4 Marks Questions

Question 14 Marks

Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Answer

Let us consider a quadrilateral ABCD in which the diagonals AC and BD intersect each other at O. It is given that the diagonals of ABCD are equal and bisect other at right angles. Therefore, AC = BD, OA = OC, OB = OD, and $\angle\text{AOB}=\angle\text{BOC}=\angle\text{COD}=\angle\text{AOD}=90^{0}$. To prove ABCD is a square, we have prov that ABCD is a parallelogram, AB = BC = CD = AD, and one of its interior angles is 90º.
In $\triangle\text{AOB}$ and $\triangle\text{COD}$
AO = CO (Diagonals bisect each other)
OB = OD (Diagonals bisect each other)
$\angle\text{AOB}=\angle\text{COD}$ (Vertically opposite angles)
$\therefore \triangle\text{AOB}\cong\triangle\text{COD}$ (SAS congruence rule)
AB = CD (By CPCT) ... (i)
However, these are alternate interior angles for line AB and CD and alternate interior angles are equal to each other only when the two lines are parallel.
AB || CD ... (ii)
From equations (i) and (ii), we obtain
ABCD is a parallelogram.
In $\triangle\text{AOD}$ and $\triangle\text{COD}$
AO = CO (Diagonals bisect each other)
$\angle\text{AOD}=\angle\text{COD}$ (Given that each is 90º)
OD = OD (Common)
$\therefore \triangle\text{AOD}\cong\triangle\text{COD}$ (SAS congruence rule)
AD = DC ... (iii)
However, AD = BC and AB = CD (Opposite sides of parallelogram ABCD)
AB = BC = CD = DA
Therefore, all the sides of quadrilateral ABCD are equal to each other.
In $\triangle\text{ADC}$ and $\triangle\text{BCD}$
AD = BC (Already proved)
AC = BD (Given)
DC = CD (Common)
$\therefore \triangle\text{ADC}\cong\triangle\text{BCD}$ (SSS Congruence rule)
However, $\therefore \angle\text{ADC}+\angle\text{BCD}=180^{0}$(Co-interior angles)
$ \angle\text{ADC}+\angle\text{ADC}=180^{0}$
$2 \angle\text{ADC}=180^{0}$
$ \angle\text{ADC}=90^{0}$
One of the interior angles of quadrilateral ABCD is a right angle.
Thus, we have obtained that ABCD is a parallelogram, AB = BC = CD = AD and one of its interior angles is 90º. Therefore, ABCD is a square.

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Question 24 Marks

ABCD is a rhombus and P, Q, R and S are the mid - points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.

Answer

Given, ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. To Prove, PQRS is a rectangle. Construction, AC and BD are joined. Proof, In $\angle\text{DRS}$ and $\angle\text{PBQ}$ DS = BQ (Halves of the opposite sides of the rhombus)

$∠\text{SDR} = ∠\text{QBP}$ (Opposite angles of the rhombus)

DR = BP (Halves of the opposite sides of the rhmbus) Thus, $Δ\text{DRS} ≅ Δ\text{BPQ} $ by SAS congruence condition. RS = PQ by CPCT --- (i) In $\angle\text{DRS}$ and $\angle\text{PBQ}$ RC = PA (Halves of the opposite sides of the rhombus) CQ = AS (Halves of the opposite sides of the rhombus) Thus, by SAS congr$Δ\text{QCR} ≅ Δ\text{SAP}$uence condition. RQ = SP by CPCT --- (ii) Now, In $\angle\text{CDE}$ R and Q are the mid points of CD and BC respectively. ⇒ QR || BD also, P and S are the mid points of AD and AB respectively. ⇒ PS || BD ⇒ QR || PS Thus, PQRS is a parallelogram. Also,$ ∠\text{PQR} = 90°$ Now, In $\text{PQRS},$ RS = PQ and RQ = SP from (i) and (ii) ∠Q = 90° Thus, PQRS is a rectangle.

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Question 34 Marks

ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that,

D is the mid - point of AC
$\text{MD}\bot\text{AC}$
$\text{CM}=\text{MA}=\frac{1}{2}\text{AB}$

Answer

In $\angle\text{ABC},$ It is given that M is the mid - point of AB and MD || BC. Therefore, D is the mid-point of AC. (Converse of mid-point theorem) As DM || CB and AC is a transversal line for them, therefore, $∠\text{MDC} + ∠\text{DCB} = 180º$ (Co-interior angles) $∠\text{MDC} + 90º = 180º$ $∠\text{MDC} = 90º$ $\text{MD}\bot\text{AC}$

In $\angle\text{ADM}$ and $Δ\text{CMD,}$ AD = CD (D is the mid - point of side AC) $∠\text{ADM} = ∠\text{CDM}$ (Each 90º) DM = DM (Common) $Δ\text{AMD} ≅ Δ\text{CMD}$ (By SAS congruence rule) Therefore, AM = CM (By CPCT) However, AM = AB (M is the mid-point of AB) Therefore, it can be said that, $\text{CM}=\text{MA}=\frac{1}{2}\text{AB}$.

