Oops! It appears that you have disabled your Javascript. In order for you to see this page as it is meant to appear, we ask that you please re-enable your Javascript!

                                                                                                                                                                                                                                                 

Areas of Parallelogram and Triangles : Exercise 9.4 (Optional)(Mathematics NCERT Class 9th)


Q.1    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.
Sol.

Given : A || gm ABCD and a rectangle ABEF with same base AB and equal areas.
To prove : Perimeter of || gm ABCD > Perimeter of rectangle ABEF.

31
Proof : Since opposite sides of a|| gm and rectangle are equal.
Therefore AB = DC               [Since ABCD is a || gm]
and,           AB = EF               [Since ABEF is a rectangle]
Therefore  DC = EF                                                                     ... (1)
 \Rightarrow          AB + DC = AB + EF  (Add AB in both sides)                   ... (2)
Since, of all the segments that can be drawn to a given line from a point not lying on it, the perpendicular segment is the shortest.
Therefore BE < BC and AF < AD
 \Rightarrow BC > BE and AD > AF
 \Rightarrow BC + AD > BE + AF                   ... (3)
Adding (2) and (3), we get
AB + DC + BC + AD > AB + EF + BE + AF
 \Rightarrow AB + BC + CD + DA > AB + BE + EF + FA
 \Rightarrow   perimeter of || gm ABCD > perimeter of rectangle ABEF.
Hence,the perimeter of the parallelogram is greater than that of the rectangle.


Q.2     In figure , D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC).
Can you know 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?

32
Sol.

Let AL be perpendicular to BC. So, AL is the height of \Delta s ABD, ADE and, AEC.
Therefore ar (ABD)  = {1 \over 2} × BD × AL
ar (ADE)  = {1 \over 2} ×DE × AL
and, ar (AEC)  = {1 \over 2} × EC × AL
Since BD = DE = EC
Therefore ar (ABD) = ar (ADE) = ar(AEC)
33 
Yes , altitudes of all triangles are same. Budhia has use the result of this question in dividing her land in three equal parts.


Q.3     In figure , ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF)

34
Sol.

Since opposite sides of a|| gm are equal
Therefore , AD = DC                         [Since ABCD is a || gm]
DE = CF                                                [Since DCFE is a || gm]
and AE = BF                                       [Since ABFE is a || gm]
Consider \Delta s ADE and BCF, in which AE = BF , AD = BC and DE = CF
Therefore by SSS criterion of congruence
\Delta \,ADE\, \cong \,\Delta BCF
 \Rightarrow ar(ADE) = ar (BCF)


Q.4      In figure, 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).

35
Sol.

Join AC.
Since \Delta s APC and BPC are on the same base PC and between the same parallels PC and AB.

37
Therefore , ar(APC) = ar (BPC) ... (1)
Since AD = CQ
and ,  AD || CQ [Given]
Therefore in the quadrilateral ADQC, one pair of opposite sides is equal and parallel.
Therefore ADQC is a parallelogram.
 \Rightarrow AP = PQ and CP = DP [Since diagonals of a|| gm bisect each other]
In \Delta s APC and DPQ we have
AP = PQ [Proved above]
\angle APC = \angle DPQ [Vertically opp. \angle s]
and , PC = PD [Proved above]
Therefore by SAS criterion of congruence,
\Delta \,APC \cong \Delta DPQ
 \Rightarrow ar (APC) = ar (DPQ) ... (2)
[Since congruent \Delta s have equal area]
Therefore ar (BPC) = ar (DPQ)        [From (1)]
Hence , ar (BPC) = ar (DPQ)


Q.5      In figure, 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) = {1 \over 4} ar (ABC)
           (ii) ar (BDE) = {1 \over 2} ar (BAE)
           (iii) ar (ABC) = 2 ar (BEC)
           (iv) ar (BFE) = ar (AFD)
           (v) ar (BFE) = 2 ar (FED)
           (vi) ar (FED) = {1 \over 8} ar (AFC).

42

Sol.

Join EC and AD. Let a be the side of \Delta ABC. then,
ar\left( {ABC} \right) = {{\sqrt 3 } \over 4}{a^2} = \Delta \left( {say} \right)

36
(i) ar\left( {BDE} \right) = {{\sqrt 3 } \over 4}{\left( {{a \over 2}} \right)^2}
\left[ {Since\,\,BD = {1 \over 2}BC = {a \over 2}} \right]
 = {{\sqrt 3 } \over {16}}{a^2} = {\Delta \over 4}
 \Rightarrow ar\left( {BDE} \right) = {1 \over 4}ar\left( {ABC} \right)

(ii) We have , ar (BDE)  = {1 \over 2} ar (BEC) ... (1)
[Since DE is a median of \Delta BEC and each median divides a triangle
in two other \Delta s of equal area]
Now, \angle EBC = 60^o
and \angle BCA = 60^o
 \Rightarrow \angle EBC = \angle BCA
But these are alterante angles with respect to the line- segments BE and CA and their transversal BC.
Hence BE || AC.
Now, \Delta s BEC and BAE stand on the same base BE and lie between the same parallels BE and AC.
Therefore ar(BEC) = ar(BAE)
Therefore From (1) ar (\Delta BDE) = {1 \over 2} ar (BAE).

