# NCERT Solutions for Class 9 Maths Chapter 9 - Areas of Parallelograms and Triangles

## Chapter 9 - Areas of Parallelograms and Triangles Exercise Ex. 9.1

Solution 1

(i) Yes. Trapezium ABCD and triangle PCD are having a common base CD and these are lying between the same parallel lines AB and CD.

(ii)

No. The parallelogram PQRS and trapezium MNRS are having a common base RS but their vertices (i.e. opposite to common base) P, Q of parallelogram and M, N of trapezium are not lying on a same line.

(iii)

Yes. The Parallelogram PQRS and triangle TQR are having a common base QR and they are lying between the same parallel lines PS and QR.

(iv)

No. We see that parallelogram ABCD and triangle PQR are lying between same parallel lines AD and BC but these are not having any common base.

(v)

Yes. We may observe that parallelogram ABCD and parallelogram APQD have a common base AD and also these are lying between same parallel lines AD and BQ.

(vi)

No. We may observe that parallelogram PBCS and PQRS are laying on same base PS, but these are not between the same parallel lines.

## Chapter 9 - Areas of Parallelograms and Triangles Exercise Ex. 9.2

Solution 1

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

We know that,

Area of parallelogram = Base corresponding attitude

Area of parallelogram ABCD = CD AE = AD CF

16 cm 8 cm = AD 10 cm

We know that,

Area of parallelogram = Base corresponding attitude

Area of parallelogram ABCD = CD AE = AD CF

16 cm 8 cm = AD 10 cm

AD = cm = 12.8 cm.

Thus, the length of AD is 12.8 cm.

Solution 2

Construction: Join HF.

In parallelogram ABCD

AD = BC and AD || BC (Opposite sides of a parallelogram are equal and parallel)

AB = CD [Opposite sides of a parallelogram are equal]

In parallelogram ABCD

AD = BC and AD || BC (Opposite sides of a parallelogram are equal and parallel)

AB = CD [Opposite sides of a parallelogram are equal]

AH = BF and AH || BF

Therefore, ABFH is a parallelogram.

Since HEF and parallelogram ABFH are on the same base HF and between the same parallel lines AB and HF.

Since HEF and parallelogram ABFH are on the same base HF and between the same parallel lines AB and HF.

area (HEF) = area (ABFH) ...(1)

Similarly, we can prove

area (HGF) = area (HDCF) ...(2)

On adding equations (1) and (2), we have

Solution 3

Here BQC and parallelogram ABCD lie on same base BC and these are between same parallel lines AD and BC.

Similarly, APB and parallelogram ABCD lie on the same base AB and between same parallel lines AB and DC

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

area (BQC) = area (APB)

area (BQC) = area (APB)

Solution 4

(i) Let us draw a line segment EF, passing through point P and parallel to line segment AB.

In parallelogram ABCD we find that

AB || EF (By construction) ...(1)

ABCD is a parallelogram

In parallelogram ABCD we find that

AB || EF (By construction) ...(1)

ABCD is a parallelogram

From equations (1) and (2), we have

AB || EF and AE || BF

So, quadrilateral ABFE is a parallelogram

Now, we may observe that APB and parallelogram AB || FE are lying on the same base AB and between the same

AB || EF and AE || BF

So, quadrilateral ABFE is a parallelogram

Now, we may observe that APB and parallelogram AB || FE are lying on the same base AB and between the same

parallel lines AB and EF.

Similarly, for PCD and parallelogram EFCD

area (PCD) = area (EFCD) ... (4)

Adding equations (3) and (4), we have

(ii)

Draw a line segment MN, passing through point P and parallel to line segment AD.

In parallelogram ABCD we may observe that

MN || AD (By construction) ...(6)

ABCD is a parallelogram

In parallelogram ABCD we may observe that

MN || AD (By construction) ...(6)

ABCD is a parallelogram

From equations (6) and (7), we have

MN || AD and AM || DN

So, quadrilateral AMND is a parallelogram

MN || AD and AM || DN

So, quadrilateral AMND is a parallelogram

Now, APD and parallelogram AMND are lying on the same base AD and between the same parallel lines AD and MN.

Similarly, for PCB and parallelogram MNCB

area (PCB) = area (MNCB) ... (9)

Adding equations (8) and (9), we have

area (PCB) = area (MNCB) ... (9)

Adding equations (8) and (9), we have

On comparing equations (5) and (10), we have

Area (APD) + area (PBC) = area (APB) + area (PCD)

Solution 5

(i) Parallelogram PQRS and ABRS lie on the same base SR

and also these are in between same parallel lines SR and PB.

and also these are in between same parallel lines SR and PB.

(ii) Now consider AXS and parallelogram ABRS

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

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

Solution 6

From figure it is clear that point A divides the field into three parts. These parts are triangular in shape - PSA, PAQ and QRA

Area of PSA + Area of PAQ + Area of QRA = area of PQRS ... (1)

We know that if a parallelogram and triangle are on the same base and between the same parallels, the area of triangle is half the area of the parallelogram.

Area of PSA + Area of PAQ + Area of QRA = area of PQRS ... (1)

We know that if a parallelogram and triangle are on the same base and between the same parallels, the area of triangle is half the area of the parallelogram.

