Areas Of Parallelograms And Triangles - Class 9 Maths - Extra Questions

$$\dfrac{ar(\triangle ADE)}{ar(\triangle ABC)}= \dfrac{(DE)^2}{(BC)^2}$$

$$\implies \dfrac{20}{ar(\triangle ABC)}= \dfrac{(5)^2}{(10)^2}$$  [given]

$$\implies \dfrac{20}{ar(\triangle ABC)}= \dfrac{25}{100}$$  [given]

$$\implies ar(\triangle ABC)= \dfrac{20 \times 100}{25}$$

$$\implies ar(\triangle ABC)= 20 \times 4$$

$$\implies (\triangle ABC)= 80 \ cm^2$$

Area of parallelogram $$=$$ Base $$\times$$ Height

$$\triangle BAE$$ and $$\triangle BDE$$ are on the same base BE and between the same parallel BE and AD.
$$\therefore$$ Area of $$\triangle BAQ$$=Area of $$\triangle BDE$$ [two triangle on the same base and between the same parallel are equal in areas]  (I)
$$\triangle BCE$$ and $$\triangle BEF$$ are on the same base BE and between the same parallel BE and CF.
$$\therefore$$ Area of $$\triangle BCE$$=Area of $$\triangle BEF$$ [two triangle on the same base and between the same parallel are equal in areas] (II)
Adding the corresponding sides (i) and (ii) we get,
Area of $$\triangle BAQ$$+Area of $$\triangle BCE$$=Area of $$\triangle BDE$$+Area of $$\triangle BEF$$
=>$$\triangle AEC$$=$$\triangle DBF$$

Consider $$\triangle DGA$$ and $$\triangle AED$$

We know that both the triangles have the same base $$AD$$ and lie between the parallel lines $$AD$$ and $$EG$$.

So we get

Area of $$\triangle DGA$$$$=$$ Area of $$\triangle AED$$.....(1)

Consider $$\triangle DBC$$ and $$\triangle BFD$$

We know that both the triangles have the same base $$DB$$ and lie between the parallel lines $$BD$$ and $$CF$$.

So we get 

Area of $$\triangle DBF$$ $$=$$ Area of $$\triangle BCD$$......(2)

By adding both the questions

Area of $$\triangle DGA$$ $$+$$ Area of $$\triangle DBF$$$$=$$ Area of $$\triangle AED$$ $$+$$Area of $$\triangle BCD$$

By adding $$\triangle ABD$$ both sides

Area of $$\triangle DGA$$$$+$$ Area of $$\triangle DBF$$ $$+$$ Area of $$\triangle ABD$$$$=$$Area of $$\triangle AED$$$$+$$Area of $$\triangle BCD$$

By adding $$\triangle ABD$$ both sides

Area of $$\triangle DGA$$ $$+$$ Area of $$\triangle DBF$$ $$+$$ Area of $$\triangle ABD$$ $$=$$ Area of $$\triangle AED$$$$+$$Area of $$\triangle BCD$$$$+$$ Area of $$\triangle ABD$$

So we get

Area of $$\triangle DGF$$ $$=$$ Area of pentagon $$ABCDE$$

Therefore, it is proved that $$ar( \ pentagon\ ABCDE)$$=$$ar ( \triangle DGF)$$.

Let us draw $$DN\bot AC$$ and $$BM\bot AC$$ as shown in above figure.

$$\triangle APB$$ and ||gm $$ABCD$$ are on the same base $$AB$$ and between same parallel $$AB$$ and $$DC$$.
Using the property,

FD is diagonal of the parallelogram BDEF
$$\therefore \quad Area\left( \triangle FBD \right) =area\left( \triangle DEF \right) $$
$$Similarly\quad Area\left( \triangle DEF \right) =Area\left( \triangle DCE \right) $$
$$Therefore\quad Area\left( \triangle FBD \right) =Area\left( \triangle DEF \right) =Area\left( \triangle FAE \right) =Area\left( \triangle DCE \right) $$
$$\therefore \triangle ABC\quad is\quad divided\quad into\quad 4\quad non-overlapping\quad triangles\quad \triangle FBD,\triangle DEF,\triangle FAE\quad and\quad \triangle DCE$$
$$Therefore\quad Area\left( \triangle ABC \right) =Area\left( \triangle FBD \right) +Area\left( \triangle DEF \right) +Area\left( \triangle FAE \right) +Area\left( \triangle DCE \right) =4\quad Area\left( \triangle DEF \right) $$
$$Area\left( \triangle DEF \right) =\frac { 1 }{ 4 } area\left( \triangle ABC \right) $$
So, as per given question
$$Area\quad of\quad \left( BDEF \right) =Area\left( \triangle FBD \right) +Area\left( \triangle DEF \right) $$
                              $$=Area\left( \triangle DEF \right) +Area\left( \triangle DEF \right) $$
                              $$=2Area\left( \triangle DEF \right) $$
                              $$=2\times \frac { 1 }{ 4 } area\left( \triangle ABC \right) $$
                              $$=\frac { 1 }{ 2 } area\left( \triangle ABC \right) $$

(i) Trapezium $$ABCD$$ and $$\triangle PDC$$ lie on the same $$DC$$ and between the same parallel lines $$AB$$ and $$DC$$.

In the following figure, DE is parallel to BC.

Given: Parallelogram $$ABCD$$ and rectangle $$ABEF$$ are on same base and have equal areas.          
$$AB=CD$$        ...opposite side of parallelogram
$$AB=EF$$         ...opposite side of rectangle

and, $$\angle F = \angle E = 90^\circ$$
$$AD> AF$$ and $$BC> BE$$     ....(Hypotenuse of right angle is greater than other sides )

Now,
Perimeter of $$\Box ABCD = AB+BC+CD+DA$$
Perimeter of $$\Box ABEF = AB+BE+EF+FA$$ 

$$A(\triangle DBC) = A(\triangle EBC)$$     ....Given

Let $$ABCD$$ be the quadrilateral plot. 

Produce $$BA$$ to meet $$CD$$ drawn parallel to $$CA$$ at $$E$$. 

Join $$EC$$. 

Triangles $$EAC$$ and $$ADC$$ lie on the same line and between same  parallels $$DE$$ and $$CA$$.

Using "Two triangles on the same base and between the same parallels are equal in area."  

$$ar(EAC) = ar (ADC) $$

$$ar(ABCD) = ar(ABC) + ar(ACD) $$

$$= ar(ABC) + ar(ADC) $$

$$= ar (ABC) + ar(EAC) = ar(EBC) $$
i.e., 

Quadrilateral $$ABCD= \triangle EBC$$ gives the explanation .

