 # Areas of Parallelograms and Triangles

## Class 9 NCERT Maths

### NCERT

1   Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and the two parallels. ##### Solution : (i)Yes.It can be observed that trapezium $ABCD$ and triangle $PCD$ have a common base $CD$ and these are lying between the same parallel lines $AB$ and $CD$.$\\$ (ii)No. It can be observed that parallelogram $PQRS$ and trapezium $MNRS$ have a common base RS. However, their vertices, (i.e., opposite to the common base) $P, Q$ of parallelogram and $M, N$ of trapezium, are not lying on the same line.$\\$ (iii)Yes. It can be observed that parallelogram $PQRS$ and triangle $TQR$ have a common base $QR$ and they are lying between the same parallel lines $PS$ and $QR.$$\\ (iv)No. It can be observed that parallelogram ABCD and triangle PQR are lying between same parallel lines AD and BC. However, these do not have any common base.\\ (v)Yes. It can be observed that parallelogram ABCD and parallelogram APQD have a common base AD and these are lying between the same parallel lines AD and BQ.\\ (vi)No. It can be observed that parallelogram PBCS and PQRS are lying on the same base PS. However, these do not lie between the same parallel lines. 2 In the given figure, ABCD is parallelogram, AE \perp DC and CF \perp AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD. ##### Solution : In parallelogram ABCD, CD = AB = 16 cm$$\\$ [Opposite sides of a parallelogram are equal] We know that$\\$ Area of a parallelogram = Base * Corresponding altitude$\\$ Area of parallelogram $ABCD = CD * AE = AD * CF$$\\ 16 cm * 8 cm = AD * 10 cm$$\\$ $AD=\dfrac{16*8}{10} cm =12.8 cm$$\\ Thus, the length of AD is 12.8 cm. 3 If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD show that ar (EFGH)=\dfrac{1}{2} ar (ABCD) ##### Solution : Let us join HF.$$\\$ In parallelogram $ABCD,$$\\ AD = BC and AD \parallel BC (Opposite sides of a parallelogram are equal and parallel)\\ AB = CD (Opposite sides of a parallelogram are equal)\\ \Rightarrow \dfrac{1}{2}AD=\dfrac{1}{2}BC and AH\parallel BF$$\\$ $\Rightarrow AH = BF$ and $AH \parallel BF$ ( $H$ and $F$ are the mid-points of $AD$ and $BC$)$\\$ Therefore, $ABFH$ is a parallelogram.$\\$ Since $\Delta HEF$ and parallelogram $ABFH$ are on the same base $HF$ and between the same parallel lines $AB$ and $HF,$$\\ \therefore Area (\Delta HEF) = \dfrac{1}{2} Area (ABFH)....(1)$$\\$ Similarly, it can be proved that$\\$ Area $(\Delta HGF) =\dfrac{1}{2}$ Area $(HDCF) ... (2)$$\\ On adding Equations (1) and (2), we obtain\\ Area (\Delta HEF) + Area (\Delta HGF)$$ \\$ $=\dfrac{1}{2}$Area $(ABFH) + \dfrac{1}{2}$Area $(HDCF)$$\\ =\dfrac{1}{2}[Area (ABFH) + Area (HDCF)]$$\\$ $\Rightarrow$ Area $(EFGH) =\dfrac{1}{2}$Area $(ABCD)$

4   $P$ and $Q$ are any two points lying on the sides $DC$ and $AD$ respectively of a parallelogram $ABCD$. Show that ar $(APB) = ar (BQC).$

##### Solution : It can be observed that $\Delta BQC$ and parallelogram $ABCD$ lie on the same base $BC$ and these are between the same parallel lines $AD$ and $BC$.$\\$ $\therefore$ Area$(\Delta BQC)=\dfrac{1}{2}$Area$(ABCD)....(1)$$\\ Similarly, \Delta APB and parallelogram ABCD lie on the same base AB and between the same parallel lines AB and DC.\\ \therefore Area (\Delta APB)=\dfrac{1}{2}Area (ABCD)...(2)$$\\$ From Equations (1) and (2), we obtain$\\$ Area $(\Delta BQC)$ = Area $(\Delta APB)$

##### Solution :

Given: $A \Delta ABC, AD$ is the median and $E$ is the mid-point of median $AD.$

To prove:$ar(\Delta BED) = 1/4 ar(\Delta ABC)$ $\\$ Proof : In $\Delta ABC, AD$ is the median.$\\$ $\therefore ar (\Delta ABD) = ar (\Delta ADC)$ $\\$ [$therefore$ Median divides a $\Delta$ into two $\Delta s$ of equal area]$\\$ $ar(\Delta ABD) = \dfrac{1}{2} ar(ABC)$ .....(i)$\\$ In $\Delta ABD, BE$ is the median.$\\$ $ar (\Delta BED) = ar (\Delta BAE)$ $\\$ $\therefore ar (\Delta BED)= = \dfrac{1}{2}ar (\Delta BED)$ $\\$ $=ar(\Delta ABD) =\dfrac{1}{2}[ \dfrac{1}{2}ar (\Delta ABC)] = \dfrac{1}{2 }ar (\Delta ABC)$ $\\$

