NCERT Solutions for Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles

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Written by Team Trustudies
Updated at 2021-02-15


NCERT solutions for class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles Exercise 9.1

Q1 ) Which of the following figures lie on the same base and between the same parallels.
In such a case, write the common base and the two parallels.

i)imageii)imageiii)image
iv)imagev)imagevi)image



NCERT Solutions for Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles


Answer :


i) image

Yes. It can be observed that trapezium ABCD and triangle PCD have a common base CD and these are lying between the same parallel lines AB and CD.

ii)image

No. It can be observed that parallelogram PQRS and trapezium MNRS have a common base RS.

However, their vertices, (i.e., opposite to the common base) P, Q of parallelogram and M, N of trapezium, are not lying on the same line.

iii)image

Yes. It can be observed that parallelogram PQRS and triangle TQR have a common base QR and they are lying between the same parallel lines PS and QR.

iv)image
No. It can be observed that parallelogram ABCD and triangle PQR are lying between same parallel lines AD and ac. However, these do not have any common base.

v)image

Yes. It can be observed that parallelogram ABCD and parallelogram APQD have a common base AD and these are lying between the same parallel lines AD and BQ.

vi)image

No. It can be observed that parallelogram PBCS and PQRS are lying on the same base PS.

However, these do not lie between the same parallel lines.

NCERT solutions for class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles Exercise 9.2

Q1 ) In Figure, ABCD is a parallelogram, AE?DC and CF?AD.
If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.
image



NCERT Solutions for Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles


Answer :

In parallelogram ABCD, CD = AB = 16 cm [Opposite sides of a parallelogram are equal]

We know that,
Area of a parallelogram
= Base x Corresponding altitude
?Area of parallelogram ABCD
= CD x AE ...(i)

Also, Area of parallelogram
= Base x Corresponding altitude
? Area of parallelogram ABCD
= AD x CF ...(ii)

Thus, from (i) and (ii), we get,
CD x AE = AD x CF
? 16cm x 8cm = AD x 10cm
? AD = 16cmx8cm10cm
? AD = 12.8cm
? the length of AD is 12.8 cm.

Q2 ) If E,F,G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar (EFGH) = 1/2 ar (ABCD)



NCERT Solutions for Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles


Answer :

Given: E,F,G and H are respectively the mid-points of the sides of a parallelogram ABCD
image

Construction: Let us join HF.

To prove: ar(EFGH) = 12 ar(ABCD)

Proof: In parallelogram ABCD,
AD = BC and AD || BC
( Opposite sides of a parallelogram are equal and parallel)
Multiplying both sides by 12 we get,
? 12 x AD = 12 x BC
? AH = BF and AH || BF
? ABFH is a parallelogram.

Since, ?HEF and parallelogram ABFH are on the same base HF and between the same parallel lines AB and HF,
? Area (?HEF) = 12 Area (ABFH) ...(i)
Similarly, it can be proved that,

Area (?HGF) = ? Area (HDCF) ...(ii)
On adding equations (i) and (ii), we get,
? Area (?HEF) + Area (?HGF) = 12 Area (ABFH) + 12 Area (HDCF)
? Area [(?HEF) + (?HGF)] = [Area (ABFH) + Area (HDCF)]
? Area (EFGH) = 12 Area (ABCD)
Hence, proved.

Q3 ) P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD.
Show that ar(APB) = ar(BQC).



NCERT Solutions for Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles


Answer :

It can be observed that ?BQC and paralleogram ABCD lie on the same base BC and these are between the same parallel lines AD and BC.
image
?APB and parallelogram ABCD lie on the same base AB and in-between same parallel AB and DC.

ar(?APB) = 12 ar(parallelogram ABCD) ...... (i)

Similarly,

ar(?BQC) = 12 ar(parallelogram ABCD) ...... (ii)

From (i) and (ii), we have

ar(? APB) = ar(?BQC)
Hence, proved

Q4 ) In Figure, P is a point in the interior of a parallelogram ABCD. Show that
(i) ar (?APB + ?PCD) = 12 ar (ABCD)
(ii) ar (?APD + ?PBC) = ar(APB) + ar(PCD)
image
[Hint : Through P, draw a line parallel to AB.]



NCERT Solutions for Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles


Answer :

image
Construction: Let us draw a line segment EF, passing through point P and parallel to line segment AB.

