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Maths 4 Marks Questions

Quadrilaterals · Jharkhand Board (JAC) STD 9 (Jharkhand - English Medium). 19 questions.

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Question 14 Marks
In the adjoining figure,$ABCD$ is a parallelogram and $E$ is the midpoint of side $BC$. If $DE$ and $AB$ when produced meet at $F$, prove that $AF = 2AB$.
Answer


Given: $A$ parrallelogram $ABCD$ in which $E$ is the mid point of side $BC, DE$ and $AB$ when produced meet at $F$.
To Prove: $\text{AF}=2\text{AB}$
Proof: In $\triangle\text{DEC}$ and $\triangle\text{FEB}$
$\angle\text{DEC}=\angle\text{FEB}$ [vertically opposite angles]
$\angle\text{DCE}=\angle\text{FBE}$ [alternate angles]
$\text{CE = EB}$ [Given]
Thus by Angle-Angle-Side criterion of congruence, we have
$\triangle\text{DEC}\cong\triangle\text{FEB}$ $[$By $AAS]$
The corresponding parts of the congruent triangle are equal.
$\therefore\text{DC = FB}$ $[$By $C.P.C.T.]$
So, $\text{AF = AB + BF}$
$=\text{AB + DC}$
$=\text{AB + AB}$
$=2\text{AB}$
$\therefore\text{AF}=2\text{AB}$
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Question 24 Marks
In the adjoining figure, $ABCD$ is a parallelogram in which $AB$ is produced to $E$ so that $BE = AB$. Prove that $ED$ bisects $BC$.
Answer


Given: $ABCD$ is a parralegram in which $AB$ is produced to $E$ such that $BE = AB$. $DE$ is joined which cuts $BC$ at $O$.
To Prove: $\text{OB = OC}$
Proof: In $\triangle\text{OCD}$ and $\triangle\text{OBE},$ we have,
$\angle\text{DOC}=\angle\text{EOB}$ [vertically opposite angles are equal]
$\angle\text{OCD}=\angle\text{OBE}$ [AB || CD, BC is a transversal thus, alternate angles are equal]
$\text{DC = BE}$ $[AB = CD$ and $BE = AB]$
Thus, by Angle-Angle-Side criterion of congruence, we have
$\therefore\triangle\text{OCD}\cong\triangle\text{OBE}$ [by AAS]
The corresponding parts of the congruent triangles are equal.
$\therefore\text{OC = OB}$
Hence, $ED$ bisect $BC$.
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Question 34 Marks
In the adjoining figure, $ABCD$ is a parallelogram whose diagonals intersect each other at $O$. A line segment $EOF$ is drawn to meet $AB$ at $E$ and $DC$ at $F$. Prove that $OE = OF.$
Answer


Given: A parallelogram $ABCD$, in which diagonals intersect at $O$. $E$ and $F$ are the points on $AB$ and $CD$
To Prove: $\text{OE = OF}$
Proof: In $\triangle\text{AOE}$ and $\triangle\text{COF},$ we have
$\angle\text{CAE}=\angle\text{DCA}$ [Alternate angles]
$\text{AO = CO}$ [diagonals are equal and bisect each other]
and, $\angle\text{AOE}=\angle\text{COF}$ [Vertically opposite angles]
Thus by Angle-Side-Angle criterion of congruence, we have,
$\therefore\triangle\text{AOE}\cong\triangle\text{COF}$ [By ASA]
The corresponding parts of the congruent triangles are equal.
$\therefore\text{OE = OF}$ $[$By $C.P.C.T.]$
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Question 44 Marks
In each of the figures given below, $ABCD$ is a rhombus. Find the value of $x$ and $y$ in each case.
Answer
$ABCD$ is a rhombus, so its all sides are equal.

In $\triangle\text{ABC},$ we have
$\text{AB = BC}$
$\Rightarrow\angle\text{CAB}=\angle\text{ACB}=\text{x}^{\circ}$
As, $\angle\text{CAB}+\angle\text{ABC}+\angle\text{ACB}=180^{\circ}$
$\Rightarrow\text{x}+110^{\circ}+\text{x}=180^{\circ}$
$\Rightarrow2\text{x}=180^{\circ}-110^{\circ}=70^{\circ}$
$\Rightarrow\text{x}=\frac{70^{\circ}}{2}=35^{\circ}$
$\therefore\text{x}=35^{\circ}$ and $\text{y}=35^{\circ}$
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Question 54 Marks
$P , Q, R$ and $S$ are respectively the midpoints of the sides $AB, BC, CD$ and $DA$ of a quadrilateral $\text{ABCD}$. Show that:
$i. PQ \| AC$ and $\text{PQ}=\frac{1}{2}\text{AC}$
$ii. PQ \| SR$
$iii. \text{PQRS}$ is a parallelogram.
Answer

