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#### Page No 312:

Given: Three angles of a quadrilateral are 75°, 90° and 75°.

Let the fourth angle be x.

Using angle sum property of quadrilateral,

$75°+90°+75°+x=360°\phantom{\rule{0ex}{0ex}}⇒240°+x=360°$

$⇒x=360°-240°\phantom{\rule{0ex}{0ex}}⇒x=120°$

So, the measure of  the fourth angle is 120$°$.

#### Page No 312:

Let $\angle$A = 2x​.
Then $\angle$B = (4x)$\angle$C = (5x) and $\angle$D = (7x)
Since the sum of the angles of a quadrilateral is 360o, we have:
2x + 4x + 5x + 7x = 360
⇒ 18 x = 360 ​
x = 20
$\angle$A = 40; $\angle$B = 80$\angle$C = 100; $\angle$D = 140

#### Page No 312:

We have AB || DC.

$\angle$ A  and  $\angle$ D are the interior angles on the same side of transversal line AD, whereas $\angle$ B and  $\angle$ C are the interior angles on the same side of transversal line BC.
Now,  $\angle$A + $\angle$D = 180
⇒  $\angle$D = 180$\angle$A
$\angle$ D = 180 55 = 125

Again , $\angle$ B + $\angle$C = 180
⇒  $\angle$C  = 180 $\angle$B
$\angle$ C = 180 −  70 = 110

#### Page No 312:

Given:  ABCD is a square in which AB = BC = CD = DA. ∆EDC is an equilateral triangle in which ED = EC = DC and
$\angle$ EDC  =   $\angle$ DEC = $\angle$​DCE =  60.
To prove:  AE = BE and
$\angle$DAE = 15
Proof: In ADE and ∆BCE, we have:
AD = BC              [Sides of a square]

DE = EC​             [Sides of an equilateral triangle]
$\angle$ADE $\angle$BCE = 90 +  60 = 150​
i.e., AE =  BE

Now,
$\angle$ADE = 150
DA = DC     [Sides of a square]
DC = DE      [Sides of an equilateral triangle]
So, DA = DE
ADE and BCE are isosceles triangles.
i.e., $\angle$DAE = $\angle$DEA =

#### Page No 312:

Given: A quadrilateral ABCD, in which BM ​⊥ AC and DN ⊥ AC and BM = DN.
To prove: AC bisects BD; or DO = BO

Proof:
Let AC and BD intersect at O.
Now, in ∆OND and ∆OMB, we have:
∠OND = ∠OMB                 (90o each)
∠DON = ∠ BOM                  (Vertically opposite angles)
Also, DN = BM                         (Given)
i.e.,
∆OND ≅ ∆OMB             (AAS congurence rule)
∴ OD = OB                          (CPCT)
​Hence, AC bisects BD.

#### Page No 312:

Given:  ABCD is a quadrilateral in which AB = AD and BC = DC
(i)

BC = DC                                                 (Given)
AC is common.
i.e., ∆ABC ≅ ∆ADC                                    (SSS congruence rule)
∴ ∠BAC = ∠DAC and ∠BCA = ∠D​CA        (By CPCT)
Thus, AC bisects ∠A and ∠ C.

(ii)
Now, in
∠BAE = ∠DAE​                               (Proven above)
AE is common.
∴ ∆ABE ≅  ∆ADE                          (SAS congruence rule)
⇒ BE = DE                                                 (By CPCT)

(iii)

#### Page No 312:

Given: ABCD is a square and ∠PQR = 90°.
Also, PB = QC = DR

(i) We have:
BC = CD                (Sides of square)
CQ = DR                   (Given)
BC = BQ + CQ
⇒ CQ = BC − BQ
∴ DR = BC − BQ           ...(i)

Also, CD = RC+ DR
∴ DR = CD −  RC = BC − RC            ...(ii)
From (i) and (ii), we have:
BC − BQ = ​BC − RC
∴ BQ = RC

(ii) In ∆RCQ and ∆QBP, we have:
PB = QC   (Given)
BQ = RC  (Proven above)
∠RCQ = ∠QBP   (90o each)
i.e., ∆RCQ ≅ ∆QBP       (SAS congruence rule)
∴ QR =  PQ                (By CPCT)

(iii) ∆RCQ ≅ ∆QBP and QR = PQ (Proven above)
∴ In ∆RPQ, ∠QPR = ∠QRP =

#### Page No 312: Let ABCD be a quadrilateral whose diagonals are AC and BD and O is any point within the quadrilateral.
Join O with A, B, C, and D.
We know that the sum of any two sides of a triangle is greater than the third side.
So, in ∆AOC, OA + OC > AC

Also, in ∆ BOD, OB + OD > BD
(OA + OC) + (OB + OD) > (AC + BD)
OA + OB + OC + OD > AC + BD

#### Page No 313:

Given: ABCD is a quadrilateral and AC is one of its diagonal.