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Question 44 Marks

ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

Answer

Given, ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Construction, AC and BD are joined. To Prove, PQRS is a rhombus. Proof, In $\angle\text{ABC}$ P and Q are the mid-points of AB and BC respectively Thus, PQ || AC and PQ $=\frac{1}{2}\text{AC}\ ...(\text{i})$ In $\angle\text{ADC}$ SR || AC and SR $=\frac{1}{2}\text{AD}...(\text{ii})$ So, PQ || SR and PQ = SR As in quadrilateral PQRS one pair of opposite sides is equal and parallel to each other, so, it is a parallelogram. PS || QR and PS = QR (Opposite sides of parallelogram) ... (iii) Now, In $\angle\text{BDC}$ Q and R are mid points of side BC and CD respectively. Thus, QR || BD and QR $=\frac{1}{2}\text{BD}...(\text{iV})$ AC = BD (Diagonals of a rectangle are equal) ... (v) From equations (i), (ii), (iii), (iv) and (v), PQ = QR = SR = PS So, PQRS is a rhombus.

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Question 54 Marks

In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ Show that:

$\triangle\text{APD}\cong\triangle\text{CQB}$
AP = CQ
$\triangle\text{AQD}\cong\triangle\text{CPB}$
APCQ is a parallelogram.

Answer

In $\triangle\text{APD}$ and $\triangle\text{CQB}$

$\angle\text{ADP}=\angle\text{CBQ}$ (Alternate interior angles for BC || AD)

AD = CB (Opposite sides of parallelogram ABCD)

DP = BQ (Given)

$\Rightarrow \triangle\text{APQ}\cong\triangle\text{CQB}$ (Using SAS congruence rule)

As we had observed that $\Rightarrow \triangle\text{APQ}\cong\triangle\text{CQB}$

AP = CQ (CPCT)

In $\triangle\text{AQB}$ and $\triangle\text{CQB}$

$\angle\text{ABQ}=\angle\text{CPD}$ (Alternate interior angles for AB || CD)

AB = CD (Opposite sides of parallelogram ABCD)

BQ = DP (Given)

$\Rightarrow \triangle\text{APQ}\cong\triangle\text{CQB}$ (Using SAS congruence rule)

As we had observed that $\Rightarrow \triangle\text{APQ}\cong\triangle\text{CQB}$

AQ = CP (CPCT)

From the result obtained in (ii) and (iv),

AQ = CP and

AP = CQ

Since opposite sides in quadrilateral APCQ are equal to each other, APCQ is a parallelogram.

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Question 64 Marks

Diagonal AC of a parallelogram ABCD bisects $\angle\text{A}$ . Show that

It bisects $\angle\text{C}$ also,
ABCD is a rhombus.

Answer

In $\angle\text{ADC}$ and $\angle\text{CBA}$ AD = CB (Opposite sides of a ||gm) DC = BA (Opposite sides of a ||gm) AC = CA (Common) Therefore, $\triangle\text{ADC}\cong\triangle\text{CBA}$ by SSS congruence condition. Thus, $\angle\text{ACD}=\angle\text{CAD}$ by CPCT $\angle\text{CAB}=\angle\text{BCA}$ (Given) $\Rightarrow \angle\text{ACD} = \angle\text{BCA}$ Thus, AC bisects also, $\angle\text{ACD}=\angle\text{CAD}$ (Proved) ⇒ AD = CD (Opposite sides of equal angles of a triangle are equal) Also, AB = BC = CD = DA (Opposite sides of a || gm) Thus, ABCD is a rhombus.

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Question 74 Marks

ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F. Show that, F is the mid - point of BC.

Answer

By converse of mid-point theorem, we know that a line drawn through the mid-point of any side of a triangle and parallel to another side, bisects the third side. In $\angle\text{ABD}$ EF || AB and E is the mid-point of AD. Therefore, G will be the mid-point of DB. As EF || AB and AB || CD, EF || CD (Two lines parallel to the same line are parallel to each other) In $\angle\text{ABD}$ GF || CD and G is the mid-point of line BD. Therefore, by using converse of mid - point theorem, F is the mid-point of BC.

