔOCB)=Area(ΔAOB)+Area(ΔOCB)  Area(ΔDCB)=Area(ΔACB)

(iii)We got ,

 Area(ΔDCB)=Area(ΔACB)

If two triangles have the same base and equal areas , then these will lie between the same parallels.

 DA||CB ……(4)

In quadrilateral ABCD , one pair of opposite sides is equal  (AB=CD)and the other pair of opposite sides is parallel  (DA||CB).

 thereforeABCDisaparallelogram.

 

Q.7. D and E are points on sides AB and AC respectively of ∆ABC such that  Area(ΔDBC)=Area(ΔEBC) Prove that DE || BC.

Solution:

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Given : D and E are points on sides AB and AC respectively of ∆ABC such that  Area(ΔDBC)=Area(ΔEBC).

To Prove: DE || BC

Proof: As  (ΔDBC) and  (ΔEBC) have equal area and are on the same base.

 therefore  (ΔDBC) and  (ΔEBC) will lie between the same parallel lines.

 therefore DE || BC

 

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

Solution:

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Given: XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and E respectively.

To Prove:  Area(ΔABE)=Area(ΔACF)

Proof: XY||BC(given)

And CF||BX ( because CF||AB(given))

 thereforeBCFX is a parallelogram.

BC= XF

BC=XY+YF …..(1)

Again ,

XY||BC   ( because BE||AC)

And BE||CY

 thereforeBCYE is a parallelogram

 thereforeBC=YE  ( becauseopposite sides of a parallelogram are equal)

 BC=XY+XE  …..(2)

From (1) and (2)  ,

XY+YF=XY+XE

 YF=XEXE=YF

 thereforeΔAXEandΔAYF have equal bases(XE=YF) on the same line EF and have common vertex A.

 thereforeTheir altitudes are also the same.

 thereforeArea(ΔAXE)=Area(ΔAFY) …….(4)

 therefore  thereforeΔBXEandΔBXE have equal bases(XE=YF) on the same line EF and are between the same parallels EF and BC(XY||BC).

 therefore  Area(ΔBEX)=Area(ΔCFY)  ( becauseTwo triangles on the same base(or equal bases) and between the same parallels are equal in area)

Add the corresponding sides of (4) and (5), we get

 Area(ΔAXE)+Area(ΔBXE)=Area(ΔAFY)+Area(ΔCFY)Area(ΔABE)=Area(ΔACF)

 

Q.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 the following figure). Show that  Area(ParallelogramABCD)=Area(ParallelogramPBQR).

[Hint: Join AC and PQ. Now compare  Area(ΔACQ)andArea(ΔAPQ)]

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Solution:

Given: 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.

To Prove: Area(Parallelogram ABCD) = Area(Parallelogram PBQR).

Construction: Join AC and PQ.

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Proof: AC is a diagonal of parallelogram ABCD

 thereforeArea(ΔABC)=12Area(parallelogramABCD)  ……………(1)

PQ  is a diagonal of parallelogram BQRP

 thereforeArea(ΔBPQ)=12Area(parallelogramBQRP)  ………(2)

 ΔACQandΔAPQ are on the same base AQ and between the same parallels AQ and CP.

 thereforeArea(ΔACQ)=Area(ΔAPQ)

Now , substract  Area(ΔABQ) from both the sides.

 Area(ΔACQ)Area(ΔABQ)=Area(ΔAPQ)Area(ΔABQ)  Area(ΔABC)=Area(ΔBPQ)12Area(parallelogramABCD)=12Area(parallelogramPBQR)

 Area(parallelogramABCD)=Area(parallelogramPBQR)     [From(1) and (2)]

 

Q.10. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that  Area(ΔAOD)=Area(ΔBOC).

Solution:

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Given: Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O.

To Prove:  Area(ΔAOD)=Area(ΔBOC).

Proof:  ΔABDandΔABC are on the same base AB and between the same parallels AB and DC.

 thereforeArea(ΔABD)=Area(ΔABC)

Now  ,substract  Area(ΔAOB) from both the sides;

 Area(ΔABD)Area(ΔAOB)=Area(ΔABC)Area(ΔAOB)Area(ΔAOD)=Area(ΔBOC)

Q.11. In the given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that

(i) Area(ΔACB)=Area(ΔACF)

 (ii)  Area(AEDF)=Area(ABCDE)

Solution:

Given: ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F.

To Prove: (i) Area(ΔACB)=Area(ΔACF)

(ii)  Area(AEDF)=Area(ABCDE)

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Proof:

 (i) ΔACBandΔACF lie on the same base AC and are between the same parallels AC and BF.