(i)  We know that the sum of any two sides of a triangle is greater than the third side.
In ∆ABC, AB + BC > AC            ...(1)

In ∆ACD, CD + DA > AC            ...(2)
​Adding inequalities (1) and (2), we get:
AB + BC + CD + DA > 2AC

(ii) In ∆ABC, we have :
​  AB + BC > AC            ...(1)
We also know that the length of each side of a triangle is greater than the positive difference of the length of the other two sides.
In ∆ACD, we have:​
AC > |DA − CD|​        ...(2)
From (1) and (2), we have:
AB + BC > |DA − CD|​
⇒ AB + BC + CD > DA

(iii) In ∆ABC, AB + BC > AC
In ∆ACD, CD + DA > AC
In ∆ BCD, BC CD > BD
In ∆ ABD, DAAB > BD
2(AB + BC + CD + DA) > 2(AC + BD)
⇒ (AB + BC + CD + DA) > (AC + BD)

#### Page No 313: Let ABCD be a quadrilateral and ∠1, ∠2, ∠3 and ∠4  are its four angles as shown in the figure.
Join BD which divides ABCD in two triangles, ∆ABD and ∆BCD.
In ∆ABD, we have:
∠1 + ∠2 + ​∠A = 180o      ...(i)
In ​∆BCD, we have:
∠3 + ∠4 + ∠C = 180o     ...(ii)
On adding (i) and (ii), we get:

(∠1 + ∠3) + ∠A + ∠C + (∠4 + ∠2) ​= 360o
⇒ ∠A + ∠C  + ∠B + ∠D = 360o                         [ ∵ ∠1 + ∠3 = ∠B; ∠4 + ∠2 = ∠D]
∴ ∠A + ∠C  + ∠B + ∠D = 360o

#### Page No 328:

ABCD is parallelogram and ∠A = 72°​.
We know that opposite angles of a parallelogram are equal.
∴∠A ​= ∠C and B ​= ∠D ​ ​
∴ ∠C = 72o
A and ∠B are adajcent angles.
i.e., ∠A ​+ ∠B​ = 180o
⇒ ∠B = 180o   ∠A
⇒ ∠B​ = 180o 72o = 108o
∴​ ∠B​ =​ ∠D108o
Hence, ∠B​ =​ ∠D = 108o​ and ∠C​ = 72o

#### Page No 328:

Given:  ABCD is parallelogram and ∠DAB = 80°​ and ∠DBC = 60°
To find: Measure of ∠CDB and ∠ADB

In parallelogram ABCD, AD ||​ BC
∴ ∠DBC = ∠ ADB = 60o     (Alternate interior angles)     ...(i)

∴ ∠ADC = 180o  − ∠DAB
​    ⇒∠ADC​ = 180o − 80o = 100o

⇒ ∠ADB + ∠C​​DB = 100o              ...(ii)
From (i) and (ii), we get:
60o + ∠C​​DB = 100o
⇒ ∠C​​DB = 100o − 60o = 40o
Hence, ∠CDB​ =​ 40o and ∠ADB​ = 60o

#### Page No 329: Given: parallelogram ABCD, M is the midpoint of side BC and ∠BAM = ∠DAM.

Proof:

Since, $AD\parallel BC$ and AM is the transversal.

So, $\angle DAM=\angle AMB$      (Alternate interior angles)

But, $\angle DAM=\angle BAM$ (Given)

Therefore, $\angle AMB=\angle BAM$

$⇒AB=BM$             (Angles opposite to equal sides are equal.)    ...(1)

Now, AB = CD       (Opposite sides of a parallelogram are equal.)

$⇒2AB=2CD\phantom{\rule{0ex}{0ex}}⇒\left(AB+AB\right)=2CD$

$⇒BM+MC=2CD$      (AB BM and MC = BM)

#### Page No 329:

ABCD is a parallelogram.
∴ ​∠A = ​∠C and B =​ ∠D (Opposite angles)
And ​∠A + ​∠B = 180o                (Adjacent angles are supplementary)
∴ ​∠B = 180o − ∠
180o − 60o = 120o             ( ∵∠A = 60o)
∴ ​∠A = ​∠C = 60o and B =​ ∠D​ = 120o

(i) In ∆ APB, ∠​PAB = $\frac{60°}{2}=30°$ and ∠PBA = $\frac{120°}{2}=60°$
∴​ ∠​APB​ = 180o − (30o + 60o) = 90o

∴ ∠APB = 180o − (30o + 120o) = 30o
Thus, ∠​PAD = ​∠APB = ​30o

In ∆ PBC, ∠​ PBC = 60o ∠​ BPC = 180o − (90o +30o) = 60o and ∠​ BCP  = 60o (Opposite angle of ∠A)
∴ ∠ PBC = ∠​ BPC = ∠​ BCP
Hence, ∆PBC is an equilateral triangle and, therefore, PB = PC = BC.​

(iii) DC = DP + PC
From (ii), we have:
DC = AD + BC                     [AD = BC, opposite sides of a parallelogram]

#### Page No 329:

ABCD is a parallelogram.
∴ AB ∣∣​ DC and BC ​∣∣​ AD

(i) In ∆AOB, ∠BAO = 35°, ​∠AOB = ∠COD = 105°  (Vertically opposite angels)
∴ ​∠ABO = 180o − (35o + 105o) = 40o

(ii)∠ODC and ∠ABO are alternate interior angles.
∴ ∠ODC = ∠ABO = 40o

(iii) ∠ACB = ∠​CAD = 40o                             (Alternate interior angles)

(iv) ∠CBDABC  ABD            ...(i)

⇒∠ABC = 180o − 75o = 105o
⇒∠CBD = 105o ABD                         (∠ABDABO)
⇒∠CBD = 105o 40o =  65o

#### Page No 329:

ABCD is a parallelogram.
i.e., ∠A = C and B∠D                  (Opposite angles)
Also, ∠A + ∠B = 180o                            (Adjacent angles are supplementary)   ​
∴​ (2x + 25)°​ + (3x − 5)°​ = 180
⇒ ​5x +20 = 180
⇒​ 5x = 160
⇒​ x = 32o
∴​∠A = 2 ⨯ 32 + 25 = 89o and ∠B = 3 32 − 5 = 91o
Hence, x = 32o, ∠AC89o and ∠BD = 91o