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Question 84 Marks

If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Answer

Let ABCD be a parallelogram. To show that ABCD is a rectangle, we have to prove that one of its interior angles is 90º.
In $\triangle \text{ABC}$ and $\triangle \text{DCB}$
AB = DC (Opposite sides of a parallelogram are equal)
BC = BC (Common)
AC = DB (Given)
$\therefore \triangle\text{ABC}\cong\triangle\text{DCB}$
$\Rightarrow \angle\text{ABC}=\angle\text{DCB}$
It is known that the sum of the measures of angles on the same side of transversal is 180º.
$\angle\text{ABC}+\angle\text{DCB}=180^{0}$
$\Rightarrow \angle\text{ABC}+\angle\text{ABC}=180^{0}$
$\Rightarrow 2\angle\text{ABC}=180^{0}$
$\Rightarrow \angle\text{ABC}=90^{0}$
Since ABCD is a parallelogram and one of its interior angles is 90º, ABCD is a rectangle.

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Question 94 Marks

ABCD is a rectangle in which diagonal AC bisects $\angle\text{A}$ as well as $\angle\text{C}.$ Show that:

ABCD is a square.
Diagonal BD bisects $\angle\text{B}$ as well as $\angle\text{D}.$

Answer

i. $\angle\text{DAC}=\angle\text{DCA}$ (AC bisects as well as)
⇒ AD = CD (Sides opposite to equal angles of a triangle are equal) also, CD = AB (Opposite sides of a rectangle)
Therefore, AB = BC = CD = AD
Thus, ABCD is a square.
ii. In $\triangle\text{BCD}$
BC = CD
Angles opposite to equal sides are equal.
also, $ \angle\text{CDB}=\angle\text{ABD}$ (Alternate interior angles)
Thus, BD bisects. Now,
$ \angle\text{CDB}=\angle\text{ABD}$
$ \angle\text{CDB}=\angle\text{ADB}$
Thus, BD bisects $\angle\text{D}.$

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Question 104 Marks

Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

Answer

Let ABCD is a quadrilateral in which P, Q, R, and S are the mid-points of sides AB, BC, CD, and DA respectively. Join PQ, QR, RS, SP, and BD. In S and P are the mid - points of AD and AB respectively. Therefore, by using mid-point theorem, it can be said that SP || BD and SP $=\frac{1}{2}\text{BD}...(\text{i})$ Similarly in $\angle\text{ABD}$ QR || BD and QR $=\frac{1}{2}\text{BD}...(\text{ii})$ From equations (i) and (ii), we obtain SP || QR and SP = QR In quadrilateral SPQR, one pair of opposite sides is equal and parallel to each other. Therefore, SPQR is a parallelogram. We know that diagonals of a parallelogram bisect each other. Hence, PR and QS bisect each other.

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Question 114 Marks

In $\triangle\text{ABC}$ and $\triangle\text{DEF},$ AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively. Show that

Quadrilateral ABED is a parallelogram.
Quadrilateral BEFC is a parallelogram.
AD || CF and AD = CF
quadrilateral ACFD is a parallelogram.
AC = DF
$\triangle\text{ABC}\cong\triangle\text{DEF}.$

Answer

It is given that AB = DE and AB || DE.

If two opposite sides of a quadrilateral are equal and parallel to each other, then it will be a parallelogram.

Therefore, quadrilateral ABED is a parallelogram.

Again, BC = EF and BC || EF

Therefore, quadrilateral BCEF is a parallelogram.

As we had observed that ABED and BEFC are parallelograms, therefore

AD = BE and AD || BE

(Opposite sides of a parallelogram are equal and parallel)

And, BE = CF and BE || CF

(Opposite sides of a parallelogram are equal and parallel)

AD = CF and AD || CF

As we had observed that one pair of opposite sides (AD and CF) of quadrilateral ACFD are equal and parallel to each other, therefore, it is a parallelogram.
As ACFD is a parallelogram, therefore, the pair of opposite sides will be equal and parallel to each other.

AC || DF and AC = DF

$\therefore \triangle\text{ABC}\cong\triangle\text{DEF}$

AB = DE (Given)

BC = EF (Given)

AC = DF (ACFD is a parallelogram)

$\therefore \triangle\text{ABC}\cong\triangle\text{DEF}$ (By SSS congruence rule).

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Question 124 Marks

ABCD is a rhombus. 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

Let us join AC. In $\triangle\text{ABC,}$ BC = AB (Sides of a rhombus are equal to each other) $\therefore \angle1=\angle2$ (Angles opposite to equal sides of a triangle are equal) However, $ \angle1=\angle3$ (Alternate interior angles for parallel lines AB and CD) $ \Rightarrow \angle2=\angle3$ Therefore, $ \Rightarrow \angle2=\angle4$ AC bisects $\angle\text{C}.$ Also, (Alternate interior angles for || lines BC and DA) $ \Rightarrow \angle1=\angle4$ Therefore, AC bisects $\angle\text{A}.$ Similarly, it can be proved that BD bisects $\angle\text{B}$ and $\angle\text{D}$ as well.

