[5 marks sum] - MATHEMATICS STD 9 Questions - Vidyadip

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Question 15 Marks

Prove that the quadrilateral formed by joining the mid$-$points of consecutive sides of a rhombus is a rectangle.

Answer

In $\triangle A B C, P$ and $Q$ are mid points of sides $A B$ and $B C$ respectively.
Therefore, $PQ \| AC$ and $PQ =\frac{1}{2} AC ($using mid$-$point theorem$) ...(1)$
In $\triangle ADC$
$R$ and $S$ are the mid points of $C D$ and $A D$ respectively
Therefore, $RS \| AC$ and $RS = \frac{1}{2} A C ($using mid$-$point theorem$) ...(2)$
From equations $(1)$ and $(2),$ we have
$P Q \| R S$ and $P Q=R S$
As in quadrilateral $\text{PQRS}$ one pair of opposite sides are equal and parallel to each other, so, it is a parallelogram.
Let diagonals of rhombus $\text{ABCD}$ intersects each other at point $O$.
Now in quadrilateral $\text{OMQN}$
$MQ \| ON \ldots( PQ \| AC )$
$QN \| OM \ldots( QR \| BD )$
So, $\text{OMQN}$ is parallelogram
$\angle MQN =\angle NOM$
$\angle PQR =\angle NOM$
But, $\angle NOM =90^{\circ}...($diagonals of a rhombus are perpendicular to each other$)$
$\angle PQR =90^{\circ}$
Clearly $\text{PQRS}$ is a parallelogram having one of its interior angle as $90^{\circ}$.
Hence, $\text{PQRS}$ is rectangle.

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Question 25 Marks

Prove that the quadrilateral formed by joining the mid$-$points of consecutive sides of a rectangle is a rhombus.

Answer

Let us join $AC$ and $BD$
In $\triangle ABC$
$P$ and $Q$ are the mid$-$point of $A B$ and $D C$ respectively
Therefore, $PQ \| AC$ and $PQ =\frac{1}{2} AC ($mid$-$point theorem$)$
Similarly in $\triangle ADC$
$S R \| A C$ and $S R=\frac{1}{2} A C ($mid$-$point theorem$)$
Clearly, $P Q \| S R$ and $P Q=S R$
As in quadrilateral $\text{PQRS}$ one pair of opposite sides is equal and parallel to each other, so, it is a parallelogram.
Therefore, $PS \| QR$ and P$S = QR ($opposite sides of parallelogram$)...$
Now, in $\triangle B C D, Q$ and $R$ are mid points of side $B C$ and $C D$ respectively.
Therefore, $Q R \| B D$ and $Q R=\frac{1}{2} B D ($mid$-$point theorem$)..$
But diagonals of a rectangle are equal
$\Rightarrow AC = BD \text {...(5) }$
Now, by using equation $(1). (2), (3), (4), (5)$ we can say that
$P Q=Q R=S R=P S$
So, $\text{PQRS}$ is a rhombus.

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Question 35 Marks

$\text{ABCD}$ is a quadrilateral $P, Q, R$ and $S$ are the mid$-$points of $AB, BC, CD$ and $AD$. Prove that $\text{PQRS}$ is a parallelogram.

Answer

Join $A C$ and $B D$
In $\triangle ABC$,
$P$ and $Q$ are mid$-$point of $A B$ and $B C$ respectively.
Therefore, $P Q \| A C$ and $P Q=\frac{1}{2} A C........(i)$
In $\triangle A D C$,
$S$ and $R$ are mid$-$point of $A D$ and $D C$ respectively.
Therefore, $S R \| A C$ and $S R=\frac{1}{2} A C.........(ii)$
From $(i)$ and $(ii)$
$P Q \| S R$ and $P Q=S R$
Therefore, $\text{PQRS}$ is a parallelogram.

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Question 45 Marks

The diagonals $PR$ and $QS$ of a quadrilateral $\text{PQRS}$ are perpendicular to each other.$ A, B, C$ and $D$ are mid$-$point of $PQ, QR, RS$ and $SP$ respectively. Prove that $\text{ABCD}$ is a rectangle.