In parallelogram ABCD, AB || EF ...(i)(By construction)

Also, ABCD is a parallelogram
So, AD || BC
(Opposite sides of parallelogram)
Also, AE || BF ...(ii)
(Since, 12 AD = AE and 12 BC = BF)
From equations (i) and (ii), we get,
AB || EF and AE || BF

Therefore, quadrilateral ABFE is a parallelogram.

It can be observed that ?APB and parallelogram ABFE are lying on the same base AB and between the same parallel lines AB and EF.

Therefore, we can say that,
Area ?APB = 12 Area (ABFE) ...(iii)
Similarly,
for ?PCD and parallelogram EFCD,
? Area ?PCD = 12 Area (EFCD) ...(iv)
Now, adding eq. (iii) and (iv), we get,
? Area ?APB + Area ?PCD = 12 Area (ABFE) + 12 Area (EFCD)
? Area ?APB + Area ?PCD = 12
[Area (ABFE) + Area (EFCD)]
? Area ?APB + Area ?PCD = 12 Area (ABCD) ...(v)

Hence, part i) is proved.

image

Construction: Let us draw a line segment MN, passing through point P and parallel to line segment AD.

In parallelogram ABCD, MN || AD ...(vi)(By construction)

Also, ABCD is a parallelogram
Thus, AB || DC
(Opposite sides of parallelogram)

Also, AM || DN ...(vii)
(Since, 12 AB = AM and 12 DC = DN)
From equations (vi) and (vii), we get,
MN || AD and AM || DN
? quadrilateral AMND is a parallelogram.


It can be observed that ?APD and parallelogram AMND are lying on the same base AD and between the same parallel lines AD and MN.

Therefore, we can say that,
Area ?APD = 12 Area (AMND) ...(viii)

Similarly,
for ?PCB and parallelogram MNCB,
? Area ?PCB = 12 Area (MNCB) ...(ix)

Now, adding eq. (viii) and (ix), we get,

Area ?APD + Area ?PCB = 12 Area (AMND) + 12 Area (MNCB)
? Area ?APD + Area ?PCB = 12 [Area (AMND) + Area (MNCB)]
? Area ?APD + Area ?PCB = 12 Area (ABCD) ...(x)

Thus, comparing eq. (v) and (x), we get,

Area ?APB + Area ?PCD = Area ?APD + Area ?PCB

Hence, proved.

Q5 ) In Figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that
i)ar(PQRS) = ar(ABRS)
ii) ar(AXS) = 12 ar(PQRS)
image



NCERT Solutions for Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles


Answer :

i) It can be observed that paralleogram PQRS and ABRS lie on the same base SR and also, these lie in between the same parallel lines SR and PB.

? Area (PQRS) = Area (ABRS) ...(i)

ii) Consider ?AXS and paralleogram ABRS.

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

? Area (?AXS) = 12 Area (ABRS) ...(ii)
From equations (i) and (ii), we get,
ar(AXS) = 12 ar(PQRS)
Hence, proved.

Q6 ) A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q.
In how many parts the fields 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 she do it?



NCERT Solutions for Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles


Answer :

image

From the figure, we can say that it is that point A which divides the field into three parts.

These parts are triangular in shape i.e., ?PSA, ?PAQ and ?QRA

Thus, Area ?PSA + Area ?PAQ + Area ?QRA = Area of Parallelogram PQRS ...(i)

Also, Area ?PAQ = (12) Area of Parallelogram PQRS ...(ii)

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

Area ?PSA + Area ?QRA = 12 Area of Parallelogram PQRS ...(iii)

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

NCERT solutions for class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles Exercise 9.3

Q1 ) In Figure, E is any point on median AD of a ?ABC. Show that ar (ABE) = ar (ACE)
image



NCERT Solutions for Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles


Answer :

Given,
AD is median of ? ABC.

? it will divide ?ABC into two triangles of equal area.

? ar(ABD) = ar(ACD) .....(i)

also,

ED is the median of ?ABC.

? ar(EBD) = ar(ECD) — (ii)

Subtracting (ii) from (i),

ar(ABD) – ar(EBD) = ar(ACD) – ar(ECD)

? ar(ABE) = ar(ACE)

Q2 ) In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) = 14 ar (ABC).