#### Page No 329:

Let ABCD be a parallelogram.
∴ ∠​A = ∠C and B = ∠D           (Opposite angles)
Let A = xo and ∠B${\left(\frac{4x}{5}\right)}^{°}$
Now, ∠​A + ∠B = 180o                 (Adjacent angles are supplementary)

Hence, AC = 100o; B = ∠D = 80o

#### Page No 329:

Let ABCD be a parallelogram.
∴ ∠​A = ∠​and B = ∠D          (Opposite angles)
Let A be the smallest angle whose measure is xo.
∴​ ∠​B = (2x − 30)o
Now, ∠​A + ∠B = 180o                (Adjacent angles are supplementary)
⇒ x + 2x − 30o = 180o
⇒ 3x = 210o
⇒ x = 70o
∴ ​∠​B = 2 ⨯ 70o − 30o = 110o
HenceAC = 70o; ∠B = ∠D = 110o

#### Page No 329:

ABCD is a parallelogram.
The opposite sides of a parallelogram are parallel and equal.
∴ AB = DC = 9.5 cm
Let BC = AD = x
∴​ Perimeter of ABCD = AB + BC + CD + DA = 30 cm
⇒ 9.5 + x + 9.5 + x = 30
⇒ 19 + 2x = 30
⇒ 2x = 11
x = 5.5 cm
Hence, AB = DC = 9.5 cm and BC = DA = 5.5 cm

#### Page No 329:

ABCD is a rhombus and a rhombus is also a parallelogram. A rhombus has four equal sides.
(i)​ In ∆ABC, ∠​BAC = ∠BCA =
i.e., x = 35o
Now, ∠​B + ∠C = 180o                 (Adjacent angles are supplementary)  ​
But ∠​C​ = x + y = 70o

⇒​ y = 70o    x

⇒​y =  70o − 35o = 35o
Hence, x = 35o; y = 35o

(ii) The diagonals of a rhombus are perpendicular bisectors of each other.
So, in ∆​AOB, ​∠OAB = 40o, ∠AOB = 90o and ∠ABO = 180o − (40o + 90o) = 50o
∴ ​x = 50o

So, ∠ABD = ​∠ADB = ​50o
Hence, x = 50o;  y = 50o

​(iii) ∠​BAC = ∠​DCA                   (Alternate interior angles)​
i.e., x  = 62o
In ∆BOC∠​BCO =  62o              [In ∆​ ABC, AB = BC, so ∠​BAC = ∠​ACB]
Also, ∠​BOC = 90o
∴ ∠​OBC = 180o − (90o + 62o) = 28o
Hence, x = 62o; y = 28o

#### Page No 329: Let ABCD be a rhombus.
AB = BC = CD = DA
Here, AC and BD are the diagonals of ABCD, where AC = 24 cm and BD = 18 cm.
Let the diagonals intersect each other at O.
We know that the diagonals of a rhombus are perpendicular bisectors of each other.
∴​ ∆AOB is a right angle triangle in which OA = AC/2 = 24/2 = 12 cm and OB = BD/2 = 18/2 = 9 cm.
Now, AB2= OA2 + OB2              [Pythagoras theorem]
⇒​ AB2=​ (12)2 + (9)2
⇒​ AB2=​ 144 + 81 = 225
⇒​ AB=​ 15 cm

Hence, the side of the rhombus is 15 cm.

#### Page No 329: Let ABCD be a rhombus.
∴ AB = BC = CD = DA = 10 cm
Let AC and BD be the diagonals of ABCD. Let AC = x and BD = 16 cm and O be the intersection point of the diagonals.
We know that the diagonals of a rhombus are perpendicular bisectors of each other.
∴​ ∆AOB is a right angle triangle, in which OA = AC $÷$ 2 = $÷$ 2 and OB = BD $÷$2 = 16 $÷$ 2 = 8 cm.

Now, AB2= OA2 + OB2              [Pythagoras theorem]
2

x2 =144
x = 12 cm

Hence, the other diagonal of the rhombus is 12 cm.
∴ Area of the rhombus =

#### Page No 330:

(i) ABCD is a rectangle.
The diagonals of a rectangle are congruent and bisect each other. Therefore, in​ ∆ AOB, we have:
OA = OB
∴​ ∠​OAB = ∠​OBA = 35o
∴​ x = 90o − 35o = 55o
And ∠AOB = 180o − (35o + 35o) = 110o
∴​ y = ∠AOB​ = 110o​                     [Vertically opposite angles]
Hence, x = 55o and y = 110o​​

(ii) In ∆AOB, we have:
OA = OB
Now, ∠​OAB = ∠OBA =
∴​ y = ∠BAC = 35o                 [Interior alternate angles]
Also, x = 90oy                          [ ​∵∠C = 90o =  x + y ]
⇒​ x = 90o − 35o = 55o
Hence, x = 55o and y = 35o​​

#### Page No 330: Given: ABCD is a rhombus, DE is altitude which bisects AB i.e. AE = EB

$DE=DE$                         (Common side)

$\angle DEA=\angle DEB=90°$     (Given)

$AE=EB$                          (Given)

(By SAS congruence Criteria)

$⇒AD=BD$                    (CPCT)

Also, $AD=AB$            (Sides of rhombus are equal)

$⇒AD=AB=BD$

Thus, is an equilateral triangle.

Therefore, $\angle A=60°$

$⇒\angle C=\angle A=60°$                         (Opposite angles of rhombus are equal)

(Adjacent angles of rhombus are supplementary.)