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Question 134 Marks

In the given figure, ABCD and AEFG are two parallelograms. If $\angle\text{C}= 58^\circ,$ find $\angle\text{F}.$

Answer

ABCD and AEFG are two parallelograms as shown below:

Since ABCD is a parallelogram, with $\angle\text{C}=58^\circ$ We know that the opposite angles of a parallelogram are equal. Therefore, $\angle\text{A}=\angle\text{C}$ $\angle\text{A}= 58^\circ,$ Similarly, AEFG is a parallelogram, with $\angle\text{A}= 58^\circ,$ We know that the opposite angles of a parallelogram are equal. Therefore, $\angle\text{F}=\angle\text{C}$ $\angle\text{F}= 58^\circ$ Hence, the required measure for $\angle\text{F}$ is 58º.

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Question 144 Marks

In a $\triangle\text{ABC},$ BM and CN are perpendiculars from B and C respectively on any line passing through A. If L is the mid-point of BC, prove that ML = NL.

Answer

Given that, In $\triangle\text{BLM}$ and $\triangle\text{CLN}$ $\angle\text{BML}=\angle\text{CNL}=90^\circ$ $\text{BL}=\text{CL}$ [L is the mid-point of BC] $\angle\text{MLB}=\angle\text{NLC}$ [Vertically opposite angle] $\therefore\triangle\text{BLM}=\triangle\text{CLN}$ $\therefore\text{LM}=\text{LN}$ [corresponding parts of congruent triangles]

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Question 154 Marks

In the adjoining figure, ABCD is a parallelogram and E is the midpoint of side BC. If DEand 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 164 Marks

ABCD is a kite having AB = AD and BC = CD. Prove that the figure found by joining the mid points of the sides, in order, is a rectangle.

Answer

Given,
A kite ABCD having AB = AD and BC = CD. P, Q, R, S are the mid-points of sides AB, BC, CD and DA respectively. PQ, QR, RS and SP are joined.
To prove:
PQRS is a rectangle.

Proof:
In $\triangle\text{ABC},$ P and Q are the mid-points of AB and BC respectively.
$\therefore\text{PQ}||\text{AC}$ and $\text{PQ}=\frac{1}{2}\text{AC}\ ...(\text{i})$
In $\triangle\text{ADC}$ R and S are the mid-points of CD and AD respectively.
$\therefore\text{RS}||\text{AC}$ and $\text{RS}=\frac{1}{2}\text{AC}\ ...(\text{ii})$
From (i) and (ii) we have
PQ || RS and PQ = RS
Thus, in quadrilateral PQRS, a pair of opposite sides is equal and parallel. So, PQRS is a parallelogram. Now, we shall prove that one angle of parallelogram PQRS is a right angle.
Since AB = AD
$\Rightarrow12\text{B}=12\text{AD}$
$\Rightarrow\text{AP}=\text{AS}\ ...(\text{iii})$ $[\therefore$ P and S are midpoints of AB and AD$]$
$\Rightarrow\angle1=\angle2\ ...(\text{iv})$
Now, in $\triangle\text{PQB}$ and $\triangle\text{SDR},$ we have
PB = SD $\Big[\therefore\text{AD}=\text{AB}\Rightarrow\Big(\frac{1}{2}\Big)\text{AB}\Big]$
BQ = DR [Since PB = SD]
And PQ = SR [Since, PQRS is a parallelogram]
So, by SSS criterion of congruence, we have
$\triangle\text{PBQ}\cong\triangle\text{SDR}$
$\Rightarrow\angle3=\angle4$ [CPCT]
Now, $\Rightarrow\angle3+\angle\text{SPQ}+\angle2=180^\circ$
And $\angle1+\angle\text{PSR}+\angle4=180^\circ$
$\therefore\angle3+\angle\text{SPQ}+\angle2=\angle1+\angle\text{PSR}+\angle4$
$\Rightarrow\angle\text{SPQ}=\angle\text{PSR}$ $[\angle1=\angle2\text{and}\angle3=\angle4]$
Now, transversal PS cuts parallel lines SR and PQ at S and P respectively.
$\therefore\angle\text{SPQ}+\angle\text{PSR}=180^\circ$
$\Rightarrow\angle2\text{SPQ}=180^\circ$
$\Rightarrow\angle\text{SPQ}=90^\circ$ $[\therefore\angle\text{PSR}=\angle\text{SPQ}]$
Thus, PQRS is a parallelogram such that $\angle\text{SPQ}=90^\circ.$
Hence, PQRS is a parallelogram.

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Question 174 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 184 Marks

In a triangle, P, Q and R are the mid points of sides BC, CA and AB respectively. If AC = 21cm, BC = 29cm and AB = 30cm, find the perimeter of the quadrilateral ARPQ.

