(i) Area of triangular plot \(=\frac { 1 }{ 2 } \times b\times h\)
\(=\frac { 1 }{ 2 } \times 120\times 90\)
= 5400 m²

(ii) In \(\Delta\)ABC they draw median AD on base BC and divide it into two equal areas ABD and ACD. Take any point E on AD and join BE and CE.
The brothers get areas
ar(\(\Delta\)ABE) and ar(\(\Delta\)ACE)
ar(\(\Delta\)BCE) is donated to school.
(iii) Any positive value is accepable. Both brothers know the impotance of education, love their community.

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)
\(\therefore \ ar(POS)+ar(QOR)=\frac { 1 }{ 2 } [ar(PABS)+ar(QABR)]\)
\(=\frac { 1 }{ 2 } ar(PQRS)\)

Given: BD is one of the diagonals of a quadrilateral ABCD. AM and CN are perpendiculars from A and C respectively on BD.
To Prove: ar(quad. ABCD)
 \(\frac { 1 }{ 2 } BD.(AM+CN)\)
Proof: ar(quad. ABCD)
 
\(ar(\Delta ABD)+ar(\Delta BCD)\)
\(\\ =\frac { BD.AM }{ 2 } +\frac { BD.CN }{ 2 } \)
Area of a triangle = \(\frac { 1 }{ 2 } \times Base\times Corresponding\ altitude\)
 \(\frac { 1 }{ 2 } BD(AM+CN)\)

Given: Two points A and B and a positive real number k.
To find: The locus of a point P such that ar(\(\Delta \)  PAB) = k.
Construction: Draw \(PM\bot AB\)
Determination: Let PM = h
ar(\(\Delta \)PAB) = k        |Given
\(\Rightarrow \frac { 1 }{ 2 } (AB)(PM)=k\)
\(\\ \Rightarrow \frac { 1 }{ 2 } (AB)(h)=k\)
\(\\ h=\frac { 2k }{ AB } \)

Points A and B are given. AB is fixed.
Also, k being a positive real number k is fixed.
h is a fixed positive real number.
The locus of P is a line parallel to the line
AB at a fixed distance of \(h=\frac { 2k }{ AB } \)  on either side of it.

Let ABCD be a rhombus whose diagonals are AC and BD. I
Then,
Area of rhombus ABCD
= Area of \(\Delta \)ABO + Area of \(\Delta \)CBD
\(=\frac { (BD)(AO) }{ 2 } +\frac { (BD)(OC) }{ 2 } \)
|Diagonals of a rhombus are perpendicular to each other
 
\(\frac { (BD) }{ 2 } (AO+OC)=\frac { (BD)(AC) }{ 2 } \)
\(\frac { 1 }{ 2 } \) Product of the lengths of its diagonals.

Given: ABCD is a parallelogram in which DE=EC
To Prove: area (ABF) = area (BEC)
Proof: AB = DE      .....(1)
I Opposite sides of a parallelogram
DE = EC     I Given     ...(2)
AB = EC     I From (1) and (2)
Also,  AB || DE
I Opposite sides of a parallelogram
\(\Rightarrow \)  AB || DC
Now, \(\Delta ABF\) and \(\Delta BEC\)are on equal bases AB and EC and between the same parallels AB and DC
 \(\therefore \ ar(\Delta ABF)=ar(\Delta BEC)\) 

Let ABC and A'BC be two triangles on the same base BC and between the same parallels PQ and RS.
 
Draw \(BM\bot PQ\),then,
\(ar(\Delta ABC)\)
 \( =\frac { 1 }{ 2 } \times Base\times Corresponding\ altitude\)
\(\\ =\frac { 1 }{ 2 } \times BC\times BM\quad \quad \quad \quad .....(i)\)
\(\\ ar(\Delta A'BC)\)
\(\\ =\frac { 1 }{ 2 } \times Base\times Corresponding\ altitude\)
\(\\ =\frac { 1 }{ 2 } \times BC\times BM\quad\quad \quad \quad \quad .....(ii)\)
Form (i) and (ii),we get
 \(ar(\Delta ABC)=ar(\Delta ABC)\)
Hence, \(\Delta ABC\) and \(\Delta A'BC\)  are equal in area.

Given: AB || DC in the given figure.
To Prove: ar(BDE) = ar(ACED)
Proof:\(\Delta \)  ADC and \(\Delta \) BDC are on the same base DC and between the same parallels AB and DC
ar(\(\Delta \) ADC) = ar(\(\Delta \) BDC)
\(\Rightarrow ar(\Delta \ ADC)+ar(\Delta \ DCE)\)
\( =ar(\Delta \ BDC)+ar(\Delta \ DCE)\)
|Adding ar(\(\Delta \quad DCE\)) to both sides
\(\Rightarrow \ ar(ACED)=ar(BDE)\)
\(\\ \Rightarrow \ ar(BDE)\ =ar(ACED)\)

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,
 \(ar(||gm\ ABCD)=base\times Corresponding\ altitude\)
\(\\ =3\times 4\)
\(\\ =12{ cm }^{ 2 }\)