Answer

In $\triangle PQS , A$ and $D$ are mid$-$points of sides $QP$ and $PS$ respectively.
Therefore, $A D \| Q S$ and $A D=\frac{1}{2} Q S$
In $\triangle QRS$
$B$ and $C$ are the mid$-$points of $Q R$ and $R S$ respectively
Therefore, $B C \| Q S$ and $B C=\frac{1}{2} Q S$
From equations $(i)$ and $(ii),$
$A D \| B C$ and $A D=B C$
As in quadrilateral $\text{ABCD}$ one pair of opposite sides are equal and parallel to each other, so it is parallelogram.
The diagonals of quadrilateral $\text{PQRS}$ intersect each other at point $O$.
Now in quadrilateral $\text{OMDN}$
$ND \| O M \ldots(A D \| Q S)$
$DM \| ON ...(DC \| PR)$
So, $\text{OMDN}$ is parallelogram
$ \angle MDN =\angle NOM$
$\angle ADC =\angle NOM $
But, $\angle NOM =90^{\circ}...($doagonals are perpendicular to each other$)$
$\Rightarrow \angle ADC =90^{\circ}$
Clearly $\text{ABCD}$ is a parallelogram having one of its interior angle as $90^{\circ}$.
Hence, $\text{ABCD}$ is rectangle.

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Question 55 Marks

Prove that the quadrilateral formed by joining the mid$-$points of a square is also a square.

Answer

Join $A C$ and $B D$
In $\triangle A B C, P$ and $Q$ are the mid$-$points of sides $A B$ and $B C$ respectively.
Therefore, $PQ \| AC$ and $PQ =\frac{1}{2} AC$
In $\triangle A D C, R$ and $S$ are the mid$-$points of sides $C D$ and $A D$ respectively.
Therefore, $RS \| AC$ and $RS =\frac{1}{2} AC$
From $(i)$ and $(ii)$
$P Q \| R S$ and $P Q=R S$
Thus, in a quadrilateral $\text{PQRS}$ one pair of opposite sides are equal and parallel.
Hence, $\text{PQRS}$ is a parallelogram.
Since $\text{ABCD}$ is a square
$ AB = BC = CD = DA$
$=\frac{1}{2} AB =\frac{1}{2} CD ; \frac{1}{2} AB =\frac{1}{2}$
$\Rightarrow PB = RC ; BQ = CQ $
Thus in $\triangle PBQ$ and $\triangle RCQ$
$ PB = RC$
$BQ = CQ$
$\angle PBQ =\angle RCQ =90^{\circ} $
Therefore, $\triangle PBQ \cong \triangle RCQ$
Hence, $P Q=Q R$
From $(iii)$ and $(iv)$
$P Q=Q R=R S$
But $\text{PQRS}$ is a parallelogram
$ \Rightarrow QR = PS$
$\Rightarrow PQ = QR = RS = PS $
Now, $PQ \|AC$
$\Rightarrow PM \| NO (vi)$
Since $P$ and $S$ are the mid$-$points of $A B$ and $A D$ respectively $PS \| BD$
$\Rightarrow PN \| MO$
Thus in quadrilateral $\text{PMON}$
$PM \| NO ...($from $(vi))$
$PN \| MO ...($from $(vii))$
So, $\text{PMON}$ is a parallelogram
$\Rightarrow \angle MPN =\angle MON$
$\Rightarrow \angle MPN =\angle BOA$
$\Rightarrow \angle MPN =90^{\circ} \ldots (\perp \because$ diagonals$)$
$\Rightarrow \angle QPS =90^{\circ} $
Thus, $\text{PQRS}$ is a quadrilateral such that $PQ = QR = RS = PS$ and $\angle QPS =90^{\circ}$
Hence, $\text{PQRS}$ is a square.

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Question 65 Marks

Prove that the diagonals of a square are equal and perpendicular to each other.

