# 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.$PE \perp SR\, and \, RF\perp PS.$.If PQ = 16 cm, PE = 8 cm & RF = 10 cm. Calculate AD.

Solution:

Area of the parallelogram PQRS=PQRS=RS×PE=16×8cm2( becausePQ=RS,PQRSisaparallelogram)=128cm2$PQRS = PQRS = RS \times PE\\ =16\times 8\, cm^{2}(\ because PQ = RS,PQRS \: is \: a\: parallelogram)\\ =128 \, cm^{2}$ ————(1)

Now, area of parallelogram PQRS=PS×RF=PS×10cm2$PS\times RF\\ =PS\times 10 \, cm^{2}$ ————(2)

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

PS×10=128PS=12810PS=12.8cm$PS\times 10 = 128\\ \Rightarrow PS=\frac{128}{10}\\ \Rightarrow PS = 12.8 \: cm$

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)$Area(EFGH)\, =\, \frac{1}{2}Area(ABCD)$

Solution:

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$AB = CD$ (opposite sides are equal)

12AD=12BCAndAH||BF$\Rightarrow \frac{1}{2}AD = \frac{1}{2}BC\\ And \: AH\, ||\, BF$

AH=BFandAH||BF$\Rightarrow AH=BF \, and\, AH||BF$(  because$\ because$The mid point of AD and BC are H and F)

therefore$\ therefore$ ABFH is a parallelogram.

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

therefore$\ therefore$ Area (ΔHEF$\Delta HEF$)= 12Area(ABFH)$\frac{1}{2} Area(ABFH)$ … (1)

Similarly, we can prove that,

Area(ΔHGF)=12Area(HDFC)$Area(\Delta HGF)= \frac{1}{2}Area(HDFC)$ … (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)$Area(\Delta HEF)+Area(\Delta HGF )=\frac{1}{2}Area(ABFH)+\frac{1}{2}Area(HDCF)\\ =\frac{1}{2}[Area(ABFH)+Area(HDCF)]\\ \Rightarrow Area(EFGH)=\frac{1}{2}(ABCD)$

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)$area(APB) = area (BQC)$

Solution:

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)$\ therefore Area(\Delta BQC)= \frac{1}{2}Area(ABCD)$ …(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)$\ therefore Area(\Delta APB)= \frac{1}{2}Area(ABCD)$ … (2)

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

Area(ΔBQC)=Area(ΔAPB)$Area (∆BQC) = Area (∆APB)$

Hence, proved.

Q.4. In the given figure,  intheinteriorofaparallelogramABCD,thereexistapointP.$\, in \,the\, interior\, of\, a\, parallelogram\, ABCD\,, there\, exist\, a\, point\, P.$Show that

(i) area(APB)+area(PCD)=12area(ABCD)$area (APB) + area (PCD) = \frac{1}{2}area (ABCD)$

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

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

Solution:

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

In the parallelogram ABCD,

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

thereforeAD||BC(oppositesidesofaparallelogram)$\ therefore AD \, ||\, BC (opposite \, sides \, of \, a \, parallelogram)$

AE||BF$\Rightarrow AE\, ||\, BF$ …(2)

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

AB||EFandAE||BF$AB\, ||\, EF\, and\, AE\, ||\, BF$

Therefore, quad ABFE is a parallelogram.

ItcanbesaidthatΔAPBandparallelogramABFEarebetweenthesameparallellinesABandEFandlyingonthesamebaseAB.$\  It \, can\, be \, said\, that\, ∆APB \, and \, parallelogram \, ABFE\, are\, between \, the\, same\, parallel \, lines\, AB\, and \, EF \, and\, lying\, on\, the\, same \, base\, AB.$  thereforeArea(ΔAPB)=12Area(ABFE)$\ therefore Area(\Delta APB)=\frac{1}{2}Area(ABFE)$ …(3)

