Question 14 Marks
The median $BE$ and $CF$ of a triangle $ABC$ intersect at $G.$ Prove that the area of $GBC =$ area of the quadrilateral $AFGE.$
AnswerGiven: In $\triangle\text{ABC}$ and we know that a median of a triangle divides it into two parts of equal area.
So, $\text{ar}(\triangle\text{ABE})=\text{ar}(\triangle\text{CBE})$
$\Rightarrow\text{ar}(\triangle\text{ABE})=\frac{1}{2}\text{ar}(\triangle\text{ABC})$
Similarly, $CF$ is the median of $\triangle\text{ABC}.$
Then, $\text{ar}(\triangle\text{ABE})=\text{ar}(\triangle\text{CBE})$
$\Rightarrow\text{ar}(\triangle\text{ABE})=\frac{1}{2}\text{ar}(\triangle\text{ABC})$ From Eqs. $(i)$ and $(ii)$ $\text{ar}(\triangle\text{ABE})=\text{ar}(\triangle\text{BCF})$
On substracting $\text{ar}(\triangle\text{GBF})$ from both sides of Eq. $(iii)$
we get $\text{ar}(\triangle\text{ABE})-\text{ar}(\triangle\text{BGF})=\text{ar}(\triangle\text{BCF})-\text{ar}(\triangle\text{GBF})$
$\Rightarrow\text{ar}(\text{AFGE})=\text{ar}(\text{GBC})$ Hence, proved
View full question & answer→Question 24 Marks
In Fig. $X$ and $Y$ are the mid-points of $AC$ and $AB$ respectively, $QP || BC$ and $CYQ$ and $BXP$ are straight lines. Prove that $ar\ (ABP) = ar\ (ACQ).$
AnswerGiven $X$ and $Y$ are the mid-points of $AC$ and $AB$ respectively. Also, $QP|| BC$ and $CYQ, BXP$ are straight lines.
To prove: $\text{ar}(\triangle\text{ABP})=\text{ar}(\triangle\text{ACQ})$
Proof: Since, $X$ and $Y$ are the mid-points”of $AC$ and $AB$ respectively.
So, $XY || BC$
We know that, triangles on the same base and between the same parallels are equal in area. Here, $\triangle\text{BYC}\ \text{and}\ \triangle\text{BXC}$ lie on same base BC and between the same parallels $BC$ and $XY.$
So, $\text{ar}(\triangle\text{BYC})=\text{ar}(\triangle\text{BXC})$
On subtracting ar $(\triangle\text{BOC})$ from both sides, we get
$\text{ar}(\triangle\text{BYC})-\text{ar}(\triangle\text{BOC})=\text{ar}(\triangle\text{BXC})-\text{ar}(\triangle\text{BOC})$
$\Rightarrow\text{ar}(\triangle\text{BOY})=\text{ar}(\triangle\text{COX})$
On adding ar $(\triangle\text{XOY})$ both sides, we get
$\text{ar}(\triangle\text{SOY})+\text{ar}(\triangle\text{XOY})=\text{ar}(\triangle\text{COX})+\text{ar}(\text{XOY})$
$\Rightarrow\text{ar}(\triangle\text{BYX})=\text{ar}(\triangle\text{CXY})\ ...(1)$
Hence, we observe that quadrilaterals $XYAP$ and $YXAQ$ are on the same base $XY$ and between the same parallels $XY$ and $PQ.$
$ar\ (XYAP) = ar\ (YXAQ) …(2)$
On adding Eqs. $(1)$ and $(2),$ we get
$\text{ar}(\triangle\text{BYX})+\text{ar}(\text{XYAP})=\text{ar}(\triangle\text{CXY})+\text{ar}(\text{YXAQ})$
$\Rightarrow\text{ar}(\triangle\text{ABP})=\text{ar}(\triangle\text{ACQ})$ Hence proved.