Answer

Given:
$\text{ABCD}$ is a square with diagonals $A C$ and $B D$ intersecting each other at point $O$.
To prove: $A C=B D$ and $\angle A O B=90^{\circ}$
Proof:
In $\triangle DAB$ and $\triangle CBA$,
$A D=B C \ldots($Sides of a square$)$
$\angle DAB =\angle CBA \ldots($ Each $90)$
$AB = AB \ldots($Common$)$
$\Rightarrow \triangle DAB \cong \triangle CBA \ldots(\text{SAS}$ Congruence$)$
$\Rightarrow A C=B D \ldots(\text{C . P . C . T})$
In $\triangle A O B$ and $\triangle B O C$,
$A D=B C \ldots($Sides of a square$)$
$O B=O B \ldots($Common$)$
$O A=O C...($Diagonals of a square bisect each other$)$
$\Rightarrow \triangle AOB \cong \triangle BOC...(\text{SSS}$ Congruence$)$
$\Rightarrow \angle A O B=\angle B O C...(\text{C.P.C.T})$
But, $\angle A O B+\angle B O C=180^{\circ} \ldots($Linear pair$)$
$\Rightarrow \angle A O B=\angle B O C=\frac{180^{\circ}}{2}=90^{\circ}$
$\therefore AC = BD$
$\Rightarrow$ Diagonals are equal
And, $\angle A O B=90^{\circ}$
$\Rightarrow$ Diagonals are perpendicular to each other.

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Question 75 Marks

Prove that the diagonals of a kite intersect each other at right angles.

Answer

Consider $ABCD$ is a kite.
Then, $AB = AD$ and $BC = DC$

In $\triangle ABC$ and $\triangle ADC,$
$AB = AD$
$BC = CD$
$AC = AC$
$\therefore \triangle ABC \times \triangle ADC ...(\text{SSS}$ Congruence$)$
$\Rightarrow \angle BCA = \angle DCA ...(\text{C.P.C.T})$
$\Rightarrow \angle BCO = \angle DCO ...(i)$
Now,
In $\triangle OBC$ and $\triangle ODC,$
$BC = CD$
$\angle BCO = \angle DCO ...[$From $(i)]$
$OC = OC ...($common$)$
$\therefore \triangle OBC \times \triangle ODC ...(\text{SAS}$ congruence$)$
$\angle COB - \angle COD ...(\text{C.P.C.T})$
But, $\angle COB + \angle CID = 180^\circ ...($Linear pair$)$
$\Rightarrow \angle COB + \angle COB = 180^\circ $
$\Rightarrow 2\angle COB = 180^\circ $
$\Rightarrow \angle COB = 90^\circ $
Hence, diagonals a kite intersect each other at right angles.

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Question 85 Marks

Prove that if the diagonals of a quadrilateral bisect each other at right angles then it is a rhombus.

Answer

Let $\text{ABCD}$ be a quadrilateral, whose diagonals $AC$ and $BD$ bisect each other at right angle.
i.e. $OA = OC, OB = OD$
And, $\angle AOB = \angle BOC = \angle COD = \angle AOD = 90^\circ $
To prove $\text{ABCD}$ a rhombus, we need to prove $\text{ABCD}$ is a parallelogram and all sides of $\text{ABCD}$ are equal.
Now, in $\triangle AOD$ and $\text{DCOD}$
$OA = OC ($Diagonal bisects each other$)$
$\angle AOD = \angle COD ...($Each $90^\circ )$
$OD = OD ...($common$)$
$\therefore \triangle AOD ≅\triangle COD ...($By $\text{SAS}$ congruence rule$)$
$\therefore AD = CD ….(i)$
Similarly, we can prove that
$AD = AB$ and $CD = BC ….(ii)$
From equations $(i)$ and $(ii)$, we can say that
$AB = BC = CD = AD$
Since opposite sides of quadrilateral $\text{ABCD}$ are equal, so, we can say that $\text{ABCD}$ is a parallelogram.
Since all sides of a parallelogram $\text{ABCD}$ are equal, so, we can say that $\text{ABCD}$ is a rhombus.

