4 Marks Questions - MATHS STD 9 Questions - Vidyadip

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Question 14 Marks

Prove that the quadrilateral formed by the bisectors of the angles of a parallelogram is a rectangle.

Answer

Given Let ABCD be a parallelogram and AP, BR, CR, be are the bisectors of $\angle\text{A},\ \angle\text{B},\ \angle\text{C}$ and $\angle\text{D},$ respectively.To prove Quadrilateral PQRS is a rectangle.
Proof Since, ABCD is a parallelogram, then DC || AB and DA is a transversal.
$\angle\text{A}+\angle\text{B}=180^\circ$
[sum of cointerior angles of a parallelogram is 180°]
$\Rightarrow\ \frac{1}{2}\angle\text{A}+\frac{1}{2}\angle\text{D}=90^\circ$ [dividing both sides by 2]
$\angle\text{PAD}+\angle\text{PDA}=90^\circ$
$\angle\text{APD}=90^\circ$ [since,sum of all angles of a triangle is 180°]
$\Rightarrow\angle\text{SPQ}=90^\circ$ [vertically opposite angles]
$\angle\text{PQR}=90^\circ$
$\angle\text{QRS}=90^\circ$
and
$\angle\text{PSR}=90^\circ$
Thus, PQRS is a quadrilateral whose each angle is 90°.
Hence, PQRS is a rectangle.

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Question 24 Marks

Show that the quadrilateral formed by joining the mid-points of the consecutive sides of a square is also a square.

Answer

Given in a square ABCD, P Q R and S are the mid-points of AB, BC, CD and DA, respectively.
To show PQRS is a square.
Construction Join AC and BD.
Proof Since, ABCD is a square.

$\therefore$ AB = BC = CD = AD
Also, P. Q. R and S are the mid-points of AB, BC, CD and DA
respectively.
Then, in $\Delta\text{ADC}$
and $\text{SR}=\frac{1}{2}=\text{AC}$ [by mid-point theorem]...(i)
In $\Delta\text{ABC},$ PQ || AC
and $\text{PQ}=\frac{1}{2}\text{AC}\ ...(\text{ii})$
From Eqs. (i) and (ii).
SR || PQ and $\text{SR}=\text{PQ}=\frac{1}{2}\text{AC}\ ...(\text{iii})$
Similarly, SP || BD and BD || RQ
$\therefore$ SP || RQ and $\text{SP}=\frac{1}{2}\text{BD}$
and $\text{RQ}=\frac{1}{2}\text{BD}$
$\therefore\ \text{SP}=\text{RQ}=\frac{1}{2}\text{BD}$
Since, diagonals of a square bisect each other at right angle.
$\therefore$ AC = BD
$\text{SP}=\text{RQ}=\frac{1}{2}\text{AC}\ ...(\text{iv})$
From Eqs. (i) and (iv). $\text{SR}=\text{PQ}=\text{SP}=\text{RQ}$ [all side are equal]
Now, in quadrilateral OERF,
OE || FR and OF || ER
$\angle\text{EOF}=\angle\text{ERF}=90^\circ$
Hence, PQRS is a square. Hence proved.

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Question 34 Marks

E is the mid-point of a median AD of $\Delta\text{ABC}$ and BE is produced to meet AC at F. Show that $\text{AF}=\frac{1}{3}\text{AC}.$

Answer

Given: $\Delta\text{ABC}$ in which E is the mid-point of a medien AD of $\Delta\text{ABC}$ and BE is produced to meet AC at F.
show that $\text{AF}=\frac{1}{3}\text{AC}$

to prove: $\text{AF}=\frac{1}{3}\text{AC}$
Construction: Draw DG || BF intersecting AC at G.
Proof: In $\Delta\text{ADG},$ E is the mid-point of AD and EF || DG.
$\therefore$ AF = FG ...(1) [Converse of mid-point theorem]
In $\Delta\text{FBC},$ D is the mid-point of BC and DG || BF.
$\therefore$ FG = GC ...(2) (Converse of mid-point theorem)
From equation (1) and (2), we get
AF = FG = GC ...(3)
But, AC = AF + FG + GC
$\Rightarrow$ AC = AF + AF + AF[Using (3)]
$\Rightarrow$ AC = 3AF
$\Rightarrow\ \text{AF}=\frac{1}{3}\text{AC}$
Hence, proved.

