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Exercise 9.4

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. Here, AB = CD = EF.                                                             …(i)
Also in triangle BEC,

BC > BE                                                                                        [Side opposite to greater angle is larger]

Similarly, AD > AF                                                                        [Reason same as above]

∴  BC + AD > BE + AF

⇒     BC + AD + AB + DC > BE + AF + AB + EF                         [From (i)]

⇒     Perimeter of parallelogram ABCD > Perimeter of rectangle ABEF.

2. In figure, D and E are two points on BC such that BD = DE = EC. Show that
Can you now answer the question that you have in the ‘Introduction’ of this chapter in NCERT, whether the field of Budhia has been actually divided into three parts of equal areas?

[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 ΔABC into n triangles of equal areas.]

Sol. In triangle ABE, BD = DE                                                       [Given]

[Median of a triangle divides the triangle into two parts of equal area]

From (i) and (ii), we have

Hence, we answer the question in the ‘Introduction’ of this chapter in NCERT that Budhia has actually divided her field into three parts of equal areas.

3. In figure, ABCD, DCFE and ABFE are parallelograms.

Sol. As ABCD, DCEF and ABFE are parallelograms.

∴   AD = BC, DE = CF and AE= BF                                                   ….(i)

∴         ΔADE  ≅  ΔBCF.                                                                       [SSS] [From (i)]

[If triangles are congruent, their areas are equal]

4. In figure, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersects DC at P, show that ar(BPC) = ar(DPQ).
[Hint: Join AC.]

Sol. Construction: Join AC.

⇒   ar(APD) + ar(DPQ) = ar(ADP) + ar(APC)
⇒   ar(DPQ) = ar(APC)
Triangles APC and BPC are on the same base PC and between the same parallels PC and AB.
∴   ar(APC) = ar(BPC)
From (i) and (ii), we get
ar(DPQ) = ar(BPC).

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) = 2ar(FED)
(vi) ar(FED) = ${1 \over 8}$ ar(AFC)
[Hint: Join EC and AD. Show that $BE\parallel AC$ and $DE\parallel AB$, etc.]

Sol. Let AB = BC = AC  = x, then BD = BE = ED =${x \over 2}$

(i) ar(ABC) = ${{\sqrt 3 } \over 4}{x^2}$

and ar(BDE) = ${{\sqrt 3 } \over 4}{\left( {{x \over 2}} \right)^2} = {{\sqrt 3 } \over {16}}{x^2}$

$\Rightarrow ar(BDE) = {1 \over 4}ar(ABC)$
(ii) ∠ACB =60°  and    ∠DBE = 60°  or  ∠CBF = 60°

$\Rightarrow BE\parallel AC$                                          [  Alternate angles are equal.]

∴   ar(BAE) = ar(BCE)                                                                  …(i)

[Triangles on the same base and between the same parallels are equal in areas.]
Since DE is median of ABEC.
$\Rightarrow ar(BED) = {1 \over 2}ar(BEC)$
[Median divides triangle into two parts of equal areas]

$\Rightarrow ar(BED) = {1 \over 2}ar(BAE)$                                        [ From (i)]
(iii) ar(BCE) = 2 ar(BDE)
[Median divides triangle into two parts of equal areas.]

$= 2 \times {1 \over 4}ar(ABC)$                                                  [From result of part (i)]

⇒       2 ar(BCE) =${1 \over 2}$ ar(ABC).
(iv) Proceeding  as result of part (ii), we get $ED\parallel AB$.

[Triangles on the same base and between the same parallels are equal in area. ]

⇒  ar(ADE) — ar(EFD) = ar(BED) — ar(EFD)

⇒    ar(AFD) = ar(BEF).
(v) Draw EL ⊥ BD. Also AD ⊥ BC

Then $EL = \sqrt {{{\left( {{x \over 2}} \right)}^2} - {{\left( {{x \over 4}} \right)}^2}} = {{\sqrt 3 } \over 4}x$

Also      $AD = {{\sqrt 3 } \over 2}x$

$ar(\Delta AFD) = {1 \over 2}.FD.AD$

$ar(\Delta FED) = {1 \over 2}.FD.EL$

∴       ${{ar(\Delta AFD)} \over {ar(\Delta FED)}} = {{AD} \over {EL}} = {{\sqrt 3 } \over 2}x \times {4 \over {\sqrt {3x} }} = 2$

$\Rightarrow ar(\Delta AFD) = 2ar(\Delta FED)$

$\Rightarrow ar(\Delta BFE) = 2ar(\Delta FED)$                           [From result (iv) ]
(vi) From result (v),
ar(ΔBFE) = 2ar(ΔFED)

⇒        BF = 2FD

∴        FC = FD + DC = FD + BD = FD + 3FD = 4FD

∴    $ar(\Delta FED) = {1 \over 2}.FD.EL = {1 \over 2}.FD.{{\sqrt 3 } \over 4}x$   [ From result (v) ]
$= {{\sqrt 3 } \over 8}FD.x$                                                                           …….(i)

$ar(\Delta AFC) = {1 \over 2}.FD.AD = {1 \over 2}4FD.{{\sqrt 3 } \over 2}x$

= $\sqrt 3 .FD.x$

$= 8\left[ {{{\sqrt 3 } \over 8}FD.x} \right]$                         [ From (i)]

=  8  or  (Δ FED)

∴    ar( ΔFED) =  ${1 \over 8}$ ar (ΔAFC).