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Question 95 Marks

In the given figure, $\text{PQRS}$ is a trapezium in which $PQ \| SR$ and $PS = QR$. Prove that: $\angle PSR = \angle QRS$ and $\angle SPQ = \angle RQP$

Answer

Construction:
Draw $SM \perp PQ$ and $RN \perp PQ$

$a$. In $\triangle PMS$ and $\triangle QNR,$
$PS = QR ....($given$)$
$\angle PMS = \angle QNR ....($Each $90^\circ )$
$SM = RN ....($Distance between parallel lines$)$
$\therefore \triangle PMS ≅ \triangle QNR ....(\text{RHS}$ congruence$)$
$\Rightarrow \angle PM = \angle RQN$
$\Rightarrow \angle SPQ = \angle RQP$
$\Rightarrow \angle PSR = \angle QRS ....($Supplement of each $\angle SPQ$ and $RQP)$
$b$. In $\triangle PQS$ and $\triangle QPR,$
$PS = QR ....($given$)$
$\angle SPQ = \angle RQP ....($proved$)$
$PQ = PQ ....($common$)$
$\therefore \triangle PQS \times \triangle QPR ....(\text{SAS}$ congruence$)$
$\Rightarrow QS = PR ....($proved$)$
$\Rightarrow PSQ = \angle QRP ....(i)$
$c$. In $\triangle TPS$ and $\triangle TQR,$
$PS = QR ....($given$)$
$\angle STP = \angle RTQ ....($Vertically opposite angles$)$
$\angle PST = \angle QRT ....[$From $(i)]$
$\therefore \triangle TPS \cong \triangle TQR ....(\text{SAS}$ congruence$)$
$\Rightarrow TP = TQ ....($proved$)$
$\Rightarrow TS = TR. ....($proved$)$

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Question 105 Marks

In the given figure, $\text{PQRS}$ is a parallelogram in which $PA = AB$.Prove that: $SA\| QB$ and $SA = QB.$

Answer

Construction:
Join $BS$ and $AQ.$
Join diagonal $QS.$

Since diagonals of a parallelogram bisect each other.
$\therefore OP = OR$ and $OQ = OS$
Also, $PA = AB = BR$
Now, $OP = OR$ and $PA = PB$
$\Rightarrow OP - PA = OR - PB$
$\Rightarrow OA = OB$
Thus, in quadrilateral $\text{SAQB}$, we have
$OQ = OS$ and $OA = OB$
$\Rightarrow $ Diagonals of a quadrilateral $\text{SAQB}$ bisect each other.
$\Rightarrow \text{SAQB}$ is a parallelogram.
$\Rightarrow SA \| QB$
$\Rightarrow SA = QB. ...($Opposite sides are equal$)$

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Question 115 Marks

$\text{ABCD}$ is a trapezium in which side $A B$ is parallel to side $D C . P$ is the mid$-$point of side $A D$. IF $Q$ is a point on the Side $B C$ such that the segment $P Q$ is parallel to $D C$, prove that $P Q=\frac{1}{2}(A B+D C)$.

Answer

$P Q\|D C$
$ \Rightarrow O Q\| D C \| A B$
Therefore, $Q$ and $O$ are mid$-$points of $B C$ and $B D$ respectively.
In $\triangle ABD$,
$P$ and $O$ are mid$-$points of $A D$ and $B D$ respectively
$\Rightarrow OP =\frac{1}{2} AB$
In $\triangle B C D$,
$Q$ and $O$ are mid$-$points of $B C$ and $B D$ respectively
$\Rightarrow OQ =\frac{1}{2} CD$
Adding $(i)$ and $(ii)$
$ OP + OQ =\frac{1}{2} AB +\frac{1}{2} CD$
$\Rightarrow PQ =\frac{1}{2}(AB + CD)$

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Question 125 Marks

In a parallelogram $\text{PQRS, M}$ and $N$ are the midpoints of the opposite sides $PQ$ and $RS$ respectively. Prove that;$RN$ and $RM$ trisect $QS.$