9   Show that the diagonals of a parallelogram divide it into four triangles of equal area.

##### Solution :

We know that diagonals of parallelogram bisect each other.$\\$ Therefore,$O$ is the mid-point of $AC$ and $BD.$ $\\$ $BO$ is the median in $\Delta ABC.$ Therefore, it will divide it into two triangles of equal areas.$\\$ $\therefore$ Area ($\Delta AOB) =$ Area ($\Delta BOC)$ ... (1)$\\$ In $\Delta BCD, CO$ is the median.$\\$ $\therefore$ Area ($\Delta BOC)$ = Area ($\Delta COD)$ ... (2)$\\$ Similarly, Area ($\Delta COD)$ = Area ($\Delta AOD)$ ... (3)$\\$ From Equations (1), (2), and (3), we obtain $\\$ Area ($\Delta AOB) =$ Area ($\Delta BOC)$ = Area ($\Delta COD)$ = Area ($\Delta AOD)$ $\\$ Therefore, it is evident that the diagonals of a parallelogram divide it into four triangles of equal area.

10   In the given figure, $ABC$ and $ABD$ are two triangles on the same base $AB$. If line-segment $CD$ is bisected by $AB$ at $O,$ show that $ar (ABC) = ar (ABD).$

##### Solution :

Consider $\Delta ACD.$ $\\$ Line-segment $CD$ is bisected by $AB$ at $O.$ Therefore, $AO$ is the median of $\Delta ACD.$ $\\$ $\therefore$ Area ($\Delta ACO) =$ Area ($\Delta ADO)$ ... (1)$\\$ Considering $\Delta BCD, BO$ is the median.$\\$ $\therefore$ Area ($\Delta BCO)$ = Area ($\Delta BDO)$ ... (2)$\\$ Adding Equations (1) and (2), we obtain$\\$ Area ($\Delta ACO)$+ Area $(\Delta BCO)$ = Area ($\Delta ADO)$ + Area ($\Delta BDO)$ $\\$ $\Rightarrow$ Area ($\Delta ABC)$ = Area ($\Delta ABD)$

11   $D, E$ and $F$ are respectively the mid-points of the sides $BC, CA$ and $AB$ of a $\Delta ABC$ . Show that: $\\$ (i)$BDEF$ is a parallelogram.$\\$ (ii) $ar (DEF) = \dfrac{1}{4}ar (ABC)$ $\\$ (iii) $ar (BDEF) = \dfrac{1}{2}ar (ABC)$

##### Solution :

(i) $F$ is the mid-point of $AB$ and $E$ is the mid-point of $AC.$ $\\$ $\therefore FE||BC$ and $FE =\dfrac{1}{2} BD$ $\\$ Line joining the mid-points of two sides of a triangle is parallel to the third and half of It$\\$ $\therefore FE||BD [BD$ is the part of $BC]$ $\\$ And $FE = BD$ $\\$ Also, $D$ is the mid-point of $BC.$$\\ B D = \dfrac{1}{2} B C \\ And FE||BC and FE = BD \\ Again E is the mid-point of AC and D is the mid-point of BC. \therefore DE||AB and DE = \dfrac{1}{2} AB \\ DE||AB [BF is the part of AB] And DE = BF \\ Again F is the mid-point of AB.$$\\$ $\therefore B F = \dfrac{1}{2} A B$ But$DE =\dfrac{ 1}{2} AB$ $\\$ $\therefore DE = BF$ $\\$ Now we have $FE||BD$ and $DE||BF$ And $FE = BD$ and $DE = BF$ $\\$ Therefore, $BDEF$ is a parallelogram.$\\$

(ii) $BDEF$ is a parallelogram.$\\$ $\therefore ar(\Delta BDF) = ar(\Delta DEF) ...........(i)$ $\\$ [diagonals of parallelogram divides it in two triangles of equal area] $DCEF$ is also parallelogram.$\\$ $\therefore ar (\Delta DEF) = ar (\Delta DEC) ..........(ii)$ $\\$ Also, $AEDF$ is also parallelogram.$\\$ $\therefore ar (\Delta AFE) = ar (\Delta DEF) ..........(iii)$ $\\$ From eq. (i), (ii) and (iii),$\\$ $ar (\dfrac DEF) = ar (\Delta BDF) = ar (\Delta DEC) = ar (\Delta AFE) ..........(iv)$ $\\$ Now, $ar (\delta ABC) = ar (\Delta DEF) + ar (\Delta BDF) + ar (\Delta DEC) + ar (\Delta AFE) ..........(v)$ $\\$ $ar (\Delta ABC) = ar (\Delta DEF) + ar (\Delta DEF) + ar (\Delta DEF) + ar (\Delta DEF)$ $\\$ [Using (iv) & (v)]$\\$ $ar (\Delta ABC) = 4 × ar (\Delta DEF)$ $\\$ $ar (\Delta DEF) = \dfrac{1}{4} ar (\Delta ABC)$ $\\$ (iii) $ar (||gm BDEF) = ar (\Delta BDF) + ar (\Delta DEF) = ar (\Delta DEF) + ar (\Delta DEF)$[Using (iv)]$\\$ $ar (||gm BDEF) = 2 ar (\Delta DEF)$ $\\$ $ar (||gm BDEF) = 2× \dfrac{1}{4} ar (\Delta ABC)$ $\\$ $ar (||gm BDEF) = \dfrac{1}{2} ar (\Delta ABC)$

