# Ncert Solutions For Class 9 Maths Ex 9.1

## Ncert Solutions For Class 9 Maths Chapter 9 Ex 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.$.$