$⇒\angle ABC+60°=180°\phantom{\rule{0ex}{0ex}}⇒\angle ABC=180°-60°\phantom{\rule{0ex}{0ex}}⇒\angle ABC=120°\phantom{\rule{0ex}{0ex}}⇒\angle ADC=\angle ABC=120°$

Hence, the angles of rhombus are .

#### Page No 330:

The angles of a square are bisected by the diagonals.
∴ ∠​OBX = 45o                        [∠​ABC = 90o and BD bisects ∠​ABC​]
And ∠​BOX = ∠COD = 80o           [Vertically opposite angles]
∴​ In ∆BOX, we have:
∠AXO = ∠OBX + ​∠BOX        [Exterior angle of ∆BOX]
⇒​ ∠AXO = 45o + 80o = 125o
∴ ​x =125o

#### Page No 330: Given:
A rhombus ABCD.

To prove: Diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.

Proof:

In $∆ABC$,

(Sides of rhombus are equal.)

$\angle 4=\angle 2$          (Angles opposite to equal sides are equal.)     ...(1)

Now,

$AD\parallel BC$          (Opposite sides of rhombus are parallel.)

$\phantom{\rule{0ex}{0ex}}$AC is transversal.

So, $\angle 1=\angle 4$        (Alternate interior angles)          ...(2)

From (1) and (2), we get

$\angle 1=\angle 2$

Thus, AC bisects $\angle A$.

Similarly,

Since, $AB\parallel DC$ and AC is transversal.

So, $\angle 2=\angle 3$     (Alternate interior angles)     ...(3)

From (1) and (3), we get

$\angle 4=\angle 3$

Thus, AC bisects ∠C.

Hence, AC bisects

In $∆DAB$,

(Sides of rhombus are equal.)

$\angle ADB=\angle ABD$          (Angles opposite to equal sides are equal.)     ...(4)

Now,

$DC\parallel AB$                        (Opposite sides of rhombus are parallel.)

$\phantom{\rule{0ex}{0ex}}$BD is transversal.

So, $\angle CDB=\angle DBA$        (Alternate interior angles)          ...(5)

From (4) and (5), we get

$\angle ADB=\angle CDB$

Thus, DB bisects $\angle D$.

Similarly,

Since, $AD\parallel BC$ and BD is transversal.

So, $\angle CBD=\angle ADB$     (Alternate interior angles)     ...(6)

From (4) and (6), we get

$\angle CBD=\angle ABD$

Thus, BD bisects ∠B.

Hence, BD bisects

#### Page No 330: Given:
In a parallelogram ABCD, AM CN.

To prove: AC and MN bisect each other.

Construction: Join AN and MC.

Proof:
Since, ABCD is a parallelogram.

$⇒AB\parallel DC\phantom{\rule{0ex}{0ex}}⇒AM\parallel NC$

Also, AM = CN           (Given)

Thus, AMCN is a parallelogram.

Since, diagonals of a parallelogram bisect each other.

Hence, AC and MN bisect each other.

#### Page No 330:

We have:
B = ∠D                        [Opposite angles of parallelogram ABCD]
AD = BC and AB = DC      [Opposite sides of parallelogram ABCD]
Also, AD || BC and AB|| DC
It is given that .
∴ ​AP = CQ                                   [∵ AD = BC]
In ∆​DPC and ∆​BQA, we have:
AB = CD, ∠B = ∠D and DP = QB                        [∵DP$\frac{2}{3}$AD and QB = $\frac{2}{3}$BC
i.e., ∆​DPC ≅ ∆​BQA
∴​ PC  = QA

Thus, in quadrilatreal AQCP, we have:
AP = CQ                   ...(i)
PC  = QA                   ...(ii)
∴ ​AQCP is a parallelogram.

#### Page No 330:

In ∆​ODF and ∆​OBE, we have:
OD = OB                                  (Diagonals bisects each other)
DOF = ∠BOE                         (Vertically opposite angles)
∠FDO = ∠OBE                         (Alternate interior angles)
i.e., ∆​ODF ≅ ∆​OBE
∴​ OF = OE                                 (CPCT)
Hence, proved.

#### Page No 330: Given: In parallelogram ABCDDP ABAQ  BC and ∠PDQ = 60°

In quadrilateral DPBQ, by angle sum property, we have

$\angle PDQ+\angle DPB+\angle B+\angle BQD=360°\phantom{\rule{0ex}{0ex}}⇒60°+90°+\angle B+90°=360°\phantom{\rule{0ex}{0ex}}⇒\angle B=360°-240°\phantom{\rule{0ex}{0ex}}⇒\angle B=120°$

Therefore,

Now,
$\angle B=\angle D=120°$           (Opposite angles of a parallelogram are equal.)

$\angle A+\angle B=180°$           (Adjacent angles of a parallelogram are supplementary.)

$⇒\angle A+120°=180°\phantom{\rule{0ex}{0ex}}⇒\angle A=180°-120°\phantom{\rule{0ex}{0ex}}⇒\angle A=60°$

Also,
$\angle A=\angle C=60°$         (Opposite angles of a parallleogram are equal.)

So, the angles of a parallelogram are

#### Page No 330: Given: In rectangle ABCDAC bisects ∠A, i.e. ∠1 = ∠2 and AC bisects ∠C, i.e. ∠3 = ∠4.

To prove:
(i) ABCD is a square,
(ii) diagonal BD bisects ∠B as well as ∠D.

