Chapter 9 : Areas of Parallelograms and Triangles - Ncert Solutions for Class 9 Maths CBSE

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Chapter 9 - Areas of Parallelograms and Triangles Excercise 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 Excercise 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
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]
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.
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)  

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

      

      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.
      
(ii)   Now consider AXS and parallelogram ABRS
       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.
From equations (1) and (2), we have
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 Excercise 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
 
(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.
      
 
       Now,
       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
   
   
 
    On adding equation (2) and (3), we have
    Area (DON) + area (DNC) = area (BOM) + area (BMA)
    So, area (DOC) = area (AOB)
 
(ii) We have
     Area (DOC) = area (AOB)
     Area (DOC) + area (OCB) = area (AOB) + area (OCB)
      (Adding area (OCB) to both sides)
       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
     
       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  
    So, BCFX is a parallelogram.
    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
   
     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)

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
      
(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)
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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