Areas of Parallelograms and Triangles – Exercise 9.2 – (MATHEMATICS) – 9th Class

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Exercise 9.2 

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


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Sol. We have AD × CF = AE × CD                                                        …(i)

But AB = CD = 16 cm                                                         [Opposite sides of a parallelogram ]

∴     AD × 10 = 8 × 16  ⇒    AD = 12.8 cm.


2. If E, F G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that
ar(EFGH) = {1 \over 2}ar(ABCD)

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Sal. Construction: Join E and G.
Proof: In parallelogram ABCD,
AB = CD mid AB\parallel CD .


⇒     {1 \over 2}AB = {1 \over 2}CD

⇒   DG = AE                                                     [ E and G are mid-points of AB and CD]

We have DG = AE and AG\parallel AE.

∴  AEGD is a parallelogram.

Now parallelogram AEGD and triangle HEG are on the same base and between the same parallels.

∴     ar(HEG) = {1 \over 2}ar(AEGD)

Similarly; we can show that

ar(FGE) = {1 \over 2}ar(EBCG)

From (i) and (ii) , we get

ar(HEG) + ar(FGH) = {1 \over 2}[ar(AEGD) + ar(EBCG)]

 \Rightarrow ar(EFGH) = {1 \over 2}ar(ABCD)


3. 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).


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Sol.
Triangle PAB and parallelogram ABCD are on the same base AB and between the same parallels AB and DC.

∴   ar(APB) = {1 \over 2}ar(ABCD)

Similarly, triangle QBC and parallelogram ABCD are on the same base BC and between the same parallels AD and BC.

∴ ar(QBC) = {1 \over 2}ar(ABCD)                                                    …..(ii)

Hence, ar(APB) = ar(QBC)                                                         [ From (i) and (ii)]


4. In Figure, P is a point in the interior of a parallelogram ABCD. Show that


(i) ar(APB) + ar(PCD)= {1 \over 2} ar(ABCD) D

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

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

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Sol. (i) Construction: Through P draw LM\parallel AB, meeting AD at  L and BC at M.
Proof: AD\parallel BC


[Opposite side of a parallelogram]
⇒    AL\parallel BM                                                               ……(i)

Also AB\parallel LM                                                                …(ii) [Construction]

⇒     ABML is a parallelogram.                                                          [From (i), (ii)]

Now parallelogram ABML and triangle ABP are on the same base AB and between the same parallels AB and LM.

∴             ar(APB){1 \over 2}ar(ABPL)                                      .,…(i)

Simlarly we can show that

ar(PDC) = {1 \over 2}ar(LMCD)                                         ……(ii)

Adding (iii) and (iv), we get

ar(APB) + ar(PDC) = {1 \over 2}ar(ABML) + {1 \over 2}ar(LMCD)

 = {1 \over 2}[ar(ABML) + ar(LMCD)]

 \Rightarrow ar(APB) + ar(PDC) = {1 \over 2}ar(ABCD)

(ii) Similarly, by drawing XY\parallel AD, through P, we can show that
ar(APD) + ar(BPC) ={1 \over 2} ar(ABCD)

⇒   ar(APD) + ar(BPC) ar(APB) + ar(PDC).
[From result of part(i) ]


5. In PQRS and ABRS are paradllelgroms and X is any point on side BR. Show that


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

(ii)    ∴   ar(AXS) = {1 \over 2}ar(ABRS)

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Sol.
(i) As parallelograms PQRS and ABRS are on the same base SR and between the same parallels SR and PB.
Hence, ar(PQRS) = ar(ABRS).

(ii) Triangle XAS and parallelogram ABRS are on the same base AS and between the same parallels AS and BB.

∴    ar(AXS)=  {1 \over 2} ar(ABRS)

But ar(ABRS)= ar(PQRS)                                                             [From result of part (i)]

∴        ar(AXS) ={1 \over 2} ar(PQRS).


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 fainter wants to sow wheat and pulses in equal portions of the field separately. How should she do it?


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Sol. The field is divided into three parts as: triangle PSA; triangle PAQ; triangle ARQ.
As triangle PAQ and parallelogram PQRS are on the same base PQ and between the same parallels PQ and RS.
∴          ar(PAQ) = {1 \over 2}ar(PQRS)                              …..(i)
From above result, we can say that
ar(PSA) + ar(QAR) = {1 \over 2}ar(PQRS)                         ……(ii)
From (1) and (ii), farmer has two options:
Option I: Wheat in ar(PAQ) and pulses ar(QAR),
Option II: Pulses in ar(PAQ) and wheat ar(QAR).


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