Proof:

(i)
Since, $AD\parallel BC$        (Opposite sides of a rectangle are parallel.)

So, $\angle 1=\angle 4$         (Alternate interior angles)

But, $\angle 1=\angle 2$        (Given)

So,

In $∆ABC,$

Since, $\angle 2=\angle 4$

So, $BC=AB$                            (Sides opposite to equal angles are equal.)

But these are adjacent sides of the rectangle ABCD.

Hence, ABCD is a square.

(ii)
Since, the diagonals of a square bisects its angles.
So, diagonals BD bisects B as well as ∠D.

#### Page No 331:

In ∆ODC and ∆​OEB, we have:
DC = BE                                   (∵ DC = AB)
∠COD = ∠BOE                        (Vertically opposite angles)
OCD = ∠OBE                        ( Alternate interior angles)

i.e., ∆​ODC ≅ ∆​OEB
⇒ OC = OB                                 (CPCT)
We know that BC = OC + OB.
∴ ED bisects BC.

#### Page No 331:

Given: ABCD is a parallelogram.
BE = CE    (E is the mid point of BC)
DE and AB when produced meet at F.
To prove: AF = 2AB

Proof:
In parallelogram ABCD, we have:
AB || DC
∠DCE = ∠EBF            (Alternate interior angles)
In ∆DCE and ∆BFE, we have:
∠DCE = ∠EBF              (Proved above)

∠DEC = ∠BEF              (Vertically opposite angles)
Also, BE = CE           (Given)
∴ ∆DCE ≅​ ∆BFE  (By ASA congruence rule)
∴ DC = BF         (CPCT)
But DC = AB, as ABCD is a parallelogram.
∴ DC = AB =  BF                   ...(i)

Now, AF = AB + BF                ...(ii)
From (i), we get:
AF = AB + AB = 2AB
Hence, proved.

#### Page No 331: Given: l || m and the bisectors of interior angles intersect at and D.

To prove: ABCD is a rectangle.

Proof:

Since,
l || m                       (Given)

So, $\angle PAC=\angle ACR$            (Alternate interior angles)

$⇒\frac{1}{2}\angle PAC=\frac{1}{2}\angle ACR$

$⇒\angle BAC=\angle ACD$

but, these are a pair of alternate interior angles for AB and DC.

$⇒AB\parallel DC$

Similarly, $BC\parallel AD$

So, ABCD is a parallelogram.

Also,
$\angle PAC+\angle CAS=180°$      (Linear pair)

$⇒\frac{1}{2}\angle PAC+\frac{1}{2}\angle CAS=90°\phantom{\rule{0ex}{0ex}}⇒\angle BAC+\angle CAD=90°\phantom{\rule{0ex}{0ex}}⇒\angle BAD=90°$

But, this an angle of the parallleogram
ABCD.

Hence, ABCD is a rectangle.

#### Page No 331: Given: In square ABCD, AK BL CM DN

To prove: KLMN is a square.

Proof:

In square
ABCD,

AB = BC = CD = DA             (All sides of a square are equal.)

And, AK BL CM DN     (Given)

So, AB $-$ AK = BC $-$ BL = CD $-$ CM = DA $-$ DN

$⇒$ KB = CL = DM = AN        ...(1)

In $∆NAK$ and $∆KBL$,

$\angle NAK=\angle KBL=90°$    (Each angle of a square is a right angle.)

$AK=BL$                        (Given)

$AN=KB$                        [From (1)]

So, by SAS congruence criteria,

$∆NAK\cong ∆KBL$

$⇒NK=KL$     (CPCT)           ...(2)

Similarly,

$⇒$MN = NK  and $\angle DNM=\angle KNA$     (CPCT)       ...(3)
MN = JM
and $\angle DNM=\angle CML$        (CPCT)        ...(4)
ML = LK and $\angle CML=\angle BLK$            (CPCT)        ...(5)

From (2), (3), (4) and (5), we get

NK = KL = MN = ML      ...(6)

And, $\angle DNM=\angle AKN=\angle KLB=LMC$

Now, In $∆NAK$,

Let

So, $\angle DNK=90°+x°$ (Exterior angles equals sum of interior opposite angles.)

$⇒\angle DNM+\angle MNK=90°+x°\phantom{\rule{0ex}{0ex}}⇒x°+\angle MNK=90°+x°\phantom{\rule{0ex}{0ex}}⇒\angle MNK=90°$

Similarly,
$\angle NKL=\angle KLM=\angle LMN=90°$       ...(7)

Using (6) and (7), we get

All sides of quadrikateral KLMN are equal and all angles are 90$°$.

So, KLMN is a square.

#### Page No 331:

BC || QA and CA || QB
i.e., BCQA is a parallelogram.
∴ BC = QA                           ...(i)
Similarly, BC || AR and AB || CR.
i.e., BCRA is a parallelogram.
BC = AR                         ...(ii)
But QR = QA + AR
From (i) and (ii), we get:
QR = BC + BC
QR = 2BC
∴ BC$\frac{1}{2}QR$

#### Page No 331:

Perimeter of ∆​ABC = AB + BC + CA                         ...(i)
Perimeter of ∆PQR  =​ PQ + QR + PR                   ...(ii)

BC || QA and CA || QB
i.e., BCQA is a parallelogram.
∴ BC = QA                                  ...(iii)
Similarly, BC || AR and AB || CR
i.e., BCRA is a parallelogram.
∴ BC = AR                                    ...(iv)
But, QR = QA + AR
From (iii) and (iv), we get:
QR = BC + BC
QR = 2BC
∴ BC = $\frac{1}{2}QR$
Similarly, CA = ​$\frac{1}{2}PQ$ and AB = ​$\frac{1}{2}PR$

From (i) and (ii), we have:
Perimeter of ∆​ABC  = $\frac{1}{2}QR$ + $\frac{1}{2}PQ$ + $\frac{1}{2}PR$
=

i.e., Perimeter of ∆​ABC  = $\frac{1}{2}$ (Perimeter of ∆​PQR)
∴ Perimeter of ∆PQR = 2 ⨯ Perimeter of ∆ABC

#### Page No 345: Given: In quadrilateral ABCD, P, Q, R and S are respectively the midpoints of the sides AB, BC, CD and DA.