 Area(ΔACB)=Area(ΔACF)

(ii)  becauseArea(ΔACB)=Area(ΔACF)

Add  Area(AEDC) on both the sides, we get

 Area(ΔACB)+Area(AEDC)=Area(ΔACF)+Area(AEDC)Area(ABCDE)=Area(AEDF)Area(AEDF)=Area(ABCDE)

 

Q.12. A villager Ram has a plot of land in 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. Ram agrees to the above proposal with the condition that he should be given an 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:

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

Let the portion ADE be taken over by the Gram Panchayat of the village from one corner D to construct a Health Centre.

Join AC. Draw a line through D parallel to AC to meet BC produced in P.

Then Ram must given the land ECP adjoining his plot so as to form a triangular plot ABP as then.

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Proof:  ΔDAPandΔDCP are between the same parallels DP and AC.

 thereforeArea(ΔDAP)=Area(ΔDCP)(Two triangles on the same base (or equal bases) and between the same parallels are equal in area)

Substract  Area(ΔDEP) from both the sides.

 Area(ΔDAP)Area(ΔDEP)=Area(ΔDCP)Area(ΔDEP)  Area(ΔADE)=Area(ΔPCE)

Now  , add  Area(ABCE) on the both the sides.

 Area(ΔADE)=Area(ΔPCE)Area(ΔADE)+Area(ABCE)=Area(ΔPCE)+Area(ABCE)  Area(ABCD)=Area(ΔABP)

 

Q.13. ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that  Area(ΔADX)=Area(ΔACY). [Hint: Join CX.]

Solution:

Given: ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y.

To Prove:  Area(ΔADX)=Area(ΔACY).

Construction: Join CX

Proof:  ΔADXandΔACX are on the same base AX and between the same parallels AB and DC.

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 thereforeArea(ΔADX)=Area(ΔACX) …….(1)

 ΔACXandΔACY are on the same base AC and between the same parallels AC and XY.

 thereforeArea(ΔACX)=Area(ΔACY) …….(2)

From equation(1) and (2)  , we get

 Area(ΔADX)=Area(ΔACY)

 

Q.14. In the given figure, AP || BQ || CR. Prove that  Area(ΔAQC)=Area(ΔPBR).

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Solution:  AP||BQ||CR

To Prove:  Area(ΔAQC)=Area(ΔPBR).

Proof:  ΔBAQandΔBQR are between the same parallels BQ and AP and on the same base BQ.

 

 Area(ΔBCQ)=Area(ΔBPQ) …….(1)

 ΔBCQandΔBQR are between the same parallels BQ and CR and on the same base BQ.

 Area(ΔBCQ)=Area(ΔBQR) …….(2)

Now , add equation(1) and (2)  , we get

 

 Area(ΔBAQ)+Area(ΔBCQ)=Area(ΔBPQ)+Area(ΔBQR)Area(ΔAQC)=Area(ΔPBR)

 

Q.15. Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that  Area(ΔAOD)=Area(ΔBOC).Prove that ABCD is a trapezium.

Solution:

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Given: Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that  Area(ΔAOD)=Area(ΔBOC).

To Prove: ABCD is a trapezium.

Proof:  Area(ΔAOD)=Area(ΔBOC)

Now , add  Area(ΔAOB) on both the sides.

 Area(ΔAOD)+Area(ΔAOB)=Area(ΔBOC)+Area(ΔAOB)Area(ΔABD)=Area(ΔABC)

But  ΔABDandΔABC are on the same base AB.

 thereforeΔABDandΔABCwill have equal corresponding altitudes

and  thereforeΔABDandΔABC will lie between the same parallels

 AB||DC

 therefore ABCD is a trapezium.(  becauseA quadrilateral is a trapezium if exactly one pair of opposite sides is parallel)

 

Q.16. In the given figure ,  Area(ΔDRC)=Area(ΔDPC)andArea(ΔBDP)=Area(ΔARC).Show that both the quadrilaterals ABCD and DCPR are trapeziums.

Solution:

Given:  Area(ΔDRC)=Area(ΔDPC)andArea(ΔBDP)=Area(ΔARC)

To Prove: both the quadrilaterals ABCD and DCPR are trapeziums.

Proof:  Area(ΔDRC)=Area(ΔDPC) (given) …..(1)

But  ΔDRCandΔDPC are on the same base DC.

 thereforeΔDRCandΔDPC will have equal corresponding altitudes.

And  thereforeΔDRCandΔDPC will lie between the same parallels.

 thereforeDC||RP ( becauseA quadrilateral is a trapezium if exactly one pair of opposite sides is parallel.)

 

 thereforeDCPR is a trapezium.

Again ,  Area(ΔBDP)=Area(ΔARC)Area(ΔBDC)+Area(ΔDPC)=Area(ΔADC)+Area(ΔDRC)Area(ΔBDC)=Area(ΔADC) ….(using(1))

But  ΔBDCandΔADC