Similarly ,$,$for ΔPCD$\Delta PCD$ and parallelogram EFCD,

Area(ΔPCD)=12Area(EFCD)$Area(\Delta PCD)=\frac{1}{2}Area(EFCD)$ …(4)

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

Area(ΔAPB)+Area(ΔPCD)=12[Area(ABEF)+Area(EFCD)]$Area(\Delta APB)+Area(\Delta PCD)=\frac{1}{2}[ Area(ABEF)+Area(EFCD)]$

Area(ΔAPB)+Area(ΔPCD)=12Area(ABCD)$Area(\Delta APB)+Area(\Delta PCD)=\frac{1}{2}Area(ABCD)$ …(5)

(ii)

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

In the parallelogram ABCD,

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

thereforeAB||DC(oppositesidesofaparallelogram)$\ therefore AB \, ||\, DC (opposite \, sides \, of \, a \, parallelogram)$

AM||DN$\Rightarrow AM\, ||\, DN$ …(7)

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

MN||ADandAM||DN$MN\, ||\, AD\, and\, AM\, ||\, DN$

Therefore ,$\ ,$ quad AMND AMNDisaparallelogram.$AMND \, \, is\,\, a \, \, parallelogram.$

It can be said that ∆APD and parallelogram AMND are  between  thesameparallellinesADandMN$\ the \: same\: parallel \: lines\: AD \: and\: MN$ and lying on the same base AD.

thereforeArea(ΔAPD)=12Area(AMND)$\ therefore Area(\Delta APD)=\frac{1}{2}Area(AMND)$ …(8)

Similarly ,$,$for ΔPCB$\Delta PCB$ and parallelogram MNCB,

Area(ΔPCB)=12Area(MNCB)$Area(\Delta PCB)=\frac{1}{2}Area(MNCB)$ …(9)

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

Area(ΔAPD)+Area(ΔPCB)=12[Area(AMND)+Area(MNCB)]$Area(\Delta APD)+Area(\Delta PCB)=\frac{1}{2}[ Area(AMND)+Area(MNCB)]$

Area(ΔAPD)+Area(ΔPCB)=12Area(ABCD)$Area(\Delta APD)+Area(\Delta PCB)=\frac{1}{2}Area(ABCD)$ …(10)

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

Area(ΔAPD)+Area(ΔPBC)=Area(ΔAPB)+Area(ΔPCD)$Area(\Delta APD)+Area(\Delta PBC)= Area(\Delta APB)+Area(\Delta PCD)$

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)$\ area (∆PXS) = \frac{1}{2}area (PQRS)$

Solution:

(i)It can be said that  parallelogramPQRS$parallelogram \, PQRS$ and the parallelogram ABRS lie inbetweenthesameparallellinesSRandPB$in \, between\, the\, same\, parallel\, lines\, SR \, and\, PB$ and also,$,$ on the same base SR.

thereforeArea(PQRS)=Area(ABRS)$\ therefore Area (PQRS) = Area (ABRS)$ …(1)

(ii)Consider ΔAXS$\Delta AXS$ and parallelogram ABRS.

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

therefore12Area(ΔAXS)=Area(ABRS)$\ therefore \frac{1}{2}Area (\Delta AXS) = Area (ABRS)$ … (2)

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

Area(ΔAXS)=12Area(PQRS)$Area(\Delta AXS)=\frac{1}{2} Area(PQRS)$

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:

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$\Delta PSA, \Delta PAQ, and\, \Delta QRA$

AreaofΔPSA+AreaofΔPAQ+AreaofΔQRA=AreaofparallelogramPQRS$Area of \Delta PSA + Area of \Delta PAQ + Area of \Delta QRA = Area\, of\, parallelogram\, PQRS$ …(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)$\ therefore Area (\Delta PAQ) = \frac{1}{2}Area (PQRS)$ … (ii)

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

Area(ΔPSA)+Area(ΔQRA)=12Area(PQRS)$Area (\Delta PSA) + Area (\Delta QRA) = \frac{1}{2}Area (PQRS)$ … (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$\Delta ABC$. Prove that

Area(ABE)=Area(ACE)$Area(ABE) = Area (ACE)$

Solution:

Given: AD is the median of ΔABC$\Delta ABC$.

