Chapter 9. Areas of parallelograms and Triangles

Important Note :

1. Two figures are called congruent, if they have the same shape and the same size.

2. If A and B are two congruent figures, then ar(A) = ar(B);and

3. If a planar region formed by a figure T is made up of two non-overlapping planar regions formed by figures P and Q, then ar(T) = ar(P) + ar(Q).

4. Two figures are said to be on the same base and between the same parallels, if they have a common base (side) and the vertices, (or the vertex) opposite to the common base of each figure lie on a line parallel to the base.

5. Parallelograms on the same base (or equal bases)and between the same parallels are equal in area.

6. Two triangles on the same base (or equal bases) and between the same parallels are equal in area.

7. Two triangles having the same base (or equal bases) and equal areas lie between the same parallels.
8. Area of a parallelogram is the product of its base and the corresponding altitude.
9. Parallelograms on the same base (or equal bases) and having equal areas lie between the same parallels.
10. If a parallelogram and a triangle are on the same base and between the same parallels, then area of the triangle is half the area of the parallelogram.
11. Triangles on the same base (or equal bases) and between the same parallels are equal in area.
12. Area of a triangle is half the product of its base and the corresponding altitude.
13. Triangles on the same base (or equal bases) and having equal areas lie between the same parallels.
14. A median of a triangle divides it into two triangles of equal areas .

EXERCISE 9.1

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 .

Solution:  (i) In figure :

Parallelogram ABCD and triangle DPC lie on the same base DC and between the same parallels DC and AB . So, the common base is DC and two parallels are DC and AB .

(ii) In figure :

Parallelogram PQRS and trapezium MNRS lie on the same base SR . but different parallels .

(iii) In figure:

Parallelogram PQRS and triangle QTR lie on the same base QR and between the same parallels QR and PS . So, the common base is QR and two parallels are QR and PS .

(iv) In figure:

Parallelogram ABCD and triangle PRQ lie on the same base BC . but different parallels .

(v) In figure:

Parallelogram ABCD and parallelogram APCD  lie on the same base AD and between the same parallels AD and BQ . So, the common base is AD and two parallels are AD and BQ .

(vi) In figure:

Parallelogram PQRS and parallelogram ABCD do not lie on the same base AD and between the same parallel PQ and SR .

 

EXERCISE 9.2

1. In Fig. 9.15, ABCD is a parallelogram, AE⊥DC and CF⊥AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.

 

Solution:  Given, AB = 16 cm, AE = 8 cm and CF = 10 cm .

ABCD is a parallelogram .

So,  AD = BC   and AB = DC .

Since, ABCD is a parallelogram and  ,then

Again , ABCD is a parallelogram and  ,then

 

From (i) and (ii) , we get

       

2. If E,F,G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that   .

Solution: Given,  E,F,G and H are respectively the mid-points of the sides of a parallelogram ABCD. Then we show that : .

Construction : we join HF .

Proof:  Since, ABCD is a parallelogram .

So, AB = CD and AD = BC

  and  

Now,   and

So, ABFH is a parallelogram .

Again,  and

So, CFHD is a parallelogram .

In  and parallelogram ABFH be on the same base HF and between the same parallels HF and AB, then

In  and parallelogram CFHD be on the same base HF and between the same parallels HF and CD, then

From (i) and (ii) , we get

 

     Proved.

3. P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that

Solution:  Given, P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Then we show that  

Proof: In figure :

     

Since,  and parallelogram ABCD be on the same base AB and between the same parallels AB and CD, then

Again,  and parallelogram ABCD be on the same base BC and between the same parallels BC and AD, then

 

From (i) and (ii) , we get

   Proved.

4. In Fig. 9.16, P is a point in the interior of a parallelogram ABCD. Show that : (i)     (ii) 

 
[Hint : Through P, draw a line parallel to AB.]

Solution: Given, P is a point in the interior of a parallelogram ABCD. Then we show that
(i) 
(ii)

Construction: Through P, we draw  and  .

Proof: Since, ABCD is a parallelogram .

So,  and  

In given figure :

(i) Since,  and parallelogram ABNM be on the same base AB and between the same parallels AB and MN, then

 

Again,  and parallelogram MNCD be on the same base DC and between the same parallels DC and MN, then

 

 Adding (i) and (ii) , we get

 

                                                Proved.

(ii) Since,  and parallelogram ADFE be on the same base AD and between the same parallels AD and EF, then

 

Again,   and parallelogram BCFE be on the same base BC and between the same parallels BC and EF, then

 

 Adding (i) and (ii) , we get

 

From (1) and (2) , we have

  

                                                Proved.

