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Edited By Ramraj Saini | Updated on Sep 28, 2023 10:17 PM IST

**Parallelograms and Triangles Class 9 Questions And Answers **are discusses here. These NCERT solutions are developed by expert team at Careers360 team keeping in mind latest CBSE syllabus 2023-24. These solutions are simple, easy to understand and cover all the concepts step by step thus, ultimately help the students. In this particular NCERT book chapter, you will learn about the areas of different triangles and parallelograms.

This Story also Contains

- NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles
- Areas of Parallelograms and Triangles Class 9 Solutions - Important Formulae
- Areas of Parallelograms and Triangles Class 9 NCERT Solutions (Intext Questions and Exercise)
- Summary Of Class 9 Maths Chapter 9 NCERT Solutions
- NCERT Solutions For Class 9 Maths - Chapter Wise
- NCERT Solutions For Class 9 - Subject Wise
- NCERT Books and NCERT Syllabus

NCERT solutions for class 9 maths chapter 9 Areas Of Parallelograms And Triangles is also covering the solutions to the application based questions as well. There are several interesting problems in the class 9 maths NCERT syllabus . Solving these problems will help you in improving the concepts of the chapter and also in exams like the Olympiads. In total there are 4 practice exercises having 51 questions. Areas of parallelograms and triangles class 9 NCERT solutions have covered all the exercises including the optional ones in a detailed manner. Here you will get NCERT solutions for class 9 Maths also.

**Also Read| **

- NCERT Exemplar Solutions For Class 9 Maths Chapter Areas Of Parallelograms And Triangles
- Areas Of Parallelograms And Triangles Class 9 Chapter Notes

Areas of Parallelograms and Triangles Class 9 Questions And Answers PDF Free Download

Area of Parallelogram = Base × Height

Area of Triangle = (1/2) × Base × Height

Alternatively, it can be expressed as half the area of the parallelogram containing the triangle:

Area of Triangle = (1/2) × Area of Parallelogram

Area of Trapezium = (1/2) × (Sum of Parallel Sides) × Distance between Parallel Sides

Area of Rhombus = (1/2) × (Product of Diagonals)

Free download **NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles **for CBSE Exam.

** Class 9 maths chapter 9 question answer - Exercise: 9.1 **

** Answer: **

In figure (i), (iii) and (v) we can see that. they lie on the same base and between the same parallel lines.

In figure (i) figure (iii) figure (v)

Common base DC QR AD

Two parallels DC and AB QR and PS AD and BQ

** Class 9 areas of parallelograms and triangles NCERT solutions - Exercise : 9.2 **

** Q1 ** In Fig. , ABCD is a parallelogram, and . If , and , find AD.

**Answer: **

We have,

AE DC and CF AD

AB = 16 cm, AE = 8 cm and CF = 10 cm

Since ABCD is a parallelogram,

therefore, AB = DC = 16 cm

We know that, area of parallelogram (ABCD) = base . height

= CD AE = (16 8 )

SInce, CF AD

therefore area of parallelogram = AD CF = 128

= AD = 128/10

= ** AD = 12.8 cm **

Thus the required 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

** Answer: **

Join GE and HE,

Since F, F, G, H are the mid-points of the parallelogram ABCD. Therefore, GE || BC ||AD and HF || DC || AB.

It is known that if a triangle and the parallelogram are on the same base and between the same parallel lines. then the area of the triangle is equal to the half of the area of the parallelogram.

Now, EFG and ||gm BEGC are on the same base and between the same parallels EG and BC.

Therefore, ar ( EFG) = ar (||gm BEGC)...............(i)

Similarly, ar ( EHG) = 1/2 . ar(||gm AEGD)..................(ii)

By adding eq (i) and eq (ii), we get

ar ( EFG) + ar ( EHG) = 1/2 (ar (||gm BEGC) + ar(||gm AEGD))

ar (EFGH) = 1/2 ar(ABCD)

Hence proved

** Answer: **

We have,

ABCD is a parallelogram, therefore AB || CD and BC || AD.

Now, APB and ||gm ABCD are on the same base AB and between two parallels AB and DC.

Therefore, ar ( APB) = 1/2 . ar(||gm ABCD)...........(i)

Also, BQC and ||gm ABCD are on the same base BC and between two parallels BC and AD.