To Prove: Area(ΔABC)=Area(ΔACE)$Area(\Delta ABC)=Area(\Delta ACE)$

Proof: In ΔABC$\Delta ABC$,

thereforeArea(ΔABD)=Area(ΔACD)$\ therefore Area(\Delta ABD)=Area(\Delta ACD)$ ….(i)

In ΔEBC$\Delta EBC$,

ED is a median.

thereforeArea(ΔEBD)=Area(ΔECD)$\ therefore Area(\Delta EBD)=Area(\Delta ECD)$ ….(ii)

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

Area(ΔABD)Area(ΔEBD)=Area(ΔACD)Area(ΔECD)Area(ΔABE)=Area(ΔACE)$Area(\Delta ABD)-Area(\Delta EBD)=Area(\Delta ACD)-Area(\Delta ECD)\\ \Rightarrow Area(\Delta ABE)=Area(\Delta ACE)$

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

Area(ΔBED)=14Area(ΔABC)$\  Area(\Delta BED)= \frac{1}{4}Area(\Delta ABC)$.

Solution:

Given: E is the mid-point of median AD in ΔABC$\Delta ABC$.

To Prove: Area(ΔBED)=Area(ΔABC)$Area(\Delta BED)=Area(\Delta ABC)$

Proof: In ΔABC$\Delta ABC$,

thereforeArea(ΔABD)=Area(ΔABC)$\ therefore Area(\Delta ABD)=Area(\Delta ABC)$ ….(i) ( because$\ because$ median of a triangle divides it into two triangles of equal area.)

In ΔABD$\Delta ABD$,

BE is a median.

thereforeArea(ΔBED)=Area(ΔBEA)$\ therefore Area(\Delta BED)=Area(\Delta BEA)$ =  12Area(ΔABD)$\ \frac{1}{2}Area(\Delta ABD)$

Area(ΔBED)=12Area(ΔABD)=1212Area(ΔABC)(From(i)))=14Area(ΔABC)$\ Area(\Delta BED)=\frac{1}{2}Area(\Delta ABD)=\frac{1}{2}\cdot \frac{1}{2}Area(\Delta ABC) (From (i)))\\=\frac{1}{4}Area(\Delta ABC)$  Area(ΔBED)=14Area(ΔABC)$\ Area(\Delta BED)=\frac{1}{4}Area(\Delta ABC)$

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

Solution:

We know that diagonals of parallelogram bisect each other.

therefore$\ therefore$O 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$\ therefore$  Area(ΔAOB)=Area(ΔBOC)$\ Area(\Delta AOB)= Area(\Delta BOC)$ …(i)

In  ΔBCD,$\ \Delta BCD,$CO is the median.

therefore$\ therefore$  Area(ΔBOC)=Area(ΔCOD)$\ Area(\Delta BOC)= Area(\Delta COD)$ …(ii)

Similarly ,$\ ,$  Area(ΔCOD)=Area(ΔAOD)$\ Area (\Delta COD) = Area (\Delta AOD)$ …(iii)

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

Area(ΔAOB)=Area(ΔBOC)=Area(ΔCOD)=Area(ΔAOD)$\ Area(\Delta AOB)=Area(\Delta BOC)=Area(\Delta COD)=Area(\Delta AOD)$

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).$\ Area(\Delta ABC) = Area(\Delta ABD).$

Solution:

Given:  ΔABCandΔDBC$\ \Delta ABC \, and \, \Delta DBC$ are on the same base AB. Line segment CD is bisected  by AB at O.

To Prove:  Area(ΔABC)=Area(ΔABD)$\ Area(\Delta ABC) =Area(\Delta ABD)$

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

therefore$\  therefore$ AO is the median of ∆ACD.