Answer

In $\triangle\text{ABC},$ R and P are mid points of AB and BC $\text{RP}||\text{AC},\text{RP}=\Big(\frac{1}{2}\Big)\text{AC}$ [By Midpoint Theorem]

In a quadrilateral, [A pair of side is parallel and equal] RP || AQ, RP = AQ Therefore, RPQA is a parallelogram $\Rightarrow\text{AR}=\frac{1}{2}\text{AB}=\frac{1}{2}\times30=15\text{cm}$ $\text{AR}=\text{QP}=15\text{cm}$ [Opposite sides are equal] $\Rightarrow\text{RP}=\frac{1}{2}\text{AC}=\frac{1}{2}\times21=10.5\text{cm}$ $\text{RP}=\text{AQ}=10.5\text{cm}$ [Opposite sides are equal] Now, Perimeter of ARPQ = AR + QP + RP + AQ = 15 + 15 + 10.5 + 10.5 = 51cm

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Question 194 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 204 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 214 Marks

If an angle of a parallelogram is two-third of its adjacent angle, find the angles of the parallelogram.

Answer

Let one of the angle of the parallelogram as x°
Then the adjacent angle becomes $\frac{2}{3}\text{x}^\circ$
We know that the sum of adjacent angles of the parallelogram is supplementary.
Therefore,
$\text{x}+\frac{2}{3}\text{x}=180$
$\frac{5}{3}\text{x}=180$
$\text{x}=180\Big(\frac{3}{5}\Big)$
$\text{x}=108^\circ$
Thus, the angle adjacent to 108°
$=\frac{2}{3}(108)^\circ$
$=72^\circ$
Since, opposite angles of a parallelogram are equal.
Therefore, the four angles in sequence are 108°, 72°, 108° and 72°.

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Question 224 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 quadrilatera 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 234 Marks

In figure, Triangle ABC is a right angled triangle at B. Given that AB = 9cm, AC = 15cm and D, E are the mid-points of the sides AB and AC respectively, calculate

The length of BC
The area of $\triangle\text{ADE}.$

Answer

I. In $\triangle\text{ABC},\angle\text{B}=90^\circ,$
By using Pythagoras theorem
AC² = AB² + BC²$\Rightarrow15^2=9^2+\text{BC}^2$
$\Rightarrow\text{BC}=\sqrt{15^2-9^2}$
$\Rightarrow\text{BC}=\sqrt{225-81}$
$\Rightarrow\text{BC}=\sqrt{144}=12\text{cm}$

II. In $\triangle\text{ABC},$
D and E are mid-points of AB and AC
$\therefore\text{DE}||\text{BC},=\frac{1}{2}\text{BC}$ [By mid−point theorem]
$\text{AD}=\text{DB}=\frac{\text{AB}}{2}=\frac{9}{2}=4.5\text{cm}$ $[\therefore$ D is the mid−point of AB$]$
Area of $\triangle\text{ADE}=\frac{1}{2}\times\text{AD}\times\text{DE}$
$=\frac{1}{2}\times4.5\times6$
$=13.5\text{cm}^2$

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Question 244 Marks

ABCD is a parallelogram, AD is produced to E so that DE = DC and EC produced meets AB produced in F.Prove that BF = BC.

Answer

Draw a parallelogram ABCD with AC and BD intersecting at O.
Produce AD to E such that DE = DC.
Join EC and produce it to meet AB produced at F.
In $\triangle\text{DCE},$
$\therefore\angle\text{DCE}=\angle\text{DEC}\dots(1)$ (In a triangle, equal sides have equal angles opposite to them)
AB || CD (Opposite sides of the parallelogram are parallel)
$\therefore$ AF || CD (AB lies on AF)
AF || CD and EF is the transversal,
$\therefore\angle\text{DCE}=\angle\text{BFC}\dots(2)$ (Pair of corresponding angles)
From (1) and (2), we get
$\angle\text{DCE}=\angle\text{BFC}$
In $\triangle\text{AFE},$
$\angle\text{AFE}=\angle\text{AEF}$ $(\angle\text{DEC}=\angle\text{BFC})$
$\therefore$ AE = AF (In a triangle, equal angles have equal sides opposite to them)
⇒ AD + DE = AB + BF
⇒ BC + AB = AB + BF $(\because$ AD = BC, DE = CD AND CD = AB, AB = DE$)$
⇒ BC = BF

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Question 254 Marks

In ABCD is a parallelogram in which P is the mid-points of DC and Q is a point on AC such that $\text{CQ}=\frac{1}{4}\text{AC}.$ if PQ produced meets BC at E, prove R is a mid-points of BC.