Therefore, ar( BQC) = 1/2 . ar(||gmABCD)...........(ii)

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

ar ( APB) = ar( BQC)

Hence proved.

** Q4 (i) ** In Fig. , P is a point in the interior of a parallelogram ABCD. Show that

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

**Answer: **

We have a ||gm ABCD and AB || CD, AD || BC. Through P, draw a line parallel to AB

Now, APB and ||gm ABEFare on the same base AB and between the same parallels EF and AB.

Therefore, ar ( APB) = 1/2 . ar(ABEF)...............(i)

Similarly, ar ( PCD ) = 1/2 . ar (EFDC) ..............(ii)

Now, by adding both equations, we get

Hence proved.

** Q4 (ii) ** In Fig. , P is a point in the interior of a parallelogram ABCD. Show that

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

**Answer: **

We have a ||gm ABCD and AB || CD, AD || BC. Through P, draw a line parallel to AB

Now, APD and ||gm ADGHare on the same base AD and between the same parallels GH and AD.

Therefore, ar ( APD) = 1/2 . ar(||gm ADGH).............(i)

Similarily, ar ( PBC) = 1/2 . ar(||gm BCGH)............(ii)

By adding the equation i and eq (ii), we get

Hence proved.

** Q5 ** In Fig. 9.17, PQRS and ABRS are parallelograms and X is any point on side BR. Show that

(i)

(ii)

**Answer: **

(i) Parallelogram PQRS and ABRS are on the same base RS and between the same parallels RS and PB.

Therefore, ............(i)

Hence proved

(ii) AXS and ||gm ABRS are on the same base AS and between same parallels AS and RB.

Therefore, ar ( AXS) = 1/2 . ar(||gm ABRS)............(ii)

Now, from equation (i) and equation (ii), we get

Hence proved.

** Answer: **

We have a field in the form of parallelogram PQRS and a point A is on the side RS. Join AP and AQ. The field is divided into three parts i.e, APS, QAR and PAQ.

Since APQ and parallelogram, PQRS is on the same base PQ and between same parallels RS and PQ.

Therefore, ............(i)

We can write above equation as,

ar (||gm PQRS) - [ar ( APS) + ar( QAR)] = 1/2 .ar(PQRS)

from equation (i),

Hence, she can sow wheat in APQ and pulses in [ APS + QAR] or wheat in [ APS + QAR] and pulses in APQ.

** Class 9 maths chapter 9 NCERT solutions - exercise : 9.3 **

** Q1 ** In Fig. , E is any point on median AD of a . Show that.

**Answer: **

We have ABC such that AD is a median. And we know that median divides the triangle into two triangles of equal areas.

Therefore, ar( ABD) = ar( ACD)............(i)

Similarly, In triangle BEC,

ar( BED) = ar ( DEC)................(ii)

On subtracting eq(ii) from eq(i), we get

ar( ABD) - ar( BED) =

Hence proved.

** Q2 ** In a triangle ABC, E is the mid-point of median AD. Show that .

** Answer: **

We have a triangle ABC and AD is a median. Join B and E.

Since the median divides the triangle into two triangles of equal area.

ar( ABD) = ar ( ACD) = 1/2 ar( ABC)..............(i)

Now, in triangle ABD,

BE is the median [since E is the midpoint of AD]

ar ( BED) = 1/2 ar( ABD)........(ii)

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

ar ( BED) = 1/2 . (1/2 ar(ar ( ABC)) ** ar ( BED) = 1/4 .ar( ABC) **

Hence proved.

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

** Answer: **

Let ABCD is a parallelogram. So, AB || CD and AD || BC and we know that Diagonals bisects each other. Therefore, AO = OC and BO = OD

Since OD = BO

Therefore, ar ( BOC) = ar ( DOC)...........(a) ( since OC is the median of triangle CBD)

Similarly, ar( AOD) = ar( DOC) ............(b) ( since OD is the median of triangle ACD)

and, ar ( AOB) = ar( BOC)..............(c) ( since OB is the median of triangle ABC)

From eq (a), (b) and eq (c), we get

ar ( BOC) = ar ( DOC)= ar( AOD) = ( AOB)

Thus, the diagonals of ||gm divide it into four equal triangles of equal area.