6. 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.]

Sol. ar(APB) × ar(CPD)
$= \left( {{1 \over 2} \times BP \times AL} \right)\left( {{1 \over 2} \times PD \times CM} \right)$

$= {1 \over 4} \times BP \times DP \times AL \times CM$
And ar(APD) × ar(BPC)

$= \left( {{1 \over 2} \times PD \times AL} \right)\left( {{1 \over 2} \times BP \times CM} \right)$

$= {1 \over 4} \times BP \times DP \times AL \times CM$                           ……(ii)
From (ii), we get

ar(APB) × ar(CPD) = ar(APD) × ar(BPC).

7. 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
(i) $ar(PRQ) = {1 \over 2}ar(ARC)$
(ii) ar(RQC) = ${3 \over 8}$ ar(ABC)
(iii) ar(PBQ) = ar(ARC)

Sol.

(i) Construction:
Join A and Q; P and C.
Proof: QP is median of triangle BQA.
∴  ar(BQP) = ar(PQA)                              …..(i)

Similarly, ar(QPR)

=  ar(RQA)                                             [QR is median of triangle PQA]

=  ar(PQA) = 2 ar(QPR)                      [From above result ]

∴    2 ar(QPR) = ar(BQP)                    [From (i)]

= ${1 \over 2}$ ar(BPC)                [PQ is median]

= ${1 \over 2}$ ar(APC)                [CP is median]

= ar (ARC)                                             [RC is median]

⇒                 ar(QPR) = ${1 \over 2}$ar(ARC).
(ii) ar(ARC) = ${1 \over 2}$ ar(CAP)                             [ CR is median of ΔACP]

${1 \over 2}\left\{ {{1 \over 2}ar(ABC)} \right\}$     [CP is median]

As RQ is median of ΔBRC

$ar(RQC) = {1 \over 2}ar(RBC) = {1 \over 2}\left\{ {ar(ABC) - ar(ARC)} \right\}$

$= {1 \over 2}\left\{ {ar(ABC) - {1 \over 4}ar(ABC)} \right\} = {3 \over 8}ar(ABC)$
(iii) ar(ARC) = ${1 \over 2}$ ar(CAP)                                    …(ii) [CR is median]

And ar(CAP) = ar(CPB)                                                                 …(iii) [CP is median]

From equations (ii) and (iii), we have

$ar(ARC) = {1 \over 2}ar(CPB)$

Further, ar(PBQ)=${1 \over 2}$ ar(PBC)                                  [PQ is median]

∴     ar(ARC) = ar(PBQ).

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 ⊥ DE meets BC at Y Show that:

(i)  ΔMBC  ≅  ΔABD
(ii) ar(BYXD) = 2ar(MBC)
(iii) ar(BYXD) = ar(ABMN)
(iv) ΔFCB ≅ ΔACE
(v) ar(CYXE) = 2ar(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) Consider triangles MBC and ABD,

∠MBA = ∠CBD                                                             [Each 90°]

Adding ∠ABC to both sides, we have

∠MBA + ∠ABC = ∠CBD + ∠ABC

⇒    ∠MBC = ∠ABD

MB = BA                                                                      [Sides of a square]

and BC = BD                                                               [Sides of a square]

∴  ΔMBC   ≅  ΔABD.                                                 [SAS]

(ii) Triangle ABD and rectangle BDXY are on the same base BD and between the same parallels BD and AX.

∴  ar(ABD) = ${1 \over 2}$ ar(BDXY)

As ΔABD  ≅  ΔMBC                                               [Using result of part (i)]

⇒   ar(ΔABD) = ar(ΔMBC)

∴  ar(MBC) = ${1 \over 2}$ ar(BDXY)

⇒   ar(BDXY) = 2 ar(MBC).
(iii) Triangle MBC and square MBAN are on the same base MB and between the same parallels MB and NC.
∴      2 ar(MBC) = ar(MBAN)

∴      ar(MBAN) = ar(BDXY).                                                   [Using result from part (ii)]

(iv) ∠ACF = ∠BCE                                                                     [90° each]

Adding ∠ACB to both sides, we get

∠ACF + ∠ACB = ∠BCE + ∠ACB

⇒    ∠BCF = ∠ ACE

Consider triangles BCF and ACE,

∠BCF = ∠ACE                                                                         [Proved above]

AC = CF                                                                                    [Sides of a square]

and BC = CE                                                                           [ Sides of a square]

∴       ΔFCB  ≅    ΔACE.                                                        [ SAS]
(v) Since rectangle CYXE and ΔACE are on the same base CE and between the same parallels CE and AX.
∴   ar(CYXE) = 2ar(ACE)

Now, use the result from part (iv), we get

ar(ACE) = ar(FCB)                                                               [Congruent triangles have equal areas]

Hence,   ar(CYXE)   =     2ar(FCB)
(vi) As triangle FCB and square ACFG are on the same base CF and between the same parallels CF and BG,

∴     ar(ACFG) = 2ar(FCB)

Comparing this result and the result from part (v),
we get

ar(CYXE) ar(ACFG)
(vii) Adding the results from parts (iii) and (vi), we get

ar(BYXD) + ar(CYXE) = ar(ABMN) + ar(ACFG)

⇒  ar(BCED) = ar(ABMN) + ar(ACFG).