Area(ΔACO)=Area(ΔADO)$\ Area(\Delta ACO) =Area(\Delta ADO)$

BO is the median of  ΔBCD$\ \Delta BCD$.

thereforeArea(ΔBCO)=Area(ΔBDO)$\ therefore Area(\Delta BCO)=Area(\Delta BDO)$ …(ii)

Adding (i) and (ii), we get

Area(ΔACO)+Area(ΔBCO)=Area(ΔADO)+Area(ΔBDO)Area(ΔABC)=Area(ΔABD)$\ Area(\Delta ACO)+Area(\Delta BCO)=Area(\Delta ADO)+Area(\Delta BDO)\\ \Rightarrow Area(\Delta ABC)=Area(\Delta ABD)$

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)$\ Area(\Delta DEF)=\frac{1}{4}Area(\Delta ABC)$

(iii)  Area(BDEF)=12Area(ΔABC)$\ Area(BDEF)=\frac{1}{2}Area(\Delta ABC)$

Solution:

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)$\ Area(\Delta DEF)=\frac{1}{4}Area(\Delta ABC)$

(iii)  Area(BDEF)=12Area(ΔABC)$\ Area(BDEF)=\frac{1}{2}Area(\Delta ABC)$

Proof:  In ΔABC,$,$

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

EF||BD$\Rightarrow EF||BD$ ……(1)

Similarly,$,$ EF||BD$EF||BD$ ……(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$\ therefore$ Area(ΔFBD)=Area(ΔDEF)$Area(\Delta FBD) =Area(\Delta DEF)$ ….(3)

Similarly ,$,$ Area(ΔDEF)=Area(ΔFAE)$Area(\Delta DEF) =Area(\Delta FAE)$ …..(4)

Area(ΔDEF)=Area(ΔDCE)$Area(\Delta DEF) =Area(\Delta DCE)$ …..(5)

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

Area(ΔFBD)=Area(ΔDEF)=Area(ΔFAE)=Area(ΔDCE)$Area(\Delta FBD) =Area(\Delta DEF)=Area(\Delta FAE)=Area(\Delta DCE)$ …..(6)

therefore$\ therefore$ ΔABC$\Delta ABC$ is divided into four non-overlapping triangles ΔFBD,ΔDEF,ΔFAEandΔDCE$\Delta FBD, \Delta DEF, \Delta FAE\, and \, \Delta DCE$

therefore$\ therefore$ Area(ΔABC)=Area(ΔFBD)+Area(ΔDEF)+Area(ΔFAE)+Area(ΔDCE)$Area(\Delta ABC)=Area(\Delta FBD)+Area(\Delta DEF)+Area(\Delta FAE)+Area(\Delta DCE)$

=4Area(ΔDEF)$4Area(\Delta DEF)$          (From equation(6))

Area(ΔDEF)=14Area(ΔABC)$\Rightarrow Area(\Delta DEF)=\frac{1}{4}Area(\Delta ABC)$ ……(7)

(iii) Area(BDEF)=Area(ΔFBD)+Area(ΔDEF)$Area( BDEF)=Area(\Delta FBD)+Area(\Delta DEF)$ (From equation(3))

==Area(ΔDEF)+Area(ΔDEF)=2Area(ΔDEF)=2.14Area(ΔABC)(Fromequation(7))=12Area(ΔABC)$=Area(\Delta DEF)+Area(\Delta DEF)\\ =2Area(\Delta DEF)\\ =2.\frac{1}{4}Area(\Delta ABC) (From \: equation(7))\\ =\frac{1}{2}Area(\Delta ABC)$

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)$Area(\Delta DOC)=Area(\Delta AOB)$

(ii) Area(ΔDCB)=Area(ΔACB)$Area(\Delta DCB)=Area(\Delta ACB)$

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

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

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)$Area(\Delta DOC)=Area(\Delta AOB)$

(ii) Area(ΔDCB)=Area(ΔACB)$Area(\Delta DCB)=Area(\Delta ACB)$

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

Construction: Draw DEACandBFAC$DE\perp AC and BF\perp AC$.