12   In the given figure, diagonals $AC$ and $BD$ of quadrilateral $ABCD$ intersect at $O$ such that $OB = OD$. If $AB = CD$, then show that:$\\$ (i) $ar (DOC) = ar (AOB)$ $\\$ (ii) $ar (DCB) = ar (ACB)$ $\\$ (iii) $DA || CB$ or $ABCD$ is a parallelogram.$\\$ [Hint: From $D$ and $B$, draw perpendiculars to $AC.$]

15   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 (see the following figure). Show that$\\$$ar (ABCD) = ar (PBQR). \\[Hint: Join AC and PQ. Now compare area (ACQ) and area (APQ)] ##### Solution : Let us join AC and PQ.$$\\$ $\Delta ACQ$ and $\Delta AQP$ are on the same base $AQ$ and between the same parallels $AQ$ and $CP.$ $\\$ Area $(\Delta ACQ)$ = Area $(\Delta APQ$)$\\$ Area$(\Delta ACQ) -$ Area $(\Delta ABQ)$ = Area $(\Delta APQ) -$ Area $(\Delta ABQ)$ Area $(\Delta ABC)$ = Area $(\Delta QBP)$... (1)$\\$ Since $AC$ and $PQ$ are diagonals of parallelograms $ABCD$ and $PBQR$ respectively, $\\$ Area$(\Delta ABC) = \dfrac{1}{2}$ Area $(ABCD)$... (2)$\\$ Area $(\Delta QBP) = \dfrac{1}{2}$ Area $(PBQR) ... (3)$ $\\$ From Equations (1), (2), and (3), we obtain$\\$ $\dfrac{1}{2}$ Area $(ABCD) = \dfrac{1}{2}$ Area $(PBQR)$ $\\$ Area$(ABCD)$= Area $(PBQR)$

16   Diagonals $AC$ and $BD$ of a trapezium $ABCD$ with $AB || DC$ intersect each other at $O.$ Prove that $ar (AOD) = ar (BOC).$

##### Solution :

It can be observed that $\Delta DAC$ and $\Delta DBC$ lie on the same base $DC$ and between the same parallels $AB$ and $CD.$ $\\$ Area $(\Delta DAC)$= Area $(\Delta DBC)$ $\\$ Area $(\Delta DAC) -$ Area $(\Delta DOC)$= Area $(\Delta DBC) -$ Area $(\delta DOC)$$\\ Area (\Delta AOD) = Area (\Delta BOC) 17 In the given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that\\ (i) ar (ACB) = ar (ACF)$$\\$ (ii) $ar (AEDF) = ar (ABCDE)$

(i) $\Delta ACB$ and $\Delta ACF$ lie on the same base $AC$ and are between The same parallels $AC$ and $BF.$$\\ Area (\Delta ACB) = Area (\Delta ACF)$$\\$ (ii) It can be observed that$\\$ Area $(\Delta ACB)$= Area $(\Delta ACF)$$\\ Area (\Delta ACB) + Area (ACDE) = Area (\Delta ACF) + Area (ACDE) \\ Area (ABCDE) = Area (AEDF) 18 A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented. ##### Solution : Let quadrilateral ABCD be the original shape of the field.\\ The proposal may be implemented as follows.\\ Join diagonal BD and draw a line parallel to BD through point A.$$\\$ Let it meet the extended side $CD$ of $ABCD$ at point $E.$$\\ Join BE and AD. Let them intersect each other at O.$$\\$ Then, portion $\Delta AOB$ can be cut from the original field so that the new shape of the field will be $\Delta BCE.$ (See figure).$\\$ We have to prove that the area of $\Delta AOB$ (portion that was cut so as to construct Health Centre) is equal to the area of $\Delta DEO$ (portion added to the field so as to make the area of the new field so formed equal to the area of the original field).

It can be observed that $\Delta DEB$ and $\Delta DAB$ lie on the same base $BD$ and are between the same parallels $BD$ and $AE.$$\\ Area (\Delta DEB) = Area (\Delta DAB) \\ Area (\Delta DEB) - Area (\Delta DOB) = Area (\Delta DAB) - Area (\Delta DOB) \\ Area (\Delta DEO) = Area (\Delta AOB) 19 ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).$$\\$ [Hint: Join $CX.$]

##### Solution :

It can be observed that $\Delta ADX$ and $\Delta ACX$ lie on the same base $AX$ and are between the same parallels $AB$ and $DC.$$\\ Area (\Delta ADX) = Area (\Delta ACX) ... (1)\\ \Delta ACY and \Delta ACX lie on the same base AC and are between the same parallels AC and XY.\\ Area (\Delta ACY) = Area (ACX) ... (2)$$\\$ From Equations (1) and (2), we obtain $\\$ Area $(\Delta ADX)$ = Area $(\Delta ACY)$

