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).

ar(parallelogram ABCD) = AB x AE
= 16 x 8 cm²
= 128 cm²             ...(1)
ar(parallelogram ABCD) = AD x CF
= AD x 10 cm²   .......(2)
From (1) and (2), we get
AD x 10 = 128
AD = \(\frac { 128 }{ 10 } \)
AD = 12.8 cm.

(i) \(\Delta \)PDC and quadrilateral ABCD lie on the same base DC and between the same parallels DC and AB.
(ii) \(\Delta \) TRQ and parallelogram SRQP lie on the same base RQ and between the same parallels RQ and SP.
(iii) Quadrilaterals APCD and ABQD lie on the same base AD and between the same parallels AD and BQ.

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)

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: ABCD is a parallelogram whose diagonals intersect at O. P is any point on BO.
To prove:
(i) ar(\(\Delta \)ADO) = ar(\(\Delta \) CDO)
(ii) ar(\(\Delta \)ABP) = ar(\(\Delta \) CBP)

Proof:
i)  Diagonals of a parallelogram bisect each other
AO = OC
O is the midpoint of AC
DO is a median of  \(\Delta \)DAC
ar(\(\Delta \) ADO) = ar(\(\Delta \) CDO)
|A median of a triangle divides it into two triangles of equal areas
(ii) BO is a median of  \(\Delta \)BAC
ar(\(\Delta \) BOA) = ar(\(\Delta \) BOC)     ..........(1)
|A median of a triangle divides it into two triangles of equal areas
PO is a median of  \(\Delta \)PAC
ar(\(\Delta \) POA) = ar(\(\Delta \) POC)      .........(2)
|A median of a triangle divides it into two triangles of equal areas
Subtracting (2) from (1), we get
ar(\(\Delta \) BOA) - ar(\(\Delta \) POA) = ar(\(\Delta \) BOC) - ar(\(\Delta \) POC)
ar(\(\Delta \) ABP) = ar(\(\Delta \) CBP)

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
 \(\therefore \ ar(\Delta ABF)=ar(\Delta BEC)\) 

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: BE = 14 cm, AD = 8 cm
\(\therefore \ Area(\triangle ADB)=\frac { 1 }{ 2 } \times 8\times 14\)
= 56 cm²
\(\because\) ABCD is a parallelogram.
\(\therefore\) ar(\(\Delta \)DBC) = ar(\(\Delta \)ADB) = 56 cm².

\(\Delta \)PSR and \(\Delta \)PSQ are on the same base PS and between the same parallels PS and QR.
ar(\(\Delta \)PSR) = ar(\(\Delta \)PSQ)
ar(\(\Delta \)PSR) - ar(\(\Delta \)PSO) = ar(\(\Delta \)PSQ) - ar(\(\Delta \)PSO)
ar(\(\Delta \)ROS) = ar(\(\Delta \)POQ).

10 cm²

40 cm²

6.4cm

A rectangle is essentially a parallelogram. A square is essentially a parallelogram. Parallelograms on the same base and between the same parallels are equal in area

Parallelograms on the same base and between the same parallels are equal in area

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