Proof: In ΔDONandΔBOM,$\Delta DON \: and \: \Delta BOM,$

DNO=BMO$\perp DNO = \perp BMO$(By Construction)

DON=BOM$\perp DON = \perp BOM$ (vertically opposite angles)

OD = OB (Given)

By AAS congruence rule,$,$

ΔDONΔBOM$\Delta DON\perp \Delta BOM$

DN=BM$\perp DN = \perp BM$ ….(1)

We know that congruent triangles have equal areas.

Area(ΔDON)=Area(ΔBOM)$Area(\Delta DON)=Area(\Delta BOM)$  …..(2)

In ΔDNCandΔBMA,$\Delta DNC\: and \: \Delta BMA,$

DNC=BMA$\perp DNC =\perp BMA$ (By Construction)

CD=AB$CD = AB$ (Given)

DN=BM$DN = BM$ [using equation(1)]

ΔDNCΔBMA$\ \Delta DNC\sim \Delta BMA$ (RHS congruence rule)

Area(ΔDNC)=Area(ΔBMA)$\ Area(\Delta DNC)=Area(\Delta BMA)$ …..(3)

Add equation(2) and (3)  ,$\ ,$ we get

Area(ΔDON)+Area(ΔDNC)=Area(ΔBOM)+Area(ΔBMA)$\ Area(\Delta DON)+Area(\Delta DNC)=Area(\Delta BOM)+Area(\Delta BMA)$  thereforeArea(ΔDOC)=Area(ΔAOB)$\ therefore Area(\Delta DOC)=Area(\Delta AOB)$

(ii)We got ,$\ ,$

Area(ΔDOC)=Area(ΔAOB)$\ Area(\Delta DOC)=Area(\Delta AOB)$

Now, add  Area(ΔOCB)$\ Area(\Delta OCB)$ on both the sides.

Area(ΔDOC)+Area(ΔOCB)=Area(ΔAOB)+Area(ΔOCB)$\ \Rightarrow Area(\Delta DOC)+Area(\Delta OCB)=Area(\Delta AOB)+Area(\Delta OCB)$  Area(ΔDCB)=Area(ΔACB)$\ \Rightarrow Area(\Delta DCB)=Area(\Delta ACB)$

(iii)We got ,$\ ,$

Area(ΔDCB)=Area(ΔACB)$\ Area (\Delta DCB) = Area (\Delta ACB)$

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

DA||CB$\ DA || CB$ ……(4)

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

thereforeABCDisaparallelogram.$\ therefore ABCD is a parallelogram.$

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

Solution:

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

To Prove: DE || BC

Proof: As  (ΔDBC)$\ (\Delta DBC)$ and  (ΔEBC)$\ (\Delta EBC)$ have equal area and are on the same base.

therefore$\ therefore$  (ΔDBC)$\ (\Delta DBC)$ and  (ΔEBC)$\ (\Delta EBC)$ will lie between the same parallel lines.

therefore$\ therefore$ 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)$\ Area (\Delta ABE) = Area(\Delta ACF)$

Solution:

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)$\ Area (\Delta ABE) = Area(\Delta ACF)$

Proof: XY||BC(given)

And CF||BX ( because$\ because$ CF||AB(given))

therefore$\ therefore$BCFX is a parallelogram.