View full question & answer→Question 34 Marks
In $\triangle\text{ABC}, D$ is the mid-point of $AB$ and $P$ is any point on $BC.$ If $CQ || PD$ meets $AB$ in $Q ($shown in figure$),$ then prove that $\text{ar}(\triangle\text{BPQ})=\frac{1}{2}\text{ar}(\triangle\text{ABC})$
AnswerGiven In $\triangle\text{ABC} D$ is the mid-point of $AB$ and $P$ is any point on $BC.$
$CQ || PD$ means $AB$ in $Q.$
To prove $\text{ar}(\triangle\text{BPQ})=\frac{1}{2}\text{ar}(\triangle\text{ABC})$
Construction: Join $PQ$ and $CD.$
Proof: Since, $D$ is the mid-point of $AB.$ So $CD$ is the median of $\triangle\text{ABC}$
We know that, a median of a triangle divides it into two triangles of equal areas.
$\therefore\text{ar}(\triangle\text{BCD})=\frac{1}{2}\text{ar}(\triangle\text{ABC})$
$\Rightarrow\text{ar}(\triangle\text{BPD})+\text{ar}(\triangle\text{DPC})=\frac{1}{2}\text{ar}(\triangle\text{ABC})$
Now, $\triangle\text{DPQ}$ and $\triangle\text{ADPC}$ are on the same base $DP$ and between the same parallel lines $DP$ and $CQ.$
So, $\text{ar}(\triangle\text{DPQ})=\text{ar}(\triangle\text{DPC})$
On putting the value from Eq. $(1)$ in Eq.$ 2$ we get
$\text{ar}(\triangle\text{BPD})+\text{ar}(\triangle\text{DPQ})=\frac{1}{2}\text{ar}(\triangle\text{ABC})$
$\Rightarrow\text{ar}(\triangle\text{BPQ})=\frac{1}{2}\text{ar}(\triangle\text{ABC} )$
View full question & answer→Question 44 Marks
In Fig.$9\ PSDA$ is a parallelogram. Points $Q$ and $R$ are taken on $PS$ such that $PQ = QR = RS$ and $PA || QB || RC.$ Prove that $ar\ (PQE) = ar\ (CFD). $
AnswerGiven In a parallelogram $PSDA,$ points $0$ and $R$ are on $PS$
such that $PQ = QR = RS$ and $PA || QB || RC.$
To prove $ar\ (PQE) = ar\ (CFD)$ Proof In parallelogram $PABQ,$ and $PA || QB$ [given]
So, $PABQ$ is a parallelogram. $PQ = AB …(i)$
Similarly, $QBCR$ is also a parallelogram.
$QR = BC …(ii)$ and $RCDS$ is a parallelogram. $RS = CD …(iii)$
Now, $PQ = QR = RS …(iv)$ From Eqs. $(i), (ii) (iii)$ and $(iv), $
$PQ || AB [$$\therefore$ in parallelogram $PSDA, PS || AD]$
In $\triangle\text{PQE}$ and $\triangle\text{DCF},\angle\text{QPE}=\angle\text{FDC} [$since, $PS || AD$ and $PD$ is transversal, then alternate interior angles are equal]
$PQ=CD$ [from Eq. $(v)$] and $\angle\text{PQE} =\angle\text{FCD}$ [ $\therefore \angle\text{PQE}=\angle\text{PRC}$ corresponding angles and $\angle\text{PRC}=\angle\text{FCD}$ alternate interior angles]
$\triangle\text{PQE}=\triangle\text{DCF} [$by $ASA$ congruence rule$]$
$\therefore$ ar $(\triangle\text{PQE}) =\text{ar}(\triangle\text{CFD})$ [since,congruent figures have equal area] Hence proved.
View full question & answer→Question 54 Marks
A point $E$ is taken on the side $BC$ of a parallelogram $ABCD. AE$ and $DC$ are produced to meet at $F.$ Prove that $ar\ (ADF) = ar\ (ABFC).$
AnswerGiven: $ABCD$ is a parallelogram.