To prove:

(i) PQ || AC and PQ $\frac{1}{2}$AC

(ii) PQ || SR

(iii) PQRS is a parallelogram.

Proof:

(i)
In $∆ABC$,

Since, P and Q are the mid points of sides AB and BC, respectively.      (Given)

(Using mid-point theorem.)

(ii)
In $∆ADC$,

Since, S and R are the mid-points of AD and DC, respectively.        (Given)

(Using mid-point theorem.)            ...(1)

From (i) and (1), we get

PQ || SR

(iii)
From (i) and (ii), we get

$PQ=SR=\frac{1}{2}AC$

So, PQ and SR are parallel and equal.

Hence, PQRS is a parallelogram.

#### Page No 345: Given: In an isosceles right ∆ABC, CMPN is a square.

To prove: P bisects the hypotenuse AB i.e., AP = PB.

Proof:

In square CMPN,

∴ CM = MP = PN = CN            (All sides are equal.)

Also, ∆ABC is an isosceles with AC = BC.

⇒ AN + NC = CM + MB

⇒ AN = MB          (∵ CN = CM)       ...(i)

Now,

In ∆ANP and ∆PMB,

AN = MB                [From (i)]

ANP = ∠PMB = 90°

PN = PM             (Sides of square CMPN)

∴ By SAS congruence criteria,

∆ANP  BMP

Hence, AP = PB         (By CPCT)

#### Page No 346:

In parallelogram ABCD, we have:
AD || BC and AB || DC

AD = BC and AB = DC
AB = AE
+ BE and DC = DF + FC
AE = BE = DF = FC
Now, DF = AE and DF || AE.
i.e., AEFD is a parallelogram​​.

Similarly, ​BEFC is also a parallelogram.
∴​ EF || BC
∴​ AD || EF || BC
Thus, AD, EF and BC are three parallel lines cut by the transversal line DC at D, F and C, respectively such that DF = FC.
These lines AD, EF and BC​ are also cut by the transversal AB at A, E and B, respectively such that​ AE =  BE. ​
Similarly, they are also cut by​ GH.
∴ GP = PH            (By intercept theorem)

#### Page No 346: Given: A parallelogram ABCD

To prove: MN is bisected at O

Proof:

I$∆$OAM and $∆$OCN,

OA = OC                 (Diagonals of parallelogram bisect each other)

AOM = CON      (Vertically opposite angles)

MAO = OCN       (Alternate interior angles)

$\therefore$ By ASA congruence criteria,

(CPCT)

Hence, MN is bisected at O.

#### Page No 346: Given: In trapezium PQRS,
PQ || SR, is the midpoint of PS and MN || PQ.

To prove: N is the midpoint of QR.

Construction: Join QS.

Proof:

In SPQ,

Since, M is the mid-point of SP and MO || PQ.

Therefore, O is the mid-point of SQ.      (By Mid-point theorem)

Similarly, in SRQ,

Since, O is the mid-point of SQ and ON || SR           (SR || PQ and MN || PQ)

Therefore, N is the mid-point of QR.      (By Mid-point theorem)

#### Page No 346: Given: In parallelogram PQRS, PQ = 12 cm and PS = 9 cm. The bisector of ∠SPQ meets SR at M.

Let ∠SPQ = 2x.

⇒ ∠SRQ = 2x and ∠TPQ = x.

Also, PQ SR

⇒ ∠TMR = ∠TPQ = x.

In △TMR, SRQ is an exterior angle.

⇒ ∠SRQ = ∠TMR + ∠MTR

⇒ 2= + ∠MTR

⇒ ∠MTR = x

⇒ △TPQ is an isosceles triangle.

TQ = PQ = 12 cm

Now,

RT = TQ QR

= TQ PS

= 12 − 9

= 3 cm

#### Page No 346:

Given: AB || DC, AP = PD and BQ = CQ

(i) In ∆QCD and ∆QBE, we have:
DQC = ​BQE   (Vertically opposite angles)

DCQ = ∠EBQ     (Alternate angles, as AE || DC)
BQ = CQ               (P is the midpoints)
∴ ∆QCD ≅ ∆QBE
​Hence, DQ = QE                     (CPCT)

(ii) Now, in ∆ADE, P and Q are the midpoints of AD and DE, respectively.
∴ PQ || AE
PQ || AB || DC
⇒​ AB || PR || DC

(iii) PQAB and DC are the three lines cut by transversal AD at P such that AP = PD.
These lines
PQABDC are also cut by transversal BC at Q such that BQ = QC.
​ Similarly, lines PQAB and DC are also cut by AC at R.
∴ AR = RC                   (By intercept theorem)

#### Page No 346:

AD is a median of  ∆ABC.
∴ BD = DC
We know that the line drawn through the midpoint of one side of a triangle and parallel to another side bisects the third side.
Here, in ∆ABC, D is the mid point of BC and DE || BA (given). Then DE bisects AC.
i.e., AE =  EC
E is the midpoint of AC.
BE is the median of ∆ABC.