20   In the given figure, $AP || BQ || CR.$ Prove that $ar (AQC) = ar (PBR).$

##### Solution :

Since $\Delta ABQ$ and $\delta PBQ$ lie on the same base $BQ$ and are between the same parallels $AP$ and $BQ,$$\\ \therefore Area (\Delta ABQ) = Area (\Delta PBQ) ... (1)\\ Again, \Delta BCQ and \Delta BRQ lie on the same base BQ and are between the same parallels BQ and CR.$$\\$ $\therefore$ Area $(\Delta BCQ)$ = Area $(\Delta BRQ)$ ... (2)$\\$ On adding Equations (1) and (2), we obtain$\\$ Area $(\Delta ABQ)$ + Area $(\Delta BCQ)$ = Area $(\Delta PBQ)$ + Area $(\Delta BRQ)$ $\\$ $\therefore$ Area $(\Delta AQC)$ = Area $(\Delta PBR)$

21   Diagonals $AC$ and $BD$ of a quadrilateral $ABCD$ intersect at $O$ in such a way that $ar (AOD) = ar (BOC)$. Prove that $ABCD$ is a trapezium.

##### Solution :

It is given that$\\$ Area $(\Delta AOD)$= Area $(\Delta BOC)$ $\\$ Area $(\Delta AOD)$+ Area $(\Delta AOB)$= Area $(\Delta BOC)$ + Area $(\Delta AOB)$ $\\$ Area $(\Delta ADB)$= Area $(\Delta ACB)$ $\\$ We know that triangles on the same base having areas equal to each other lie between the same parallels.$\\$ Therefore, these triangles, $\Delta ADB$ and $\Delta ACB,$ are lying between the same parallels.$\\$ i.e., $AB || CD$ $\\$ Therefore, $ABCD$ is a trapezium.

22   In the given figure, $ar (DRC) = ar (DPC)$ and $ar (BDP) = ar (ARC)$. Show that both the quadrilaterals $ABCD$ and $DCPR$ are trapeziums.

##### Solution :

It is given that$\\$ Area $(\Delta DRC)$ = Area $(\Delta DPC)$ $\\$ As $\Delta DRC$ and $\delta DPC$ lie on the same base $DC$ and have equal areas, therefore, they must lie between the same parallel lines.$\\$ $\therefore DC || RP$ $\\$ Therefore, $DCPR$ is a trapezium.$\\$ It is also given that$\\$ Area $(\Delta BDP)$= Area $(\Delta ARC)$ $\\$ Area $(\Delta BDP) -$ Area $(\Delta DPC)$ = Area $(\Delta ARC) -$ Area $(\Delta DRC)$ $\\$ $\therefore$ Area $(\delta BDC)$ = Area $(\Delta ADC)$ $\\$ Since $\Delta BDC$ and $\Delta ADC$ are on the same base $CD$ and have equal areas, they must lie between the same parallel lines.$\\$ $\therefore AB || CD$ $\\$ Therefore, $ABCD$ is a trapezium.

23   Parallelogram $ABCD$ and rectangle $ABEF$ are on the same base $AB$ and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.

##### Solution :

As the parallelogram and the rectangle have the same base and equal area, therefore, these will also lie between the same parallels. Consider the parallelogram $ABCD$ and rectangle $ABEF$ as follows.