$i.$ In $\triangle\text{ABC,}$
$P$ and $Q$ are the mid$-$points of sides $AB$ and $BC$ respectively.
$\Rightarrow\text{PQ }\|\text{ AC}$ and $\text{PQ}=\frac{1}{2}\text{AC}...(\text{i})$
$ii.$ In $\triangle\text{ADC,}$
$R$ and $S$ are the mid$-$points of sides $CD$ and $AD$ respectively.
$\Rightarrow SR \| AC$ and
$\Rightarrow\text{SR}=\frac{1}{2}\text{AC}...(\text{ii})$
From $(i)$ and $(ii),$ we have
$\text{PQ = SR }$ and $PQ \| SR$
$iii.$ Thus, in quadrilateral $\text{PQRS}$, one pair of opposite sides are equal and parallel.
Hence, $\text{PQRS}$ is a parallelogram.
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Question 64 Marks
In the adjoining figure, $\text{BM}\perp\text{AC}$ and $\text{DN}\perp\text{AC}.$ If $BM = DN$, prove that $AC$ bisects $BD$.
Answer
Given: $A$ quadrilateral $ABCD$, in which $\text{BM}\perp\text{AC}$ and $\text{DN}\perp\text{AC}$ and $BM = DN$.
To prove: $AC$ bisects $BD$; or $DO = BO$
 Proof: Let $AC$ and $BD$ intersect at $O$.
Now, in $\triangle\text{OND}$ and $\triangle\text{OMB},$
we have: $\angle\text{OND}=\angle\text{OMB}$ (90° each) $\angle\text{DON}=\angle\text{BOM}$ (Vertically opposite angles)
Also, $\text{DN = BM}$ (Given) i.e., $\triangle\text{OND}\cong\triangle\text{OMB}$ (AAS congrurence rule)
$\therefore\text{OD = OB}$ $(C.P.C.T.) $
Hence, $AC$ bisects $BD$.
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Question 74 Marks
In the adjoining figure, $AD$ is a median of $\triangle\text{ABC}$ and $DE || BA$. Show that $BE$ is also a median of $\angle\text{ABC}.$
Answer
Given: A $\triangle\text{ABC}$ in which $AD$ is its median and $DE || AB$

To Prove: $BE$ is a median of $\triangle\text{ABC}.$
Proof: In $\triangle\text{ABC},$
$DE || AB$ [Given]
$D$ is the mid-point of $BC$.
The line drawn through the midpoint of one side of a triangle, parallel to another side, intersects the third side at its midpoint.
So, by Mid point Theorem, $E$ is the mid-point of $AC$.
$\therefore$ $BE$ is the median of $\triangle\text{ABC}$ drawn through $B$.
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Question 84 Marks
In the adjoining figure,$\text{ABCD}$ is a square and $\triangle\text{EDC}$ is an equilateral triangle. Prove that:
$i. AE = BE$,
$ii. \angle\text{DAE}=15^{\circ}.$
Answer
Given: $\text{ABCD}$ is a square in which $AB = BC = CD = DA$.
$\triangle\text{EDC}$ is an equilateral triangle in which $ED = EC = DC$ and $\angle\text{EDC}=\angle\text{DEC}=\angle\text{DCE}=60^{\circ}.$
To prove: $\text{AE = BE}$ and $\angle\text{DAE}=15^{\circ}$
Proof: in $\triangle\text{ADE}$ and $\triangle\text{BCE},$
we have: $\text{AD = BC} [$Sides of a square$]$
$\text{DE = EC} [$Sides of an equilateral triangle$]$
$\angle\text{ADE}=\angle\text{BCE}=90^{\circ}+60^{\circ}=150^{\circ}$
$\therefore\triangle\text{ADE}\cong\triangle\text{BCE}$
i.e., $\text{AE = BE}$
Now, $\angle\text{ADE}=150^{\circ}$
$\text{DA = DC} [$Sides of a square$] $
$\text{DC = DE} [$Sides of an equilateral triangle$]$
So, $\text{DA = DE}$
$\triangle\text{ADE}$ and $\triangle\text{BCE}$ are isosceles triangle.
i.e.,$\angle\text{DAE}=\angle\text{DEA}=\frac{1}{2}(180^{\circ}-150^{\circ})=\frac{30^{\circ}}{2}=15^{\circ}$
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Question 94 Marks
In the adjoining figure, $AD$ and $BE$ are the medians of $\triangle\text{ABC}$ and $DF || BE$. Show that $\text{CF}=\frac{1}{4}\text{AC}.$
Answer
Given: A $\triangle\text{ABC}$ in which $AD$ and BE are the medians. $DF$ is drawn parallel to $BE$.

