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

Page / Exercise

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

Solution 1
(i)      Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesYes. Trapezium ABCD and triangle PCD are having a common base CD and these are lying between the same parallel lines AB and CD.
 
(ii)  Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
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) Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
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)  Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
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) Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
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) Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
 
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 Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles corresponding attitude
Area of parallelogram ABCD = CD Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles AE = AD Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles CF  
16 cm Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles 8 cm = AD Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles 10 cm
AD = Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles cm = 12.8 cm.
Thus, the length of AD is 12.8 cm.
Solution 2
 
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
 
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]
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles AH = BF and AH || BF    Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Therefore, ABFH is a parallelogram.
Since Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesHEF and parallelogram ABFH are on the same base HF and between the same parallel lines AB and HF.
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Trianglesarea (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesHEF) = Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles area (ABFH)            ...(1)
Similarly, we can prove
area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesHGF) = Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles area (HDCF)               ...(2)
On adding equations (1) and (2), we have
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Solution 3
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesHere Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBQC and parallelogram ABCD lie on same base BC and these are between same parallel lines AD and BC.
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Similarly, Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesAPB and parallelogram ABCD lie on the same base AB and between same parallel lines AB and DC
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
From equation (1) and (2), we have
area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBQC) = area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesAPB)  

Solution 4
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
(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
    Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
    From equations (1) and (2), we have
    AB || EF and AE || BF    
    So, quadrilateral ABFE is a parallelogram
    Now, we may observe that Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesAPB and parallelogram AB || FE are lying on the same base AB and between the same 
    parallel lines AB and EF.
    Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Similarly, for Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles PCD and parallelogram EFCD

area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesPCD) = area (EFCD)                                     ... (4) 

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

Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

 

(ii) 

Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
     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
     Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
     From equations (6) and (7), we have
     MN || AD and AM || DN    
     So, quadrilateral AMND is a parallelogram
     Now, Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesAPD and parallelogram AMND are lying on the same base AD and between the same parallel lines AD and MN.
     Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
     Similarly, for Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesPCB and parallelogram MNCB
     area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesPCB) = Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles area (MNCB)                            ... (9)
      Adding equations (8) and (9), we have

       Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

      On comparing equations (5) and (10), we have
      Area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesAPD) + area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesPBC) = area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesAPB) + area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesPCD)  

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.
       Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
(ii)   Now consider Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesAXS and parallelogram ABRS
       As these lie on the same base and are between same parallel lines AS and BR
       Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Solution 6
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
From figure it is clear that point A divides the field into three parts. These parts are triangular in shape - Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesPSA, Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesPAQ and Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesQRA

Area of Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesPSA + Area of Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesPAQ + Area of Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesQRA = 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.
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
From equations (1) and (2), we have
Area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesPSA) + area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesQRA) =  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 Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesABC. So, it will divide Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesABC into two triangles of equal areas.
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Solution 2
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Here, Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesDAC and Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesDBC lie on same base DC and between same parallels AB and CD
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Solution 3
(i)  Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesACB and Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesACF lie on the same base AC and are between
      the same parallels AC and BF
      Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
(ii)    Area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesACB) = area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesACF)   
        Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles  
 
 
Solution 4
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
 
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 Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesAOB can be cut from the original field so that new shape of field will be Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBCE.
Now we have to prove that the area of Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesAOB (portion that was cut so as to construct Health Centre) is equal to the area of the Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesDEO (portion added to the field so as to make the area of new field so formed equal to the area of original field)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesDEB and Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesDAB lie on same base BD and are between same parallels BD and AE.
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Solution 5
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Solution 6
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Solution 7
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Solution 8
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
 
Therefore, ABCD is a trapezium

Solution 9
 
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
AD is median of Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesABC. So, it will divide Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesABC into two triangles of equal areas.
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Solution 10
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Solution 11
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Solution 12
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
(i)  In Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesABC
     E and F are mid points of side AC and AB respectively
     So, EF || BC and EF = Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles BC          (mid point theorem)
      But BD = Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles 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
 
(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.
       Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
 
       Now,
       Area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesAFE) + area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBDF) + area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesCDE) + area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesDEF) = area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesABC)
       Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
 
(iii)   Area (parallelogram BDEF) = area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesDEF) + area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBDE)
       Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesArea (parallelogram BDEF) = area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesDEF) + area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesDEF)
       Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesArea (parallelogram BDEF) = 2 area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesDEF)
       Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles Area (parallelogram BDEF) = 2 Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesNcert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesABC)
       Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesArea (parallelogram BDEF) = Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesABC)
Solution 13
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
     Let us draw DN Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles AC and BM Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles AC
(i)  In Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesDON and Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBOM
     Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesDNO = Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBMO         (By construction)
     Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesDON = Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBOM         (Vertically opposite angles)
    OD = OB            (Given)
    By AAS congruence rule
   Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesDON Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBOM
    Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
    Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
 
    On adding equation (2) and (3), we have
    Area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesDON) + area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesDNC) = area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBOM) + area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBMA)
    So, area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesDOC) = area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesAOB)
 
(ii) We have
     Area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesDOC) = area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesAOB)
     Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles Area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesDOC) + area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesOCB) = area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesAOB) + area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesOCB)
      (Adding area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesOCB) to both sides)
       Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesArea (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesDCB) = area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesACB)
(iii)  Area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesDCB) = area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesACB)
      Now if two triangles are having same base and equal areas, these will be between same parallels
      Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
       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 14
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Since, Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBCE and Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBCD are lying on a common base BC and also having equal areas, so, Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBCE and Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBCD will lie between the same parallel lines.
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Solution 15
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
    Given that
    XY || BC   Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles    EY || BC    
    BE || AC   Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles   BE || CY    
    So, EBCY is a parallelogram.
    It is given that
    XY || BC   Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles    XF || BC    
    FC || AB   Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles    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
    Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
    Consider parallelogram EBCY and Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesAEB
    These are on same base BE and are between same parallels BE and AC
    Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
     Also parallelogram BCFX and Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesACF are on same base CF and between same parallels CF and AB
     Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
      From equations (1), (2), and (3), we have
      Area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesABE) = area (Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesACF)

Solution 16
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Ncert Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles