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Areas of Parallelogram and Triangles : Exercise 9.2 (Mathematics NCERT Class 9th)


Q.1     In figure ABCD is a parallelogram, AE  \bot DC and CF  \bot AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.
7Sol.

We have,
Area of a || gm = Base × Height

Therefore, Area of || gm ABCD = AB × AE
 = \left( {16 \times 8} \right)c{m^2} = 128{\mkern 1mu} c{m^2}            ...(1) 
Also area of || gm ABCD = AD × CF
 = \left( {AD \times 10} \right)c{m^2}                                                     ... (2)
From (1) and (2) we get

128 = AD × 10
 \Rightarrow AD = {{128} \over {10}}cm = 12.8\,cm


Q.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\left( {ABCD} \right)
Sol.        \Delta HGF and || gm HDCF stand on the same base HF and lie between the same parallels HF and DC. 

8Therefore,   ar(\Delta HGF) = {1\over 2}ar(HDCF)          ... (1)
Similarly, \Delta HEF and ||gm ABFH stand on the same base HF and lie between the same parallels
HF and AB.
Therefore ar(\Delta HEF) = {1\over 2}ar(ABFH)          ... (2)
Therefore Adding (1) and (2), we get
ar(\Delta HGF) + ar(\Delta HEF) = {1 \over 2}ar( HDCF ) + ar(ABFH)
 \Rightarrow ar\left( {EFGH} \right) = {1 \over 2}ar\left( {ABCD} \right)


Q.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).
Sol.        \Delta APB and || gm ABCD stand on the same base AB and lie between the same parallels AB and DC.

9Therefore ar\left( {\Delta APB} \right) = {1 \over 2}ar\left( {ABCD} \right).... (1)
Similarly , \Delta BQC and || gm ABCD stand on the same base BC and lie between the same parallels BC and AD.
Therefore ar\left( {\Delta BQC} \right) = {1 \over 2}ar\left( {ABCD} \right) .... (2)
From (1) and (2) , we have
ar(\Delta APB) = ar (\Delta BQC)


Q.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\left( {ABCD} \right)
             (ii) ar (APD) + ar(PBC) = ar (APB) + ar (PCD).

10
Sol.       Draw EPF parallel to AB or DC and GPH parallel to AD or BC.

11Now AGHD is a|| gm
[Since GH || DA and AG|| DH]
Similarly, HCBG, EFCD and ABFE are parallelograms.
(i) \Delta APB and || gm ABFE stand on the same base AB and lie between the same parallels AB and EF.

Therefore  ar\left( {APB} \right) = {1 \over 2}ar\left( {ABFE} \right) ... (1)
Similarly,  ar\left( {PCD} \right) = {1 \over 2}ar\left( {EFCD} \right) ... (2)
Adding (1) and (2) , we get
ar\left( {APB} \right) + ar\left( {PCD} \right) = {1 \over 2}\left[ {ar\left( {ABFE} \right) + ar\left( {EFCD} \right)} \right]
ar\left( {APB} \right) + ar\left( {PCD} \right) = {1 \over 2}ar\left( {ABCD} \right)\,\,\,\,\,\,\,\,\,\,....\left( 3 \right)

(ii) \Delta APD and || gm AGHD are on the same base AD and lie between the same parallels AD and HG.
Therefore ar\left( {APD} \right) = {1 \over 2}ar\left( {AGHD} \right)\,\,\,\,\,\,\,\,\,\,\,....\left( 4 \right)
Similarly, ar\left( {PCB} \right) = {1 \over 2}ar\left( {GBCH} \right)\,\,\,\,\,\,\,\,\,\,\,....\left( 5 \right)
Adding (4) and (5), we get
ar\left( {APD} \right) + ar\left( {PCB} \right) = {1 \over 2}\left[ {ar\left( {AGHD} \right) + ar\left( {GBCH} \right)} \right]
ar\left( {APD} \right) + ar\left( {PCB} \right) = {1 \over 2}ar\left( {ABCD} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 6 \right)
From (3) and (6) we get
ar (APD) + ar (PBC) = ar (APB) + ar(PCD).


Q.5      In figure PQRS and ABRS are parallelograms and X is any point on side BR. Show that 
               (i) ar (PQRS) = ar (ABRS)
               (ii) ar (AXS)  = {1 \over 2} ar (PQRS).

12Sol.        

(i) || gm PQRS and || gm ABRS stand on the same base RS and lie between the same parallels SR and PAQB.
Therefore ar(PQRS) = ar (ABRS) ... (1)

(ii) \Delta AXS and || gm ABRS stand on the same base AS and lie between the same parallels AS and RB.
Therefore ar (AXS)  = {1 \over 2} ar (ABRS)
 \Rightarrow ar (AXS)  = {1 \over 2} ar (PQRS)                [Using (1)]


Q.6      A farmer was having a field in the form of a parallelogram PQRS. He 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 farmer wants to sow wheat and pulses in equal portions of the field separately. How should he do it?
Sol.       Clearly, the field i.e., || gm PQRS is divided into 3 parts. Each part is of the shape of triangle.

13Since \Delta APQ and || gm PQRS stand on the same base PQ and lie between the same parallels PQ and SR.
Therefore ar (APQ)  = {1 \over 2} ar (PQRS) ... (1)
Clearly, ar (APS) + ar (AQR) = ar (PQRS) – ar (APQ)
= ar (PQRS)  - {1 \over 2}ar\left( {PQRS} \right)                 [Using (1)]
 = {1 \over 2}ar\,\left( {PQRS} \right)\,\,\,\,\,\,\,\,\,\,....\left( 2 \right)
From (1) and (2), we get
ar (APS) + ar (AQR) = ar (APQ)
Thus , the farmer should sow wheat and pulses either as [(\Delta s APS and AQR) or \Delta APQ] or as
[\Delta APQ or (\Delta s APS and AQR )]



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