** Answer: **

We have, ABC and ABD on the same base AB. CD is bisected by AB at point O.

OC = OD

Now, in ACD, AO is median

ar ( AOC) = ar ( AOD)..........(i)

Similarly, in BCD, BO is the median

ar ( BOC) = ar ( BOD)............(ii)

Adding equation (i) and eq (ii), we get

ar ( AOC) + ar ( BOC) = ar ( AOD) + ar ( BOD)

Hence proved.

** Answer: **

We have a triangle ABC such that D, E and F are the midpoints of the sides BC, CA and AB respectively.

Now, in ABC,

F and E are the midpoints of the side AB and AC.

Therefore according to mid-point theorem, the line segment connecting the midpoints of two sides of a triangle is parallel to the third side and half of the third side.

EF || BC or EF || BD

also, EF = 1/2 (BC)

[ D is the midpoint of BC]

Similarly, ED || BF and ED = FB

Hence BDEF is a parallelogram.

** Q5 (ii) ** D, E and F are respectively the mid-points of the sides BC, CA and AB of a . Show that

** Answer: **

We already proved that BDEF is a ||gm.

Similarly, DCEF and DEAF are also parallelograms.

Now, ||gm BDEF and ||gm DCEF is on the same base EF and between same parallels BC and EF

Ar (BDEF) = Ar (DCEF)

Ar( BDF) = Ar ( DEF) .............(i)

It is known that diagonals of ||gm divides it into two triangles of equal area.

Ar(DCE) = Ar (DEF).......(ii)

and, Ar( AEF) = Ar ( DEF)...........(iii)

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

Ar( BDF) = Ar(DCE) = Ar( AEF) = Ar ( DEF)

Thus, Ar ( ABC) = Ar( BDF) + Ar(DCE) + Ar( AEF) + Ar ( DEF)

Ar ( ABC) = 4 . Ar( DEF)

Hence proved.

** Q5 (iii) ** D, E and F are respectively the mid-points of the sides BC, CA and AB of a . Show that

** Answer: **

Since we already proved that,

ar( DEF) = ar ( BDF).........(i)

So, ar(||gm BDEF) = ar( BDF) + ar ( DEF)

= 2 . ar( DEF) [from equation (i)]

Hence proved.

** Q6 (i) ** In Fig. , diagonals AC and BD of quadrilateral ABCD intersect at O such that. If , then show that:

[ ** Hint: ** From D and B, draw perpendiculars to AC.]

**Answer: **

We have ABCD is quadrilateral whose diagonals AC and BD intersect at O. And OB = OD, AB = CD

Draw DE AC and FB AC

In DEO and BFO

DOE = BOF [vertically opposite angle]

OED = BFO [each ]

OB = OD [given]

Therefore, by AAS congruency

DEO BFO

DE = FB [by CPCT]

and ar( DEO) = ar( BFO) ............(i)

Now, In DEC and ABF

DEC = BFA [ each ]

DE = FB

DC = BA [given]

So, by RHS congruency

DEC BFA

1 = 2 [by CPCT]

and, ar( DEC) = ar( BFA).....(ii)

By adding equation(i) and (ii), we get

Hence proved.

** Q6 (ii) ** In Fig. , diagonals AC and BD of quadrilateral ABCD intersect at O such that . If , then show that:

[ ** Hint: ** From D and B, draw perpendiculars to AC.]

**Answer: **

We already proved that,

Now, add on both sides we get

Hence proved.

** Q6 (iii) ** In Fig. , diagonals AC and BD of quadrilateral ABCD intersect at O such that . If , then show that:

or ABCD is a parallelogram.

[ ** Hint ** : From D and B, draw perpendiculars to AC.]

** Answer: **

Since DCB and ACB both lie on the same base BC and having equal areas.

Therefore, They lie between the same parallels BC and AD

CB || AD

also 1 = 2 [ already proved]

So, AB || CD

Hence ABCD is a || gm

** Q7 ** D and E are points on sides AB and AC respectively of such that . Prove that .

** Answer: **

We have ABC and points D and E are on the sides AB and AC such that ar( DBC ) = ar ( EBC)

Since DBC and EBC are on the same base BC and having the same area.