(i) Since triangles $$ACB$$ and $$ACF$$ are on the same base $$AC$$ and between the same parallels $$AC$$ and $$BF$$.
Therefore $$ar(ACB) = ar (ACF)$$

(ii) Adding $$ar(ACDE)$$ on both sides, we get
$$ar(ACF) + ar(ACDE) = ar(ACB) + ar(ACDE)$$
$$\Rightarrow ar(AEDF) = ar(ABCDE)$$

Given: $$AP\parallel BQ$$. 

Given: $$ABCD$$ is a quadrilateral, $$AC$$ is diogonal And $$DE \ || \ AC$$ and also $$DE$$ meats $$BC$$ produced at $$E.$$

Given : $$ABE$$ and $$BDE$$ are two equilateral triangles.

Considering $$a$$ as the side of $$\triangle ABC.$$

Then, ar$$(ABC) =\dfrac{\sqrt{3}}{4}a^2$$

(i)  Since $$D$$ is the midpoint of $$BC.$$  

So, $$BD= \dfrac{a}{2}$$

ar$$(BDE) =  \dfrac{\sqrt{3}}{4}\left(\dfrac{a}{2}\right)^2=\dfrac{\sqrt{3}}{16}a^2$$ 

$$\Rightarrow   ar(BDE)=\dfrac{1}{4}ar(ABC) $$

(ii) Since $$DE$$ is a median of $$\triangle BEC$$ and each median divides a triangle in two other  $$\triangle$$s of equal area, 

So, ar$$ (BDE)  =\dfrac{1}{2}ar(BEC)$$ ... (1) 

$$∠EBC=∠BCA=60^o$$ 

This suggests that they are alterante angles between the lines $$BE$$ and $$AC$$ with $$BC$$ as the transversal.

So, $$BE\parallel AC.$$ 

Since $$\triangle BEC$$ and $$BAE$$ stand on the same base $$BE$$ and lie between the same parallel lines $$BE$$ and $$AC.$$ 

So, $$ar(BEC) = ar(BAE)  $$

$$ ar (BDE)  =\dfrac{1}{2}ar(BEC)$$ (from (1))

ar $$( Δ  BDE) =  \dfrac{1}{2}  ar (BAE).$$ 

(iii) $$ar(BDE) =  \dfrac{1}{2}  ar (BEC) $$ 

($$ED$$ is a median of $$ \triangle BEC$$ and median divides the triangle in two other $$\triangle s$$ of equal area. )

$$ar (BDE) =  \dfrac{1}{4}  ar (ABC)$$  (Using result from (i))

So, $$\dfrac{1}{4}  ar (ABC) =   \dfrac{1}{2}  ar (BEC) $$

$$\Rightarrow ar (ABC) = 2 ar (BEC)$$

(iv) $$∠ABD=∠BDE=60^∘ $$ (Angles of equilateral triangle)

These are alternate angles between the lines $$AB$$ and $$ED$$ with $$BD$$ as transversal. 

So, $$BA$$ || $$ ED.$$ 

So, $$ar (BDE) = ar (AED) $$ ( Δs  on the same base $$ED$$ and between the same parallel lines $$AB$$ and $$DE$$) 

Subtracting $$ar(FED)$$ from both sides,

$$\Rightarrow ar(BDE) – ar(FED) = ar (AED) – ar (FED) $$

$$\Rightarrow  ar (BEF) = ar (AFD)$$

(v) In  Δ  $$ABC,$$

By Appolonius theorem,   

$$AB^2+AC^2=2AD^2+2BD^2 $$

Since $$AB = AC,$$

$$AD^2=AB^2−BD^2 $$

$$AD^2=(a)^2−\left(\dfrac{a}{2}\right)^2 $$

$$AD^2=a^2−\dfrac{a^2}{4}=\dfrac{3a^2}{4} $$

$$AD=\dfrac{\sqrt{3}}{2}a $$

In  Δ $$BED ,$$  

By using Appolonius theorem,

Drop perpendicular from $$EP$$ to $$BD, $$

So , $$EP$$ is the median $$BED$$

$$EP^2=DE^2−DP^2 $$

$$EP^2=\left(\dfrac{a}{2}\right)^2−\left(\dfrac{a}{4}\right)^2$$

$$EP^2=\dfrac{a^2}{4}−\dfrac{a^2}{16}=\dfrac{3a^2}{16}  $$  

$$\Rightarrow  EP=\dfrac{\sqrt{3}a}{4} $$

$$ar(AFD) =  \dfrac{1}{2}  \times FD  \times AD $$

$$ar(AFD) =  \dfrac{1}{2}   \times FD  \times \dfrac{\sqrt{3}}{2}a $$  

$$ar(EFD) =  \dfrac{1}{2} \times FD \times EP $$

$$ar(EFD) = \dfrac{\sqrt3}{4}a$$

Now, 

$$ar(AFD) =  \dfrac{\sqrt{3}}{2}a  $$

$$ar(EFD) =  \dfrac{\sqrt{3}}{4}a$$

$$ar (AFD) = 2 ar (EFD)$$ 

Also, $$  ar (BEF) = ar (AFD)$$

So, $$ar (BFE) = ar (AFD) = 2ar (EFD)$$

(vi) $$ar (BDE) =  \dfrac{1}{4}  ar (ABC) $$

So, $$\Rightarrow ar (BEF) + ar (FED) =  \dfrac{1}{4} \times 2ar (ADC) $$

$$  2ar (FED) + ar (FED) =  \dfrac{1}{2}  (ar(AFC) – ar (AFD)$$ [Using part (v)]

$$  3ar (FED) =  \dfrac{1}{2}  ar (AFC) –  \dfrac{1}{2}  × 2 ar (FED)$$ 

$$  4ar (FED) =  \dfrac{1}{2}  ar (AFC) $$

$$ar(FED) = \dfrac18 ar(AFC)$$

Given ABCD is a trapezium the diagonal AC and BD with $$AB\parallel CD$$ intersect each other at O.

Given, $$\Delta ABE$$ and $$\Delta BAD$$ have the same base.

