In \(\Delta\)GHK; F and E are the mid-points of HG and GK respectively.
\(\therefore\) By mid-point theorem,
\(FE=\frac { 1 }{ 2 } KH\quad and\quad FE||KH\)  ...........(i)
In quadrilateral EFHK,
EF || HK        (by (i))
\(\therefore\) EFHK is a trapezium.
Also,
ar(EFHK) = ar(FHD) + ar(DEF) + ar(DEK)   .............(ii)
We have,    FE || HD and FE = HD
\(\therefore\) FEDH is a ||^gm,
So, ar(FHD) = ar(DEF)   ..........(iii)
Similarly, DFGE is a ||^gm, 
\(\therefore\) ar(DEF) = ar(GEF)   ...........(iv)
Also, DFEK is a ||^gm,
\(\therefore\) ar(DEF) = ar(DEK)    ...........(v)
Using (iii), (iv) & (v) we get,
ar(GEF) = ar(FHD)
= ar(DEK) = ar(DEF)
\(=\frac { 1 }{ 4 } \) ar(GHK)                ...........(vi)
Using (vi) and (ii), we get
ar(EFHK) = \(\frac { 3 }{ 4 } ar(GHK)\)
Hence proved

Given: AB CD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y.

To Prove: ar(\(\Delta \)ADX) = ar(\(\Delta \)ACY).
Construction: Join CX.
Proof: \(\Delta \)ADX and \(\Delta \)ACX are on the same base AX and between the same parallels AB and DC.
ar(\(\Delta \)ADX) = ar(\(\Delta \)ACX)        ...(1)
|Two triangles on the same base (or equal bases) and between the same parallels are equal in area
\(\Delta \)ACX and \(\Delta \)ACY are on the same base AC and between the same parallels AC and XY.
ar(\(\Delta \)ACX) = ar(\(\Delta \)ACY)        ...(2)
|Two triangles on the same base (or equal bases) and between the same parallels are equal in area
From (1) and (2), we get
ar(\(\Delta \)ADX) = ar(\(\Delta \)ACY).

Given: P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD.
  
To Prove: ar( \(\Delta \)APB) = ar(\(\Delta \) BQC). 
Proof:  \(\Delta \)APB and || gm ABCD are on the same base AB between the same parallels AB and DC.
ar(\(\Delta \) APB) = \(\frac { 1 }{ 2 } ar(||\ gm\ ABCD)\)   .......(i)
(\(\Delta \) BQC) and || gm ABCD are on the same parallels BC and AD.
ar(\(\Delta \) BQC) = \(\frac { 1 }{ 2 } ar(||\ gm\ ABCD)\)   .......(ii)
from (1) and (2),
ar(\(\Delta \) APB) = ar(\(\Delta \) BQC)

(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 triangles PEA and QFD, we have
\(\angle \)APE = \(\angle \)DQF
(Corresponding angles)
AE = DF
(Oppoosite sides of 8 ||^gm AEFD)
\(\angle \)AEP=\(\angle \)DFQ
(Corresponding angles)
\(\therefore\) \(\Delta\)PEA\(\cong \)\(\Delta\)QFD     (by AAS)
As congruent triangles have equal area.
\(\therefore\) ar(\(\Delta\)PEA)=ar(\(\Delta\)QFD)

Construction: Join BX.

In \(\Delta\)XDB and \(\Delta\)ADC,
1. XD = CD (given)
2. \(\angle XDB=\angle ADC\) (vertically opposite angles)
3. BD = DA(CD is median of triangle ABC)
So, \(\Delta\)s are congruent by SAS rule
\(\therefore\) XB = AC (By c.p.c.t.)
Now, add XDA to both sides,
\(\Delta\)(XDB) + \(\Delta\)(XDA) = \(\Delta\)(ADC) + \(\Delta\)(XDA)\(\Delta\)(XBA) = \(\Delta\)(XAC)
(since the two triangles are on the same bae and have equal areas, they lic between same parallels XA and BC)
Similarly, we can prove \(\Delta\)s BAY and CAY and then AY||BC.
Hence, XABC and BCAY are parallelograms, where XA=BC and AY = BC. thus, XA = AY (Proved)
Now, since XA = AY and points XAY lie on the equal lines, they are collinear.
Hence Proved.

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) = \(\frac { 1 }{ 2 } ar(\triangle ABC)\)

DB is the transversal because DC||AB

because \(\angle CDB=\angle ABD=90°\)(form a pair of alternate angles)
DC || AB and DC = AB
\(\therefore\) ABCD is a ||^gm
ar(ABCD) = b\(\times\)h
= 2.5 \(\times\)4 = 10 cm².

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 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
\(\Rightarrow \ ar(ACED)=ar(BDE)\)
\(\\ \Rightarrow \ ar(BDE)\ =ar(ACED)\)

40 cm²

6.4cm

If a parallelogram and a triangle are on the same base and between the same parallels, then area of the triangle is half the area of the parallelogram.

Diagonal=\(\sqrt { 2 }\) side;Area=(side)²

Area of parallelelogram=\(Base\times Corresponding\quad altitude\)

Theorem