They must lie between the same parallels DE and BC

Hence DE || BC.

**Answer: **

We have a ABC such that BE || AC and CF || AB

Since XY || BC and BE || CY

Therefore, BCYE is a ||gm

Now, The ||gm BCEY and ABE are on the same base BE and between the same parallels AC and BE.

ar( AEB) = 1/2 .ar(||gm BEYC)..........(i)

Similarly, ar( ACF) = 1/2 . ar(||gm BCFX)..................(ii)

Also, ||gm BEYC and ||gmBCFX are on the same base BC and between the same parallels BC and EF.

ar (BEYC) = ar (BCFX).........(iii)

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

ar( ABE) = ar( ACF)

Hence proved.

**Answer: **

Join the AC and PQ.

It is given that ABCD is a ||gm and AC is a diagonal of ||gm

Therefore, ar( ABC) = ar( ADC) = 1/2 ar(||gm ABCD).............(i)

Also, ar( PQR) = ar( BPQ) = 1/2 ar(||gm PBQR).............(ii)

Since AQC and APQ are on the same base AQ and between same parallels AQ and CP.

ar( AQC) = ar ( APQ)

Now, subtracting ABQ from both sides we get,

ar( AQC) - ar ( ABQ) = ar ( APQ) - ar ( ABQ)

ar( ABC) = ar ( BPQ)............(iii)

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

Hence proved.

** Q10 ** Diagonals AC and BD of a trapezium ABCD with intersect each other at O. Prove that .

** Answer: **

We have a trapezium ABCD such that AB || CD and it's diagonals AC and BD intersect each other at O

ABD and ABC are on the same base AB and between same parallels AB and CD

ar( ABD) = ar ( ABC)

Now, subtracting AOB from both sides we get

ar ( AOD) = ar ( BOC)

Hence proved.

** Q11 ** In Fig. , ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that

(i)

(ii)

** Answer: **

We have a pentagon ABCDE in which BF || AC and CD is produced to F.

(i) Since ACB and ACF are on the same base AC and between same parallels AC and FB.

ar( ACB) = ar ( ACF)..................(i)

(ii) Adding the ar (AEDC) on both sides in equation (i), we get

ar( ACB) + ar(AEDC) = ar ( ACF) + ar(AEDC)

Hence proved.

** Answer: **

We have a quadrilateral shaped plot ABCD. Draw DF || AC and AF || CF.

Now, DAF and DCF are on the same base DF and between same parallels AC and DF.

ar ( DAF) = ar( DCF)

On subtracting DEF from both sides, we get

ar( ADE) = ar( CEF)...............(i)

The portion of ADE can be taken by the gram panchayat and on adding the land CEF to his (Itwaari) land so as to form a triangular plot.( ABF)

** We need to prove that ar( ABF) = ar (quad. ABCD) **

Now, adding ar(quad. ABCE) on both sides in eq (i), we get

ar ( ADE) + ar(quad. ABCE) = ar( CEF) + ar(quad. ABCE)

ar (ABCD) = ar( ABF)

** Answer: **

We have a trapezium ABCD, AB || CD

XY ||AC meets AB at X and BC at Y. Join XC

Since ADX and ACX lie on the same base CD and between same parallels AX and CD

Therefore, ar( ADX) = ar( ACX)..........(i)

Similarly ar( ACX) = ar( ACY).............(ii) [common base AC and AC || XY]

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

ar( ADX) = ar ( ACY)

Hence proved.

** Answer: **

We have, AP || BQ || CR

BCQ and BQR lie on the same base (BQ) and between same parallels (BQ and CR)

Therefore, ar ( BCQ) = ar ( BQR)........(i)

Similarly, ar ( ABQ) = ar ( PBQ) [common base BQ and BQ || AP]............(ii)

Add the eq(i) and (ii), we get

ar ( AQC) = ar ( PBR)

Hence proved.

** Answer: **

We have,

ABCD is a quadrilateral and diagonals AC and BD intersect at O such that ar( AOD) = ar ( BOC) ...........(i)

Now, add ar ( BOA) on both sides, we get

ar( AOD) + ar ( BOA) = ar ( BOA) + ar ( BOC)

ar ( ABD) = ar ( ABC)

Since the ABC and ABD lie on the same base AB and have an equal area.

