NCERT Class 9 Maths Solutions Chapter - 9 Areas of Parallelograms and Triangles, Ex - 9.3

Ex - 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

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

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

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

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

(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

 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

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

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

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

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

(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

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

Question 14.  In the given figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).

Solution

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

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

Therefore, ABCD is a trapezium

Ex-9.2

Ex-10.1