A point $E$ is taken on the side $BC.$
$AE$ and $DC$ are produced to meet at $F.$
Proof: since $ABCD$ is a parallelogram and diagonal $AC$ divides it into two triangles of equal area.
we have $\text{ar}(\triangle\text{ADC})=\text{ar}(\triangle\text{ABC})\ ...(1)$
As $\text{DC}\parallel\text{AB},$
So $\text{CF}\parallel\text{AB}$
Since triangle on the same base and between the same parallel are equal in area,
so we have $\text{ar}(\triangle\text{ACF})=\text{ar}(\triangle\text{BCF})\ ...(2)$ Adding $(1)$ and $(2),$
we get $\text{ar}(\triangle\text{ADC})+\text{ar}(\triangle\text{ACF})=\text{ar}(\triangle\text{ABC})+\text{ar}(\triangle\text{BCF})$
$\Rightarrow\text{ar}(\triangle\text{ADF})=\text{ar}({\text{ABFC}})$ Hence, proved.
View full question & answer→Question 64 Marks
In trapezium $ABCD, AB || DC$ and $L$ is the mid-point of $BC.$ Through $L,$ a line $PQ || AD$ has been drawn which meets $AB$ in $P$ and $DC$ produced in $Q ($Fig$)$. Prove that $ar\ (ABCD) = ar\ (APQD).$
AnswerAs $AB || DC,$ so $AB || DQ$
In $\triangle\text{CLQ}$ and $\triangle\text{BLP}$, we have
$\therefore$ $\angle\text{QCL}=\angle\text{LBP}$ [Alt, $\angle\text{S}$]
$CL = LB$ [$\because L$ is the mid point of $BC$]
$\angle\text{CLQ}=\angle\text{BLP}$ [vertically opposite $\angle\text{S}$ ]
$\therefore\triangle\text{QCL}\cong\angle\text{BLP}$ [By $ASA$ congruence rule]
$\Rightarrow\text{ar}(\triangle\text{CLQ})=\text{ar}(\triangle\text{BLP})\ ...(1)$ [congruence $\triangle\text{s}$ are equal in area]
Adding ar $(APLCD) = \text{ar}(\triangle\text{BLP})+\text{ar}(\text{APLCD})$
$\Rightarrow\text{ar}(\text{APQD})=\text{ar}(\text{ABCD})$
Hence, $ar(ABCD) = ar(APQD).$
View full question & answer→Question 74 Marks
In $\triangle\text{ABC}$ if $L$ and $M$ are the points on $AB$ and $AC,$ respectively such that $LM || BC.$ Prove that $ar (LOB) = ar (MOC).$
AnswerSince triangle on the same base and between the same parallel are equal in area, so, We have, $\therefore\text{ar}(\triangle\text{LBM})=\text{ar}(\triangle\text{LCM})$ $\triangle\text{LBM}$ and $\triangle\text{LCM}$ are on the same base $LM$ and between the same parallel $LM$ and $BC$] $\therefore\text{ar}(\triangle\text{LBM})=\text{ar}(\triangle\text{LCM})$ $\text{ar}(\triangle\text{LOB})=\text{ar}(\triangle\text{MOC})$ [Cancelling $(\triangle\text{LOM})$ from both sides]
View full question & answer→Question 84 Marks
$ABCD$ is a trapezium in which $AB || DC, DC = 30\ cm$ and $AB = 50\ cm.$ If $X$ and $Y$ are, respectively the mid-points of $AD$ and $BC,$ prove that $\text{ar}(\text{DCYX})=\frac{7}{9}\ \text{ar}(\text{XYBA})$
Answerjoin $DY$ and extend $DY$ intersecting $AB$ produced at $M.$
Now, In $\triangle\text{MBY}$ and $\triangle\text{DCY}$,
we have $\angle1=\angle2$ [vertically opposite $\angle\text{S}$ ]
$\angle3=\angle4$ [$\because AB || DC$ and alt. $\angle\text{S}$ are equal]
$BY = CY[\because$ $Y$ is the mid point of $BC]$
$\therefore\triangle\text{MBY}\cong\triangle\text{DCY}$ [By ASA cong. Rule]
So, $MB = DC = 30\ cm [CPCT]$
Now, $AM = AB + BM = 50\ cm + 30\ cm = 80\ cm$
In $\triangle\text{ADM},$