Therefore, AB || CD

Hence ABCD is a trapezium.

** Q16 ** In Fig. , and . Show that both the quadrilaterals ABCD and DCPR are trapeziums. ** **

** Answer:**

Given,

ar( DPC) = ar( DRC) ..........(i)

and ar( BDP) = ar(ARC)............(ii)

from equation (i),

Since DRC and DPC lie on the same base DC and between same parallels.

CD || RP (opposites sides are parallel)

Hence quadrilateral DCPR is a trapezium

Now, by subtracting eq(ii) - eq(i) we get

ar( BDP) - ar( DPC) = ar( ARC) - ar( DRC)

ar( BDC) = ar( ADC) (Since theya are on the same base DC)

AB || DC

Hence ABCD is a trapezium.

** Areas of parallelograms and triangles class 9 solutions - Exercise : 9.4 **

** Answer: **

We have ||gm ABCD and a rectangle ABEF both lie on the same base AB such that, ** ar(||gm ABCD) = ar(ABEF) **

for rectangle, AB = EF

and for ||gm AB = CD

CD = EF

AB + CD = AB + EF ...........(i)

SInce BEC and AFD are right angled triangle

Therefore, AD > AF and BC > BE

(BC + AD ) > (AF + BE)...........(ii)

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

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

(AB + BC + CD + DA) > (AB + BE + EF + FA)

Hence proved, perimeter of ||gm ABCD is greater than perimeter of rectangle ABEF.

** Q2 ** In Fig. , D and E are two points on BC such that . Show that .

**Answer: **

In ABC, D and E are two points on BC such that BD = DE = EC

AD is the median of ABE, therefore,

area of ABD= area of AED...................(1)

AE is the median of ACD, therefore,

area of AEC= area of AED...................(2)

From (1) and (2)

area of ABD=area of AED= area of AEC

Hence proved.

** Q3 ** In Fig. , ABCD, DCFE and ABFE are parallelograms. Show that .

**Answer: **

Given,

ABCD is a parallelogram. Therefore, AB = CD & AD = BC & AB || CD .......(i)

Now, ADE and BCF are on the same base AD = BC and between same parallels AB and EF.

Therefore, ar ( ADE) = ar( BCF)

Hence proved.

**Answer: **

Given,

ABCD is a ||gm and AD = CQ. Join AC.

Since DQC and ACD lie on the same base QC and between same parallels AD and QC.

Therefore, ar( DQC) = ar( ACD).......(i)

Subtracting ar( PQC) from both sides in eq (i), we get

ar( DPQ) = ar( PAC).............(i)

Since PAC and PBC are on the same base PC and between same parallel PC and AB.

Therefore, ar( PAC) = ar( PBC)..............(iii)

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

Hence proved.

[ ** Hint: ** Join EC and AD. Show that and , etc.]

** Answer: **

Let join the CE and AD and draw . It is given that ABC and BDE is an equilateral triangle.

So, AB =BC = CA = and D i sthe midpoint of BC

therefore,

(i) Area of ABC = and

Area of BDE =

Hence,

[ ** Hint: ** Join EC and AD. Show that and , etc.]

**Answer: **

Since ABC and BDE are equilateral triangles.

Therefore, ACB = DBE =

BE || AC

BAE and BEC are on the same base BE and between same parallels BE and AC.

Therefore, ar ( BAE) = ar( BEC)

ar( BAE) = 2 ar( BED) [since D is the meian of BEC ]

Hence proved.

[ ** Hint ** : Join EC and AD. Show that and , etc.]

**Answer: **

We already proved that,

ar( ABC) = 4.ar( BDE) (in part 1)

and, ar( BEC) = 2. ar( BDE) (in part ii )

ar( ABC) = 2. ar( BEC)

Hence proved.

**Answer: **

Since ABC and BDE are equilateral triangles.

Therefore, ACB = DBE =

BE || AC

BDE and AED are on the same base ED and between same parallels AB and DE.

Therefore, ar( BED) = ar( AED)

On subtracting EFD from both sides we get

Hence proved.

[ ** Hint ** : Join EC and AD. Show that and , etc.]