From equations (1) and (2), we have

Area (PSA) + area (QRA) = area (PQRS) ... (3)

Area (PSA) + area (QRA) = area (PQRS) ... (3)

Clearly, farmer must sow wheat in triangular part PAQ and pulses in other two triangular parts PSA and QRA or wheat in triangular part PSA and QRA and pulses in triangular parts PAQ.

## Chapter 9 - Areas of Parallelograms and Triangles Exercise Ex. 9.3

Solution 1

AD is median of ABC. So, it will divide ABC into two triangles of equal areas.

Solution 2

AD is median of ABC. So, it will divide ABC into two triangles of equal areas.

Solution 3

Solution 4

Solution 5

(i) In ABC

E and F are mid points of side AC and AB respectively

So, EF || BC and EF = BC (mid point theorem)

But BD = BC (D is mid point of BC)

So, BD = EF

Now the line segments BF and DE are joining two parallel lines EF and BD of same length.

So, line segments BF and DE will also be parallel to each other and also these will be equal in length.

Now as each pair of opposite sides are equal in length and parallel to each other.

Therefore BDEF is a parallelogram

Now the line segments BF and DE are joining two parallel lines EF and BD of same length.

So, line segments BF and DE will also be parallel to each other and also these will be equal in length.

Now as each pair of opposite sides are equal in length and parallel to each other.

Therefore BDEF is a parallelogram

(ii) Using the result obtained as above we may say that quadrilaterals BDEF, DCEF, AFDE are parallelograms.

We know that diagonal of a parallelogram divides it into two triangles of equal area.

We know that diagonal of a parallelogram divides it into two triangles of equal area.

Now,

Area (AFE) + area (BDF) + area (CDE) + area (DEF) = area (ABC)

Area (AFE) + area (BDF) + area (CDE) + area (DEF) = area (ABC)

(iii) Area (parallelogram BDEF) = area (DEF) + area (BDE)

Area (parallelogram BDEF) = area (DEF) + area (DEF)

Area (parallelogram BDEF) = 2 area (DEF)

Area (parallelogram BDEF) = 2 area (ABC)

Area (parallelogram BDEF) = area (ABC)

Solution 6

Let us draw DN AC and BM AC

(i) In DON and BOM

DNO = BMO (By construction)

DON = BOM (Vertically opposite angles)

OD = OB (Given)

By AAS congruence rule

DON BOM

DNO = BMO (By construction)

DON = BOM (Vertically opposite angles)

OD = OB (Given)

By AAS congruence rule

DON BOM

On adding equation (2) and (3), we have

Area (DON) + area (DNC) = area (BOM) + area (BMA)

So, area (DOC) = area (AOB)

Area (DON) + area (DNC) = area (BOM) + area (BMA)

So, area (DOC) = area (AOB)

(ii) We have

Area (DOC) = area (AOB)

Area (DOC) = area (AOB)

Area (DOC) + area (OCB) = area (AOB) + area (OCB)

(Adding area (OCB) to both sides)

Area (DCB) = area (ACB)

Area (DCB) = area (ACB)

(iii) Area (DCB) = area (ACB)

Now if two triangles are having same base and equal areas, these will be between same parallels

Now if two triangles are having same base and equal areas, these will be between same parallels

For quadrilateral ABCD, we have one pair of opposite sides are equal (AB = CD) and other pair of opposite sides are parallel

(DA || CB).

Therefore, ABCD is parallelogram

Solution 7

Since, BCE and BCD are lying on a common base BC and also having equal areas, so, BCE and BCD will lie between the same parallel lines.

Solution 8

Given that

XY || BC EY || BC

BE || AC BE || CY

So, EBCY is a parallelogram.

It is given that

XY || BC XF || BC

FC || AB FC || XB

XY || BC EY || BC

BE || AC BE || CY

So, EBCY is a parallelogram.

It is given that

XY || BC XF || BC

FC || AB FC || XB

So, BCFX is a parallelogram.

Now parallelogram EBCY and parallelogram BCFX are on same base BC and between same parallels BC and EF

Now parallelogram EBCY and parallelogram BCFX are on same base BC and between same parallels BC and EF

Consider parallelogram EBCY and AEB

These are on same base BE and are between same parallels BE and AC

These are on same base BE and are between same parallels BE and AC

Also parallelogram BCFX and ACF are on same base CF and between same parallels CF and AB

From equations (1), (2), and (3), we have

Area (ABE) = area (ACF)

Area (ABE) = area (ACF)

Solution 9

Solution 10

Here, DAC and DBC lie on same base DC and between same parallels AB and CD

Solution 11

(i) ACB and ACF lie on the same base AC and are between

the same parallels AC and BF

the same parallels AC and BF

(ii) Area (ACB) = area (ACF)

Solution 12

Let quadrilateral ABCD be original shape of field. 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. Now portion AOB can be cut from the original field so that new shape of field will be BCE.

Now 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 the DEO (portion added to the field so as to make the area of new field so formed equal to the area of original field)

Let it meet the extended side CD of ABCD at point E. Join BE and AD. Let them intersect each other at O. Now portion AOB can be cut from the original field so that new shape of field will be BCE.

Now 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 the DEO (portion added to the field so as to make the area of new field so formed equal to the area of original field)

DEB and DAB lie on same base BD and are between same parallels BD and AE.

Solution 13

Solution 14

Solution 15

Solution 16

Therefore, ABCD is a trapezium

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