BC= XF

BC=XY+YF …..(1)

Again ,$\ ,$

XY||BC   ( because$\ because$ BE||AC)

And BE||CY

therefore$\ therefore$BCYE is a parallelogram

therefore$\ therefore$BC=YE  ( because$\ because$opposite sides of a parallelogram are equal)

BC=XY+XE$\ \Rightarrow BC=XY+XE$  …..(2)

From (1) and (2)  ,$\ ,$

XY+YF=XY+XE

YF=XEXE=YF$\ \Rightarrow YF=XE\\ \Rightarrow XE=YF$

thereforeΔAXEandΔAYF$\ therefore \Delta AXE\: and\: \Delta AYF$ have equal bases(XE=YF) on the same line EF and have common vertex A.

therefore$\ therefore$Their altitudes are also the same.

thereforeArea(ΔAXE)=Area(ΔAFY)$\ therefore Area(\Delta AXE)=Area(\Delta AFY)$ …….(4)

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

therefore$\ therefore$  Area(ΔBEX)=Area(ΔCFY)$\ Area(\Delta BEX)=Area(\Delta CFY)$  ( because$\ because$Two 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)$\ Area(\Delta AXE)+Area(\Delta BXE)=Area(\Delta AFY)+Area(\Delta CFY)\\ \Rightarrow Area(\Delta ABE)=Area(\Delta ACF)$

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).$\ Area(Parallelogram ABCD) = Area(Parallelogram PBQR).$

[Hint: Join AC and PQ. Now compare  Area(ΔACQ)andArea(ΔAPQ)$\ Area (\Delta ACQ) \: and \: Area (\Delta APQ)$]

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.

Proof: AC is a diagonal of parallelogram ABCD

thereforeArea(ΔABC)=12Area(parallelogramABCD)$\ therefore Area(\Delta ABC)=\frac{1}{2}Area(parallelogram ABCD)$  ……………(1)

PQ  is a diagonal of parallelogram BQRP

thereforeArea(ΔBPQ)=12Area(parallelogramBQRP)$\ therefore Area(\Delta BPQ)=\frac{1}{2}Area(parallelogram BQRP)$  ………(2)

ΔACQandΔAPQ$\ \Delta ACQ \:and \: \Delta APQ$ are on the same base AQ and between the same parallels AQ and CP.

thereforeArea(ΔACQ)=Area(ΔAPQ)$\ therefore Area(\Delta ACQ)=Area(\Delta APQ)$

Now ,$\ ,$ substract  Area(ΔABQ)$\ Area(\Delta ABQ)$ from both the sides.

Area(ΔACQ)Area(ΔABQ)=Area(ΔAPQ)Area(ΔABQ)$\ \Rightarrow Area(\Delta ACQ)-Area(\Delta ABQ)=Area(\Delta APQ)-Area(\Delta ABQ)$  Area(ΔABC)=Area(ΔBPQ)12Area(parallelogramABCD)=12Area(parallelogramPBQR)$\ \Rightarrow Area(\Delta ABC)=Area(\Delta BPQ)\\ \Rightarrow \frac{1}{2}Area(parallelogram\: ABCD)= \frac{1}{2}Area(parallelogram\: PBQR)$

Area(parallelogramABCD)=Area(parallelogramPBQR)$\ \Rightarrow Area(parallelogram\: ABCD)= Area(parallelogram\: PBQR)$     [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).$\ Area (\Delta AOD) = Area(\Delta BOC).$

Solution:

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

To Prove:  Area(ΔAOD)=Area(ΔBOC)$\ Area (\Delta AOD) = Area(\Delta BOC)$.

Proof:  ΔABDandΔABC$\ \Delta ABD\: and \: \Delta ABC$ are on the same base AB and between the same parallels AB and DC.

thereforeArea(ΔABD)=Area(ΔABC)$\ therefore Area(\Delta ABD)=Area(\Delta ABC)$

Now  ,$\ ,$substract  Area(ΔAOB)$\ Area(\Delta AOB)$ from both the sides;

Area(ΔABD)Area(ΔAOB)=Area(ΔABC)Area(ΔAOB)Area(ΔAOD)=Area(ΔBOC)$\ \Rightarrow Area(\Delta ABD)-Area(\Delta AOB)=Area(\Delta ABC)-Area(\Delta AOB)\\ \Rightarrow Area(\Delta AOD)=Area(\Delta BOC)$