The ratio of the areas of two triangle with common base $$= 4 : 3$$
Height of the larger triangle $$(h_{1}) = 20$$ $$cm$$
Height of the smaller triangle $$(h_{2}) = ?$$
$$\dfrac {\text {Area of larger triangle}}{\text {Area of smaller triangle}} = \dfrac {h_{1}}{h_{2}}$$ ....$$ [\because$$ triangles have common base]
$$\Rightarrow \dfrac {4}{3} = \dfrac {20}{h_{2}}$$
$$\Rightarrow h_{2} = \dfrac {3\times 20}{4} = 15 $$ $$cm$$
$$\therefore$$ height of smaller triangle is $$15$$ $$cm$$.

Given In the figure, triangle ABC and triangle ABD are two triangles on the same base AB. If line segment CD is bisected by AB at O

$${\textbf{Hint: Recall the properties of parallelogram.}}$$

$${\textbf{Step -1: Form a pictorial representation of given data.}}$$

                 $${\text{Let }}ABCD{\text{ be the plot of the land of the shape of quadrilateral}}{\text{.}}$$

            

                 $${\text{Now, join the diagonal }}BD.$$

                 $${\text{Draw }}AE{\text{ parallel to }}BD,{\text{ join }}BE,{\text{ that intersect }}AD{\text{ at }}O.$$

             

                 $${\text{So, we get }}\vartriangle BCE{\text{ is the shape of the new field}}{\text{. }}$$

                 $$\vartriangle AOB{\text{ is the area of constructing health centre}}{\text{.}}$$

                 $${\text{And }}\vartriangle DEO{\text{ is the land joined to the plot}}{\text{.}}$$

$$\boldsymbol{{\textbf{Step -2: To prove, ar}}\left(\mathbf {\vartriangle DEO} \right) = {\textbf{ar}}\left(\mathbf {\vartriangle AOB} \right)}$$

                 $$\vartriangle DEB{\text{ and }}\vartriangle DAB{\text{ lie on the same base }}BD$$ $${\text{in between the same parallels }}BD{\text{ and }}AE.$$

                 $$ \Rightarrow {\text{ar}}\left( {\vartriangle DEB} \right) = {\text{ar}}\left( {\vartriangle DAB} \right)$$

                 $$ \Rightarrow ar\left( {\vartriangle DEB} \right) - ar\left( {\vartriangle BOD} \right) = ar\left( {\vartriangle DAB} \right) - ar\left( {\vartriangle BOD} \right)$$           

                 $$\left( {{\textbf{Subtracting the same areas from the both sides}}} \right)$$

                 $$ \Rightarrow {\text{ar}}\left( {\vartriangle DEO} \right) = {\text{ar}}\left( {\vartriangle AOB} \right)$$

                 $${\text{So, Rammaya can dispose ar}}\left( {\vartriangle DEO} \right){\text{ from his plot }}ABCE{\text{ and get ar}}\left( {\vartriangle AOB} \right)$$ 

                 $${\text{in adjacent and then he will have a triangle land}}$$ $$\vartriangle AOE.$$

$${\textbf{Hence, Rammaya can dispose ar}}\left(\mathbf {\vartriangle DEO} \right){\textbf{ from his plot ABCE and get ar}}\left(\mathbf {\vartriangle AOB} \right)$$ 

$${\textbf{in adjacent and then he will have a triangle land}}$$ $$\mathbf{\vartriangle AOE.}$$

Let 'n' be the number of shots in a side of the base, then we have,

$$PQ\parallel XT$$                            [ Since, $$XT$$ is a part of $$XY$$ ]

$$\Delta TRP$$ and $$\Delta TPK$$ have their bases on the same line and have a common vertex.
$$\therefore$$ The heights of these triangles are equal.
$$\therefore$$ Their areas are proportional to their corresponding bases.
$$RP : PK = 3 : 2$$   (Given)
$$\therefore \dfrac{RP}{PK}  = \dfrac{3}{2}$$       ....... $$(i)$$
$$\therefore \dfrac{A(\Delta TRP)}{A (\Delta TPK)} = \dfrac{RP}{PK} = \dfrac{3}{2}$$      $$[$$From $$(i)]$$
$$\therefore A(\Delta TRP) : A(\Delta TPK) = 3 : 2$$

In figure ABCDE is a pentagon . A line through B parallel to AC meet DC produced at F.

Given: $$ABCD$$ is a parallelogram. $$A$$ point $$E$$ is taken on the side $$BC,AE$$ and $$DC$$ are produced to meet at $$F$$.

We have,

In $$\Delta ABC$$

$$D,E,F$$ are mid point of the side $$AB,\,BC,\,CA$$ respectively.

Then,

$$ ar\left( CEDF \right)=24c{{m}^{2}} $$

$$ ar\left( BEFD \right)=? $$

But According to given question

$$ar\left( CEDF \right)=ar\left( BEFD \right)=24c{{m}^{2}}$$

Hence, this is the answer.

Since $$X$$ and $$Y$$ are the mid points $$AC$$ and $$AB$$ respectively.

Area of parallelogram $$base \times height $$ or $$\dfrac 12d_1\times d_2$$

Let $$l$$ and $$h$$ be the length and height respectively (in cm)

The area of Rhombus is

Base is $$5m$$

$$BP||AC$$ and $$AD||EQ$$

Join $$SQ$$, bisect the diagonal $$PM$$ at $$M$$. Since diagonals of a parallelogram bisect each other, so $$SM=MQ$$

It does not lie on the same base and between the same parallels.

$$\textbf{Step-1: Apply the concept of parallelogram}$$

It does not lie on the same base and between the same parallels.

$$\textbf{Step-1: Apply the concept of parallelogram}$$

Clearly, $$ABCQ$$ and $$ARBC$$ are parallelograms.

The given figure lies on the same base and between the same parallels. The common base is $$CD$$ and the two parallels are $$CD$$ and $$BP$$.

It does not lie on the same base and between the same parallels.

The given figure, lies on the same base and between the same parallels. The common base is $$BC$$ and the two parallels are $$BC$$ and $$AD$$.