we have $\text{XY}=\frac{1}{2}\text{AM}=\frac{1}{2}\times80\text{cm}=40\text{cm}$
As $AB || XY || DC$ and $X$ and $Y$ are the mid point of $AD$ and $BC,$
so height of trapezium $DCYX$ and $XYBA$ are equal. let the equal height be $h\ cm.$
$\frac{\text{ar}(\text{DCXY})}{\text{ar}(\text{XYBA})}=\frac{\frac{1}{2}(30+40)\times\text{h}}{\frac{1}{2}\times(40+50)\times\text{h}}=\frac{70}{90}=\frac{7}{9}$
Hence, ar $(DCXY)=\frac{7}{9}\text{ar}(\text{XYBA})$
View full question & answer→Question 94 Marks
In figure, $CD || AE$ and $CY || BA.$ Prove that $\text{ar}(\triangle\text{CBX}) =\text{ar}(\triangle\text{AXY})$
Answer$CD || AE$ and $CY || BA$. we have to prove that $\text{ar}(\triangle\text{CBX})=\text{ar}(\triangle\text{AXY}).$
Since triangle on the same base and between the same parallel are equal in area,
so we have $\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{ABY})$
$\Rightarrow\text{ar}(\triangle\text{CBX})+\text{ar}(\triangle\text{ABX})=\text{ar}(\text{ABX})+\text{ar}(\triangle\text{AXY})$
Hence, $\text{ar}(\triangle\text{CBX})=\text{ar}(\triangle\text{AXY})$ [cancelling $\text{ar}(\triangle\text{ABX)}$ from both sides.
View full question & answer→Question 104 Marks
$ABCD$ is a parallelogram in which $BC$ is produced to $E$ such that $CE = BC$ (Fig.) $AE$ intersects $CD$ at $F.$ If ar $(DFB) = 3\ cm^2$ , find the area of the parallelogram $ABCD$.
AnswerGiven, $ABCD$ is a parallelogram and $CE = BC$
i.e., $C$ is the mid-point of $BE.$
Also, $\text{ar}(\triangle\text{DFB})=3\text{cm}^2$
Now, $\triangle\text{ADF}$ and $\triangle\text{DFB}$
are on the same base DF and between parallel $CD$ and $AB.$
Then $\text{ar}(\triangle\text{ADF})=\text{ar}(\triangle\text{DFB})=3\text{cm}^2\ ...(\text{i})$ In
$\triangle\text{ADF}$ by the converse of of mid point theorem,
$EF = AF [$since, $C$ is mid-point of $BE] ...(ii)$ In $\triangle\text{ADF}$
and $\triangle\text{ECF},$
$\angle\text{AFD}=\angle\text{CFG}$ [vertically opposite angle]
$AF = EF \angle\text{DAF}=\angle\text{CEF} [$since $BE || AD$ and $AE$ is transversal,
then alternate interior angles are equal]
$\triangle\text{ADF}\cong\triangle\text{ECF} [$by $ASA$ congurance rule$]$
Then, $\text{ar}(\triangle\text{ADF})=\text{ar}(\triangle\text{CFE})$ [since, congurant figure have equal area]
$\therefore\text{ar}(\triangle\text{CEF})=\text{ar}(\triangle\text{ADF})=3\text{cm}^2 [$From EQ. $(i)] ...(iii)$
Now, In $\triangle\text{BFE}, C$ is the mid point of $BE$ then $CF$ is median of
$\triangle\text{BFE},$
$\therefore\text{ar}(\triangle\text{CEF})=\text{ar}(\triangle\text{BFC})$ [since, median of a triangle divides it into two triangle of equal area]
$\Rightarrow\text{ar}(\triangle\text{CEF})=\text{ar}(\triangle\text{BFC})$ [since, median of a triangle divides it into two triangles of equal area]
$\Rightarrow\text{ar}(\triangle\text{BFC})=3\text{cm}^2$
Now, $\text{ar}(\triangle\text{BDC})=\text{ar}(\triangle\text{CFB})+\text{ar}(\triangle\text{BFC})$
$=3+3=6\ \text{cm}^2 [$from $EQS. (i)$ and $(iv)]$
We know that, diagonal of a parallelogram divides it into two congurant triangle of equal area.
