NCERT Class 9 Maths Solutions Chapter - 9 Areas of Parallelograms and Triangles, Ex - 9.2

Ex - 9.2

Question 1.  In the given figure, ABCD is parallelogram, AE  DC and CF  AD. If AB = 16 cm. AE = 8 cm and CF = 10 cm, find AD.


Solution

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.

Question 2.  If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD show that  ar (EFGH) = ar (ABCD)

Solution

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

Question 3.  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).

Solution

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)

Question 4.  In the given figure, P is a point in the interior of a parallelogram ABCD. Show that

(i)  ar (APB) + ar (PCD) =   ar (ABCD) 
(ii)  ar (APD) + ar (PBC) = ar (APB) + ar (PCD)

Solution

(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)

Question 5.  In the given figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that 

    
(i)    ar (PQRS) = ar (ABRS)    

(ii)    ar (PXS) =   ar (PQRS)

Solution

(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

       

Question 6.  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 field 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?

Solution

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.

Ex-9.1

Ex-9.3