#### Page No 346:

In  ABC, we have:
AC = AE + EC     ...(i)
AE = EC               ...(ii)       [BE is the median of  ABC]
∴ AC = 2EC          ...(iii)

​In ∆BECDF || BE.
∴ EF = CF        (By midpoint theorem, as D is the midpoint of BC)
But EC = EF + CF
EC =​ 2 ⨯ ​CF       ...(iv)
From (iii) and (iv), we get:
AC = 2 ⨯ (2 ⨯​ CF)
∴​ CF =  $\frac{1}{4}AC$

#### Page No 346:

∆ABC is shown below.  D, E and F are the midpoints of sides AB, BC and CA, respectively. As, D and  E are the mid points of sides AB, and BC of ∆ ABC.
DE ∣∣ AC   (By midpoint theorem)
Similarly, DF ​∣∣ BC and EF ​∣∣ AB.
Therefore, ADEF, BDFE and DFCE are all parallelograms.
Now, DE is the diagonal of the parallelogram BDFE.
∴ ​∆BDE ≅ ​∆FED
Simiilarly, DF is the ​diagonal of the parallelogram ADEF.

∴ ∆DAF ≅ ∆FED
And, EF is the ​diagonal of the parallelogram DFCE.
∴ ∆EFC ≅ ∆FED
So, all the four triangles are congruent.

#### Page No 346:

∆ 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 AB and AC of ∆ ABC.
∴ 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:
A = ∠EDF          (Opposite angles are equal)
In parallelogram BDEF, we have:
B = DEF          (Opposite angles are equal)
In parallelogram DCEF, we have:
C = ∠​ DFE        (Opposite angles are equal)

#### Page No 347:

Let ABCD be the rectangle and P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively.
Join AC, a diagonal of the rectangle.
In ∆ ABC, we have:
∴ PQ ∣∣ AC and PQ$\frac{1}{2}$AC                    [By midpoint theorem]
Again, in ∆ DAC, the points S and R are the mid points of AD and DC, respectively.
∴ SR ∣∣ AC and SR = $\frac{1}{2}$AC                    [By midpoint theorem]
Now, PQ ∣∣ AC and SR ∣∣ AC
⇒ PQ ∣∣ SR

Also, PQ = SR                  [Each equal to $\frac{1}{2}$ AC ]         ...(i)
So, PQRS is a parallelogram.
Now, in ∆SAP and QBP, we have:
AS = BQ
A = ∠B = 90o
AP = BP
i.e., ∆SAP ≅ ∆QBP​
∴ PS = PQ                ...(ii)
Similarly, ∆SDR ≅ ∆QCR​
SR = RQ                ...(iii)
From (i), (ii) and (iii), we have:
PQ = PQ = SR = RQ
Hence, PQRS is a rhombus.

#### Page No 347: Let ABCD be a rhombus and P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively.
Join the diagonals, AC and BD.
In ∆ ABC, we have:
PQ ∣∣ AC and PQ = $\frac{1}{2}$AC                    [By midpoint theorem]
Again, in ∆DACthe points S and R are the midpoints of AD and DC, respectively.
∴ SR ∣∣ AC and SR = $\frac{1}{2}$AC                    [By midpoint theorem]
Now, PQ ∣∣ AC and SR ∣∣ AC  ⇒ PQ ∣∣ SR

Also, PQ = SR                  [Each equal to $\frac{1}{2}$ AC ]            ...(i)
So, PQRS is a parallelogram.
We know that the diagonals of a rhombus bisect each other at right angles.
∴ ∠EOF = 90o
​Now, RQ∣∣ DB
⇒RE ∣∣ FO
Also, SR∣∣ AC
FR ∣∣ OE
∴ OERF is a parallelogram.
So, ∠FRE = ∠EOF = 90o​         (Opposite angles are equal)
Thus, PQRS is a parallelogram with ∠R = 90o.
∴​ PQRS is a rectangle.

#### Page No 347: Let ABCD be a square and P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively.
Join the diagonals AC and BD. Let BD cut SR at F and AC cut RQ at E. Let O be the intersection point of AC and BD.

In ∆ ABC, we have:
∴ PQ ∣∣ AC and PQ = $\frac{1}{2}$AC                    [By midpoint theorem]
Again, in ∆DACthe points S and R are the midpoints of AD and DC, respectively.
∴ SR ∣∣ AC and SR = $\frac{1}{2}$AC                    [By midpoint theorem]
Now, PQ ∣∣ AC and SR ∣∣ AC  ⇒ PQ ∣∣ SR

Also, PQ = SR                  [ Each equal to $\frac{1}{2}$ AC ]             ...(i)
So, PQRS is a parallelogram.

Now, in ∆SAP and QBP, we have:
AS = BQ
A = ∠B = 90o
AP = BP
i.e.,  ∆SAP ≅ ∆QBP​
∴ PS = PQ                  ...(ii)
Similarly, ∆SDR ≅ ∆RCQ​
∴ SR = RQ                 ...(iii)
From (i), (ii) and (iii), we have:
PQ = PS = SR = RQ           ...(iv)

We know that the diagonals of a square bisect each other at right angles.
∴ ∠​EOF = 90o
​Now, RQ ∣∣ DB
RE ∣∣ FO
Also, SR ∣∣ AC
FR ∣∣ OE
∴ OERF is a parallelogram.
So, ∠FRE = ∠EOF = 90o​         (Opposite angles are equal)
Thus, PQRS is a parallelogram with ∠R = 90o and PQ = PS = SR = RQ.