5. In Fig. 9.17, PQRS and ABRS are parallelograms and X is any point on side BR.

 

Show that : (i)     (ii) 

Solution:  Given, PQRS and ABRS are parallelograms and X is any point on side BR. Then we show that : (i)     (ii)

Proof: In given figure :

    

Since, two parallelograms PQRS and ABRS are on the same base SR and between the same parallels SR and PB , then

 

                              Proved.

(ii) Since,  and parallelogram ABRS be on the same base AS and between the same parallels AS and BR , then

 

From (1) and (2) , we get

  

                                   Proved.

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 fields 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: We draw a parallelogram  PQRS by the above statement .

The field is divided in three parts . The shape of the parts is triangular . The parts are APQ , APS and AQR .

Since,  and parallelogram PQRS be on the same base PQ and between the same parallels PQ and SR , then

 

 

Again, 

From (i) and (ii) , we get

  

  

Therefore, the farmer wants to sow wheat or pulse in triangular shape APQ or either in triangular shape APS and AQR respectively .

EXERCISE 9.3

1. In Fig.9.23, E is any point on median AD of a ∆ ABC. Show that   .

 

Solution: Given, E is any point on median AD of a ∆ABC. Then we show that   .
Proof:  We know that a median of a triangle divides it into triangles of equal areas.

So, AD is a median of a triangle ABC , then

 

 

And  DE is a median of a triangle BEC , then

 

Subtracting  (i) from (ii) , we get

 

  Proved.

2. In a triangle ABC, E is the mid-point of median AD. Show that .

Solution: Given, a triangle ABC, E is the mid-point of median AD. Then we show that

Proof:  We know that a median of a triangle divides it into triangles of equal areas.

So, AD is a median of a triangle ABC , then

 

 

Since, E is the mid-point of median AD .

So, BE and CE are the median of the triangles ABD and ACD respectively .

  BE is a median of a triangle ABD , then

 

  

From (i) and (ii) , we get

 

    Proved.

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

Solution:  let ABCD is a parallelogram and the diagonals are AC and BD . Then we show that

Proof:  Since,  and are on the same base AC and   .

So,

 

  

We know that the diagonals of a parallelogram bisect each other .

i.e., OA = OC and OB = OD

So, OA , OB , OC and OD are median of the triangles ABD , ABC , BCD and ACD respectively .  

Again, OB is a median of a triangle ABC , then

 

 

From (i) and (ii) , we have

      Proved.

4. In Fig. 9.24, ABC and ABD are two triangles on the same base AB. If line- segment CD is bisected by AB at O, show that .

 

Solution: Given, ABC and ABD are two triangles on the same base AB. If line- segment CD is bisected by AB at O. Then we show that .

Proof:  Since, CD is bisected by AB at O .

i.e., OC = OD

 So, OA and OB are the median of the triangles ACD and CBD respectively .

Since, OA is median of the triangle ACD, then

 

And  OB is median of the triangle CBD, then

  

Adding (i) and (ii) , we get

 

   Proved.

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)   (iii) 

Solution: Given, D, E and F are respectively the mid-points of the sides BC, CA and AB of a ∆ABC.

Then we show that :  (i) BDEF is a parallelogram.

(ii) 

(iii) 

Proof:  (i)  Since, D and E are the mid-point of the sides BC and AC respectively .

 and

So,  and [F is mid-point of AB]

   

Again, E and F are the mid-point of the sides AC and AB respectively .

and 

So,  and

Therefore, BDEF is a parallelogram .

(ii) Since, D and E are the mid-point of the sides BC and AC respectively .

 and

So,  and  [F is mid-point of AB]

Again, E and F are the mid-point of the sides AC and AB respectively .

and 

So, and

Thus,  BDEF is a parallelogram .

Similarly , We show that DCEF and AEDF are two parallelogram .

For parallelogram AEDF :

andare on the same base EF and between the same parallels AE and DE .

So,   

For  parallelogram BDEF

andare on the same base EF and between the same parallels AE and DE .

So, 

For parallelogram DCEF

andare on the same base EF and between the same parallels AE and DE .

So,    

Now,

 

 

 

   Proved.

(iii)  We have,

      Proved.

6. In Fig. 9.25, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that :

     

(i)   
(ii) 
(iii)  or ABCD is a parallelogram.
[Hint : From D and B, draw perpendiculars to AC.]

Solution:  Given, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD and AB = CD. Then we show that :
(i) 
(ii) 
(iii)  or ABCD is a parallelogram.

Construction: We draw and.

Proof: In given figure :

  (i) In and , we have

        [given]

   

 [Vertically opposite angles]

   [ASA]

    [CPCT]

So,

In and

   [given]

    [given]

    [SAS]

 

Adding (1) and (2) , we get

  Proved.

(ii) In and , we have

        CD = AB

        OD = OB

   [ Vertically opposite angles]

 [SAS]

  

Adding  on both sides , we get

 

 Proved. 

(iii)  Since, 

So,  and are on the same base CB and between the same parallels CB and DA .

