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 $$CD$$