Answer

Since $M$ and $N$ are the mid$-$point of $P Q$ and $RS$ respectively.
$\therefore PM =\frac{1}{2} PQ$ and $RN =\frac{1}{2} RS \text {...(i) }$
But $\text{PQRS}$ is a parallelogram,
$\therefore PQ = RS$ and $PQ \| RS$
$\Rightarrow \frac{1}{2} PQ =\frac{1}{2} RS$ and $PQ \| RS$
$\Rightarrow PM = RN$ and $PQ \| RS$
$\Rightarrow \text{PMRN}$ is a parallelogram.
$\Rightarrow PN \| RM$
$\Rightarrow NY \| RX . . . \text { (ii) }$
We know that the segment drawn through the mid$-$point of one side of a triangle and parallel to the other side bisects the third side.
In $\triangle S R X, N$ is the mid$-$point of $RS$ and $NY \| R X....[$From $(ii)]$
$\therefore Y$ is the mid$-$point of $Q Y$
$\Rightarrow X Y=Y S \text {....(iii) }$
Similarly, in $\triangle P Q Y, M$ is the mid$-$point of $P Q$ and $M X \| P Y....[$From $(ii)]$
$\Rightarrow Q X=X Y \text {....(iv) }$
From $(iii)$ and $(iv)$, we get
$Q X=X Y=Y S$
$\Rightarrow X$ and $Y$ trisect $Q S$
$\Rightarrow PN$ and $RM$ trisect $QS.$

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Question 135 Marks

Prove that the line segment joining the mid$-$points of the diagonals of a trapezium is parallel to each of the parallel sides, and is equal to half the difference of these sides.

Answer

Join $A C$ and $B D .$
$M$ and $N$ are mid$-$points of $A C$ and $B D$ respectively.
Join $M N$.
Draw a line $C N$ cutting $A B$ at $E$.
Now, in $\triangle SBC$ and $\text{BNE}$,
$D N=N B ... (N$ is the mid$-$point of $B D$, given $)$
$\angle CDB =\angle EBN...($Alternate angles as $DC \|AB)$
$\angle DNC = BNE...($Vertically opposite angles$)$
$\Rightarrow \triangle DNC \cong \triangle BNE...($By $A-S-A$ Test$)$
$\Rightarrow DC = BE$
By Mid$-$point Theorem, in $\triangle A C E, M$ and $N$ are mid$-$points
$MN =\frac{1}{2} AE$ and $MN \| AE$ or $MN \| AB$
Also, $A B \| C D$,
$\therefore M N \| C D$
$ \Rightarrow MN =\frac{1}{2}[ AB = BE]$
$\Rightarrow MN =\frac{1}{2}[ AB = CD] \quad \ldots($since $BE = CD)$

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Question 145 Marks

$\text{PQRS}$ is a parallelogram. $M$ and $N$ are the mid$-$points of the adjacent sides $QR$ and $RS. O$ is the mid$-$point of the diagonal $PR$. Prove that $\text{MONR}$ is a rectangle and $MN$ is half of $PR.$

Answer

In $\triangle S R Q$
$N$ and $O$ are the mid$-$points of $S R$ and $P R$ respectively.
Therefore, $O N \| Q R$ and $O N=\frac{1}{2} Q R$
i.e. $O N=M R$
$P R=S Q...($diagonals of rectangle are equal and bisect each other$)$
$\Rightarrow O$ is mid$-$point of $SQ$
In $\triangle RQS$,
$M$ and $O$ are the mid$-$points of $Q R$ and $S Q$ respectively.
Therefore, $OM \| SR$ and $OM =\frac{1}{2} QR$
i.e. $OM = SQ$
$\angle MRN =\angle QRS =90^{\circ}..(iii) (\text{PQRS}$ is a rectangle$)$
From $(i) , (ii)$ and $(iii)$
Therefore, quadrilateral $\text{MONR}$ has two opposite pairs of sides equal and parallel and an interior angle as right angle, so it is a rectangle.
In $\triangle S Q R$
$M$ and $N$ are the mid$-$points of $Q R$ and $S R$ respectively.
Therefore, $MN \| SQ$ and $MN =\frac{1}{2} SQ$
But $S Q=P R$

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Question 155 Marks

$\text{PQR}$ is a triangle formed by the adjacent sides $PQ$ and $QR$ and diagonal $PR$ of a parallelogram $\text{PQRS}$. If in $\triangle PQR, \angle P : \angle Q : \angle R = 3 : 8 : 4$, Calculate the measures of all the angles of parallelogram $\text{PQRS}.$