NCERT Solutions for Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles


Answer :

image

ar(BED) = 12×BD×DE
(Since, E is the mid-point of AD,)
AE = DE
(Since, AD is the median on side BC of triangle ABC,)
BD = DC

DE = 12 AD — (i)
BD = 12BC — (ii)
From (i) and (ii), we get,
ar(BED) = 12×12BC × 12AD
? ar(BED) = 12×12ar(ABC)
? ar(BED) = 14 ar(ABC)

Hence, proved.

Q3 ) Show that the diagonals of a parallelogram divide it into four triangles of equal area.



NCERT Solutions for Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles


Answer :

image

O is the mid point of AC and BD. (diagonals of bisect each other)

In ? ABC, BO is the median.
? ar(AOB) = ar(BOC) .... (i)
also,
In ? BCD, CO is the median.
? ar(BOC) = ar(COD) ...... (ii)
In ? ACD, OD is the median.
? ar(AOD) = ar(COD) ....... (iii) In ? ABD, AO is the median.
? ar(AOD) = ar(AOB) — (iv)

From equations (i), (ii), (iii) and (iv), we get,
ar(BOC) = ar(COD) = ar(AOD) = ar(AOB)

Hence, we get, the diagonals of a parallelogram divide it into four triangles of equal area.

Q4 ) 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).
image



NCERT Solutions for Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles


Answer :

In ?ACD,
Line-segment CD is bisected by AB at O.

So, AO is the median of ?ACD.

? Area (?ACO) = Area (?ADO) ...(i)

Similarly,
In ?BCD, BO is the median.
? Area (?BCO) = Area (?BDO) ...(ii)

Now, adding equations (i) and (ii), we get,

Area (?ACO) + Area (?BCO) = Area (?ADO) + Area (?BDO)
? Area (?ABC) = Area (?ABD)
Hence, proved.

Q5 ) 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) = 14 ar(ABC)
iii) ar(BDEF) = 12 ar(ABC)



NCERT Solutions for Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles


Answer :

image

(i) In ? ABC,
EF || BC and EF = 12 BC (by mid point theorem)
also,
BD = 12 BC (D is the mid point)
So, BD = EF
also,
BF and DE are parallel and equal to each other.
? the pair opposite sides are equal in length and parallel to each other.
? BDEF is a parallelogram.


(ii) Proceeding from the result of (i),

BDEF, DCEF, AFDE are parallelograms.
Diagonal of a parallelogram divides it into two triangles of equal area.
? ar(?BFD) = ar(?DEF) (For parallelogram BDEF) — (i)
also,
ar(? AFE) = ar(? DEF) (For parallelogram DCEF) — (ii)
ar(? CDE) = ar(? DEF) (For parallelogram AFDE) — (iii)

From (i), (ii) and (iii) ar(?BFD)=ar(?AFE)=ar(?CDE)=ar(?DEF)
?ar(?BFD)+ar(?AFE)+ar(?CDE)+ar(?DEF)=ar(?ABC)
?4ar(?DEF)=ar(?ABC)
? ar(DEF) = 14 ar(ABC)


(iii) Area (parallelogram BDEF) = ar(?DEF) +ar(? BDE)
? ar(parallelogram BDEF) = ar(? DEF) +ar(? DEF)
? ar(parallelogram BDEF) = 2× ar(? DEF)
? ar(parallelogram BDEF) = 2× 14 ar(? ABC)
? ar(parallelogram BDEF) = 14 ar(? ABC)

Hence, proved.

Q6 ) 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) DA || CB or ABCD is a parallelogram.
[Hint: From D and B, draw perpendiculars to AC.]
image



NCERT Solutions for Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles


Answer :

Construction: Let us draw DN?AC and BN?AC.

image

i) In (?DON) and (?BOM),

?DNO = ?BMO ...(By construction)
?DON = ?BOM
(Vertically opposite angles)
Also, OD = OB ...(Given)
? ?DON ? ?BOM
(By AAS congruency test)
Thus, DN = BM ...(i)(CPCT)
But, we know that, congruent triangles have equal areas.
? Area (?DON) = Area (?BOM) ...(ii)

Now, in (?DNC) and (?BMA),
?DNC = ?BMA ...(By construction)
DN = BM ...(Using Equation (i))
Also, CD = AB ...(Given)
? ?DNC ? ?BMA
(By RHS congruency test)
? Area (?DNC) = Area (?BMA) ...(iii)
On adding eq. (ii)and (iii), we get,
Area ?DON + Area ?DNC = Area ?BOM + Area ?BMA
? Area ?DOC = Area ?AOB


ii)We have, Area ?DOC = Area ?AOB

Adding Area ?OCB on both sides, we get,

? Area ?DOC + Area ?OCB = Area ?AOB + Area ?OCB
? Area ?DCB = Area ?ACB


iii)Now, we have, Area ?DCB = Area ?ACB

We also know that, if two triangles have the same base and equal areas, then these will lie between the same parallels.