$\therefore$ Area of parallelogram $ABCD = 2 \times $ Area of $\triangle\text{BDC}$
$= 2 \times 6 = 12\ cm^2$
Hence, the area of parallelogram $ABCD$ is $12\ cm^2.$
View full question & answer→Question 114 Marks
In Fig. $ABCD$ and $AEFD$ are two parallelograms. Prove that $ar\ (PEA) = ar\ (QFD).$
AnswerWe have to prove that $\text{ar}(\triangle\text{PEA})=\text{ar}(\triangle\text{QFD})$ join $PD.$
In $\triangle\text{PEA}$ $\triangle\text{QFD}$ we have
$\angle\text{APE}=\angle\text{DQF}[\because$ corresponding $\angle\text{S}$ are equal as $AB || CD]$
$\angle\text{APE}=\angle\text{DFQ}[\because$ corresponding $\angle\text{S}$ are equal as $AE || DF]$
$AE = DF [\because$ opposite sides of $a || gm$ are equal$]$
$\therefore\triangle\text{PEA}\cong\triangle\text{QFD} [$By $AAS$ Cong. Rule$]$
Hence, $\text{ar}(\triangle\text{PEA})=\text{ar}(\triangle\text{QED})$
View full question & answer→Question 124 Marks
In Fig. $ABCDE$ is any pentagon. $BP$ drawn parallel to $AC$ meets $DC$ produced at $P$ and $EQ$ drawn parallel to $AD$ meets $CD$ produced at $Q.$ Prove that $ar\ (ABCDE) = ar\ (APQ).$
$$
AnswerGiven $ABCDE$ is a pentagon.
$BP || AC$ and $EQ|| AD.$
To prove $ar (ABCDE) = ar (APQ)$
Proof: We know that, triangles on the same base and between the same parallels are equal in area.
Here, $\triangle\text{ADQ}$ and $\triangle\text{ADE}$ lies on the same base $AD$ and between the same parallel.
$AD$ and $EQ.$
So, $\text{ar}(\triangle\text{ADQ})=\text{ar}(\triangle\text{ADE})\ ...(1)$
Similarly, $\triangle\text{ACP}$ and $\triangle\text{ACB}$ lies on the same base $AC$ and base the same parallel.
$AC$ and $BP.$
So, $\text{ar}(\triangle\text{ACP})=\text{ar}(\triangle\text{ACB})\ ...(2)$
On adding Eqs. $(i)$ and $(ii),$ we get
$\text{ar}(\triangle\text{ADQ})+\text{ar}(\triangle\text{ACP})=\text{ar}(\text{ADE})+\text{ar}(\triangle\text{ACB})$
On adding $\text{ar}(\triangle\text{ACD})$ both sides, we get
$\text{ar}(\triangle\text{ADQ})+\text{ar}(\triangle\text{ACP})+\text{ar}(\text{ACD})$ $=\text{ar}(\triangle\text{ADE})+\text{ar}(\triangle\text{ACB})+\text{ar}(\triangle\text{ACD})$
$\Rightarrow\text{ar}(\triangle\text{APQ})=\text{ar}(\text{ABCDE})$ Hence proved.
View full question & answer→Question 134 Marks
If the mid-points of the sides of a quadrilateral are joined in order, prove that the area of the parallelogram so formed will be half of the area of the given quadrilateral (Fig.) $[$Hint: Join $BD$ and draw perpendicular from $A$ on $BD] $
AnswerGiven: Let $ABCD$ is a quadilateral and $P, F, R$ and $S$ are the mid point of the sides $BC, CD$ and $AD$ and $AB $ respectively and $PFRS$ is a parallelegram.
To prove: $ar ($parallelogram $PFRS) =\frac{1}{2} ar ($quadilateral $ABCD)$
Construction: join $BD$ and $BR.$
Proof: Median $BR$ divides $\triangle\text{BDA}$ into two triangles of equal area.