The given figure, lies on the same base and between the same parallels. The common base is $$AB$$ and the two parallels are $$AB$$ and $$DE$$.

$$CD \parallel AE$$ and $$CY \parallel BA.$$ We have to prove that 

Given:From fig (1) ABCD is a parallelogram and $$ \mathrm{P} $$ is any point in $$ \mathrm{BC} $$.
To prove:
Area of $$ \Delta \mathrm{ABP}+ $$ area of $$ \Delta \mathrm{DPC}= $$ Area of $$ \Delta \mathrm{APD} $$
$$ \text { Proof: }$$
$$\Delta \mathrm{APD} $$ and $$ \| \mathrm{gm} \mathrm{ABCD} $$ are on the same base AD and between the same $$ \| $$ lines $$ \mathrm{AD} $$ and $$ \mathrm{BC} $$
$$ \operatorname{ar}(\Delta \mathrm{APD})=1 / 2 \operatorname{ar}(\| \mathrm{gm} \mathrm{ABCD}) \ldots(1) $$
In parallelogram ABCD $$ \operatorname{ar}(\| g m A B C D)=\operatorname{ar}(\Delta A B P)+\operatorname{ar}(\Delta A P D)+\operatorname{ar}(\Delta D P C) $$
Now, divide both sides by $$ 2, $$ we get
 $$ 1 / 2 \operatorname{ar}(\| \mathrm{gm} \mathrm{ABCD})=1 / 2 \operatorname{ar}(\Delta \mathrm{ABP})+1 / 2 \operatorname{ar}(\Delta \mathrm{APD})+1 / 2 \operatorname{ar}(\Delta \mathrm{DPC}) \ldots(2) $$
From $$( 1) $$ and (2) $$ \operatorname{ar}(\Delta \mathrm{APD})=1 / 2 \operatorname{ar}(\| \operatorname{gm} \mathrm{ABCD}) $$
Substituting ( 2) in (1) $$ \operatorname{ar}(\Delta \mathrm{APD})=1 / 2 \operatorname{ar}(\Delta \mathrm{ABP})+1 / 2 \operatorname{ar}(\Delta \mathrm{APD})+1 / 2 \operatorname{ar}(\Delta \mathrm{DPC}) $$
$$ \operatorname{ar}(\Delta \mathrm{APD})-1 / 2 \operatorname{ar}(\Delta \mathrm{APD})=1 / 2 \operatorname{ar}(\Delta \mathrm{ABP})+1 / 2 \operatorname{ar}(\Delta \mathrm{DPC}) $$
$$ 1 / 2 \operatorname{ar}(\Delta \mathrm{APD})=1 / 2[\operatorname{ar}(\Delta \mathrm{ABP})+\operatorname{ar}(\Delta \mathrm{DPC})] $$
$$ \operatorname{ar}(\Delta \mathrm{APD})=\operatorname{ar}(\Delta \mathrm{ABP})+\operatorname{ar}(\Delta \mathrm{DPC}) $$
Or ar $$ (\Delta \mathrm{ABP})+\operatorname{ar}(\Delta \mathrm{DPC})=\operatorname{ar}(\triangle \mathrm{APD}) $$
Hence proved.

Given:From fig (2)
$$ \| \mathrm{gm} \mathrm{ABCD} $$ in which $$ \mathrm{O} $$ is any point inside it.
To prove:
(i) area of $$ \Delta \mathrm{OAB}+ $$ area of $$ \Delta \mathrm{OCD}=1 / 2 $$ area of $$ \| \mathrm{gm} \mathrm{ABCD} $$
(ii) area of $$ \Delta \mathrm{OBC}+ $$ area of $$ \Delta \mathrm{OAD}=1 / 2 $$ area of $$ \| \mathrm{gm} \mathrm{ABCD} $$
Draw POQ $$ \| $$ AB through $$ O $$. It meets $$ A D $$ at $$ P $$ and $$ B C $$ at $$ Q $$.
Proof: (i) $$ \mathrm{AB} \| \mathrm{PQ} $$ and $$ \mathrm{AP} \| \mathrm{BQ} $$
$$ \mathrm{ABQP} $$ is a $$ \| \mathrm{gm} $$
Similarly, PQCD is a $$ \| $$ gm Now, $$ \Delta \mathrm{OAB} $$ and $$ \| \mathrm{gm} $$ ABQP are on same base $$ \mathrm{AB} $$ and between same $$ \| $$ lines $$ \mathrm{AB} $$ and $$ \mathrm{PQ} $$
$$ \operatorname{ar}(\Delta \mathrm{OAB})=1 / 2 \operatorname{ar}(\| \mathrm{gm} \mathrm{ABQP}) \ldots .(1) $$
Similarly, ar $$ (\Delta \mathrm{OCD})=1 / 2 \operatorname{ar}(\| \mathrm{gm} \mathrm{PQCD}) \ldots . $$ ( 2)
Now by adding (1) and (2) $$ \operatorname{ar}(\Delta \mathrm{OAB})+\operatorname{ar}(\Delta \mathrm{OCD})=1 / 2 $$ ar $$ (\| \mathrm{gm} \mathrm{ABQP})+1 / 2 $$ ar $$ (\| \mathrm{gm} \mathrm{PQCD}) $$
$$ =1 / 2[\operatorname{ar}(\| \mathrm{gm} \mathrm{ABQP})+\operatorname{ar}(\| \mathrm{gm} \mathrm{PQCD})] $$
$$ =1 / 2 \mathrm{ar}(\| \mathrm{gm} \mathrm{ABCD}) $$
$$ \operatorname{ar}(\Delta \mathrm{OAB})+\operatorname{ar}(\Delta \mathrm{OCD})=1 / 2 \operatorname{ar}(\| \mathrm{gm} \mathrm{ABCD}) $$
Hence proved.
(ii) we know that,
$$ \operatorname{ar}(\Delta O A B)+\operatorname{ar}(\Delta O B C)+\operatorname{ar}(\Delta O C D)+\operatorname{ar}(\Delta O A D)=\operatorname{ar}(\| g m \operatorname{ABCD}) $$
$$ [\operatorname{ar}(\Delta \mathrm{OAB})+\operatorname{ar}(\Delta \mathrm{OCD})]+[\operatorname{ar}(\Delta \mathrm{OBC})+\operatorname{ar}(\Delta \mathrm{OAD})]=\operatorname{ar}(\| \mathrm{gm} \mathrm{ABCD}) $$
$$ 1 / 2 $$ ar $$ (\| \mathrm{gm} \mathrm{ABCD})+\operatorname{ar}(\Delta \mathrm{OBC})+\operatorname{ar}(\Delta \mathrm{OAD})=\operatorname{ar}(\| \mathrm{gm} \mathrm{ABCD}) $$
ar $$ (\Delta \mathrm{OBC})+\operatorname{ar}(\Delta \mathrm{OAD})=\operatorname{ar}(\| \mathrm{gm} \mathrm{ABCD})-1 / 2 \mathrm{ar}(\| \mathrm{gm} \mathrm{ABCD}) $$
$$ \operatorname{ar}(\Delta \mathrm{OBC})+\operatorname{ar}(\Delta \mathrm{OAD})=1 / 2 \mathrm{ar}(\| \mathrm{gm} \mathrm{ABCD}) $$
Hence proved.