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Question 44 Marks

AB || DE, AB = DE, AC || DF and AC = DF. Prove that BC || EF and BC = EF.

Answer

Given In figure AB || DE and AC || DF, also AB = DE and AC = DFTo prove BC || EF and BC = EF
Proof In quadrilateral ABED, AB || DE and AB = DE
So, ABED is a parallelogram. AD || BE and AD = BE
Now, in quadrilateral ACFD, AC || FD and AC = FD …(i)
Thus, ACFD is a parallelogram.
AD || CF and AD = CF …(ii)
From Eqs. (i) and (ii), AD = BE = CF and CF || BE …(iii)
Now, in quadrilateral BCFE, BE = CF
and BE || CF [from Eq. (iii)]
So, BCFE is a parallelogram. BC = EF and BC || EF. Hence proved.

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Question 54 Marks

E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. Prove that EF || AB and$\text{EF}=\frac{1}{2}(\text{AB}=\text{CD})$
[Hint: Join BE and produce it to meet CD produced at G.]

Answer

(Change D to E in mid points of AD)
Given: A trapezium ABCD in which E and F are repectively the mid -points of the mid poient of the non-parallel sides AD and BC.
to prove: EF || AB and $\text{EF}=\frac{1}{2}(\text{AB}=\text{CD})$
construction: join DF and produce it to intersect AB produced at G.
proof: in $\Delta\text{CFD}$ and $\Delta\text{BFG},$ we have
DC || AB
$\therefore\ \angle\text{C}=\therefore\ \angle\text{3}$ [Alternate interior angles]
CF = BF
$\therefore\ \angle\text{1}=\therefore\ \angle\text{2}$ [Vertically opposite angles]
[so, by ASA critertion of congruence, we have]
$\Delta\text{CFD}\cong\Delta\text{BFG}$
$\therefore$ CD || BG [CPCT]
In $\Delta\text{DAG},$ EF joins mid-points AD and GD repectively
$\therefore$ EF || AG [$\because$ mid-point therem]
$\Rightarrow$ EF || AB
so, $\text{EF}=\frac{1}{2}\text{AG}$ [mid-point therem]
$\text{EF}=\frac{1}{2}(\text{AB}+\text{BG})$
$\Rightarrow\ \text{EF}=\frac{1}{2}(\text{AB}+\text{CD})[\because\ \text{CD}=\text{BG}]$
Hence, proved.

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Question 64 Marks

P is the mid-point of side BC of a parallelogram ABCD such that $\angle\text{BAP}=\angle\text{DAP}$ Prove that AD = 2CD.

Answer

Given in a parallelogram ABCD, P is a mid-point of BC such that $\angle\text{BAP}=\angle\text{DAP}.$
To prove AD = 2CD
Proof Since, ABCD is a parallelogram.
So, AD || BC and AB is transversal, then
$\angle\text{A}+\angle\text{B}=180^\circ$ (sum of cointerior angles is 180°)
$\angle\text{B}=180^\circ-\angle\text{A}\ ...(\text{i})$
in $\Delta\text{ABP}$ $\angle\text{PAB}+2\angle\text{B}+\angle\text{BPA}=180^\circ$[by angle sum property of a triangle)
$\Rightarrow\ \frac{1}{2}\angle\text{A}+180^\circ-\angle\text{A}+\angle\text{BPA}=180^\circ$ [from Eq. (i)]
$\Rightarrow\ \angle\text{BPA}=\frac{\angle\text{A}}{2}=0$
$\Rightarrow\ \angle\text{BPA}=\frac{\angle\text{A}}{2}\ ...\text{(ii)}$
$\Rightarrow\angle\text{BPA}=\angle\text{BAP}$
$\Rightarrow\ \text{AB}=\text{BP}$ [opposite sides of equal angles are equal]
On multiplying both sides by 2, we get
$2\text{AB}=2\text{BP}$
$\Rightarrow\ 2\text{AB}=\text{BP}$ (since Pis the mid-point of BC)
$\Rightarrow\ 2\text{CD}=\text{AD}$
(since, ABCD is a parallelogram, then AB = CD and BC = AD)

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Question 74 Marks

Show that the quadrilateral formed by joining the mid-points of the consecutive sides of a square is also a square.

