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) | ii) | iii) | |||
---|---|---|---|---|---|

iv) | v) | vi) |

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

Answer :

i)

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)

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)

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)

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)

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)

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.

Q1 )
In Figure, ABCD is a parallelogram,

If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.

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

= CD x AE ...(i)

Also, Area of parallelogram

= Base x Corresponding altitude

= AD x CF ...(ii)

Thus, from (i) and (ii), we get,

CD x AE = AD x CF

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

Construction: Let us join HF.

To prove: ar(EFGH) =

Proof: In parallelogram ABCD,

AD = BC and AD || BC

( Opposite sides of a parallelogram are equal and parallel)

Multiplying both sides by

Since,

Similarly, it can be proved that,

Area (

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

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

ar(

Similarly,

ar(

From (i) and (ii), we have

ar(

Hence, proved

Q4 )
In Figure, P is a point in the interior of a parallelogram ABCD. Show that

(i) ar (

(ii) ar (

[Hint : Through P, draw a line parallel to AB.]

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

Answer :

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,

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

AB || EF and AE || BF

Therefore, quadrilateral ABFE is a parallelogram.

It can be observed that

Therefore, we can say that,

Area

Similarly,

for

Now, adding eq. (iii) and (iv), we get,

[Area (ABFE) + Area (EFCD)]

Hence, part i) is proved.

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,

From equations (vi) and (vii), we get,

MN || AD and AM || DN

It can be observed that

Therefore, we can say that,

Area

Similarly,

for

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

Area

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

Area

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) =

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.

ii) Consider

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

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

ar(AXS) =

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 :

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.,

Thus, Area

Also, Area

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

Area

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

Answer :

Given,

AD is median of

also,

ED is the median of

Subtracting (ii) from (i),

ar(ABD) – ar(EBD) = ar(ACD) – ar(ECD)

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

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

Answer :

ar(BED) =

(Since, E is the mid-point of AD,)

AE = DE

(Since, AD is the median on side BC of triangle ABC,)

BD = DC

DE =

BD =

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

ar(BED) =

Hence, proved.

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

Answer :

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

In

also,

In

In

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).

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

Answer :

In

Line-segment CD is bisected by AB at O.

So, AO is the median of

Similarly,

In

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

Area (

Hence, proved.

Q5 )
D, E and F are respectively the mid-points of the sides BC, CA and AB of a

i) BDEF is a parallelogram.

ii) ar(DEF) =

iii) ar(BDEF) =

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

Answer :

(i) In

EF || BC and EF =

also,

BD =

So, BD = EF

also,

BF and DE are parallel and equal to each other.

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

BDEF, DCEF, AFDE are parallelograms.

Diagonal of a parallelogram divides it into two triangles of equal area.

also,

ar(

ar(

From (i), (ii) and (iii)

(iii) Area (parallelogram BDEF) = ar(

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.]

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

Answer :

Construction: Let us draw

i) In (

(Vertically opposite angles)

Also, OD = OB ...(Given)

(By AAS congruency test)

Thus, DN = BM ...(i)(CPCT)

But, we know that, congruent triangles have equal areas.

Now, in (

DN = BM ...(Using Equation (i))

Also, CD = AB ...(Given)

(By RHS congruency test)

On adding eq. (ii)and (iii), we get,

Area

ii)We have, Area

Adding Area

iii)Now, we have, Area

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

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

Hence, proved.

Q7 )
D and E are points on sides AB and AC respectively of

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

Answer :

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 :

XY || BC = EY || BC ...(Given)

Also, BE || AC = BE || CY

Similarly,

XY || BC = XF || BC

Also, FC || AB = FC || XB

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

Now, considering parallelogram EBYC and

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

Also, parallelogram BCFX and

Thus, Area (

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

Area (

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).]

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

Answer :

Construction: Let us join AC and PQ.

Also, Area (

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

(

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 :

It can be observed that

Area (

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).

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

Answer :

i)

ii)

Area (

Now adding Area (ACDE) on both side,

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.

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

We have to prove that the area of

It can be observed that

Area (

Now subtracting area (

Area (

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 :

It can be observed that,

Also,

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

Area (

Hence, proved.

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

Answer :

It can be observed that,

Also,

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

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.

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

Answer :

Given:

Area (

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

Hence, proved.

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

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

Answer :

Given:

Area (

As,

Also, Area (

Hence, it is proved that ABCD is a trapezium.

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]

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

From (1) and (2), we have

(AB + CD) + (BC+AD) > (AB + EF) +( BE + AF)

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

Since , BD= DE= EC (Given )