From fig ( 2 )
Given: PQRS and ABRS are parallelograms on the same base SR. X is any point on the side BR.
Join $$ \mathrm{AX} $$ and $$ \mathrm{SX} $$.
To prove: area of $$ \Delta \mathrm{AXS}=1 / 2 $$ area of $$ \| \mathrm{gm} $$ PQRS we know that,
 $$ \| \mathrm{gm} $$ PQRS and ABRS are on the same base SR and between the same parallels.
So, ar $$ \| \mathrm{gm} $$ PQRS $$ = $$ ar $$ \| \mathrm{gm} $$ ABRS $$ \ldots $$ (1)
we know that, $$ \Delta $$ AXS and $$ \| $$ gm ABRS are on the same base AS and between the same parallels.
So, ar $$ \Delta \mathrm{AXS}=1 / 2 $$ ar $$ \| \mathrm{gm} $$ ABRS $$ =1 / 2 $$ ar $$ \| \mathrm{gm} $$ PQRS $$ [\text { From }(1)] $$
Hence proved.

From the figure it is given that, 

Given:From fig (1)
$$ \operatorname{ar} \| \mathrm{gm} \mathrm{ABCD}=29 \mathrm{cm}^{2} $$
To find:
Height of parallelogram ABEF if $$ \mathrm{AB}=5.8 \mathrm{cm} $$
Now, let us find We know that $$ \| \mathrm{gm} \mathrm{ABCD} $$ and $$ \| \mathrm{gm} $$ ABEF with equal bases and between the same parallels so that there area are same.
ar $$ (\| g m A B E F)=\operatorname{ar}(\| g m A B C D) $$
$$ \operatorname{ar}(\| \mathrm{gm} \mathrm{ABEF})=29 \mathrm{cm}^{2} \ldots $$(1) [Since, ar $$ \left.\| \mathrm{gm} \mathrm{ABCD}=29 \mathrm{cm}^{2}\right] $$
also, ar ( $$ \| \mathrm{gm} $$ ABEF $$ = $$ base $$ \times $$ height) $$ 29=\mathrm{AB} \times $$ height $$ [\text { From }(1)] $$
$$ 29=5.8 \times $$ heightHeight $$ =29 / 5.8 $$
$$ =5 $$
$$ \therefore $$ Height of parallelogram ABEF is $$ 5 \mathrm{cm} $$

PSDA is a parallelogram. Points Q and R are taken on PS such that $$PQ = RS = RS$$ and $$PA \parallel QB \parallel RC.$$

$$AD \parallel DC$$,
$$AD \parallel BC$$,
$$AP \parallel DQ$$,
$$AD \parallel PQ$$,
Parallelogram $$ABCD$$ and parallelogram $$APQD$$ are on some base $$AD$$ and in between  $$AD \parallel BC$$

Given:From fig (1)
$$ \mathrm{ABCD} $$ and $$ \mathrm{AEFG} $$ are two parallelograms as shown in the figure.
To prove:
area of $$ \| \mathrm{gm} \mathrm{ABCD}= $$ area of $$ \| \mathrm{gm} \mathrm{AEFG} $$
Proof:
let us join BG. We know that,
$$ \operatorname{ar}(\Delta \mathrm{ABG})=1 / 2(\text { ar } \| \mathrm{gm} \mathrm{ABCD}) \ldots \ldots(1) $$
Similarly, $$ \operatorname{ar}(\Delta \mathrm{ABG})=1 / 2(\operatorname{ar} \| \mathrm{gm} \mathrm{AEFG}) \ldots(2) $$
From (1) and (2)
$$ 1 / 2(\text { ar } \| \mathrm{gm} \mathrm{ABCD})=1 / 2(\text { ar } \| \mathrm{gm} \text { AEFG }) $$
So, $$ \operatorname{ar}\|\mathrm{gm} \mathrm{ABCD}=\operatorname{ar}\| \mathrm{gm} \mathrm{AEFG}) $$
Hence proved.

(c) Given:$$ \mathrm{ABCD} $$ is a parallelogram.
$$ \mathrm{P} $$ is a point on $$ \mathrm{DC} $$ such that area of $$ \Delta \mathrm{DAP}=25 \mathrm{cm}^{2} $$ and area of $$ \Delta \mathrm{BCP}=15 \mathrm{cm}^{2} $$
To Find:
(i) area of $$ \| \mathrm{gm} \mathrm{ABCD} $$
(ii) DP: PC Now let us find, From fig ( 3 )
(i) we know that, ar $$ (\Delta \mathrm{APB})=1 / 2 $$ ar $$ (\| \mathrm{gm} \mathrm{ABCD}) $$
Then,
$$\begin{aligned}1 / 2 \mathrm{ar}(\| \mathrm{gm} \mathrm{ABCD}) &={\operatorname{ar}(\Delta \mathrm{DAP})+\operatorname{ar}(\Delta \mathrm{BCP})} \\&=25+15\end{aligned}$$
$$=40 \mathrm{cm}^{2}$$
So, ar $$ (\| g m A B C D)=2 \times 40=80 c m^{2} $$
(ii) we know that, $$ \triangle \mathrm{ADP} $$ and $$ \Delta \mathrm{BCP} $$ are on the same base $$ \mathrm{CD} $$ and between same parallel lines $$ \mathrm{CD} $$ and $$ \mathrm{AB} $$.
$$ \operatorname{ar}(\Delta \mathrm{DAP}) / \operatorname{ar}(\Delta \mathrm{BCP})=\mathrm{DP} / \mathrm{PC} $$
$$ 25 / 15=\mathrm{DP} / \mathrm{PC} $$
$$ 5 / 3=\mathrm{DP} / \mathrm{PC} $$
So, DP: $$ \mathrm{PC}=5: 3 $$