Answer

Given in a square ABCD, P Q R and S are the mid-points of AB, BC, CD and DA, respectively.
To show PQRS is a square.
Construction Join AC and BD.
Proof Since, ABCD is a square.
$\therefore$ AB = BC = CD = AD
Also, P. Q. R and S are the mid-points of AB, BC, CD and DA
respectively.
Then, in $\Delta\text{ADC}$
and $\text{SR}=\frac{1}{2}=\text{AC}$ [by mid-point theorem]...(i)
In $\Delta\text{ABC},$ PQ || AC
and $\text{PQ}=\frac{1}{2}\text{AC}\ ...(\text{ii})$
From Eqs. (i) and (ii).
SR || PQ and $\text{SR}=\text{PQ}=\frac{1}{2}\text{AC}\ ...(\text{iii})$
Similarly, SP || BD and BD || RQ
$\therefore$ SP || RQ and $\text{SP}=\frac{1}{2}\text{BD}$
and $\text{RQ}=\frac{1}{2}\text{BD}$
$\therefore\ \text{SP}=\text{RQ}=\frac{1}{2}\text{BD}$
Since, diagonals of a square bisect each other at right angle.
$\therefore$ AC = BD
$\text{SP}=\text{RQ}=\frac{1}{2}\text{AC}\ ...(\text{iv})$
From Eqs. (i) and (iv). $\text{SR}=\text{PQ}=\text{SP}=\text{RQ}$ [all side are equal]
Now, in quadrilateral OERF,
OE || FR and OF || ER
$\angle\text{EOF}=\angle\text{ERF}=90^\circ$
Hence, PQRS is a square. Hence proved.

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Question 84 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

Given In isosceles triangle ABC, a square $\Delta\text{DEF}$ is inscribed.To prove CE = BE
Proof In an isosceles $\Delta\text{ABC},\ \angle\text{A}=90^\circ$
and AB=AC …(i)
Since, $\Delta\text{DEF}$ is a square.
AD = AF [all sides of square are equal] … (ii)
On subtracting Eq. (ii) from Eq. (i), we get
AB – AD = AC- AF
BD = CF ….(iii)
Now, in $\Delta\text{CFE}$ and $\Delta\text{BDE}$

BD = CF [from Eq (iii)]
DE = EF [sides of a square]
and $\angle\text{CEF}=\angle\text{BDE}$ [by SAS conguencerule]
$\therefore$ CE = BE [by CPCT]
Hence, vertex E of the square bisects the hypotenuse BC.

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Question 94 Marks

P is the mid-point of the side CD of a parallelogram ABCD. A line through C parallel to PA intersects AB at Q and DA produced at R. Prove that DA = AR and CQ = Q

Answer

Given in a perallelogram ABCD, P is the mid-point of DC.
To prove DA = AR and CQ = QR
proof ABCD is a parallelogram

$\therefore$ BC = AD and BC || AD
Also DC = AB and DC || AB
Since, P is the mid-point of DC.
$\therefore\ \text{DP}=\text{PC}=\frac{1}{2}\text{DC}$
now, QC || AP and PC || AQ
So, APCQ is a parallelogram.
$\therefore\ \text{AQ}=\text{PC}=\frac{1}{2}\text{DC} $
$=\frac{1}{2}\text{AB}=\text{BQ}$ [$\therefore$ DC = AB ...(i)]
Now, in $\Delta\text{AQR}$ and $\Delta\text{BQC},$ AQ = BQ [from eq. ...(i)]
$\angle\text{AQR}=\angle\text{BQR}$ [veetically opposite angles]
and $\angle\text{ARQ}=\angle\text{BCQ}$ [Alternate interior angaes]
$\therefore\ \Delta\text{AQR}=\Delta\text{BQR}$ [by AAS congruence rile]
$\therefore$ AR = BC [ by CPCT rule]
But AR =DA
$\therefore$ AR = DA
Also, CQ = QR [by CPCt rule]
hence proved.

