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..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.

 

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