Answer

Figure is given as follows:

ABCD is a parallelogram, where P is the mid-points of DC and Q is a point on AC such that $\text{CQ}=\frac{1}{4}\text{AC}.$ PQ produced meets BC at R. We need to prove that R is a mid-points of BC. Let us join BD to meet AC At O. It is given that ABCD is a parallelogram. Therefore, $\text{OC}=\frac{1}{2}\text{AC}$ (Because diagonals of a parallelogram bisect each other) Also, $\text{CQ}=\frac{1}{2}\text{AC}$ Therefore, $\text{CQ}=\frac{1}{2}\text{OC}$ In $\triangle\text{DCQ},$ p and Q are the mid-points of CD respectively. Theorem states, the line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it. Therefore, we get: PQ || DO Also, in $\triangle\text{COB},$ Q is the mid-points of BC. Hence proved.

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Question 264 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 274 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 284 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 294 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 304 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 314 Marks

P , Q, R and S are respectively the midpoints of the sides AB, BC, CD and DA of a quadrilateral ABCD. Show that:

PQ || AC and $\text{PQ}=\frac{1}{2}\text{AC}$
PQ || SR
PQRS is a parallelogram.

Answer

In $\triangle\text{ABC,}$ P and Q are the mid-points of sides AB and BC respectively.

$\Rightarrow\text{PQ }||\text{ AC}$ and $\text{PQ}=\frac{1}{2}\text{AC}...(\text{i})$

In $\triangle\text{ADC,}$ R and S are the mid-points of sides CD and AD respectively.

$\Rightarrow\text{SR || AC}$ and $\Rightarrow\text{SR}=\frac{1}{2}\text{AC}...(\text{ii})$From (i) and (ii), we have$\text{PQ = SR }$ and $\text{PQ ∥ SR}$

Thus, in quadrilateral PQRS, one pair of opposite sides are equal and parallel.

Hence, PQRS is a parallelogram.

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Question 324 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 334 Marks

In the given figure, ABCD is a trapezium. Find the values of x and y.

Answer

The figure is given as follows:

We know that ABCD is a trapezium with AB || DC Therefore, $\angle\text{A}+\angle\text{D}=180^\circ$ It is given that $\angle\text{A}=\text{x}+20^\circ$ and $\angle\text{D}=2\text{x}+10^\circ.$ (x + 20º) + (2x + 10º) = 180º 3x + 30º = 180º 3x = 180º - 30º 3x = 150º x = 50º Similarly, y + 92º = 180º y = 180º - 92º y = 88º Hence, the required values for x and y is 50º and 88º respectively.

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Question 344 Marks

The diagonals of a rectangle ABCD meet at O, If $\angle\text{BOC}=44^\circ,$ find $\angle\text{OAD}.$

Answer

The rectangle ABCD is given as:

We have, $\angle\text{BOC}+\angle\text{BOA}=180^\circ$ (Linear pair) $44^\circ+\angle\text{BOA}=180^\circ$ $\angle\text{BOA}=180^\circ-44^\circ$ $\angle\text{BOA}=136^\circ$ Since, diagonals of a rectangle are equal and they bisect each other. Therefore, in $\triangle\text{OAB},$ we have OA = OB (Angles opposite to equal sides are equal.) Therefore, $\angle1=\angle2$ Now,in $\triangle\text{OAB},$ we have $\angle\text{BOA}+\angle1+\angle2=180^\circ$ $\angle\text{BOA}+2\angle1=180^\circ$ $2\angle1=44^\circ$ $\angle1=22^\circ$ Since, each angle of a rectangle is a right angle. Therefore, $\angle\text{BAD}=90^\circ$ $\angle1+\angle3=90^\circ$ $22^\circ+\angle3=90^\circ$ $\angle3=68^\circ$ Thus, $\angle\text{OAD}=68^\circ$ Hence, the measure of $\angle\text{OAD}$ is 68º. $\angle\text{C}+\angle\text{D}=150^\circ$ Hence, the sum of $\angle\text{C}$ and $\angle\text{D}$ is 150º.

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Question 354 Marks

In a quadrilateral ABCD, CO and DO are the bisectors of $\angle\text{C}\ \text{and}\ \angle\text{D}$ respectively. Prove that $\angle\text{COD} = \frac{1}{2} (\angle\text{A}\ \text{and}\ \angle\text{B})$.

Answer

In $\triangle\text{DOC}$
$\angle1+\angle\text{COD}+\angle2=180^\circ$ [Angle sum property of a triangle]
$\Rightarrow\angle\text{COD}=180-(\angle1-\angle2)$
$\Rightarrow\angle\text{COD}=180-\angle\ 1+\angle2$
$\Rightarrow\angle\text{COD}=180-\Big[\frac{1}{2}\text{LC}+\frac{1}{2}\text{LD}\Big]$ $\big[\because$ OC and OD are bisectors of LC and LD respectively$\big]$
$\Rightarrow\angle\text{COD}=180-\frac{1}{2}(\text{LC+LD)}\ ...(\text{i})$
In quadrilateral ABCD
$\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}=360^\circ$ [Angle sum property of quadrilateral]
$\angle\text{C}+\angle\text{D}=360^\circ-(\angle\text{A}+\angle\text{B)}\dots(\text{ii})$
Substituting (ii) in (i)
$\Rightarrow\angle\text{COD}=180-\frac{1}{2}(360^\circ-(\angle\text{A}+\angle\text{B))}$
$\Rightarrow\angle\text{COD}=180-180+\frac{1}{2}(\angle\text{A}+\angle\text{B})$
$\Rightarrow\angle\text{COD}=\frac{1}{2}(\angle\text{A}+\angle\text{B})$