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Question 104 Marks

A diagonal of a parallelogram bisects one of its angles. Show that it is a rhombus.

Answer

ABCD ois a parallelogram and diagonal AC bisect $\angle\text{A}.$ we have to show that ABCD is a ehombus.

$\angle1=\angle2\ ...(1)$ [$\because$ AC bisect $\angle\text{A}$] $\angle2=\angle4\ ...(2)$ [Alt. interior angle] From (1) and (2), we get $\angle1=\angle4$ Noe, in $\Delta\text{ABC},$ we have $\angle1=\angle4$ [proved adove] $\therefore$ BC = AB [$\because$ side. Opp. to equal $\angle\text{S}$ are equal] Also, AB = DC and AD = BC [$\because$ Opposite sides of a parallelogram are equal] So, ABCD is a parallelogram in which its sides AB = BC = CD = AD. Hence, ABCD is arhombus.

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Question 114 Marks

P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and $\text{AC}\bot\text{BD}.$ Prove that PQRS is a square.

Answer

Given: A quadrilateral ABCD is which AC = BD and $\text{AC}\bot\text{BD}.$ P, Q, R and S respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD.

To prove: PQRS is a pquare.
proof: parallelogram PQRS is a rectangle.
[same as in Q4]
$\text{PQ}=\frac{1}{2}\text{AC}\ ...(\text{i})$ [proved as in Q4]
PS joins mid-points of sides AB and AD respectively.
$\text{PS}=\frac{1}{2}\text{BD}...(\text{ii})$ [mid-point theorem]
But, $\text{AC}=\text{BD}\ ...(\text{iii})$ [Given]
From (i), (ii) and (iii), we get
PS = PQ
since adjacent sides of a rectangle are equal.
$\Rightarrow$ Rectangle PQRS is a square.

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Question 124 Marks

Through A, B and C, lines RQ, PR and QP have been drawn, respectively parallel to sides BC, CA and AB of a $\Delta\text{ABC}$ as shown Show that $\text{BC}=\frac{1}{2}$ QR.

Answer

Given In $\Delta\text{ABC}$ PQ || AB and PR || AC and RQ || BC.
To show $\text{BC}\frac{1}{2}\text{QR}$
Proof In quadrilateral BCAR, BR || CA and BC|| RA
So, quadrilateral, BCAR is a parallelogram.
BC = AR …(i)
Now, in quadrilateral BCQA, BC || AQ
and AB||QC
So, quadrilateral BCQA is a parallelogram,
BC = AQ …(ii)
On adding Eqs. (i) and (ii), we get
2BC = AR+ AQ
$\Rightarrow$ 2 BC = RQ
$\Rightarrow\ \text{BC}=\frac{1}{2}\text{QR}$
Now, BEDF is a quadrilateral, in which $\Delta\text{BED}=\Delta\text{BFD}=90^\circ$
$\Delta\text{FSE}=360^\circ-(\Delta\text{FDE}+\Delta\text{BED})=360^\circ-(60^\circ+90^\circ+90^\circ)$
= 360° - 240° = 120°

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Question 134 Marks

In a parallelogram ABCD, AB = 10cm and AD = 6cm. The bisector of $\angle\text{A}$meets DC in E. AE and BC produced meet at F. Find the length of CF.

Answer

ABCD is a parallelogram. in wgich AB = 10cm and AD = 6cm. the bisector of$\angle\text{A}$ meets DC in E. AE and BC produced meer at F.