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

? we can say that, ABCD is a parallelogram.
Hence, proved.

Q7 ) D and E are points on sides AB and AC respectively of ?ABC such that ar (DBC) = ar (EBC). Prove that DE || BC.



NCERT Solutions for Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles


Answer :

image

? ?BCE and ?BCD are lying on a common base BC and also have equal areas.

? ?BCE and ?BCD will lie between the same parallel lines.
? DE || BC

Hence, it is proved that DE || BC

Q8 ) XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively,
show that ar (ABE) = ar (ACF)



NCERT Solutions for Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles


Answer :

image

XY || BC = EY || BC ...(Given)
Also, BE || AC = BE || CY
? we can say that, EBYC is a parallelogram.
Similarly,
XY || BC = XF || BC
Also, FC || AB = FC || XB
? BCFX is a parallelogram.


Parallelograms EBCY and BCFX are on the same base BC and between the same parallels BC and EF.

? Area (EBCY) = (12 ) Area (BCFX) ...(i)

Now, considering parallelogram EBYC and ?AEB,

We can say that, these lie on the same base BE and are between the same parallels BE and AC.

? Area (?AEB) = (12 ) Area (EBYC) ...(ii)

Also, parallelogram BCFX and ?ACF are on the same base CF and between the same parallels CF and AB.

Thus, Area (?ACF) = (12 ) Area (BCFX) ...(iii)

So, from eq.(i), (ii) and (iii), we get,

Area (?AEB) = (?ACF)
Hence, proved.

Q9 ) 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 Figure).
Show that ar (ABCD) = ar (PBQR).
[Hint : Join AC and PQ. Now compare ar (ACQ) and ar (APQ).]
image



NCERT Solutions for Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles


Answer :

Construction: Let us join AC and PQ.

image

?ACQ and ?AQP are the same base AQ and between the same parallels AQ and CP.

? Area ?ACQ = Area ?AQP
?Area(?ACQ)?Area(?ABQ)=Area(?APQ)?Area(?ABQ)
? Area (?ABC) = Area (?QBP) ...(i)


? AC and PQ are diagonals of parallelograms ABCD and PBQR, respectively
? Area (?ABC) = (12 ) Area (ABCD) ...(ii)

Also, Area (?QBP) = (12 ) Area (PBQR) ...(iii)

From equations (i),(ii) and (iii), we get,

(12 ) Area (ABCD) = (12 ) Area(PBQR)
? Area (ABCD) = Area(PBQR)
Hence, proved.

Q10 ) Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).



NCERT Solutions for Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles


Answer :

image

It can be observed that ?DAC and ?DBC lie on the same base DC and between the same parallels AB and CD.

? Area (?DAC) = Area (?DBC)
?Area(?DAC)?Area(?DOC)=Area(?DBC)?Area(?DOC)
Area (?AOD) = Area (?BOC)
Hence, proved.

Q11 ) 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).
image



NCERT Solutions for Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles


Answer :

i) ?ACB and ?ACF lie on the same base AC and are between The same parallels AC and BF.

? Area (?ACB) = Area (?ACF)


ii) ? it can be observed that
Area (?ACB) = Area (?ACF)

Now adding Area (ACDE) on both side,

? Area (?ACB) + Area (ACDE) = Area (?ACF) + Area (ACDE)

? Area (ABCDE) = Area (AEDF)
Hence, proved.

Q12 ) A villager Itwaari has a plot of land of 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.
Itwaari agrees to the above proposal 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.



NCERT Solutions for Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles


Answer :

Let quadrilateral ABCD be the original shape of the field.

image

The proposal may be implemented as follows:

Join diagonal BD and draw a line parallel to BD through point A.
Let it meet the extended side CD of ABCD at point E.

Join BE and AD. Let them intersect each other at O.

Then, portion ?AOB can be cut from the original field so that the new shape of the field will be ?BCE as shown below.

image

We have to prove that the area of ?AOB (portion that was cut so as to construct Health Centre) is equal to the area of ?DEO(portion added to the field so as to make the area of the new field so formed equal to the area of the original field).