(iii) Since ED is a median of \Delta BEC and we know that each median divides a triangle in two other \Delta s of equal area.
Therefore ar(BDE) = {1 \over 2} ar (BEC)
From part (i),
ar (BDE) = {1 \over 4} ar (ABC)

Combining these results , we get
{1 \over 4} ar (ABC) = {1 \over 2} ar (BEC)
 \Rightarrow ar (ABC) = 2 ar (BEC)

(iv) Now,\angle ABD = \angle BDE = 60^\circ [Given]
But \angle ABD\,and\,\angle BDE are alternate angles with respect to the line- segment BA and DE and their transversal BD.
Hence BA || ED.
Now, \Delta s BDE and AED stand on the same base ED and lie between the same parallels BA and DE
Therefore ar (BDE) = ar (AED)
 \Rightarrow ar(BDE) – ar(FED) = ar (AED) – ar (FED)
 \Rightarrow ar (BEF) = ar (AFD)

(v) In \Delta ABC, A{D^2} = A{B^2} - B{D^2}
 = (a)^2 -\left({{a \over 2}}\right)^2
 = {a^2} - {{{a^2}} \over 4} = {{3{a^2}} \over 4}
38
 \Rightarrow AD = {{\sqrt 3 } \over 2}a
In \Delta BED , E{L^2} = D{E^2} - D{L^2}
 = {\left( {{a \over 2}} \right)^2} - {\left( {{a \over 4}} \right)^2} = {{{a^2}} \over 4} - {{{a^2}} \over {16}} = {{3{a^2}} \over {16}}        [EL is median of\Delta BED]
 \Rightarrow EL = {{\sqrt 3 a} \over 4}
Therefore ar(AFD) = {1 \over 2} × FD × AD
= {1 \over 2} × FD × {{\sqrt 3 } \over 2}a ... (1)
and ar(EFD) = {1 \over 2} × FD × EL
= {1 \over 2} × FD = {{\sqrt 3 } \over 4}a ... (2)
From (1) and (2), we have ar (AFD) = 2 ar (EFD)
Combining this result with part (iv).
We have ar (BFE) = ar (AFD) = 2ar (EFD)

(vi) From part (i)
ar (BDE) = {1 \over 4} ar (ABC)
 \Rightarrow ar (BEF) + ar (FED) = {1 \over 4} × 2ar (ADC)
 \Rightarrow 2ar (FED) + ar (FED) = {1 \over 2} (ar(AFC) – ar (AFD) [Using part (v)]
 \Rightarrow 3ar (FED) = {1 \over 2} ar (AFC) – {1 \over 2} × 2 ar (FED)
 \Rightarrow 4ar (FED) = {1 \over 2} ar (AFC)
 \Rightarrow ar (FED) = {1 \over 8} ar (AFC)


Q.6     Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar (APB) × ar (CPD) = ar (APD) × ar (BPC).
Sol.       Diagonals AC and BD of a quadrilateral ABCD intersect each other at P.

39
Draw AM  \bot BD and CN  \bot BD.
Now, ar (APB) × ar(CPD)
 = \left( {{1 \over 2} \times BP \times AM} \right) \times \left( {{1 \over 2} \times DP \times CN} \right)
 = {1 \over 4} × BP × DP × AM × CN
and , ar (APD) × ar (BPC)
 = \left( {{1 \over 2} \times DP \times AM} \right) \times \left( {{1 \over 2} \times BP \times CN} \right)
 = {1 \over 4} × BP × DP × AM × CN
From (1) and (2), we have
ar (APB) × ar (CPD) = ar (APD) × ar (BPC)


Q.7     P and Q are respectively the mid- points of sides AB and BC of of a triangle ABC and R is the mid- point of AP, show that
                   (i) ar (PRQ) = {1 \over 2} ar (ARC)
           (ii) ar (RQC) = {3 \over 8} ar (ABC)

                  (iii) ar (PBQ) = ar (ARC)
Sol.

P and Q are respectively the mid- points of sides AB and BC of \Delta ABC and R is the mid- point of AP.

40
Join AQ and PC.
(i) We have,
ar (PQR) = {1 \over 2} ar (APQ)              [Since QR is a median of \Delta APQ and it divides the \Delta into two other \Delta s of equal area]
 = {1 \over 2} \times {1 \over 2} ar (ABQ)                                 [Since QP is a median of \Delta ABQ]
 = {1 \over 4} ar (ABQ) = {1 \over 4} \times {1 \over 2} ar (ABC)
[Since AQ is a median of \Delta ABC]
 = {1 \over 8} ar (ABC) ... (1)
Again , ar(ARC) = {1 \over 2} ar (APC)                  [Since CR is a median of \Delta APC]
 = {1 \over 2} \times {1 \over 2} ar (ABC)                [Since CP is a median of \Delta ABC]
 = {1 \over 4} ar (ABC) ... (2)
From (1) and (2) , we get
ar (PQR) = {1 \over 8} ar (ABC) = {1 \over 2} \times {1 \over 4} ar (ABC)
 = {1 \over 2} ar (ARC).