$\angle\text{BAE}=\angle\text{EAD}\ ...(\text{i})$ [$\therefore$ bisect of $\angle\text{A}$] $\angle\text{EAD}=\angle\text{EFB}\ ...(\text{ii})$[Alt. $\angle\text{s}$] $\Rightarrow\ \angle\text{BAE}=\angle\text{EFB}$ [From (i) and (ii)] $\Rightarrow\ \text{BF}=10\text{cm}[\because\text{AB}=10\text{cm}]$ $\Rightarrow\ \text{BF}+\text{CF}=10\text{cm}\ \Rightarrow\ 6\text{cm}$ [$\because\ \text{BC}=\text{AD}=6\text{cm}$ opposite sides of a || gm] $\Rightarrow\ \text{CF}=10-6\text{cm}=4\text{cm}$

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Question 144 Marks

E is the mid-point of the side AD of the trapezium ABCD with AB || DC. A line through E drawn parallel to AB intersect BC at F. Show that F is the mid-point of BC.
[Hint: Join AC]

Answer

Given: A trapezium ABCD in which AB || CD and E is mid-point of the side AB. Also, EF ||AB. To prove: F is the mid-point of BC.

Construction: join AC Which intersect EF at o. Proof: in $\Delta\text{ADC},$ E is the mid-point of AD and EF || DC. [$\therefore$ EF || AB and DC || AB $\Rightarrow$ AB || EF || DC] $\therefore$ O is the mid- point of AC. [Converse of mid- point theorem] Now, in $\Delta\text{CAB0},$ O is mid- point of AC and OF || AB. $\Rightarrow$ OF bisects BC. [Converse of mid-point threorem] Or F is the mid- point of bc. Hence, proved.

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Question 154 Marks

ABCD is a rectangle in which diagonal BD bisects $\angle\text{B}.$ Show that ABCD is a square.

Answer

Given: A rectangle ABCD in which diagonal BD bisects $\angle\text{B}$

To prove: ABCD is a square.
Proof: DC || AB [Opposite sides of a rectangle are parallel]
$\Rightarrow\ \angle4=\angle1\ ...(\text{i})$ [Alternate interior angles]
Similarly, $\angle3=\angle2\ ...(\text{ii})$ [Alternate interior angles]
And $\angle1=\angle2\ ...(\text{iii})$ [Given]
From equation (1), (2) and (3), we get
$\angle3=\angle4$
In $\Delta\text{BAD}$ and $\Delta\text{BDC}$ we have
$\angle1=\angle2$[Given]
BD = BD[Common side)
$\angle3=\angle4$ [proved above]
So, By ASA criterion of congruence, we have
$\Delta\text{BAD}\cong\Delta\text{BCD}$
$\therefore\ \text{AB}=\text{BC}$ [CPCT]
As, adjacent sides of rectangle are equal. So, ABCD is a square.

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Question 164 Marks

P and Q are the mid-points of the opposite sides AB and CD of a parallelogram ABCD. AQ intersects DP at S and BQ intersects CP at R. Show that PRQS is a parallelogram.

Answer

Given: A quadrilateral ABCD in which P and Q are the mid-points of the sides AB and CD respectively. AQ intersect DP at S and BQ intersect CP at R.

To prove: PRQS is a parallelogram.
Proof: DC||AB [$\therefore$ Opposite sides of a parallelogram are parallel]
$\Rightarrow$ AP || QC
DC = AB [$\because$ Opposite sides of a parallelogram are equal]
$\Rightarrow\ \frac{1}{2}\text{DC}=\frac{1}{2}\text{AB}$
$\Rightarrow$ QC = AP [$\because$ P is the mid-point of AB and Q is mid-point of CD]
$\Rightarrow$ APCQ is a parallelogram. [$\because$ AP || CQ and QC = AP]
$\therefore$ AQ||PC [$\because$ Opposite sides of a || gm are parallelogram)
$\Rightarrow$ SQ || PR
Similarly, SP || QR
$\therefore$ Qudrilateral PRQS is a parallogram.
Hence. proved

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Question 174 Marks

ABCD is a rhombus in which altitude from D to side AB bisects AB. Find the angles of the rhombus.