$\therefore\text{ar}(\triangle\text{BRA})=\frac{1}{2}\text{ar}(\triangle\text{BDA})\ ...(\text{i})$ Similarly, median $RS$ divides
$\triangle\text{BRA}$ into two triangles of equal area.
$\therefore\text{ar}(\triangle\text{ASR})=\frac{1}{2}\text{ar}(\triangle\text{BAR})\ ...(\text{ii})$ From Eqs. $(iii)$ and $(iv),$ we get
$\text{ar}(\triangle\text{ASR})+\text{ar}(\triangle\text{CFP})=\frac{1}{4}\text{ar}(\triangle\text{BDA})$
$\Rightarrow\text{ar}(\triangle\text{ASR})+\text{ar}(\triangle\text{CFP})=\frac{1}{4}\text{}(\text{quadilateral BCDA})$ Similarly,
$\text{ar}(\triangle\text{DRF})+\text{ar}(\triangle\text{BSP})=\frac{1}{4}\text{ar}(\text{quadilateral BCDA})$ On adding $EQS.$ $(v)$ and $(vi),$ we get
$\text{ar}(\triangle\text{ASR})+\text{ar}(\triangle\text{CFP })+\text{ar}(\triangle\text{DRF})+\text{ar}(\triangle\text{BSP})$
$=\frac{1}{2}\text{ar quadilateral BCDA}$ But $\text{ar}(\triangle\text{ASR})+\text{ar}(\triangle\text{CFP})+\text{ar}(\triangle\text{DRF})+\text{ar}(\triangle\text{BSP})+\text{ar}(\text{parallelogram PFS})$$= ar ($quadilateral $BCDA)$ On substracting Eq. $(vii)$ from Eq. $(viii),$ we get
$ar ($parallelogram $PFRS) =\frac{1}{2} ar ($quadilateral $BCDA).$ Hence proved.
View full question & answer→Question 144 Marks
$O$ is any point on the diagonal $PR$ of a parallelogram $PQRS$ (Fig.) Prove that $\text{ar}(\triangle\text{PSO})=\text{ar}(\triangle\text{PQO})$
Answerjoin $SQ,$ bisect the diagonal $PM$ at $M.$ since diagonal of a parallelogram bisects each other, So $SM = MQ.$
Therefore, $PM$ is a median of $\triangle\text{PQS}$
$\text{ar}(\triangle\text{PSM})=\text{ar}(\triangle\text{PQM})\ ...(1)$
[ $\because$ median divides a triangle into two triangles of equal area]
Again, as $OM$ is the median of $\triangle\text{OSQ}$, so
$\text{ar}(\triangle\text{OSM})=\text{ar}(\triangle\text{OQM})\ ...(2)$
Adding $(1)$ and $(2),$ we get
$\text{ar}(\triangle\text{PSM})+\text{ar}(\triangle\text{OSM})=\text{ar}(\triangle\text{PQM})+\text{ar}(\triangle\text{OQM})$
$\Rightarrow\text{ar}(\triangle\text{PSO})=\text{ar}(\triangle\text{PQO})$ Hence proved.
View full question & answer→Question 154 Marks
The diagonals of a parallelogram $ABCD$ intersect at a point $O$. Through $O$, a line is drawn to intersect $AD$ at $P$ and $BC$ at $Q.$ Show that $PQ$ divides the parallelogram into two parts of equal area.
AnswerGiven: In a parallelogram $ABCD,$ diagonal intersect at $O$ and draw a line $PQ,$ which intersect $AD$ and $BC,$
To prove: $PQ$ divides the parallelogram $ABCD$ into two parts of equal area.
$ar\ (ABQP) = ar\ (CDPQ)$
Proof: we have that, diagonal of a parallelogram bisect each other.
