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.

                    

Question 1:

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

 

 

 

 

 

 

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

show that ar (EFGH) = .

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).

 

In the figure, P is a point in the interior of a

parallelogram ABCD. Show that

(i) ar (APB) + ar (PCD) =

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

In the figure, PQRS and ABRS are parallelograms and X is

any point on side BR. Show that

(i) ar (PQRS) = ar (ABRS)

(ii) ar (AXS) = ar

 

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 are 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?

 

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

                                                             

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

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

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 ar (ABC) = ar (ABD).

 

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

1. BDEF is a parallelogram.

2. ar (DEF) =

3. ar (BDEF) =

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) ar (DOC) = ar (AOB)               

(ii) ar (DCB) = ar (ACB)

(iii) DA || CB or ABCD is a parallelogram.

[Hint: From D and B, draw perpendiculars to AC.]

 

 

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

Prove that DE || BC.

XY is a line parallel to side BC of a triangle ABC.

If BE || AC and CF || AB meet XY at E and F respectively,

show that ar (ABE) = ar (ACF).

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).]

 

 

 

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

Prove that ar (AOD) = ar (BOC).

In Fig. 9.27, 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)

 

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 l

and adjoining his plot so as to form a triangular plot.

Explain how this proposal will be implemented.

 

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). [Hint: Join CX.]

In Fig.9.28, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).

                                                     

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.

In Fig.9.29, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the

quadrilaterals ABCD and DCPR are trapeziums.

 

 

Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas.

Show that the perimeter of the parallelogram is greater than that of the rectangle.

In Fig. 9.30, D and E are two points on BC such that BD = DE = EC. Show that

ar (ABD) = ar (ADE) = ar (AEC)

 

 

 

 

Can you now answer the question that you have

left in the ‘Introduction’ of this chapter, whether

the field of Budhia has been actually divided into three parts of equal area?

[Remark: Note that by taking BD = DE = EC,

the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas.

In the same way, by dividing BC into n equal parts and

joining the points of division so obtained to the opposite vertex of BC,

you can divide ∆ABC into n triangles of equal areas.]

In Fig. 9.31, ABCD, DCFE and ABFE are parallelograms.

Show that: ar (ADE) = ar (BCF)

                                                    

In Fig. 9.32, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar (BPC) = ar (DPQ).     

[Hint: Join AC.]

                                                         

In Fig.9.33, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that:

(i) ar (BDE) =

(ii) ar (BDE) =

(iii) ar (ABC) = 2 ar (BEC)

(iv) ar (BFE) = ar (AFD)

(v) ar (BFE) = 2 ar (FED)

(vi) ar (FED) =                                                                                                                      

 [ Hint: Join EC and AD. Show that BE || AC and DE || AB, etc.

 

 

 

 

 

Diagonals AC and BD of a quadrilateral ABCD intersect each other at P.

Show that: ar (APB) × ar (CPD) = ar (APD) × ar (BPC). 

[Hint: From A and C, draw perpendiculars to BD.]

P and Q are respectively the mid-points of sides AB and BC of a triangle ABC

and R is the mid-point of AP, show that

1. ar (PRQ) =  

2. ar (RQC) =  

3. ar (PBQ) = ar (ARC)

In Fig. 9.34, ABC is a right triangle right angled at A.

BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively.

Line segment AX ⊥ DE meets BC at Y. Show that:

                                                  

1. ∆ MBC ≅ ∆ ABD                                  

2. ar (BYXD) = 2 ar (MBC)

3. ar (BYXD) = ar (ABMN)                      

4. ∆ FCB ≅ ∆ ACE

5. ar (CYXE) = 2 ar (FCB)                         

6. ar (CYXE) = ar (ACFG)

7. ar (BCED) = ar (ABMN) + ar (ACFG)

Note: Result (vii) is the famous Theorem of Pythagoras.

You shall learn a simpler proof of this theorem in Class X.