To prove: $\text{CF =}\frac{1}{2}\text{AC}$
Proof: In $\triangle\text{CBE},$D is the mid point of $BC$ and $DF$ is parallel to $BE$.
The line drawn through the midpoint of one side of a triangle, parallel to another side, intersects the third side at its midpoint.
So,by Mid point Theorem Fis the mid point of $EC$.
$\therefore\text{CF}=\frac{1}{2}\text{EC}$
$=\frac{1}{2}\Big(\frac{1}{2}\text{AC}\Big)$ [BE is the median through $B]$
$=\frac{1}{4}\text{AC}.$
Thus, $\text{CF}=\frac{1}{4}\text{AC}.$
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Question 104 Marks
In the given figure, $\text{ABCD}$ is a quadrilateral in which $AB = AD$ and $BC = DC$. Prove that
$i. AC$ bisects $\angle\text{A}$ and $\angle\text{C},$
$ii. BE = DE$,
$iii. \angle\text{ABC}=\angle\text{ADC}.$
Answer
Given: $\text{ABCD}$ is a quadrilateral in which $AB = AD$ and $BC = DC$
$i.$ In $\triangle\text{ABC}$ and $\triangle\text{ADC},$ we have:
$\text{AB = AD}\ ($Given$)$
$\text{BC = DC}\ ($Given$)$
$\text{AC}$ is common.
i.e., $\triangle\text{ABC}\cong\triangle\text{ADC}\ (\text{SSS}$ congruence rule$)$
$\therefore\angle\text{BAC}=\angle\text{DAC}$ and $\angle\text{BCA}=\angle\text{DCA}\ ($By $\text{C.P.C.T.})$
Thus, $AC$ bisects $\angle\text{A}$ and $\angle\text{C}.$
$ii.$ Now, in $\triangle\text{ABE}$ and $\triangle\text{ADE},$ we have:
$\text{AB = AD}\ ($Given$)$
$\angle\text{BAE}=\angle\text{DAE}\  ($Proven above$)$
$\text{AE}$ is common.
$\therefore\triangle\text{ABE}\cong\triangle\text{ADE}\  (\text{SAS}$ congruence rule$)$
$\Rightarrow\text{BE = DE}\  ($By $\text{C.P.C.T.})$
$iii. \triangle\text{ABC}\cong\triangle\text{ADC}\ ($Proven above$)$
$\therefore\angle\text{ABC}=\angle\text{ADC}\ ($By $\text{C.P.C.T.})$
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Question 114 Marks
The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is $60^\circ$. Find the angles of the parallelogram.
Answer


$\angle\text{DCM}=\angle\text{DCN}+\angle\text{MCN}$
$\Rightarrow90^{\circ}=\angle\text{DCN}+60^{\circ}$
$\Rightarrow\angle\text{DCN}=30^{\circ}$
In $\triangle\text{DCN,}$
$\angle\text{DNC}+\angle\text{DCN}+\angle\text{D}=180^{\circ}$
$\Rightarrow90^{\circ}+30^{\circ}+\angle\text{D}=180^{\circ}$
$\Rightarrow\angle\text{D}=60^{\circ}$
$\Rightarrow\angle\text{B}=\angle\text{D}=60^{\circ}$ (opposite angles of parallelogram are equal)
$\Rightarrow\angle\text{A}=180^{\circ}-\angle\text{B}=180^{\circ}-60^{\circ}=120^{\circ}$
$\Rightarrow\angle\text{C}=\angle\text{A}=120^{\circ}$
Thus, the angles of a parallelogram are $60^\circ , 120^\circ , 60^\circ $ and $120^\circ$.
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Question 124 Marks
In the adjoining figure, $ABCD$ is a parallelogram in which $\angle\text{A}=70^{\circ}.$ Calculate $\angle\text{B},\angle\text{C}$ and $\angle\text{D}.$
Answer