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)$\ Area(\Delta ACB) =Area(\Delta ACF)$

(ii)  Area(AEDF)=Area(ABCDE)$\ Area(AEDF) =Area(ABCDE)$

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)$\ Area(\Delta ACB) =Area(\Delta ACF)$

(ii)  Area(AEDF)=Area(ABCDE)$\ Area(AEDF) =Area(ABCDE)$

Proof:

(i) ΔACBandΔACF$\ \Delta ACB\: and\: \Delta ACF$ lie on the same base AC and are between the same parallels AC and BF.

Area(ΔACB)=Area(ΔACF)$\ Area(\Delta ACB) =Area(\Delta ACF)$

(ii)  becauseArea(ΔACB)=Area(ΔACF)$\ because Area(\Delta ACB) =Area(\Delta ACF)$

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

Area(ΔACB)+Area(AEDC)=Area(ΔACF)+Area(AEDC)Area(ABCDE)=Area(AEDF)Area(AEDF)=Area(ABCDE)$\ \Rightarrow Area(\Delta ACB)+Area(AEDC)=Area(\Delta ACF)+Area(AEDC)\\ \Rightarrow Area(ABCDE)=Area(AEDF)\\ \Rightarrow Area(AEDF)=Area(ABCDE)$

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.

Proof:  ΔDAPandΔDCP$\ \Delta DAP\: and \: \Delta DCP$ are between the same parallels DP and AC.

thereforeArea(ΔDAP)=Area(ΔDCP)$\ therefore Area(\Delta DAP)=Area( \Delta DCP)$(Two triangles on the same base (or equal bases) and between the same parallels are equal in area)

Substract  Area(ΔDEP)$\ Area(\Delta DEP)$ from both the sides.

Area(ΔDAP)Area(ΔDEP)=Area(ΔDCP)Area(ΔDEP)$\ \Rightarrow Area(\Delta DAP)-Area(\Delta DEP)=Area( \Delta DCP)-Area(\Delta DEP)$  Area(ΔADE)=Area(ΔPCE)$\ Area(\Delta ADE)=Area(\Delta PCE)$

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

Area(ΔADE)=Area(ΔPCE)Area(ΔADE)+Area(ABCE)=Area(ΔPCE)+Area(ABCE)$\ Area(\Delta ADE)=Area(\Delta PCE)\\ \Rightarrow Area(\Delta ADE)+Area(ABCE)=Area(\Delta PCE)+Area(ABCE)$  Area(ABCD)=Area(ΔABP)$\ \Rightarrow Area(ABCD)=Area(\Delta ABP)$

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).$\ Area(\Delta ADX) =Area(\Delta ACY).$ [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).$\ Area(\Delta ADX) =Area(\Delta ACY).$

Construction: Join CX

Proof:  ΔADXandΔACX$\ \Delta ADX\: and\: \Delta ACX$ are on the same base AX and between the same parallels AB and DC.

thereforeArea(ΔADX)=Area(ΔACX)$\ therefore Area(\Delta ADX)=Area( \Delta ACX)$ …….(1)

ΔACXandΔACY$\ \Delta ACX \: and \: \Delta ACY$ are on the same base AC and between the same parallels AC and XY.

thereforeArea(ΔACX)=Area(ΔACY)$\ therefore Area(\Delta ACX)=Area( \Delta ACY)$ …….(2)

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

Area(ΔADX)=Area(ΔACY)$\ Area(\Delta ADX)=Area(\Delta ACY)$

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

Solution:  AP||BQ||CR$\ AP||BQ||CR$

To Prove:  Area(ΔAQC)=Area(ΔPBR)$\ Area(\Delta AQC)=Area(\Delta PBR)$.

Proof:  ΔBAQandΔBQR$\ \Delta BAQ\: and\:\Delta BQR$ are between the same parallels BQ and AP and on the same base BQ.

