NCERT Solution for Class 9 Mathematics Chapter 9 - Areas of Parallelograms and Triangles RR ACADEMY MEERUT

NCERT Solution for Class 9 Mathematics Chapter 9 - Areas of Parallelograms and Triangles Page/Excercise 9.1

Question 1

Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and the two parallels.              (i)                                (ii)                                      (iii)            (iv)                                   (v)                                          (iv)

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.

NCERT Solution for Class 9 Mathematics Chapter 9 - Areas of Parallelograms and Triangles Page/Excercise 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 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.

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 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 

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 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)  

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 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)  

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 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        

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 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.

NCERT Solution for Class 9 Mathematics Chapter 9 - Areas of Parallelograms and Triangles Page/Excercise 9.3

Question 1

In the given figure, E is any point on median AD of a ABC. Show that 
ar (ABE) = ar (ACE)  

Solution 1

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

Question 2

In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) =  ar (ABC)

Solution 2

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

Question 3

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

Solution 3

Question 4

In the given figure, ABC and ABD are two triangles on the same base AB. If line-segment CD is bisected by AB at O, show that ar (ABC) = ar (ABD).

Solution 4

Question 5

D, E and F are respectively the mid-points of the sides BC, CA and AB of a ABC. Show that (i) BDEF is a parallelogram. (ii) ar (DEF) =  ar (ABC) (iii) ar (BDEF) =  ar (ABC)

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)

Question 6

In the given figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that: 
(i)    ar (DOC) = ar (AOB)
(ii)    ar (DCB) = ar (ACB)
(iii)    DA || CB or ABCD is a parallelogram.  

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

Question 7

D and E are points on sides AB and AC respectively of ABC such that 
ar (DBC) = ar (EBC). Prove that DE || BC.

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. 

Question 8

XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and E respectively, show that
ar (ABE) = ar (ACF)

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)

Question 9

The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see the following figure). Show that 
ar (ABCD) = ar (PBQR).

Solution 9

Question 10

Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).

Solution 10

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

Question 11

In the given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. show that 
(i)    ar (ACB) = ar (ACF) (ii)    ar (AEDF) = ar (ABCDE)

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)                  

Question 12

A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.

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. 

Question 13

ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).

Solution 13

Question 14

In the given figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).

Solution 14

Question 15

Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a trapezium.

Solution 15

Question 16

In the given figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

Solution 16

   Therefore, ABCD is a trapezium