Here, it can be observed that parallelogram $ABCD$ and rectangle $ABEF$ are between the same parallels $AB$ and $CF.$$\\ We know that opposite sides of a parallelogram or a rectangle are of equal lengths. Therefore,\\ AB = EF (For rectangle)\\ AB = CD (For parallelogram)\\ \therefore CD = EF \\ \therefore AB + CD = AB + EF ... (1)\\ Of all the line segments that can be drawn to a given line from a point not lying on it, the perpendicular line segment is the shortest.\\ \therefore AF < AD \\ And similarly, BE < BC$$\\$ $\therefore AF + BE < AD + BC$ ... (2)$\\$ From Equations (1) and (2), we obtain$\\$ $AB + EF + AF + BE < AD + BC + AB + CD$$\\ Perimeter of rectangle ABEF < Perimeter of parallelogram ABCD. 24 In the following figure, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC). \\ Can you now answer the question that you have left in the ’Introduction’ of this chapter, whether the field of Budhia has been actually divided into three parts of equal area?\\ [Remark: Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into n equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide \Delta ABC into n triangles of equal areas.] ##### Solution : Let us draw a line segment AL \perp BC. Let us draw a line segment AL \perp BC. \\ We know that,\\ Area of a triangle = 12 × Base × Altitude\\ Area (\Delta ADE)= 12 ×DE×AL \\ Area (\Delta ABD)= 12 ×BD×AL \\ Area(\Delta AEC)= 12 ×EC×AL \\ It is given that DE = BD = EC \\ 12 ×DE×AL= 12 ×BD×AL= 12 ×EC×AL \\ Area (\Delta ADE) = Area (\Delta ABD) = Area (\Delta AEC) It can be observed that Budhia has divided her field into 3 equal parts. 25 In the following figure, ABCD, DCFE and ABFE are parallelograms. Show that ar (\Delta ADE) = ar (\Delta BCF). ##### Solution : It is given that ABCD is a parallelogram. We know that opposite sides of a parallelogram are equal.\\ \therefore AD = BC ... (1)\\ Similarly, for parallelograms DCEF and ABFE, it can be proved that\\ DE = CF ... (2)$$\\$ And, $EA = FB ... (3)$$\\ In \Delta ADE and \Delta BCF, \\ AD = BC [Using equation (1)]\\ DE = CF [Using equation (2)]\\ EA = FB [Using equation (3)]\\ \therefore \Delta ADE \cong \Delta BCF (SSS congruence rule) \therefore Area (\Delta ADE) = Area (\Delta BCF) 26 In the following figure, ABCD is parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar (\Delta BPC) = ar (\Delta DPQ). ##### Solution : It is given that ABCD is a parallelogram. AD || BC and AB || DC(Opposite sides of a parallelogram are parallel to each other) Join point A to point C. Consider \Delta APC and \Delta BPC$$\\$ $\Delta APC$ and $\Delta BPC$ are lying on the same base $PC$ and between the same parallels $PC$ and $AB$. Therefore,$\\$ Area $(\Delta APC)$ = Area $(\delta BPC)$ ... (1)$\\$ In quadrilateral $ACDQ,$ it is given that $AD = CQ$$\\ Since ABCD is a parallelogram,\\ AD || BC (Opposite sides of a parallelogram are parallel)\\ CQ is a line segment which is obtained when line segment BC is produced. \\ \therefore AD || CQ \\ We have,\\ AC = DQ and AC|| DQ \\ Hence, ACQD is a parallelogram.\\ Consider BDCQ and BACQ$$\\$ These are on the same base $CQ$ and between the same parallels $CQ$ and $AD.$ Therefore,$\\$ Area $(\Delta DCQ)$= Area $(\Delta ACQ)$$\\ \therefore Area (\Delta DCQ) - Area (\Delta PQC) = Area (\Delta ACQ) - Area (\Delta PQC)$$\\$ $\therefore$Area $(\Delta DPQ)$ = Area $(\Delta APC)$ ... (2)$\\$ From equations (1) and (2), we obtain$\\$ Area $(\Delta BPC)$ = Area$(\Delta DPQ)$

27   In the following figure, $ABC$ and $BDE$ are two equilateral triangles such that $D$ is the mid-point of $BC$. If $AE$ intersects $BC$ at $F,$ show that$\\$ (i) $ar (BDE) = \dfrac{1}{4} ar (ABC)$$\\ (ii) ar (BDE) = \dfrac{1}{2} ar (BAE)$$\\$ (iii) $ar (ABC) = 2 ar (BEC)$$\\ (iv) ar (BFE) = ar (AFD)$$\\$ (v) $ar (BFE) = 2 ar (FED)$$\\ (vi) ar (FED) = \dfrac{1}{8} ar (AFC)$$\\$ [Hint: Join $EC$ and $AD.$ Show that $BE || AC$ and $DE || AB$ etc.]

##### Solution :

(i) Let $G$ and $H$ be the mid-points of side $AB$ and $AC$ respectively. $\\$ Line segment $GH$ is joining the mid-points and is parallel to third side. $\\$ Therefore, $BC$will be half of the length of $BC$ (mid-point theorem).

$\therefore GH=\dfrac{1}{2} BC$ and $GH || BD$ $\\$ $\therefore GH = BD = DC$ and $GH || BD (D$ is the mid-point of $BC)$ $\\$ Similarly,$\\$ $\bullet GD=HC=HA$$\\ \bullet HD=AG=BG$$\\$ Therefore, clearly $\Delta ABC$ is divided into $4$ equal equilateral triangles viz $\Delta BGD, \Delta AGH, \Delta DHC$ and $\delta GHD$ $\\$ In other words, $\Delta BGD = \dfrac{1}{4} \Delta ABC$ $\\$ Now consider $\Delta BDG$ and $\Delta BDE$ $\\$ $BD = BD$(Common base)$\\$ As both triangles are equilateral triangle, we can say $BG = BE$$\\ DG = DE Therefore, \delta BDG \cong \Delta BDE [By SSS congruency]\\ Thus, area (\Delta BDG) = area (\Delta BDE) \\ ar (\Delta BDE) = \dfrac{1}{4} ar (\Delta ABC) Hence proved (ii) Area (\Delta BDE) = Area (\Delta AED) (Common base DE and DE || AB)$$\\$ Area $(\Delta BDE) -$ Area $(\Delta FED)$ $\\$ = Area $(\Delta AED) -$ Area $(\Delta FED)$ $\\$ Area $(\Delta BEF)$ = Area $(\Delta AFD)$ ... (1)$\\$ Now, Area $(\Delta ABD)$ = Area $(\Delta ABF)$ + Area $(\Delta AFD)$ $\\$ Area $(\Delta ABD)$ = Area $(\Delta ABF)$ + Area $(\Delta BEF)$ [From equation (1)]$\\$ Area$(\Delta ABD)$ = Area $(\Delta ABE) AD$ is the median in $\Delta ABC$