 Given:From fig (2)
A parallelogram ABCD in which $$ A B $$ is produced to E.
A straight line through $$ A $$ is drawn parallel to $$ \mathrm{CE} $$ to meet $$ \mathrm{CB} $$ produced at $$ \mathrm{F} $$ and parallelogram BFGE is Completed.
To prove:
area of $$ \| \mathrm{gm} $$ BFGE = Area of $$ \| \mathrm{gm} \mathrm{ABCD} $$
Proof:
Let us join $$ \mathrm{AC} $$ and EF. We know that,
$$ \operatorname{ar}(\Delta \mathrm{AFC})=\operatorname{ar}(\Delta \mathrm{AFE}) \ldots \ldots(1) $$
now subtract ar ( $$ \triangle \mathrm{ABF} $$ ) on both sides, we get
$$ \operatorname{ar}(\Delta \mathrm{AFC})-\operatorname{ar}(\Delta \mathrm{ABF})=\operatorname{ar}(\Delta \mathrm{AFE})-\operatorname{ar}(\Delta \mathrm{ABF}) $$
Or ar $$ (\Delta \mathrm{ABC})=\operatorname{ar}(\Delta \mathrm{BEF}) $$
Now multiply by 2 on both sides, we get
2. $$ \operatorname{ar}(\Delta \mathrm{ABC})=2 . $$ ar $$ (\Delta \mathrm{BEF}) $$
Or ar $$ (\| \mathrm{gm} \mathrm{ABCD})=\operatorname{ar}(\| \mathrm{gm} \text { BFGE }) $$
Hence proved.

Parallelogram $$PQRS$$ is on base $$SR$$. Trapezium $$SRMN$$ is on base $$SR$$. But they are not between $$SR \parallel PQ$$

Parallelogram $$PQRS$$ and $$\triangle TRQ$$ are on the base $$RQ$$ and in between  $$PS \parallel RQ$$

Parallelogram $$ABCD$$ is on base $$BC$$ and in between  $$AD \parallel BC$$
$$\triangle PRQ$$ is on base $$PQ$$ and in between $$AD \parallel BC$$
$$\therefore$$ parallelogram$$ABCD$$ and $$\triangle RQP$$ are not on same base. But they are in between $$AD \parallel CD$$

Suppose $$ABCD$$ is a parallelogram (given)
Consider the triangle $$ABC$$ and $$ADC$$:
$$AB=CD$$  [$$ABCD$$ is a parallelogram]
$$AD=BC$$
$$AD=AD$$   [common]
By side-side-side criterion of congruence, we have, 
$$\Delta ABC\cong \Delta ADC$$
Area of congruent triangle are equal.
Therefore, Area of $$ABC=Area\ of\ ADC$$

Since $$\triangle EBC$$ and parallelogram $$ABCD$$ are on the same base $$BC$$ and between the same parallels i.e., $$BC || AD$$
$$\therefore Ar.(\triangle EBC)=\dfrac 12 \times Ar. $$ (parallelogram $$ABCD$$)
(parallelogram $$ABCD$$) $$=2\times Ar. (\triangle EBC)$$
$$=2\times 480\ cm^2$$
$$=960\ cm^2$$.

Let us consider EF, side of $$ \| \mathrm{gm} $$ BCFE meets $$ \mathrm{AC} $$ at $$ \mathrm{G} $$. We know that,
$$ \mathrm{E} $$ is the mid-point and $$ \mathrm{EF} \| \mathrm{BC} $$
$$ \mathrm{G} $$ is the mid-point of $$ \mathrm{AC} $$.
So, $$ \mathrm{AG}=\mathrm{GC} $$
Now, in $$ \triangle \mathrm{AEG} $$ and $$ \Delta \mathrm{CFG} $$,
The alternate angles are:
$$ \angle \mathrm{EAG}, \angle \mathrm{GCF} $$
Vertically opposite angles are:
$$ \angle \mathrm{EGA}=\angle \mathrm{CGF} $$
$$ \mathrm{So}, \mathrm{AG}=\mathrm{GC} $$
Proved.
$$ \therefore \Delta \mathrm{AEG} \cong \Delta \mathrm{CFG} $$
$$ \mathrm{ar}(\Delta \mathrm{AEG})=\operatorname{ar}(\Delta \mathrm{CFG}) $$
Now,
$$\begin{aligned}\text { ar }(\| g m \text { EBCF }) &=\operatorname{ar} \mathrm{BCGE}+\text { ar }(\Delta \mathrm{CFG}) \\&=\operatorname{arctg} \mathrm{E}+\operatorname{ar}(\Delta \mathrm{AEG}) \\&=\operatorname{ar}(\Delta \mathrm{ABC})\end{aligned}$$
We know that, ar $$ (\Delta \mathrm{ABC})=25 \mathrm{sq} $$. units
Hence, ar $$ (\| g m \text { EBCF })=25 $$ sq. units

$$ar.(PQRS)=ar. (\square ABRS)$$
$$\square PQRS$$ and $$\square ABRS$$ are on Base $$SR$$ and in between $$SR \parallel PB$$.
$$\therefore ar. ( \square PQRS)=ar.( \square ABRS)$$.

$$P$$ and $$Q$$ are any two points lying on the sides $$DC$$ and $$AD$$ respectively of a parallelogram $$ABCD$$.
To prove: $$ar( \triangle APB)=ar ( \triangle BQC)$$
Proof: $$ABCD$$ is a parallelogram.
$$\therefore AB | | DC\ AB=DC$$
$$AD\ | | BC\ AD=BC$$
Now $$\triangle APB$$ and $$\square ABCD$$ are on same $$AB$$ and in between $$AB | | DC$$
$$\therefore Area ( \triangle  APB)=\dfrac 12 Area \square ABCD)$$  .........(i)
Similarly, $$\triangle BQC$$ and $$\square BADC$$ are on the same base $$BC$$ and in between $$BC | | AD$$.
$$\therefore Area ( \triangle BQC)=\dfrac 12 Area ( \square ABCD)$$........(ii)
From (i) and (ii) 
$$\therefore $$ Area $$( \triangle APB)=$$ Area $$(\triangle BQC)$$.

Here, $$PQ \parallel SR,\ PS \parallel AD \parallel BC \parallel QR$$
These are on the same base and in between a pair of parallel lines.

Data: $$P$$ is the any point in the interior of a parallelogram $$ABCD. PA, PB, PC$$ and $$PD$$ are joined.
To prove: $$ar. (ABP)=ar(PCD)=\dfrac 12 ar ( \square ABCD)$$
Construction: $$AB \parallel  XY$$ is drawn through $$P$$.
Proof: $$XY \parallel  AB\parallel DC$$
$$\therefore ABYX$$ and $$XDCY$$ are parallelograms.
$$\triangle APB$$ and $$\square ABYX$$ are on base $$AB$$ and in between $$AB \parallel XY$$.
$$\therefore APB =\dfrac 12 \square ABYX$$ ......(i)
Now, $$\triangle PCD$$ and $$\square XDCY$$ are on base $$DC$$ and in between $$DC \parallel XY$$.
$$\therefore ar. ( \triangle PCD )=\dfrac 12 ar. ( \square XDCY)$$  .....(ii)
By adding (i) and (ii), 
$$ar. ( \triangle APB )+ar. (\triangle PCD)=\dfrac 12 ar. \square ABYX+\dfrac 12 ar. \square XDCY$$
$$\therefore ar. ( \triangle APB)+ar. (\triangle PCD)=\dfrac 12 ar. ( \square ABCD)$$.