(ii) We have,
ar (RQC) = ar(RQA) + ar (AQC) – ar (ARC) ... (3)
Now, ar (\Delta RQA) = {1 \over 2} ar (PQA)              [Since RQ is a median of \Delta PQA]
 = {1 \over 2} \times {1 \over 2} ar (AQB)               [Since PQ is a median of \Delta AQB]
 = {1 \over 4} ar (AQB)
 = {1 \over 4} \times {1 \over 2} ar(ABC)                               [Since AQ is a median of \Delta ABC]
 = {1 \over 8} ar (ABC) ... (4)
ar (AQC) = {1 \over 2} ar (ABC) ... (5)                [Since AQ is a median of \Delta ABC]
ar (\Delta ARC) = {1 \over 2} ar (APC)                  [Since CR is a median of \Delta APC]
 = {1 \over 2} \times {1 \over 2} ar(ABC) [Since CP is a median of \Delta ABC]
 = {1 \over 4} ar (ABC) ... (6)
From (3), (4) , (5) and (6) we have
ar (RQC) = {1 \over 8} ar (ABC) + {1 \over 2} ar (ABC) – {1 \over 4} ar (ABC)
 = \left( {{1 \over 8} + {1 \over 2} - {1 \over 4}} \right) ar (ABC)
 = {3 \over 8} ar (ABC)

(iii) We have,
ar (PBQ) = {1 \over 2} ar (ABQ)                    [Since PQ is a median of \Delta ABQ]
 = {1 \over 2} \times {1 \over 2}ar\left( {ABC} \right)                [Since AQ is a median of \Delta ABC]
 = {1 \over 4} ar (ABC)
= ar(ARC)                               [Using (6)]


Q.8     In figure, 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  \bot DE meets BC at Y. Show that :
          (i) \Delta MBC \cong \,\Delta ABD
          (ii) ar (BYXD) = 2 ar (MBC)
          (iii) ar (BYXD) = ar (ABMN)
          (iv) \Delta FCB \cong \,\Delta ACE
          (v) ar (CYXE) = 2 ar (FCB)
          (vi) ar (CYXE) = ar (ACFG)
        
(vii) 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.

41

Sol.

(i) In \Delta s MBC and ABD, we have
BC = BD [Sides of the square BCED]
MB = AB [Sides of the square ABMN]
\angle MBC = \angle ABD [Since Each = 90º + \angle ABC]
Therefore by SAS criterion of congruence, we have
\Delta MBC \cong \,\Delta ABD

(ii) \Delta ABD and square BYXD have the same base BD and are between the same parallels BD and AX.
Therefore ar(ABD)  = {1 \over 2} ar (BYXD)
But \Delta MBC \cong \,\Delta ABD [Proved in part (i)]
 \Rightarrow ar(MBC) = ar (ABD)
Therefore ar(MBC) = ar (ABD) = {1 \over 2} ar (BYXD)
 \Rightarrow ar(BYXD) = 2 ar (MBC).

(iii) Square ABMN and \Delta MBC have the same base MB and are between same parallels MB and NAC.
Therefore ar(MBC) = {1 \over 2} ar (ABMN)
 \Rightarrow ar(ABMN) = 2ar (MBC)
= ar (BYXD) [Using part (ii)]

(iv) In \Delta s ACE and BCF, we have
CE = BC [Sides of the square BCED]
AC = CF [Sides of the square ACFG]
and \angle ACE = \angle BCF [Since Each = 90º + \angle BCA]
Therefore by SAS criterion of congruence,
\Delta ACE \cong \Delta BCF

(v) \Delta ACE and square CYXE have the same base CE and are between same parallels CE and AYX.
Therefore ar(ACE) = {1 \over 2} ar (CYXE)
 \Rightarrow ar(FCB) = {1 \over 2} ar (CYXE) [Since \Delta ACE \cong \Delta BCF, part (iv)]
 \Rightarrow ar(CYXE) =2 ar (FCB) .

(vi) Square ACFG and \Delta BCF have the same base CF and are between same parallels CF and BAG.
Therefore ar(BCF) = {1 \over 2} ar(ACFG)
 \Rightarrow {1 \over 2} ar(CYXE) = {1 \over 2} ar (ACFG) [Using part (v)]
 \Rightarrow ar(CYXE) = ar (ACFG)

(vii) From part (iii) and (vi) we have
ar (BYXD) = ar (ABMN)
and ar (CYXE) = ar (ACFG)
On adding we get
ar (BYXD) + ar(CYXE) = ar (ABMN) + ar(ACFG)
ar (BCED) = ar (ABMN) + ar(ACFG)



51 Comments

Leave a Reply

Contact Us

Call us: 8287971571,0261-4890014

Or, Fill out the form & get a call back.!