$ar (\delta ABD)=\dfrac{1}{2}ar(\Delta ABC)$ $\\$ $=\dfrac{4}{2}ar(\Delta BDE) \text{(As proved earlier)}$ $\\$ $ar (\Delta ABD)=2ar (\Delta BDE) \qquad (3)$ $\\$ From(2) and (3) , we obtain $\\$ $2 ar(\Delta BDE)=ar(\Delta ABE)$ $\\$ $ar(BDE)=\dfrac{1}{2}ar(BAE)$ $\\$ (iii)$ar(\Delta ABE)=ar(\Delta BEC)$ (Common base $BE$ and $BE||AC$)$\\$ $ar(\Delta ABF)+ar(\Delta BEF)=ar(\Delta BEC)$ $\\$ Using equation (1),we obtain $\\$ $ar(\Delta ABF)+ar(\Delta AFD)=ar(\Delta BEC)$ $\\$ $\dfrac{1}{2}ar(\Delta ABC)=ar(\Delta BEC)$ $\\$ $ar (\Delta ABC)=2 ar(\Delta BEC)$

(iv) It is seen that $\Delta BDE$ and $ar \Delta AED$ lie on the same base $(DE)$ and between the parallels $DE$ and $AB$ $\\$ $\therefore ar (\Delta BDE) = ar (\Delta AED)$ $\\$ $\therefore ar (\Delta BDE) - ar (\Delta FED)$ $\\$ $= ar (\Delta AED) - ar (\Delta FED)$ $\\$ $\therefore ar (\Delta BFE) = ar (\Delta AFD)$ $\\$ (v) Let $h$ be the height of vertex $E,$ corresponding to the side $BD$ in $\Delta BDE$. $\\$ Let $H$ be the height of vertex $A$, corresponding to the side $BC$ in $\Delta ABC.$ $\\$ In (i), it was shown that $ar (BDE) = \dfrac{1}{4} ar (ABC)$ $\\$ In (iv), it was shown that $ar (\Delta BFE) = ar (\Delta AFD).$ $\\$ $\therefore ar (\Delta BFE) = ar (\Delta AFD)$ $\\$ $= 2 ar (\Delta FED)$ Hence,$\\$ (vi) $ar( \Delta AFC)=ar(\Delta AFD)+ar( \Delta ADC)$ $\\$ $=2ar(\Delta FED)+ \dfrac{1}{2 }ar( \Delta ABC)$ [using(v) =$2ar( FED)+\dfrac{ 1}{2} [4 ar( \Delta BDE)]$ [Usingresultofpart(i)]$\\$ $=2ar( \Delta FED)+2ar(\Delta BDE)=2ar(\Delta FED)+2ar(\Delta AED)$ [ $\Delta BDE$ and $\Delta AED$ are on the same base and between same parallels] $\\$ $=2ar(\Delta FED)+2[ar(\Delta AFD)+ar(\Delta FED)]$ $\\$ $=2ar( \Delta FED)+2ar(\Delta AFD)+2ar(\Delta FED)$ [Using(viii)]$\\$ $=4ar( \Delta FED)+4ar(\Delta FED)$$\\ ar(\Delta AFC)=8ar(\Delta FED) \\ ar(\Delta FED)= \dfrac{1}{8} ar( AFC) 28 Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that \\[Hint: From A and C, draw perpendiculars to BD] ##### Solution : Given : A quadrilateral ABCD, in which diagonals AC and BD intersect each other at point E. To Prove : ar(\Delta AED)* ar(\Delta BEC) \\ =ar(\Delta ABE)* ar(\Delta CDE) \\ \text{Construction:} From A,draw AM\perp BD AM BD and fromC, draw CN\perp BD \\ Proof : ar(\Delta ABE)=\dfrac{1}{2}*BE*AM..........(1) \\ ar(\Delta AED )=\dfrac{1}{2}*DE * AM...........(2) \\ Dividing eq. (ii) by (i),we get , \\ \dfrac{ar(\Delta AED)}{ar(\Delta ABE)}=\dfrac{\dfrac{1}{2}*DE*AM}{\dfrac{1}{2}* BE * AM} \\ \dfrac{ar(\Delta AED)}{ar(\Delta ABE)}=\dfrac{DE}{BE}......(iii) \\ Similarly \dfrac{ar(\Delta AED)}{ar(\Delta ABE)}=\dfrac{DE}{BE}.........(iv) \\ From eq.(iii) and (iv) , we get \\ \dfrac{ar(\Delta AED)}{ar(\Delta ABE)}=\dfrac{ar(\Delta CDE)}{ar(\Delta BEC)} \\ \implies ar(\Delta AED)*ar(\Delta BEC)=ar(\Delta ABE)*ar(\Delta CDE)$$\\$ Hence proved.