In a parallelogram, opposite angles are equal.
$\therefore\angle\text{A}=\angle\text{C}=72^{\circ}$
 The sum of all the four angles of a parallelogram is $360^\circ$
So, $\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}=360^{\circ}$
$\Rightarrow72^{\circ}+\angle\text{B}+72^{\circ}+\angle\text{D}=360^{\circ}$
$[\because\angle\text{A} = \angle\text{C}]$
$\Rightarrow2\angle\text{B}+144^{\circ}=360^{\circ}$
$[\because\angle\text{B}=\angle\text{D}]$
$\Rightarrow2\angle\text{B}=360^{\circ}-144^{\circ}=216{^\circ}$
$\Rightarrow\angle\text{B}=\frac{216}{2}=108^{\circ}$
$\therefore\angle\text{B}=108^{\circ},\angle\text{C}=72^{\circ}$ and $\angle\text{D}=108^{\circ}.$
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Question 134 Marks
A square is inscribed in an isosceles right triangle so that the square and the triangle have one angle common. Show that the vertex of the square opposite the vertex of the common angle bisects the hypotenuse.
Answer


Let $\triangle\text{ABC}$ be an isosceles right triangle, right-angled at $B$.
$\Rightarrow\text{AB = BC}$
Let $PBSR$ be a square inscribed in $\triangle\text{ABC}$ with common $\angle\text{B}.$
$\Rightarrow\text{PB = BS = SR = RP}$
Now, $\text{AB} - \text{PB = BC} -\text{BS}$
$\Rightarrow\text{AP = CS ...(i)}$
In $\triangle\text{APR}$ and $\triangle\text{CSR}$
$\text{AP = CS}$ $[$from $(i)]$
$\angle\text{APR}=\angle\text{CSR}$ $($Each $90^\circ )$
$\text{PR = SR}$ (sides of a square)
$\therefore\triangle\text{APR}\cong\triangle\text{CSR}$ $($by $SAS$ congruence criterion$)$
$\Rightarrow\text{AR = CR}$ $[C.P.C.T.]$
Thus, point $R$ bisects the hypotenuse $AC$.
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Question 144 Marks
In a rhombus $ABCD$, the altitude from $D$ to the side $AB$ bisects $AB$. Find the angles of the rhombus.
Answer


Let the altitude from $D$ to the side $AB$ bisect $AB$ at point $P$.
Join $BD$.
In $\triangle\text{AMD}$ and $\triangle\text{BMD},$
$\text{AM = BM}$ (M is the mid-point of AB)
$\angle\text{AMD}=\angle\text{BMD}$ $($Each $90^\circ )$
$\text{MD = MD}$ (common)
$\therefore\triangle\text{AMD}\cong\triangle\text{BMD}$ $($by $SAS$ congruence criterion$)$
$\Rightarrow\text{AD = BD}$ $(C.P.C.T.)$
But, $\text{AD = AB}$ (sides of a rhombus)
$\Rightarrow\text{AD = AB = BD}$
$\Rightarrow\triangle\text{ADB}$ is an equilateral triangle.
$\Rightarrow\angle\text{A}=60^{\circ}$
$\Rightarrow\angle\text{C}=\angle\text{A}=60^{\circ}$ (opposite angles are equal)
$\Rightarrow\angle\text{B}=180^{\circ}-\angle\text{A}=180^{\circ}-60^{\circ}=120^{\circ}$
$\angle\text{D}=\angle\text{B}=120^{\circ}$
Hence, in rhombus $ABCD$, $\angle\text{A}=60^{\circ},\angle\text{B}=120^{\circ},\angle\text{C}=60^{\circ}$ and $\angle\text{D}=120^{\circ}.$
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Question 154 Marks
In a rhombus $ABCD$ show that diagonal $AC$ bisects $\angle\text{A}$ as well as $\angle\text{C}$ and diagonal $BD$ bisects $\angle\text{B}$ as well as $\angle\text{D}.$
Answer


In $\triangle\text{ABC}$ and $\triangle\text{ADC},$
$\text{AB = AD}$ (sides of a rhombus are equal)
$\text{BC = CD}$ (sides of a rhombus are equal)
$\text{AC = AC}$ (common)
$\therefore\triangle\text{ABC}\cong\triangle\text{ADC}$ (by $SSS$ congruence criterion)
$\Rightarrow\angle\text{BAC}=\angle\text{DAC}$ and $\angle\text{BCA}=\angle\text{DCA}$ $(C.P.C.T.)$
$\Rightarrow\text{AC}$ bisects $\angle\text{A}$ as well as $\angle\text{C}.$
Similarly,
In $\triangle\text{BAD}$ and $\triangle\text{BCD},$
$\text{AB = BC}$ (sides of a rhombus are equal)
$\text{AD = CD}$ (sides of a rhombus are equal)
$\text{BD = BD}$ (common)
$\therefore\triangle\text{BAD}\cong\triangle\text{BCD}$ (by $SSS$ congruence criterion)
$\Rightarrow\angle\text{ABD}=\angle\text{CBD}$ and $\angle\text{ADB}=\angle\text{CDB}$ $(C.P.C.T.)$
$\Rightarrow\text{BD}$ bisects $\angle\text{B}$ as well as $\angle\text{D}.$
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Question 164 Marks
In the adjoining figure, $PQRS$ is a trapezium in which $PQ || SR$ and $M$ is the midpoint of $PS$.
$A$ line segment $MN || PQ$ meets $QR$ at $N$. Show that $N$ is the midpoint of $QR$.
Answer
Construction: Join diagonal $QS$. Let $QS$ intersect $MN$ at point $O$.