Therefore,

Or ABCD is a parallelogram .

7. D and E are points on sides AB and AC respectively of ∆ABC such that  Prove that  .

Solution: Given, D and E are points on sides AB and AC respectively of ∆ABC such that  . Then we prove that  .

Proof: In figure :

 Since,  

So, DBC and EBC are two triangles on the same base BC and equal areas , then

       Proved .

8. XY is a line parallel to side BC of a triangle ABC. If  and   meet XY at E and F respectively, show that .

Solution:  Given, XY is a line parallel to side BC of a triangle ABC. If and  meet XY at E and F respectively. Then we show that  .

Construction : let XY line passes through A .

Proof: Since,  

So, ACBE is a parallelogram .

 

Since,  and parallelogram ACBE are on the same base AE and between the same parallels AE and BC .

 

and

So, ABCF is a parallelogram .

  and parallelogram ABCF are on the same base AF and between the same parallels AF and BC .

 

Again, parallelogram ACBE and parallelogram ABCF are on the same base BC and between the same parallels BC and EF .

So,   

   

     [From (i) and (ii) ]

                               Proved.

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 Fig. 9.26). Show that
ar (ABCD) = ar (PBQR) .

  

[Hint : Join AC and PQ. Now compare ar (ACQ) and ar (APQ).]

Solution:  Given, 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 .

 Then we show that  .

Construction: We join AC and PQ .

Proof: In given figure :

  

Since, and lie on the same base AC and between the same parallels AB and DC .

So,

 

 and lie on the same base PQ and between the same parallels BQ and PR .

So,

Again,  

and lie on the same base AQ and between the same parallels AQ and CP .

So,  

Subtracting  from (iii) on the both sides , we have

 

   

From (i) , (ii) and (iv) , we have

  

     Proved.

10. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that .

Solution: Given, the diagonals AC and BD of a trapezium ABCD with  intersect each other at O. Then we prove that

Proof: In figure :

     

Since,  and lie on the same base DC and between the same parallels DC and AB , then

  

Subtracting from (i) on both sides , we have

  

   Proved.

11. In Fig. 9.27, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that  (i)       (ii)

    

Solution:  Given, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F.

Then we show that:  (i)      (ii) 

Proof: In given figure :

      

(i) Since,   .

 and lie on the same base AC and between the same parallels AC and BF , then

 

(ii) Since, and lie on the same base AC and between the same parallels AC and BF , then

 

 

[ Adding   on both sides ]

   

  Proved.

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:  In given figure ,

    

Let PQMN be the land of the shape of a quadrilateral and the Gram Panchayat of the village decided to take some portion of the plot from one of the corner P of the plot PQMN .

To prove :  

Construction: we draw and join QR and NR .

Proof: Since,  and are on the same base QN and between the same parallels QN and PR .

So, 

Subtracting   from (i) , we get

   

 

Again,   

  [from (ii)]

 

Therefore, the Gram Panchayat  must provide land NOR to itwari to meet his requirement of having a triangular plot .

13. ABCD is a trapezium with AB∥DC . A line parallel to AC intersects AB at X and BC at Y. Prove that  .    [Hint : Join CX.]

Solution: Given, ABCD is a trapezium with  . A line parallel to AC intersects AB at X and BC at Y. Then we prove that  .     

Construction: We join CX .

Proof: In figure :

Since, and lie on the same base AC and between the same parallels AC and XY , the

 

Again, and lie on the same base AX and between the same parallels AX and DC , the

 

From (i) and (ii) , we get

  Proved.

14. In Fig.9.28, . Prove that   .

   

Solution:  Given, . Then we prove that  .

Proof: In given figure :

Since,  and

Since, and lie on the same base BQ and between the same parallels BQ and AP , then

 

Again,  and lie on the same base BQ and between the same parallels BQ and CR , then

  

Adding (i) and (ii) , we get

   

  Proved.

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:  Given, Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that  . Then we prove that ABCD is a trapezium.

Proof:  In figure :

  

We have, 

Adding   on both sides in (i) , we get

  

 and lie on the same base AB and between the same parallels AB and DC , then .

So, ABCD is a trapezium .    Proved.

16. In Fig.9.29, ar(∆DRC)=ar(∆DPC) and ar(∆BDP) = ar(∆ARC) . Show that both the quadrilaterals ABCD and DCPR are trapeziums.

 

Solution: Given,  and  . Then we show that both the quadrilaterals ABCD and DCPR are trapeziums.

Proof:  In given figure :

We have, 

and  

 

Subtracting  (i) from (ii) , we get

 

 

 and  lie on the same base DC and between the same parallels DC and AB , then .

So, ABCD is a trapezium .

Again,  and  lie on the same base DC and between the same parallels DC and RP , then .

So, DCPR is a trapezium .