(i) Let the plot be ABCD.
Construction: Join AC. Draw BE||AC.
Proof: ar(\(\Delta\)ADE) = ar(Quad.ABCD)
Health centre can be constructed in triangular plot (\(\Delta\)AOB) and the farmer can have the triangular plot \(\Delta\)ADE.
Proof:
ar(\(\Delta\)ADE) = ar(\(\Delta\)ADC) + ar(\(\Delta\)ACE)
= ar(\(\Delta\)ADC) + ar(\(\Delta\)ABC)
= ar(Quad.ABCD)
(ii) Helpful, wise, and co-operating.
(iii) Yes, constructing a Health Centre is justified and essential also, because Health is Wealth.

In figure, AD is the median of \(\Delta\)ABC.

\(\therefore \ ar(\triangle ABD)=\frac { 1 }{ 2 } ar(\triangle ABC)\)      ...........(i)
Now, BE is the median of \(\Delta\)ABD
\(\therefore \ ar(\triangle BED)=\frac { 1 }{ 2 } ar(\Delta ABD)\)       ..............(ii)
From (i) and (ii), we get
ar(\(\Delta\)BED) = \(\frac { 1 }{ 2 } ar(\triangle ABC)\)

Through O, draw AB || PS
Also  PA || BS
PABS is a parallelogram.
ar(POS) =\(\frac { 1 }{ 2 } ar(PABS)\) 

(Triangle and a parallelogram are on the same base and between the same parallels)
Similarly, ar(QOR )= \(\frac { 1 }{ 2 } \)ar(QABR)
\(\therefore \ ar(POS)+ar(QOR)=\frac { 1 }{ 2 } [ar(PABS)+ar(QABR)]\)
\(=\frac { 1 }{ 2 } ar(PQRS)\)

Construction: Join HF
H and F are the mid-points of AD and BC, respectively
\(\Rightarrow\) HD = FC    [AD||BC]
\(\therefore\) HD || FC
\(\Rightarrow\) HDFC is a rectangle.

Now, \(\Delta\) HFG and rectangle HFCD are on the same base HF and lie between the same parallels HF and DC.
\(\therefore \ ar(\Delta HFG)=\frac { 1 }{ 2 } ar(\Box HFCD)\quad \quad \quad ......(i)\)
Similarly, ar(\(\Delta EHF)=\frac { 1 }{ 2 } ar(\Box ABFH)\)        ...........(ii)
Adding (i) and (ii), we get
ar(\(\Delta\)HFG) + ar(\(\Delta\)EHF) = \(\frac { 1 }{ 2 } ar(\Box HFCD)+ar(\Box ABFH)\)
\(\Rightarrow \ ar(\Box EFGH)=\frac { 1 }{ 2 } ar(\Box ABCD)\)
\(16=\frac { 1 }{ 2 } ar(\Box ABCD)\)
\(\therefore \ ar(\Box ABCD)=32\ { cm }^{ 2 }\)

Given: A rectangle ABCD.

To prove: AC= BD
Consider, \(\Delta\)DAC and \(\Delta\)CBD,
\(\angle \)D =\(\angle \)C
AD = BC
(opp. sides are equal)
DC=DC (Common)
\(\therefore\) \(\Delta\)DAC\(\cong \)\(\Delta\)CBD (By SAS)
\(\Rightarrow\) AC = BD (c.p.c.t)
Hence proved.

Given:\(\Delta \) ABC and \(\Delta \) ABD are two triangles on the same base AB. Line segment CD is bisected by AB at O.
To Prove: ar(ABC) = ar(ABD)
Proof:  Line segment CD is bisected by AB at O
CO = DO
\(\Rightarrow \) O is the midpoint of CD.
\(\Rightarrow \) AO is a median of \(\Delta \)ACD and BO is a median of \(\Delta \)BCD
AO is a median of \(\Delta \)ACD
ar(\(\Delta \) AOC) = ar(\(\Delta \) AOD) ... (1)
| A median of a triangle divides it into two triangles of equal areas
BO is a median of \(\Delta \) BCD
ar(\(\Delta \) BOC) = ar(\(\Delta \) BOD) ... (2)
| A median of a triangle divides it into two triangles of equal areas
Adding (1) and (2), we get,
\(ar(\Delta AOC)+ar(\Delta BOC)\) 
\(\\ =ar(\Delta AOD)+ar(\Delta BOD)\)
\(\\ \Rightarrow ar(\Delta ABC)=ar(\Delta ABD)\)

Given: The diagonals of a parallelogram ABCD intersect at a point O. Through O, a line is drawn to intersect AD at P and BC at Q.
 