**Answer: **

In right angled triangle ABD, we get

So, in PED,

So,

Therefore, the Area of ..........(i)

And, Area of triangle ...........(ii)

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

ar( AFD) = 2. ar( EFD)

Since ar( AFD) = ar( BEF)

[ ** Hint: ** Join EC and AD. Show that and , etc.]

** Answer: **

.....(from part (v) ar( BFE) = 2. ar( FED) ]

Hence proved.

** Q6 ** Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that

[ * Hint: * From A and C, draw perpendiculars to BD.]

** Answer: **

Given that,

A quadrilateral ABCD such that it's diagonal AC and BD intersect at P. Draw and

Now, ar( APB) = and,

Therefore, ar( APB) ar( CDP)=

....................(i)

Similarly, ar( APD) ar ( BPC) =

................(ii)

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

Hence proved.

** Answer: **

We have ABC such that P, Q and R are the midpoints of the side AB, BC and AP respectively. Join PQ, QR, AQ, PC and RC as shown in the figure.

Now, in APC,

Since R is the midpoint. So, RC is the median of the APC

Therefore, ar( ARC) = 1/2 . ar ( APC)............(i)

Also, in ABC, P is the midpoint. Thus CP is the median.

Therefore, ar( APC) = 1/2. ar ( ABC)............(ii)

Also, AQ is the median of ABC

Therefore, 1/2. ar ( ABC) = ar (ABQ)............(iii)

In APQ, RQ is the median.

Therefore, ar ( PRQ) = 1/2 .ar ( APQ).............(iv)

In ABQ, PQ is the median

Therefore, ar( APQ) = 1/2. ar( ABQ).........(v)

From eq (i),

...........(vi)

Now, put the value of ar( APC) from eq (ii), we get

Taking RHS;

(from equation (iii))

(from equation (v))

(from equation (iv))

Hence proved.

** Answer: **

In RBC, RQ is the median

Therefore ar( RQC) = ar( RBQ)

= ar (PRQ) + ar (BPQ)

= 1/8 (ar ABC) + ar( BPQ) [from eq (vi) & eq (A) in part (i)]

= 1/8 (ar ABC) + 1/2 (ar PBC) [ since PQ is the median of BPC]

= 1/8 (ar ABC) + (1/2).(1/2)(ar ABC) [CP is the medain of ABC]

= 3/8 (ar ABC)

Hence proved.

** Answer: **

QP is the median of ABQ

Therefore, ar( PBQ) = 1/2. (ar ABQ)

= (1/2). (1/2) (ar ABC) [since AQ is the median of ABC

= 1/4 (ar ABC)

= ar ( ARC) [from eq (A) of part (i)]

Hence proved.

** Answer: **

We have, a ABC such that BCED, ACFG and ABMN are squares on its side BC, CA and AB respec. Line segment meets BC at Y

(i) [each 90]

Adding on both sides, we get

In ABD and MBC, we have

AB = MB

BD = BC

Therefore, By SAS congruency

ABD MBC

Hence proved.

** Answer: **

SInce ||gm BYXD and ABD are on the same base BD and between same parallels BD and AX

Therefore, ar( ABD) = 1/2. ar(||gm BYXD)..........(i)

But, ABD MBC (proved in 1st part)

Since congruent triangles have equal areas.

Therefore, By using equation (i) we get

Hence proved.

** Answer: **

Since, ar (||gm BYXD) = 2 .ar ( MBC) ..........(i) [already proved in 2nd part]

and, ar (sq. ABMN) = 2. ar ( MBC)............(ii)

[Since ABMN and AMBC are on the same base MB and between same parallels MB and NC]

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

** Answer: **

[both 90]

By adding ABC on both sides we get

ABC + FCA = ABC + BCE ** FCB = ACE **

** In ** FCB and ACE

FC = AC [sides of square]

BC = AC [sides of square] ** FCB = ACE ** FCB ACE

Hence proved.

** Answer: **

Since ||gm CYXE and ACE lie on the same base CE and between the same parallels CE and AX.

Therefore, ** 2. ar( ACE) = ar (||gm CYXE) B ** ut, FCB ACE (in iv part)

Since the congruent triangle has equal areas. So,

Hence proved.