Answer

$\text{PQRS}$ is a parallelogram.
Let $\angle RPQ = 3x^\circ $
Then, $\angle PQR = 8x^\circ $ and $\angle QRP = 4x^\circ $
In $\triangle PQR,$
$\angle RPQ + \angle PQR + \angle QRP = 180^\circ ...($sum of angles of triangle$= 180^\circ )$
$3x^\circ + 8x^\circ + 4x^\circ = 180^\circ $
$15x^\circ = 180^\circ $
$x = 12^\circ $
$\Rightarrow \angle RPQ = 3x^\circ = 3 \times 12^\circ = 36^\circ $
$\Rightarrow \angle PQR = 8x^\circ = 8 \times 12^\circ = 96^\circ $
$\Rightarrow \angle QRP = 4x^\circ = 4 \times 12^\circ = 48^\circ $
Now,
$\angle PSR = \angle PQR = 96^\circ ...($opposite angles of a parallelogram are equal$)$
$\angle RPS = \angle QRP = 48^\circ ...($Alternate angles since $QR \| PS)$
$\angle PRS = \angle RPQ = 36^\circ ...($Alternate angles since $QR \| PS)$
Therefore,
$\angle PSR = \angle PQR = 96^\circ , \angle RPS + \angle RPQ = 84^\circ , \angle QRP = 84^\circ .$

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Question 165 Marks

The angles of a triangle formed by $2$ adjacent sides and a diagonal of a parallelogram are in the ratio $1 : 5 : 3$. Calculate the measures of all the angles of the parallelogram.

Answer

$\text{ABCD}$ is a parallelogram.
Let $\angle CAB = x^\circ $
Then, $\angle ABC = 5x^\circ $ and $\angle BCA = 3x^\circ $
In $\triangle ABC,$
$\angle CAB + \angle ABC + \angle BCA = 180^\circ ...($sum of angles of $\triangle = 180^\circ )$
$x^\circ + 5x^\circ + 3x^\circ = 180^\circ $
$9x^\circ = 180^\circ $
$x^\circ = 20^\circ $
$\Rightarrow \angle CAB = x^\circ = 20^\circ $
$\Rightarrow \angle ABC = 5x^\circ = 5 \times 20^\circ = 100^\circ $
$\Rightarrow \angle BCA = 3x^\circ = 3 \times 20^\circ = 60^\circ $
Now,
$\angle ADC = \angle ABC = 100^\circ ...($opposite angles of a parallelogram are equal$)$
$\angle ACD =\angle CAB = 20^\circ ...($Alternate angles since $BC \| AD)$
$\angle CAD = \angle BCA = 60^\circ ...($Alternate angles since $BC \| AD)$
Therefore,
$\angle ADC= \angle ABC = 100^\circ , \angle ACD + \angle BCA = 80^\circ , \angle CAD + \angle CAB = 80^\circ .$

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Question 175 Marks

In the given figure,$\text{ABCD}$ is a parallelogram, find the values of $x$ and $y.$

Answer

$\text{ABCD}$ is a parallelogram.
Opposite angles of a parallelogram are equal.
$\therefore \angle A = \angle C$
$\Rightarrow 4x + 3y - 6 = 9y + 2$
$\Rightarrow 4x - 6y = 8$
$\Rightarrow 2x - 3y = 4 ....(i)$
$AB \| CD$ and $AD$ is the transversal.
$\therefore \angle A + \angle D = 180^\circ .... ($Co $-$ interior angles are supplementary$)$
$\Rightarrow (4x + 3y - 6) + (6x + 22) = 180^\circ$
$\Rightarrow 10x + 3y + 16 = 180^\circ$
$\Rightarrow 10x + 3y = 164 ....(ii)$
Adding equations $(i)$ and $(ii),$ we get
$12x + = 168$
$\Rightarrow x = 14$
Substituting the value of $x$ in $(i),$ we get
$2 \times 14 - 3y = 4$
$\Rightarrow 28 - 3y = 4$
$\Rightarrow 3y = 24$
$\Rightarrow y = 8$
Hence, $x = 14$ and $y = 8$.

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[5 marks sum] - MATHEMATICS STD 9 Questions - Vidyadip