$\therefore OA = OC $ and $OB = OD ...(1)$
In $\triangle\text{AOB}$ and $\triangle\text{COD}$
$OA = OC$
$OB = OD [$from Eq. $(1)]$
And $\angle\text{AOB}=\angle\text{COD}$ [vertically opposite angles]
$\triangle\text{AOB}=\triangle\text{COD}$ [by $SAS$ congruence rule]
$\text{ar}(\triangle\text{AOB})=\text{ar}(\triangle\text{COD})\ ...(2)$ [since, congruent fig have equal area]
Now, in $\triangle\text{AOP}$ and $\triangle\text{COQ},$
$\angle\text{PAO}=\angle\text{OCQ}$ [alternate interior angle]
$OA = OC [$from eq. $(1)]$
and $\angle\text{AOP}=\angle\text{COQ}$ [vertically opposite angle]
$\therefore\triangle\text{AOP}\cong\triangle\text{COQ} [$by $ASA$ congurancy$]$
$\therefore\text{ar}(\triangle\text{AOP})=\text{ar}(\triangle\text{COQ})$
[since, congurance fig have equal area]
Similarliy.
$\text{ar}(\triangle\text{POD})=\text{ar}(\triangle\text{BOQ})$
Now, $\text{ar}(\text{ABQP})=\text{ar}(\triangle\text{COQ})+\text{ar}(\triangle\text{COD})+\text{ar}(\triangle\text{POD})$
$=\text{ar}(\triangle\text{AOP})+\text{ar}(\triangle\text{AOB})+\text{ar}(\triangle\text{BOQ}) [$from Eqs. $(ii), (iii)$ and $(iv)]$
Hence proved.
View full question & answer→Question 164 Marks
If the medians of a $\triangle\text{ABC}$ intersect at $G,$ show that $ar\ (AGB) = ar\ (AGC) = ar\ (BGC)$ $=\frac{1}{3}\text{ar}(\text{ABC})$
AnswerGiven: In $\triangle\text{ABC}$, $AD, BE$ and $CF$ are median and intersect at $G.$
To prove: $\text{ar}(\triangle\text{AGB})=\text{ar}(\triangle\text{AGC})=\text{ar}(\triangle\text{BGC})=\frac{1}{3}\text{ar}(\triangle\text{ABC})$
Proof: we have that, a median of a triangle divides it into two triangles divides it into two triangle of equals area.
$\therefore\text{ar}(\triangle\text{ABD})=\text{ar}(\triangle\text{ACD})$
In $\triangle\text{BGC},\text{GD}$ is a median.
$\therefore\text{ar}(\triangle\text{GBD})=\text{ar}(\triangle\text{GCD})$
On substracting Eq. $(ii)$ from Eq. $(i),$ we get $\text{ar}(\triangle\text{ABD})-\text{ar}(\triangle\text{GBD})=\text{ar}(\triangle\text{ACD})-\text{ar}(\triangle\text{GDC})$
$\Rightarrow\text{ar}(\triangle\text{AGB})=\text{ar}(\triangle\text{AGC})$
Now, $\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{AGB})+\text{ar}(\triangle\text{BGC})+\text{ar}(\triangle\text{AGC})$
$\Rightarrow\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{AGB})+\text{ar}(\triangle\text{AGB})+\text{ar}(\triangle\text{AGB}) [$from Eq. $(v)]$
$\Rightarrow\text{ar}(\triangle\text{ABC})=3\text{ar}(\triangle\text{AGB})$ From Eqs. $(iii)$ and $(iv),$
$\text{ar}(\triangle\text{AGB})=\text{ar}(\triangle\text{BGC})=\text{ar}(\triangle\text{AGC})$
Now, $\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{AGB})+\text{ar}(\triangle\text{BGC})+\text{ar}(\triangle\text{AGC})$
$\Rightarrow\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{AGB})+\text{ar}(\text{AGB})+\text{ar}(\triangle\text{AGB})$
$\Rightarrow\text{ar}(\triangle\text{ABC})=3\text{ar}(\triangle\text{AGB})$
$\Rightarrow\text{ar}(\triangle\text{AGB})=\frac{1}{3}\text{ar}(\triangle\text{ABC})$
From Eqs. $(v)$ and $(vi).$
$\text{ar}(\triangle\text{BGC})=\frac{1}{3}\text{ar}(\triangle\text{ABC})$ And
$\text{ar}(\triangle\text{AGC})=\frac{1}{3}\text{ar}(\triangle\text{ABC})$
View full question & answer→