** Answer: **

Since ar( ||gm CYXE) = 2. ar( ACE) {in part (v)}...................(i)

Also, ** FCB and ** quadrilateral ACFG lie on the same base FC and between the same parallels FC and BG.

Therefore, ar (quad. ACFG) = 2.ar( ** FCB ** ) ................(ii)

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

Hence proved.

** Answer: **

We have,

ar(quad. BCDE) = ar(quad, CYXE) + ar(quad. BYXD)

= ar(quad, CYXE) + ar (quad. ABMN) [already proved in part (iii)]

Thus,

Hence proved.

- Figures on the Same Base and Between the Same Parallels
- Parallelograms on the same Base and Between the same Parallels
- Triangles on the same Base and between the same Parallels

Maths chapter 9 class 9 - Important Points

- The area of a parallelogram is equal to the product of its base and the corresponding altitude.
- The altitude of a parallelogram is the perpendicular distance between the two parallel sides.
- The area of a triangle is half the product of its base and the corresponding altitude.
- The altitude of a triangle is the perpendicular distance between the base and the opposite vertex.
- If two parallelograms are on the same base and between the same parallels, then their areas are equal.
- If a parallelogram and a triangle are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.
- If two triangles have the same base and lie between the same parallels, then their areas are equal.
- The sum of the areas of two triangles formed by drawing a diagonal of a parallelogram is equal to the area of the parallelogram.
- The sum of the areas of four triangles formed by drawing the diagonals of a parallelogram is equal to twice the area of the parallelogram.
- The area of a trapezium is half the product of the sum of its parallel sides and the corresponding altitude.

Interested students can practice class 9 maths ch 9 question answer using the following links

- NCERT Solutions for Class 9 Maths Exercise 9.1
- NCERT Solutions for Class 9 Maths Exercise 9.2
- NCERT Solutions for Class 9 Maths Exercise 9.3
- NCERT Solutions for Class 9 Maths Exercise 9.4

Chapter No. | Chapter Name |

Chapter 1 | Number Systems |

Chapter 2 | Polynomials |

Chapter 3 | Coordinate Geometry |

Chapter 4 | Linear Equations In Two Variables |

Chapter 5 | Introduction to Euclid's Geometry |

Chapter 6 | Lines And Angles |

Chapter 7 | Triangles |

Chapter 8 | Quadrilaterals |

Chapter 9 | Areas of Parallelograms and Triangles |

Chapter 10 | Circles |

Chapter 11 | Constructions |

Chapter 12 | Heron’s Formula |

Chapter 13 | Surface Area and Volumes |

Chapter 14 | Statistics |

Chapter 15 | Probability |

** How To Use NCERT Solutions For Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles **

- Before coming to this, please cover the basic properties related to triangles and parallelograms as well as the previous chapter.
- Learn some formulas and properties given in the NCERT textbook.
- Understand the method of solution through some examples.
- Practice the approach to which solutions are solved in the practice problems given.
- While doing practice problems, if you get stuck anywhere then assist yourself with NCERT solutions for class 9 maths chapter 9 Areas Of Parallelograms and Triangles.

* Keep working hard and happy learning! *

1. What are the important topics in chapter ch 9 maths class 9?

Areas of Parallelograms and Triangles, figures with the same base and between parallel lines, are the important topics of this chapter. Students can practice NCERT solutions for class 9 maths to command these concepts which are very important for the exam.

2. How does the NCERT solutions are helpful ?

NCERT solutions are not only helpful for the students if they stuck while solving NCERT problems but also, these solutions are provided in a very detailed manner which will give them conceptual clarity.

3. Where can I find the complete area of parallelogram and triangle class 9 ?

Here you will get the detailed NCERT solutions for class 9 maths by clicking on the link. you can practice these ncert solutions for class 9 maths chapter 9 to command the concepts which give the confidence during exams.

4. Do we need to master all the exercises included in the NCERT Solutions for Class 9 Maths Chapter 9?

In order to be able to tackle various types of questions that appear in exams, it is crucial to study and practice all the questions included in the NCERT Solutions for Class 9 Maths Chapter 9. The solutions are presented in a student-friendly language that facilitates easier comprehension of complex problems. These solutions are designed by Careers360 experts who possess extensive knowledge of mathematics.

Mar 02, 2024

Mar 02, 2024

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