Area(ΔBCQ)=Area(ΔBPQ)$\ Area(\Delta BCQ)=Area(\Delta BPQ)$ …….(1)

ΔBCQandΔBQR$\ \Delta BCQ\: and\:\Delta BQR$ are between the same parallels BQ and CR and on the same base BQ.

Area(ΔBCQ)=Area(ΔBQR)$\ Area(\Delta BCQ)=Area(\Delta BQR)$ …….(2)

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

Area(ΔBAQ)+Area(ΔBCQ)=Area(ΔBPQ)+Area(ΔBQR)Area(ΔAQC)=Area(ΔPBR)$\ Area(\Delta BAQ)+Area(\Delta BCQ)=Area(\Delta BPQ)+Area(\Delta BQR)\\ \Rightarrow Area(\Delta AQC)=Area(\Delta PBR)$

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

Solution:

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

To Prove: ABCD is a trapezium.

Proof:  Area(ΔAOD)=Area(ΔBOC)$\ Area(\Delta AOD) = Area (\Delta BOC)$

Now ,$\ ,$ add  Area(ΔAOB)$\ Area(\Delta AOB)$ on both the sides.

Area(ΔAOD)+Area(ΔAOB)=Area(ΔBOC)+Area(ΔAOB)Area(ΔABD)=Area(ΔABC)$\ Area(\Delta AOD)+Area(\Delta AOB)=Area(\Delta BOC)+Area(\Delta AOB)\\ \Rightarrow Area(\Delta ABD)=Area(\Delta ABC)$

But  ΔABDandΔABC$\ \Delta ABD\: and\: \Delta ABC$ are on the same base AB.

thereforeΔABDandΔABC$\ therefore \Delta ABD\: and \: \Delta ABC$will have equal corresponding altitudes

and  thereforeΔABDandΔABC$\ therefore \Delta ABD\: and \: \Delta ABC$ will lie between the same parallels

AB||DC$\ \Rightarrow AB||DC$

therefore$\ therefore$ ABCD is a trapezium.(  because$\ because$A 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).$\ Area (\Delta DRC) =Area (\Delta DPC)\: and \:Area (\Delta BDP) = Area(\Delta ARC).$Show that both the quadrilaterals ABCD and DCPR are trapeziums.

Solution:

Given:  Area(ΔDRC)=Area(ΔDPC)andArea(ΔBDP)=Area(ΔARC)$\ Area(\Delta DRC)=Area(\Delta DPC)\: and\: Area(\Delta BDP)=Area(\Delta ARC)$

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

Proof:  Area(ΔDRC)=Area(ΔDPC)$\ Area(\Delta DRC)=Area(\Delta DPC)$ (given) …..(1)

But  ΔDRCandΔDPC$\ \Delta DRC\: and\:\Delta DPC$ are on the same base DC.

thereforeΔDRCandΔDPC$\ therefore \Delta DRC\: and \Delta DPC$ will have equal corresponding altitudes.

And  thereforeΔDRCandΔDPC$\ therefore \Delta DRC\: and \Delta DPC$ will lie between the same parallels.

therefore$\ therefore$DC||RP ( because$\ because$A quadrilateral is a trapezium if exactly one pair of opposite sides is parallel.)

therefore$\ therefore$DCPR is a trapezium.

Again ,$\ ,$  Area(ΔBDP)=Area(ΔARC)Area(ΔBDC)+Area(ΔDPC)=Area(ΔADC)+Area(ΔDRC)Area(ΔBDC)=Area(ΔADC)$\ Area(\Delta BDP)=Area(\Delta ARC)\\ \Rightarrow Area(\Delta BDC)+Area(\Delta DPC)=Area(\Delta ADC)+Area(\Delta DRC)\\ \Rightarrow Area(\Delta BDC)=Area(\Delta ADC)$ ….(using(1))