Answer

Let sides of a rhombus be AB = BC = CD = DA = x New, join DB

in $\Delta\text{ALD}$ and $\Delta\text{BLD},\angle\text{DLA}=\angle\text{DLB}=90^\circ$ [since, DL is a perpendicular bisector of AB] $\text{AL}=\text{BL}=\frac{\text{x}}{2}$ and DL = DL [common side] $\therefore\ \Delta\text{ALD}=\Delta\text{BLD}$ [by SAS cogeuence rule] AD = BD [by CPCt] Now, in $\Delta\text{ADB},$ AD = AB = DB = x Then, $\Delta\text{ADB}$ is an equilateral triangle. $\therefore\ \angle\text{C}=\angle\text{BCD}=\angle\text{DBC}=60^\circ$ Similarly, $\Delta\text{DBC}$ is an equilateral triangle. $\therefore\ \angle\text{C}=\angle\text{BDC}=\angle\text{DBC}=60^\circ$ Also, $\angle\text{A}=\angle\text{C}$ $\therefore\ \angle\text{D}=\angle\text{B}=180^\circ-60^\circ=120^\circ$ [since, sum of interior angles is 180°]

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Question 184 Marks

ABCD is a quadrilateral in which AB || DC and AD = BC. Prove that $\angle\text{A}=\angle\text{B}$ and $\angle\text{C}=\angle\text{D}.$

Answer

ABCD is a quadrilaateral such that AB || DC and AD = BC

Construction Extend AB to E and draw a line CE parallet to AD.
proos since, AD || CE and transverrsal AE cuts then at A and E repectively.
$\therefore\angle\text{A}+\angle\text{E}=180^\circ$ [since, sum of cointerios angles is 180°]
$\Rightarrow\ \angle\text{A}=180^\circ-\angle\text{E}\ ...(\text{i})$
So, quadrilateral AECD is a parallelogram.
$\Rightarrow\ \text{AD}=\text{CE}\Rightarrow\text{BC}=\text{CE}$ [$\because\ \text{AD}=\text{BC}$ given]
Now, in $\Delta\text{BCE}$ $\text{CE}=\text{BC}$ [proved above]
$\Rightarrow\ \angle\text{CBE}=\angle\text{CEB}$
[opposite angles of equal side are equal]
$\Rightarrow\ 180^\circ-\angle\text{B}=\angle\text{E}$ $[\because\angle\text{B}+\angle\text{CBE}=180^\circ]$
$\Rightarrow\ 180^\circ-\angle\text{E}=\angle\text{B}\ ...(\text{ii})$
From Eqs. (i) and (ii) $\angle\text{A}=\angle\text{B}$ Hence proved.

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Question 194 Marks

Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ Show that AC and PQ bisect each other.

Answer

Given ABCD is a parallelogram and AP = CQ
To show AC and PQ bisect each other.

proof in $\Delta\text{AMP}$ and $\Delta\text{CMQ}$
$\angle\text{MAP}=\angle\text{MCQ}$ [altemate inerior angles]
$\text{AP}=\text{CQ}$ [given]
and $\angle\text{APM}=\angle\text{CQM}$ [alternate interior angles]
$\therefore\ \Delta\text{AMP}=\Delta\text{CMQ}$ [by ASA congruence rule]
$\Rightarrow\ \text{AM}=\text{CM}$ [by CPCT rule]
and $\text{PM}=\text{MQ}$ [by CPCT rule]
Hence, AC and PQ bisech each other. [Hence proved.]

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Question 204 Marks

The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 60º. Find the angles of the parallelogram.

Answer

Let the parallelogram be ABCD, in which $\angle\text{ADC}$ and $\angle\text{ABC}$ are obtuse angles. Now, DE and DF are two altitudes of parallelogram and angle between them is 60°.