To prove:
(i) $$ar.(\triangle DOC)=ar.(\triangle AOB)$$
Construction:Draw $$DM\bot AC$$ and $$BN\bot AC$$.
Proof:
(i) $$DM\bot CO, BN\bot AO$$.
In $$\triangle DOM$$ and $$\triangle BON$$
$$DO=BO$$ (data)
$$\angle DMP=\angle BNO=90^o$$ (Construction)
$$\angle DOM=\angle BON$$ (Vertically opposite angles)
$$\therefore \triangle DOM \cong \triangle BON$$ (AAS Postulate)
$$\therefore DM=BN$$
$$\therefore ar.(\triangle DOM)=ar.(\triangle BON)$$     .........(i)
In $$\triangle DMC$$, and $$\triangle BNA$$,
$$\angle DMC=\angle BNA=90^{o}$$
$$DC=AB$$ (Data)
$$DM=BN$$ (Proved)
$$\therefore \bot \triangle DMC=\bot \triangle BNA$$
$$\therefore ar.(\triangle DMC)=ar,(\triangle BNA)$$     ........(ii)
Adding $$(i)$$ and $$(ii)$$,
$$ar.(\triangle DOM)+ar.(\triangle DMC)=ar.(\triangle BON)+ar.(\triangle BNA)$$
$$\therefore (\triangle DOC)=(\triangle AOB)$$    .........(iii)

Data: $$P$$ and $$Q$$ are respectively the mid points of sides $$AB$$ and $$BC$$ of a triangle $$\triangle ABC$$ and $$R$$ is the mid point of $$AP$$.
To prove: $$ar. (\triangle RQC)=\dfrac 38 ar. (\triangle ABC)$$
Proof: $$P$$ and $$Q$$ are respectively the mid points of sides $$AB$$ and $$BD$$
$$\therefore PQ || AC$$ and $$PQ =\dfrac 12 AC$$
($$\because $$ Mid-point theorem)
Marks $$S$$ such that it is the mid point of $$AC$$ and join $$QS$$
$$\therefore PQSA$$ is a parallelogram. Its diagonals bisects quadrilateral into two congruent triangles.
$$ar.(\triangle PAS)=ar.(\triangle PQS)=ar.(\triangle PQA)=ar.(\triangle QAS)$$
Similarly, $$PSCQ, QSCT$$ and $$PSQB$$ are parallelogram.
$$\therefore ar.(\triangle PSQ)=ar.(\triangle CQS)$$
$$ar.(\triangle QSC)=ar.(\triangle CTQ)$$
$$ar.(\triangle PSQ)=ar.(\triangle BQP)$$
$$\therefore \triangle PAS=\triangle SQP =\triangle PAQ=\triangle SQA =\triangle QSC =\triangle CTQ= \triangle QBP ....(i)$$
and $$\triangle ABC =\triangle PBQ +\triangle PAS+\triangle PQS +\triangle QSC$$
$$\triangle ABC =\triangle PBQ +\triangle PBQ +\triangle PBQ+\triangle PBQ$$
$$\triangle ABC=4\times \triangle PBQ$$
$$ar.(\triangle ABC)=4\times ar. (\triangle PBQ)$$
$$ar(\triangle PBQ)=\dfrac 14 \times ar.(\triangle ABC)....(ii)$$
In $$PACT$$ parallelogram
$$ar.(\triangle ARC)=\dfrac 12 ar.(\triangle PAC)$$ ($$\because CR$$ central line)
$$\dfrac 12 \times \dfrac 12 ar.(PACT)$$
$$=\dfrac 14 ar.(PACT)$$
$$=\dfrac 14 ar.(\triangle ABC)$$
$$\therefore ar.(\triangle ARC)=ar.(\triangle PBQ)$$.

Construction: $$AD \parallel MN$$ is drawn through $$P$$.
Proof: $$\triangle ADP$$ and $$\square ADMN$$ lie on the base $$AD$$ and in between $$AD \parallel MN$$
$$\therefore ar( \triangle ADP)=\dfrac 12 ar( \square ADMB)$$  .....(i)
Similarly, $$\triangle CPB$$ and $$\square BCNM$$ are on base $$BC$$ and in between $$BC \parallel MN$$.
$$\therefore ar. ( \triangle CPB )=\dfrac ar.( square BCNM)$$   .....(ii)
By adding (i) and (ii).
$$ar.( \triangle ADP)+ar. (\triangle CPB)=\dfrac 12r.(\square ADMN)+\dfrac 12ar. ( \square BCNM)$$
$$\therefore ar.( \triangle ADP)+ar. ( \triangle CPB)=ar. ( \triangle APB)+ar. ( \triangle PCD)$$
$$\therefore ar.(\square ABCD)=ar. (\square ABCD)$$

$$ar. (AXS)+=\dfrac 12ar.( \square PQRS)$$.
$$\triangle ACS$$ and $$\square ABRS$$ are on $$AS$$ and in between $$SR \parallel BR$$.
$$\therefore ar.( \triangle AXS)=\dfrac 12ar.( \square ABRS)$$   ......(ii)$
From (i) and (ii), 
$$ar.(\triangle AXS)=\dfrac 12 ar.(\square PQRS)$$
$$( \because \square ABRS =\square PQRS)$$

Data: Diagonals of $$\square ABCD, AC$$ and $$BD$$ intersect at $$'O'$$.
To prove: $$ar.(\triangle AOD)=ar.(\triangle DOC)=ar.(\triangle COD)=ar.(\triangle AOB)$$
Proof: $$'O'$$ is the mid-point of diagonals $$AC$$ and $$BD$$.
$$DO$$ is the median of $$\triangle ADC$$.
$$\therefore ar.(\triangle AOD)=ar.(\triangle DOC)$$   ....(i)
$$CO$$ is the median of $$\triangle DCB$$.
$$\therefore ar.(\triangle DOC)=ar.(\triangle BOC)$$    ....(ii)
$$BO$$ is the median of $$\triangle ABC$$.
$$\therefore ar.(\triangle BOC)=ar.(\triangle AOB)$$     ......(iii)
$$AO$$ is the median $$\triangle ADB$$.
$$\therefore ar.(\triangle AOB)=ar.(\triangle AOD)$$    ......(iv)
From $$(i), (ii), (iii)$$ and $$(iv)$$,
$$ar.(\triangle AOD)=ar. (\triangle DOC)=ar.(\triangle COB)=ar.(\triangle AOB)$$
$$\therefore$$ The diagonals of a parallelogram divide it into four triangles of equal area.