29   $P$ and $Q$ are respectively the mid-points of sides $AB$ and $BC$ of a triangle $ABC$ and $R$ is the mid-point of $AP,$ show that$\\$ (i) $ar (PRQ) = \dfrac{1}{2} ar (ARC)$ $\\$ (ii)$ar(RQC)=\dfrac{3}{8}ar (ABC)$ $\\$ (iii)$ar(PBQ)=ar(ARC)$

##### Solution :

(i) $PC$ is the median of $\Delta ABC.$ $\\$ $ar (\Delta BPC) = ar (\Delta APC) ..........(i)$ $\\$ $RC$ is the median of $APC.$$\\ ar (\Delta ARC) = \dfrac{1}{2} ar ( \Delta APC) ..........(ii)$$\\$ [Median divides the triangle into two triangles of equal area] $\\$ $PQ$ is the median of $\Delta BPC$.$\\$

$\therefore ar (\Delta PQC) = \dfrac{1}{2} ar (\Delta BPC) ..........(iii)$ $\\$ From eq. (i) and (iii), we get,$\\$ $ar (\Delta PQC) = \dfrac{1}{2} ar ( \Delta APC) ..........(iv)$ $\\$ From eq. (ii) and (iv), we get,$\\$ $ar ( \Delta PQC) = ar (\Delta ARC) ..........(v)$ $\\$ We are given that $P$ and $Q$ are the mid-points of $AB$ and $BC$ respectively.$\\$ $PQ || AC$ and $PA=\dfrac{1}{2}AC$ $\\$ $ar ( \Delta APQ) = ar (\Delta PQC) ..........(vi)$ $\\$[triangles between same parallel are equal in area]$\\$ From eq. (v) and (vi), we get$\\$ $ar (\Delta APQ) = ar (\Delta ARC) ..........(vii)$ $\\$ $R$ is the mid-point of $AP.$ Therefore $RQ$ is the median of $\Delta APQ.$ $\\$ $ar ( \Delta PRQ) =\dfrac{ 1}{2} ar (\Delta APQ) ..........(viii)$ $\\$ From (vii) and (viii), we get,$\\$ $ar(\Delta PRQ)=\dfrac{ 1}{2} ar(\Delta ARC)$ $\\$

(ii) $PQ$ is the median of $\Delta BPC$ $\\$ $\therefore ar(\Delta PQC)= \dfrac{1}{2 } ar(\Delta BPC)$ $\\$ $= \delta{1}{2}×\dfrac{1}{2} ar(\Delta ABC)= \dfrac{1}{4} ar( ABC)..........(ix)$$\\ Also ar ( \Delta PRC) = \dfrac{1}{2} ar (\Delta APC) [Using (iv)]\\ ar(\Delta PRC)=\dfrac{ 1}{2}×\dfrac{1}{2}ar( \Delta ABC)= \dfrac{1}{4} ar(\Delta ABC)..........(x) \\ Adding eq. (ix) and (x), we get,\\ ar( \Delta PQC)+ar( \Delta PRC)= (\dfrac{1}{4}+\dfrac{1}{4})ar(\Delta ABC) \\ \implies ar (quad. PQCR) =\dfrac{ 1}{2} ar (\Delta ABC) ..........(xi) \\ Subtracting ar (\Delta PRQ) from the both sides,\\ ar(quad.PQCR)-ar(\Delta PRQ)=\dfrac{ 1}{2} ar(\Delta ABC)-ar(\Delta PRQ) \\ ar( \Delta RQC)=\dfrac{ 1}{2} ar( \Delta ABC)-\dfrac{ 1}{2} ar(\Delta ARC)[Usingresult(i)]\\ ar(\Delta ARC)=\dfrac{ 1}{2 }ar(\Delta ABC)-\dfrac{ 1}{2}×\dfrac{1}{2} ar(\Delta APC) \\ ar( \Delta RQC)=\dfrac{ 1}{2} ar(\Delta ABC)-\dfrac{ 1}{4} ar(\Delta APC) \\ ar(\Delta RQC)=\dfrac{ 1}{2 }ar(\Delta ABC)-\dfrac{ 1}{4}×\dfrac{1}{2} ar(\Delta ABC)[PC is median of ABC] \\ ar(\Delta RQC)=\dfrac{ 1}{2} ar( \Delta ABC)- \dfrac{1}{8} ar(\Delta ABC) \\ ar( \Delta RQC)=(\dfrac{1}{2}-\dfrac{1}{8})* ar(\Delta ABC) \\ ar( \Delta RQC)=\dfrac{ 8}{3} ar(\Delta ABC) \\ (iii)ar(\Delta PRQ)= \dfrac{1}{2} ar(\Delta ARC) [Using result (i) ]\\ 2ar(\Delta PRQ) = ar(\Delta ARC)..(xii) \\ ar (\Delta PRQ) = \dfrac{1}{2} ar (\Delta APQ) [RQ is the median of \Delta APQ] ..........(xiii) \\ But ar (\Delta APQ) = ar (\Delta PQC) [Using reason of eq. (vi)]\\ ..........(xiv) From eq. (xiii) and (xiv), we get,\\ ar (\Delta PRQ) = \dfrac{1}{2} ar (\Delta PQC) ..........(xv)\\ But ar (\Delta BPQ) = ar (\Delta PQC) [PQ is the median of BPC] ..........(xvi) \\ From eq. (xv) and (xvi), we get,\\ ar (\Delta PRQ) = \dfrac{1}{2} ar (\Delta BPQ) ..........(xvii) \\ Now from (xii) and (xvii), we get,\\ 2×\dfrac{1}{2}ar(\Delta BPQ)=ar(\Delta ARC) \\ ar(\Delta BPQ)=ar(\Delta ARC) 30 In the following figure, ABC is a right triangle right angled at A.$$ BCED, ACFG$ and $ABMN$ are squares on the sides $BC, CA$ and $AB$ respectively. Line segment $AX \perp DE$ meets $BC$ at $Y.$ Show that:$\\$ (i) $\Delta MBC \cong \Delta ABD$ $\\$ (ii) $ar(BYXD) = 2ar(MBC)$ $\\$ (iii) $ar(BYXD) = 2ar(ABMN)$ $\\$ (iv) $\Delta FCB \cong \Delta ACE$ $\\$ (v) $ar(CYXE) = 2ar(FCB)$ $\\$ (vi) $ar(CYXE) = ar(ACFG)$ $\\$ (vii) $ar(BCED) = ar(ABMN) + ar(ACFG)$ $\\$ Note: Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem in class X.