Now, BEDF is a quadrilateral, in which $\angle\text{BED}=\angle\text{BFD}=90^\circ$ $\therefore\ \angle\text{FBE}=360^\circ(\angle\text{FDE}+\angle\text{BED}+\angle\text{BFD)}$ $=360^\circ-(60^\circ+90^\circ+90^\circ)$ $=360^\circ-240^\circ=120^\circ$ Since, ABCD is a parallelogram. $\therefore\ \angle\text{ADC}=120^\circ$ Now, $\angle\text{A}+\angle\text{B}=180^\circ$ [sum of two cointerior angles is 180°] $\therefore\ \angle\text{A}=180^\circ-\angle\text{B}$ $5=180^\circ-120^\circ$ $[\because\angle\text{FBE}=\angle\text{B}]$ $\Rightarrow\ \angle\text{A}=60^\circ$ Also, $\angle\text{C}=\angle\text{A}=60^\circ$ Hence, angles of the parallelogram are 60°, 120°, 60°, and 120°, respevtively.

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Question 214 Marks

P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD such that$\text{AC}\bot\text{BD}.$ Prove that PQRS is a rectangle.

Answer

Given In quadrilateral ABCD, P, O, S and S are the mid-points of the sides AB, BC, CD and DA, respectively. Also,$\text{AC}\bot\text{BD}$ To prove PQRS is a rectangle. Proof Since, $\text{AC}\bot\text{BD}$ $\angle\text{COD}=\angle\text{AOD}=\angle\text{AOB}=\angle\text{COB}=90^\circ$ in $\Delta\text{ADC},$ S and R are the mid-points od AD and DC repectively, then by mid-point theorem SR || AC and $\text{SR}=\frac{1}{2}\text{AC}\ ...(\text{i})$

in $\Delta\text{ABC},$ P and Q are the mid-points of AB and BC repectively, then by mid-point therem PQ || AC and $\text{PQ}=\frac{1}{2}\text{AC}\ ...(\text{ii})$ From eqs. (i) and (ii). PQ || SR and $\text{PQ}=\text{SR}=\frac{1}{2}\text{AC}\ ...(\text{iii})$ Similarly, SP || RQ and $\text{SP}=\text{RQ}=\frac{1}{2}\text{BD}\ ...(\text{iv})$ Now, inquadrilaterral EOFR. OE || FR, OF|| ER $\therefore\angle\text{EOF}=\angle\text{ERF}=90^\circ$ $[\because\angle\text{COD}=90^\circ]\Rightarrow\angle\text{EOF}=90^\circ]\ ...(\text{v})$ So, PQRS is a rectangle. Hence proved.

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Question 224 Marks

Prove that the line joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides of the trapezium.

Answer

Given Let ABCD be a trapezium in which AB|| DC and let M and N be the mid-points of the diagonals AC and BD, respectively.

To prove MN || AB || CD
Consttruction join CN and Produce it to meet AB at E.
DN = BN [since, N is mid-point of BD]
$\angle\text{DCN}=\angle\text{BEN}$ [alternate interior angles]
and $\angle\text{CDN}=\angle\text{EBN}$ [alternate interior angles]
$\therefore\ \Delta\text{CDN}\cong\Delta\text{EBN}$ [by CPCT rule]
Thus, in $\Delta\text{CAE},$ the points M and N are the mid-points of AC and CE, resoectivrly
$\therefore$ Mn || AE [by mid-point therem]
$\Rightarrow$ Mn || AB || CD Hence, proved

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Question 234 Marks

P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. Prove that PQRS is a rhombus.

Answer

Given In a quadrilateral ABCD, P, Q, R and S are the mid-points of sides AB, BC, CD and DA, respectively. Also, AC = BD To prove PQRS is a rhombus.