It can be observed that ?DEB and ?DAB lie on the same base BD and are between the same parallels BD and AE.

Area (?DEB) = Area (?DAB)

Now subtracting area (?DOB ) On both side

Area (?DEB) - Area (?DOB) = (?DAB) - Area (?DOB)

? Area (?DEO) = Area (?AOB)

Q13 ) ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y.
Prove that ar (ADX) = ar (ACY).
[Hint: Join CX.]



NCERT Solutions for Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles


Answer :

image

It can be observed that, ?ADX and ?ACX lie on the same base AX and are between the same parallels AB and DC.

? Area (?ADX) = Area (?ACX) ...(i)

Also, ?ADY and ?ACX lie on the same base AC and are between the same parallels AC and XY.

? Area (?ACY) = Arca (?ACX) ...(ii)

From Equations (i) and (ii), we get,

Area (?ADX) = Area (?ACY)

Hence, proved.

Q14 ) In Figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).
image



NCERT Solutions for Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles


Answer :

It can be observed that, ?ABQ and ?PBQ lie on the same base BQ and are between the same parallels AP and BQ.

? Area (?ABQ) = Area (?PBQ) ...(i)

Also, ?BCQ and ?BRQ lie on the same BQ and are between the same parallels BQ and CR.

? , Area (?BCQ) = Arca (?BRQ) ...(ii)

From Equations (i) and (ii), we get,
? Area (?ABQ) + Area(?BCQ) = Area (?PBQ) + Area (?BRQ)

? Area (?AQC) = Area (?PBR)
Hence, proved.

Q15 ) 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.
image



NCERT Solutions for Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles


Answer :

Given:
Area (?AOD) = Area (?BOC)

? Area (?AOD) + Area (?AOB) = (?BOC) + Area (?AOB)

? Area (?ADB) = Area (?ACB)

We know that triangles on the same base having area equal to each other lie between the same parallels.

? these triangles, ?ADB and ?ACB, are lying between the same parallels.i.e., AB || CD

? ABCD is a trapezium.
Hence, proved.

Q16 ) In Figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.
image



NCERT Solutions for Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles


Answer :

Given:
Area (?DRC) = Area (?DPC).

As, ?DRC and ?DPC lie on the same base DC and have equal area, therefore, they must lie between the same parallel lines.

? DC || RP
? DCPR is a trapezium.

Also, Area (?BDP) = Area (?ARC).

? Area (?BDP) - Area (?DPC) = Area (?ARC) - Area (?DRC)

? Area (?BDC) = Area(?ADC)

? ?BDC and ?ADC lie on the same base CD and have equal areas, they must lie between the same parallel lines.

? AB || CD
Hence, it is proved that ABCD is a trapezium.

NCERT solutions for class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles Exercise 9.4

Q1 ) Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.



NCERT Solutions for Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles


Answer :


We have a parallelogram ABCD and rectangle ABEF such that

ar(||gm ABCD) = ar( rect. ABEF)

AB = CD [Opposite sides of parallelogram]
and AB = EF [Opposite sides of a rectangle]
? CD = EF
? AB + CD = AB + EF … (1)

BE < BC and AF < AD [In a right triangle, hypotenuse is the longest side]

? (BC + AD) > (BE + AF) …(2)

From (1) and (2), we have

(AB + CD) + (BC+AD) > (AB + EF) +( BE + AF)
? (AB + BC + CD + DA) > (AB + BE + EF + FA)
? Perimeter of parallelogram ABCD > Perimeter of rectangle ABEF.

Q2 ) In figure, D and E are two points on BC such that BD = DE = EC. Show that ar(ABD) = ar(ADE) = ar(AEC).



NCERT Solutions for Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles


Answer :

Let us draw AF, perpendicular to BC such that AF is the height of ?ABD, ?ADE and ?AEC



? area of triangle =( 12) × base ×Height
? Area of ? ABD= (12 ) × BD × AF
? Area of ? ADE = (12 ) × DE × AF
? Area of ? AEC = (12 ) × EC × AF
Since , BD= DE= EC (Given )
? Area of ? ABD= Area of ? ADE= Area of ? AEC

Q3 ) In figure, ABCD, DCFE and ABFE are parallelograms. Show that ar(ADE) = ar(BCF).



NCERT Solutions for Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles


Answer :

Since, ABCD is a parallelogram [Given]
?Its opposite sides are parallel and equal.