Data: Diagonals $$AC$$ and $$BD$$ of a quadrilateral $$ABCD$$ intersect at $$O$$ such that $$OB=OD$$. If $$AB=CD$$,

Here 

(i) $$\Delta BAD$$ is an obtuse angled triangle and $$ABCD$$ is a trapezium.

We know that ratio of areas of two similar triangles is equal to the ratio of square of their corresponding heights.
$$\cfrac { ar.\left( \triangle ABC \right)  }{ ar.\left( \triangle DEF \right)  } =\cfrac { { \left( AL \right)  }^{ 2 } }{ { \left( DN \right)  }^{ 2 } } =\cfrac { { \left( 4 \right)  }^{ 2 } }{ { \left( 9 \right)  }^{ 2 } } =\cfrac { 16 }{ 81 } $$

Data: The side $$AB$$ of a parallelogram $$ABCD$$ is produced to any point $$P$$. A line through $$A$$ and parallel to $$CP$$ meets $$CB$$ produced at $$Q$$ and then parallelogram $$PBQR$$ is completed.
To prove: $$ar.(ABCD)=ar.(PBQR)$$
Construction: $$AC$$ and $$PQ$$ are joined.
Proof: $$\triangle ACQ$$ and $$\triangle APQ$$ are on base $$AQ$$ and in between $$AQ\parallel CP$$.
$$\therefore (\triangle ACQ)=(\triangle APQ)$$
Now, $$\triangle ACQ-\triangle ABQ=\triangle APQ-\triangle ABQ$$
$$\therefore \triangle ABC=\triangle QBP$$  .....(i)
Diagonals divides quadrilateral into two congruent triangles.
$$\therefore \triangle ABC=\dfrac{1}{2}(\square ABCD)$$     (ii)
Similarly $$\triangle QBP=\dfrac{1}{2}(\square PBQR)$$       (iii)
From $$(i), (ii)$$ and $$(iii)$$,
$$\dfrac{1}{2}ar.(\square ABCD)=\dfrac{1}{2}ar.(\square PBQR)$$
$$\therefore ar.(\square ABCD)=ar.(\square PBQR)$$

Given: $$CD || AE$$ and $$CY || BE$$ 

$$PQRS$$ and $$EFRS$$ are on the same base $$SR$$ and between the same parallels $$EF$$ and $$SR$$. So their areas are equal

In $$\triangle AOQ$$ and $$\triangle BOP$$
$$\angle A=\angle B={90}^{o}$$ (given)
$$\angle AOQ=\angle POB$$ (vertically opposite angle)
By AA similarity cirterion
$$\triangle AOQ\sim \triangle BOP$$
$$\cfrac { ar.\left( \triangle AOQ \right)  }{ ar.\left( \triangle BOP \right)  } =\cfrac { { \left( OQ \right)  }^{ 2 } }{ { \left( OP \right)  }^{ 2 } } $$
$$\Rightarrow$$ $$\cfrac{x(let)}{150}={ \left( \cfrac { 7 }{ 5 }  \right)  }^{ 2 }$$
$$\Rightarrow$$ $$\cfrac{x}{150}=\cfrac{49}{25}$$
$$\Rightarrow$$ $$x=\cfrac{150\times 49}{25}=294$$
Thus $$ar(\triangle AOQ)=294{cm}^{2}$$

Construction: From vertices $$A$$ and $$D$$ drawn $$AE\bot BC$$ and $$DE\bot BC$$ respectively
Proof: From vertices $$A$$ and $$D$$, $$AE\bot BC$$ and $$DF\bot BC$$
$$\angle AEO=\angle DFO={90}^{o}$$
$$\angle AOE=\angle DOF$$ (vertically opposite angles)
By AA similarity cirterion
$$\triangle AEO\sim \triangle DFO$$
$$\Rightarrow$$ $$\cfrac{AE}{DF}=\cfrac{AO}{DO}...(i)$$
Now area of $$\triangle ABC=\cfrac{1}{2}BC\times AE$$
area of $$\triangle DBC=\cfrac{1}{2}BC\times DF$$
$$\cfrac { ar.\left( \triangle ABC \right)  }{ ar.\left( \triangle DBC \right)  } =\cfrac { \cfrac { 1 }{ 2 } BC\times AE }{ \cfrac { 1 }{ 2 } BC\times DF } =\cfrac{AF}{DF}$$....(ii)
from equation (i) and (ii)
$$\cfrac { ar.\left( \triangle ABC \right)  }{ ar.\left( \triangle DBC \right)  } =\cfrac{AO}{DO}$$

Data: $$ABC$$ is a right angles triangles at $$A, BCED, ACFG$$ and $$ABMN$$ are square on the sides $$BC, CA$$ and $$AB$$ respectively. Line segment $$AX\bot DE$$ meets $$BC$$ at $$Y$$.
To Prove: $$ar.(BCED)=ar.(ABMN)+ar.(ACFG)$$
Proof:
$$ar.(BCED)=ar.(BYXD)+ar.(CYXE)$$
$$\therefore ar.(BCED)=ar.(ABMN)+ar.(ACFG)$$

area of trapezium $$PQMS$$
$$= \dfrac12 \times (PQ + MS) \times ST$$
$$= \dfrac12 \times 12 \times 5 = 30 \ cm^2$$
$$ar(PQRN)= 60 \ cm^2$$

Join $$BD$$ and $$EF$$.
$$ar(\triangle AEF)= ar(||gm \ ABCD)- \dfrac58 \ ar(||gm \ ABCD)$$
                          $$=\left(1-\dfrac58 \right) ar (||gm \ ABCD)$$
$$= \dfrac38 \ ar (||gm \ ABCD)$$.

$$Theorem:$$ Triangles on the same base and between same parallel lines are equal in areas.    .............(1)
$$Converse:$$ Triangles with equal areas and on same base lies between parallel line               ..............(2)

Since, $$ar(DRC)=ar(DPC)$$ and on same base $$CD$$