Proof In $\Delta\text{ADC},$ S and R are the mid-points of AD and DC respectively. Then, by mid-point theorem.
SR || AC and $\text{SR}=\frac{1}{2}\text{AC}\ ...(\text{i})$
In $\Delta\text{ABC},$ P and Q are the mid-points of AB and BC respectively. Then, by mid-point theorem.
PQ || AC and $\text{PQ}=\frac{1}{2}\ ...(\text{ii})$
From Eqs. (i) and (ii). $\text{SR}=\text{PQ}=\frac{1}{2}\text{AC}\ ...(\text{iii})$
Similarly, in $\Delta\text{BCD}$ RQ|| BD and $\text{RQ}=\frac{1}{2}\text{BD}\ ...(\text{iv})$
and in $\Delta\text{BAD}$ SP || BD and $\text{SP}=\frac{1}{2}\text{BD}\ ...(\text{v})$
From Eqs. (iv) and (v). $\text{SP}=\text{RQ}=\frac{1}{2}\text{BD}=\frac{1}{2}\text{AC}$ (given, AC = BD]...(vi)
From Egs. (ii) and (vi). SR = PQ = SP = RQ
It shows that all sides of a quadrilateral PQRS are equal. Hence, PQRS is a rhombus.
Hence proved.

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Question 244 Marks

P and Q are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. Show that PQ is bisected at O.

Answer

ABCD is a parallelogram. Its diagonal AC and BD bisect each other at O. PQ passes through the point of intersection O of its diagonal AC and BD.

In $\Delta\text{AOP}$ and $\Delta\text{COQ},$ we have
$\angle3=\angle4$ [Alternate int. $\angle\text{s}$]
OA || OC[Diagonais of a || gm bisects each other]
$\angle1=\angle2$ [vertically opposite angles]
$\therefore\ \Delta\text{AOP}\cong\Delta\text{COQ}$ [By ASA Congruence rule]
So, OP [CPCT]
Hence, PQ is bisected at O.

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Question 254 Marks

D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. Prove that by joining these mid-points D, E and F, the triangles ABC is divided into four congruent triangles.

Answer

Given In a $\Delta\text{ABC}$ D, E and F are respectively the mid-points of the sides AB, BC and CA. To prove $\Delta\text{ABC}$ is divided into four congruent triangles. Proof Since, ABC is a triangle and D, E and F are the mid-points of sides AB, BC and CA, respectively.

Then, $\text{AD}=\text{BD}=\frac{1}{2}\text{AB},\ \text{BE}=\text{EC}=\frac{1}{2}\text{BC}$
and $\text{AF}=\text{CF}=\frac{1}{2}\text{AC}$ Now, using the mid -point therem, Ef || AB and $\text{EF}=\frac{1}{2}\text{AB}=\text{AD}=\text{BD}$ ED || AC and $\text{ED}=\frac{1}{2}\text{AC}=\text{AF}=\text{CF}$ and DF || BC and $\text{DF}=\frac{1}{2}\text{BC}=\text{BE}=\text{CE}$ in $\Delta\text{ADF}$ and $\Delta\text{EFD,}$ AD = EF AF = DE and DE = FD [common] $\therefore\ \Delta\text{ADF}\cong\Delta\text{EFD}$ [by SSS congruence rule] Similarly, $\Delta\text{DEF}\cong\Delta\text{EDB}$ and $\Delta\text{DEF}\cong\Delta\text{CFE}$ So, $\Delta\text{ACB}$ is divided into four congruent triangles. Hence prodev.

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Question 264 Marks

E and F are points on diagonal AC of a parallelogram ABCD such that AE = CF. Show that BFDE is a parallelogram.

Answer

Given: A parallelogram ABCD: E and F are points of diagonal AC of parallelpgram ABCD such that AE = CF.

To prove: BFDE is parallelogram. Proof: ABCD is a parallelogram.$\therefore$ OD = OB...(1) [$\therefore$ Diagonals of parallelogram bisect each other]
OA = OC...(2)($\because$ Diagonals of parallelogram bisect each other]
AE = CF...(3)[Given]
Subtracting (3) from (2), we get
OA - AE = OC - CF
$\Rightarrow$ OE = OF...(4)
$\therefore$ BFDE is parallelogram. [$\therefore$ OD = OB and OE = OF]
Hence, proved.

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