$PQ || SR$ and $MN || PQ \Rightarrow PQ || MN || SR$ By converse of mid-point theorem a line drawn,
through the mid-point of any side of a triangle and parallel to another side bisects the third side.
Now, in $\triangle\text{SPQ}$
$MO || PQ$ and $M$ is the mid-point of $SP$ So, this line will intersect $QS$ at point $O$ and $O$ will be the mid-point of $QS$.
Also, $MN || SR$ Thus, in $\triangle\text{QRS},$ $ON || SR$ and $O$ is the midpoint of line $QS$.
So, by using converse of mid-point theorem, $N$ is the mid-point of $QR$.
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Question 174 Marks
Prove that the sum of all the angles of a quadrilateral is $360^\circ$.
Answer

Let $ABCD$ be a quadrilateral and $\angle1,\angle2,\angle3$ and $\angle4$ are its four angles as shown in the figure.
Join $BD$ which divides $ABCD$ in two triangles, $\triangle\text{ABD}$ and $\triangle\text{BCD}.$ In $\triangle\text{ABD},$ we have: $\angle1+\angle2+\angle\text{A}=180^{\circ}..(\text{i})$ In $\triangle\text{BCD},$
we have: $\angle3+\angle4+\angle\text{C}=180^{\circ}...(\text{ii})$
On adding $(i)$ and $(ii)$, we get: $(\angle1+\angle3)+\angle\text{A}+\angle\text{C}+(\angle4+\angle\text{2})=360^{\circ}$
$\Rightarrow\angle\text{A}+\angle\text{C}+\angle\text{B}+\angle\text{D}=360^{\circ}$
$[\because\angle1+\angle3=\angle\text{B, }\angle4+\angle2=\angle\text{D}]$
$\therefore\angle\text{A}+\angle\text{C}+\angle\text{B}+\angle\text{D}=360^{\circ}$
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Question 184 Marks
In the adjoining figure, $D, E, F$ are the midpoints of the sides $BC, CA$ and $AB$ respectively, of $\triangle\text{ABC}.$ Show that $\angle\text{EDE}=\angle\text{A},\angle\text{DEF}=\angle\text{B}$ and $\angle\text{DFE}=\angle\text{C}.$
Answer
$\triangle\text{ABC}$ is shown below. $D, E$ and $F$ are the midpoints of sides $BC, CA$ and $AB$, respectively.
As $F$ and $E$ are the mid points of sides $AD$ and $AC$ of $\triangle\text{ABC}.$
$\therefore$ $FE || BC$ (By mid point theorem)
Similarly, $DE || FB$ and $FD || AC$.
Therefore, $AFDE, BDEF$ and $DCEF$ are all parallelograms.
In parallelogram $AFDE$, we have:
$\angle\text{A}=\angle\text{EDF}$ (Opposite angle are equal)
In parallelogram $BDEF$, we have:
$\angle\text{B}=\angle\text{DEF}$ (Opposite angles are equal)
In parallelogram $DCEF$, we have:
$\angle\text{C}=\angle\text{DFE}$ (Opposite angles are equal)
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Question 194 Marks
The lengths of the diagonals of a rhombus are $24\ cm$ and $18\ cm$ respectively. Find the length of each side of the rhombus.
Answer
$ABCD$ is a rhombus in which diagonal $AC = 24\ cm$ and $BD = 18\ cm$.
We know that in a rhombus, diagonals bisect each other at right angles.
So in $\triangle\text{AOB}$ $\angle\text{AOB}=90^{\circ}$
$\text{AO}=\frac{1}{2}\text{AC}=\frac{1}{2}\times24=12\text{cm}$ and, $\text{BO}=\frac{1}{2}\text{BD}=\frac{1}{2}\times18=9\text{cm}$
 Now, by Pythagoras Theorem, we have $\text{AB}^{2}=\text{AO}^{2}+\text{OB}^2$
$\Rightarrow\text{AB}^2=12^2+9^2$
$\Rightarrow144+81=225$
$\Rightarrow\text{AB}=\sqrt{225}=15\text{cm}$
SO the length of each side of the rhombus is $15\ cm$.
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