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Question 364 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 374 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 384 Marks

Two opposite angles of a parallelogram are (3x - 2)° and (50 - x)°. Find the measure of each angle of the parallelogram.

Answer

We know that,
Opposite sides of a parallelogram are equal.
(3x - 2)° = (50 - x)°
⇒ 3x + x = 50 + 2
⇒ 4x = 52°
⇒ x = 13°
Therefore, (3x - 2)° = (3 × 13 - 2)° = 37°
(50 - x)° = (50 - 13) = 37°
Adjacent angles of a parallelogram are supplementary.
$\therefore$ x + 37 = 180°
$\therefore$ x = 180° - 37° = 143°
Hence, four angles are: 37°, 143°, 37°, 143°.

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Question 394 Marks

In a $\triangle\text{ABC},$ E and F are the mid-points of AC and AB respectively. The altitude AP to BC intersects FE at Q. Prove that AQ = QP.

Answer

In a $\triangle\text{ABC}$ E and F are mid points of AB and AC $\therefore\text{EF},||\text{FE},\frac{1}{2}\text{BC}=\text{FE}$ [By midpoint theorem] In $\triangle\text{ABP}$ F is the mid-point of AB and $\text{FQ}||\text{BP}$ $[\therefore\text{EF}||\text{BP}]$ Therefore, Q is the mid-point of AP [By mid-point theorem] Hence, AQ = QP.

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Question 404 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 414 Marks

ABC is a triangle and through A, B, C lines are drawn parallel to BC, CA and AB respectively intersecting at P, Q and R. Prove that the perimeter of $\triangle\text{PQR}$ is double the perimeter of $\triangle\text{ABC}.$

Answer

Clearly ABCQ and ARBC are parallelograms.
Therefore, BC = AQ and BC = AR
⇒ AQ = AR
⇒ A is the mid-point of QR
Similarly B and C are the mid points of PR and PQ respectively.
$\therefore\text{AB}=\Big(\frac{1}{2}\Big)\text{PQ},$ $\text{BC}=\Big(\frac{1}{2}\Big)\text{QR},$ $\text{CA}=\Big(\frac{1}{2}\Big)\text{PR}.$
⇒ PQ = 2AB, QR = 2BC and PR = 2CA
⇒ PQ + QR + RP = 2 (AB + BC + CA)
⇒ Perimeter of $\triangle\text{PQR}=2$ $($perimeter of $\triangle\text{ABC})$

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Question 424 Marks

In figure, M, N and P are mid-points of AB, AC and BC respectively. If MN = 3cm, NP = 3.5cm and MP = 2.5cm, calculate BC, AB and AC.

Answer

Given MN = 3cm, NP = 3.5cm and MP = 2.5cm. To find BC, AB and AC

In $\triangle\text{ABC}$ M and N are mid-points of AB and AC $\therefore\text{MN}=\frac{1}{2}\text{BC},\text{MN}||\text{BC}$ [By mid-point theorem] $\Rightarrow3=\frac{1}{2}\text{BC}$ $\Rightarrow3\times2=\text{BC}$ $\Rightarrow\text{BC}=6\text{cm}$ Similarly AC = 2MP = 2(2.5) = 5cm AB = 2 NP = 2(3.5) = 7cm

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Question 434 Marks

If the bisectors of two adjacent angles A and B of a quadrilateral ABCD intersect at a point O such that $\angle\text{C}+\angle\text{D}=\text{k}\angle\text{AOB},$ then find the value of k.

Answer

The quadrilateral can be drawn as follows:

We have AO and BO as the bisectors of angles $\angle\text{A}$ and $\angle\text{B}$ respectively. In $\triangle\text{AOB},$ We have, $\angle\text{AOB}+\angle1+\angle2=180^\circ$ $\angle\text{AOB}=180^\circ-(\angle1+\angle2)$ $\angle\text{AOB}=180^\circ-\Big(\frac{1}{2}\angle\text{A}+\frac{1}{2}\angle\text{B}\Big)$ $\angle\text{AOB}=180^\circ-\frac{1}{2}\Big(\angle\text{A}+\angle\text{B}\Big)\ ...(1)$ By angle sum property of a quadrilateral, we have: $\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}=360^\circ$ $\angle\text{A}+\angle\text{B}=360^\circ-(\angle\text{C}+\angle\text{D})$ Putting in equation (1): $\angle\text{AOB}=180^\circ-\frac{1}{2}\Big[360^\circ-(\angle\text{C}+\angle\text{D})\Big]$ $\angle\text{AOB}=180^\circ-180^\circ+\frac{(\angle\text{C}+\angle\text{D})}{2}$ $\angle\text{AOB}=\frac{1}{2}(\angle\text{C}+\angle\text{D})$ $(\angle\text{C}+\angle\text{D})=2\angle\text{AOB}\ ...(2)$ On comparing equation (2) with $(\angle\text{C}+\angle\text{D})=\text{k}\angle\text{AOB}$ We get k = 2. Hence, the value for k is 2.

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Question 444 Marks

In a $\triangle\text{ABC},$ D, E and F are, respectively the mid points of BC, CA and AB. If the lengths of sides AB, BC and CA are 7cm, 8cm and 9cm, respectively, find the perimeter of $\triangle\text{DEF}.$

Answer

Given that,

AB = 7cm, BC = 8cm, AC = 9cm In $\triangle\text{ABC},$ F and E are the mid points of AB and AC. $\therefore\text{EF}=\frac{1}{2}\text{BC}$ Similarly $\text{DF}=\frac{1}{2}\text{AC}$ and $\text{DE}=\frac{1}{2}\text{AB}$ Perimeter of $\triangle\text{DEF}=\text{DE}+\text{EF}+\text{DF}$ $=\Big(\frac{1}{2}\Big)\text{AC}+\Big(\frac{1}{2}\Big)\text{BC}+\Big(\frac{1}{2}\Big)\text{AC}$ $=\frac{1}{2}\times7+=\frac{1}{2}\times8+\frac{1}{2}\times9$ $=3.5+4+4.5$ $=12\text{cm}$

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Question 454 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 464 Marks

ABCD is a trapezium in which AB || CD and AD = BC. Show that,

$\angle\text{A}=\angle\text{B}$
$\angle\text{C}=\angle\text{D}$
$\triangle\text{ABC}\cong\triangle\text{BAD}$
Diagonal AC = diagonal BD.

Answer

Construction: Draw a line through C parallel to DA intersecting AB produced at E.

CE = AD (Opposite sides of a parallelogram)

AD = BC (Given)

Therefor, BC = CE

also,

$\angle\text{A}+\angle\text{CBE}=180^{0}$ (Angles on the same side of transversal and ∠CBE = ∠CEB)

$\angle\text{B}+\angle\text{CBE}=180^{0}$ (Linear pair)

$\angle\text{A}=\angle\text{B}$

$\angle\text{A}+\angle\text{D}=\angle\text{B}+\angle\text{C}=180^{0}$ (Angles on the same side of transversal)

$\angle\text{A}+\angle\text{D}=\angle\text{B}+\angle\text{C}$

$\angle\text{D}=\angle\text{C}$

In $\angle\text{ABC}$ and $\angle\text{BAD}$

AB = AB (Common)

$\angle\text{DBA}=\angle\text{CBA}$

AD = BC (Given)

Thus, by SAS $\angle\text{DBA}=\angle\text{CBA}$ Congruence condition.

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Question 474 Marks

ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA. AC is a diagonal. Show that :

SR || AC and SR $=\frac{1}{2}\text{AC}$
PQ = SR
PQRS is a parallelogram.

Answer

In $\triangle\text{ADC}$ and R are the mid-points of sides AD and CD respectively.

In a triangle, the line segment joining the mid-points of any two sides of the triangle is parallel to the third side and is half of it.

SR || AC and SR $=\frac{1}{2}\text{AC}\ ...(\text{i})$

In $\triangle\text{ABC}$ and Q are mid - points of sides AB and BC respectively. Therefore, by using mid - point theorem,

PQ || AC and PQ $=\frac{1}{2}\text{AC}\ ...(\text{ii})$

Using equations (i) and (ii), we obtain

PQ || SR and PQ = SR ...(iii)

⇒ PQ = SR

From equation (3), we obtained

PQ || SR and PQ = SR

Clearly, one pair of opposite sides of quadrilateral PQRS is parallel and equal.

Hence, PQRS is a parallelogram.

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Question 484 Marks

In a quadrilateral ABCD, bisectors of angles A and B intersect at O such that $\angle\text{AOB}=75^\circ,$ then write the value of $\angle\text{C}+\angle\text{D}.$

Answer

The quadrilateral can be drawn as follows:

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Question 494 Marks

In the adjoining figure, ABCD is a square and $\triangle\text{EDC}$ is an equilateral triangle. Prove that:

AE = BE,
$\angle\text{DAE}=15^{\circ}.$

Answer

Given: 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 504 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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