\(ar(\Box PDCQ)=ar(\Box PQBA)\)
Proof: AC is diagonal pf || gm ABCD
\(ar(\Delta ABC)=ar(\Delta ACD)\)     ........(i)
\(\\ =\frac { 1 }{ 2 } ar(||gm\ ABCD)\)
In \(\Delta AOP\) and \(\Delta COQ\)
AO + CO
|Diagonals of a parallelogram bisect each other
\(\angle AOP=\angle COQ\) |Vertically opposite angles
\(\angle OAP=\angle OCQ\) |Alternate interior angles
\(\Delta AOP=\Delta COQ\) |By ASA Congruence Rule
\(ar(\Delta AOP)=ar(\Delta COQ)\) |Congruent figures have equal areas
\(\Rightarrow \ ar(\Delta AOP)+ar(\Box OPDC)\) 
\(\\ ar(\Delta ACD)=ar(\Box PDCQ)\)
\(\\ \Rightarrow \frac { 1 }{ 2 } ar(||\ gm\ ABCD)=ar(\Box PDCQ)\quad \quad \ from(i)\)
\(\\ \Rightarrow ar(\Box PQBA)=ar(\Box PDCQ)\)
\(\\ \Rightarrow ar(\Box PDCQ)=r(\Box PQBA)\)           

 Given: ABCD is a trapezium with AB || CD. AM is its height.

To Prove: ar(trapezium ABCD)
\(=\frac { 1 }{ 2 } (AB+DC)(AM)\) 
Construction: Join AC
Proof:ar(trapezium ABCD)
\(ar(\Delta \ ADC)+ar(\Delta \ ABC)\)
\(\\ =\frac { 1 }{ 2 } \times DC\times AM+\frac { 1 }{ 2 } \times AB\times AM\)
\(\\ =\frac { 1 }{ 2 } \times (DC+AB)\times AM\)
\(\\ =\frac { 1 }{ 2 } (AB+Dc)(AM)\)

Given: Diagonals AC and BD of a trapezium ABCD with AB || CD intersect each other at O.
To Prove: ar(\(\Delta \)AOD) = ar(\(\Delta \) BOC)
Proof:  \(\Delta \)ADB and \(\Delta \)ACB are on the same base AB and between the same parallels AB and DC
ar(\(\Delta \)ADB) = ar(\(\Delta \) ACB)
 |Two triangles on the same base and between the same parallels are equal in area
\(\Rightarrow \)  ar(\(\Delta \) ADB) - ar(\(\Delta \) AOB)
\(\Rightarrow \)  ar(\(\Delta \) ACB) - ar(\(\Delta \)AOB)
I Subtracting ar( \(\Delta \)AOB) from both sides
\(\Rightarrow \)  ar(\(\Delta \) AOD) = ar(\(\Delta \)BOC)

Given: ABCD is a quadrilateral and BD is one of its diagonals.
To Prove: ABCD is a parallelogram and to determine its area.
Proof:\(\angle ABD=\angle BDC(=90°)\)        |Given
But these angles form a pair of equal alternate.interior angles for lines AB, DC and a transversal BD
AB || DC
Also,  AD = DC (= 3 cm) I Given
Hence, quadrilateral ABCD is a parallelogram.
I A quadrilateral is a parallelogram if its one pair of opposite sides are parallel and equal
Now,
 \(ar(||gm\ ABCD)=base\times Corresponding\ altitude\)
\(\\ =3\times 4\)
\(\\ =12{ cm }^{ 2 }\)