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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. 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?

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)

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)

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).

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

Sol.

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

$\Rightarrow$ $AD = {{\sqrt 3 } \over 2}a$
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. 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. 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.

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)

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