##### Solution :

(i) We know that each angle of a square is $90^o.$ $\\$ Hence, $\angle ABM = \angle DBC = 90^o$ $\\$ $\therefore \angle ABM + \angle ABC = \angle DBC + \angle ABC$ $\\$ $\therefore \angle MBC = \angle ABD$ $\\$ In $\Delta MBC$ and $\Delta ABD,$ $\\$ $\angle MBC = \angle ABD$ (Proved above)$\\$ $MB = AB$ (Sides of square $ABMN$)$\\$ $BC = BD$ (Sides of square $BCED$)$\\$ $\therefore \Delta MBC \cong \Delta ABD$ ($SAS$ congruence rule)$\\$ (ii) We have $\Delta MBC \cong \Delta ABD$ $\\$ $\therefore ar (\Delta MBC) = ar (\Delta ABD) ... (1)$ $\\$ It is given that $AX \perp DE$ and $BD \perp DE$ (Adjacent sides of square $BDEC$)$\\$ $\therefore BD || AX$(Two lines perpendicular to same line are parallel to each other)$\\$ $\Delta ABD$ and parallelogram $BYXD$ are on the same base $BD$ and between the same parallels $BD$ and $AX.$ $\\$ Area $(\Delta YXD) = 2$ Area $(\Delta MBC)$ [Using equation (1)] ... (2)$\\$

(iii) $\Delta MBC$ and parallelogram $ABMN$ are lying on the same base $MB$ and between same parallels $MB$ and $NC.$$\\ 2 ar (\Delta MBC) = ar (ABMN)$$\\$ $ar (\Delta YXD) = ar (ABMN)$ [Using equation (2)] ... (3)$\\$ (iv) We know that each angle of a square is $90^o$ $\\$ $\therefore \angle FCA = \angle BCE = 90^o$ $\\$ $\therefore \angle FCA + \angle ACB =\angle BCE + \angle ACB$ $\\$ $\therefore \angle FCB = \angle ACE$ $\\$ In $\Delta FCB$ and $\Delta ACE,$ $\\$ $\angle FCB = \angle ACE$ $\\$ $FC = AC$(Sides of square $ACFG$)$\\$ $CB = CE$(Sides of square $BCED$)$\\$ $\Delta FCB \cong \Delta ACE (SAS$ congruence rule)$\\$

(v) It is given that $AX \perp DE$ and $CE \perp DE$ (Adjacent sides of square $BDEC)$$\\ Hence, CE || AX(Two lines perpendicular to the same line are parallel to each other)\\ Consider BACE and parallelogram CYXE$$\\$ $BACE$ and parallelogram $CYXE$ are on the same base $CE$ and between the same parallels $CE$ and $AX.$ $\\$ $\therefore ar (\Delta YXE) = 2 ar (\Delta ACE) ... (4)$ $\\$ We had proved that $\\$ $\therefore \Delta FCB \cong \Delta ACE$ $\\$ $ar (\Delta FCB) \cong ar (\Delta ACE) ... (5)$ $\\$ On comparing equations (4) and (5), we obtain$\\$ $ar (CYXE) = 2 ar (\Delta FCB) ... (6)$$\\ (vi) Consider BFCB and parallelogram ACFG \\ BFCB and parallelogram ACFG are lying on the same base CF and between the same parallels CF and BG.$$\\$ $\therefore ar (ACFG) = 2 ar (\Delta FCB)$ $\\$ $therefore ar (ACFG) = ar (CYXE)$ [Using equation (6)] ... (7)$\\$ (vii) From the figure, it is evident that$\\$ $ar (\Delta CED) = ar (\Delta YXD) + ar (CYXE)$ $\\$ $\therefore ar (\Delta CED) = ar (ABMN) + ar (ACFG)$ [Using equations (3) and (7)]