∴​ PQRS is a square.

#### Page No 347:

Let ABCD be the quadrilateral in which P, Q, R, and S are the midpoints of sides AB, BC, CD, and DA, respectively.
Join PQ, QR, RS, SP and BD. BD is a diagonal of ABCD. In ΔABD, S and P are the midpoints of AD and AB, respectively.

∴​ SP || BD and SP BD ... (i)        (By midpoint theorem)

Similarly in Δ BCD, we have:

QR || BD and QR BD ... (ii)   (By midpoint theorem)

From equations (i) and (ii), we get:
SP ||  BD || QR

∴ SP || QR and SP = QR             [Each equal to BD]

In quadrilateral SPQR, one pair of the opposite sides is equal and parallel to each other.
∴  SPQR is a parallelogram.

We know that the diagonals of a parallelogram bisect each other.

∴ PR and QS bisect each other.

#### Page No 347: Given: In quadrilateral ABCDBD = AC and K, L, M and N are the mid-points of AD, CD, BC and AB, respectively.

To prove: KLMN is a rhombus.

Proof:

Since, K and L are the mid-points of sides AD and CD, respectively.

So, KL || AC and KL = $\frac{1}{2}$AC                   ...(1)

Similarly, in
ABC,

Since, M and N are the mid-points of sides BC and AB, respectively.

So, NM || AC and NM = $\frac{1}{2}$AC               ...(2)

From (1) and (2), we get

KL = NM and KL || NM

But this a pair of opposite sides of the quadrilateral KLMN.

So, KLMN is a parallelogram.

Now, iABD,

Since, K and N are the mid-points of sides AD and AB, respectively.

So, KN || BD and KN = $\frac{1}{2}$BD                ...(3)

But BD = AC            (Given)

$⇒\frac{1}{2}$BD = $\frac{1}{2}$AC

$⇒$KN = NM          [From (2) and (3)]

But these are a pair of adjacent sides of the parallelogram KLMN.

Hence, KLMN is a rhombus.

#### Page No 347: Given: In quadrilateral ABCDAC $\perp$ BDP, Q, R and S are the mid-points of AB, BC, CD and AD, respectively.

To prove: PQRS is a rectangle.

Proof:

In ΔABCP and Q are mid-points of AB and BC, respectively.

∴ PQ || AC and PQ = $\frac{1}{2}$AC       (Mid-point theorem)          ...(1)

Similarly, in ΔACD,

So, R and S are mid-points of sides CD and AD, respectively.

∴ SR || AC and SR = $\frac{1}{2}$AC          (Mid-point theorem)        ...(2)

From (1) and (2), we get

PQ || SR and PQ = SR

But this is a pair of opposite sides of the quadrilateral PQRS,

So, PQRS is parallelogram.

Now, in ΔBCDQ and R are mid-points of BC and CD, respectively.

∴ QR || BD and QR = $\frac{1}{2}$BD       (Mid-point theorem)          ...(3)

From (2) and (3), we get

SR || AC and QR || BD

But, AC  BD      (Given)

∴ RS  QR

Hence, PQRS is a rectangle.

#### Page No 347: AC BD and AC ⊥ BDP, Q, R and S are the mid-points of AB, BC, CD and AD, respectively.

To prove: PQRS is a square.

Construction: Join AC and BD.

Proof:

In ΔABC,

$\because$ P
and Q are mid-points of AB and BC, respectively.

$\therefore$ PQ || AC and PQ = $\frac{1}{2}$AC       (Mid-point theorem)           ...(1)

Similarly, in ΔACD,

$\because$ R and S are mid-points of sides CD and AD, respectively.

$\therefore$ SR || AC and SR = $\frac{1}{2}$AC        (Mid-point theorem)          ...(2)

From (1) and (2), we get

PQ || SR and PQ = SR

But this a pair of opposite sides of the quadrilateral PQRS.

So, PQRS is parallelogram.

Now, in ΔBCD,

$\because$ Q and R are mid-points of sides BC and CD, respectively.

$\therefore$ QR || BD and QR = $\frac{1}{2}$BD        (Mid-point theorem)          ...(3)

From (2) and (3), we get

RS || AC and QR || BD

But, AC  BD                     (Given)

∴ RS  QR

But this a pair of adjacent sides of the parallelogram PQRS.

So, PQRS is a rectangle.

Again, AC = BD                (Given)

$⇒$ $\frac{1}{2}$AC = $\frac{1}{2}$BD

$⇒$ RS = QR                   [From (2) and (3)]

But this a pair of adjacent sides of the rectangle PQRS.

Hence, PQRS is a square.

#### Page No 349:

(b) 73°​

Explanation:
Let the measure of the fourth angle be xo.
Since the sum of the angles of a quadrilateral is 360o, we have:
80o + 95o + 112ox = 360o
⇒ 287o
x = 360o
x = 73o
Hence, the measure of the fourth angle is 73o.

#### Page No 349:

(b) 60°​

Explanation:
Let ∠A = 3x​, ∠B = 4x, ∠C = 5x and ∠D = 6x.
Since the sum of the angles of a quadrilateral is 360o, we have:

3x + 4x + 5x + 6x = 360o
⇒
18x =
360o ​
⇒
x = 20
o
∴ ∠A = 60o​, ∠B = 80o, ∠C = 100o and ∠D = 120o
Hence, the smallest angle is
60$°$.

#### Page No 349:

(c) 45°

Explanation:
∠B = 180o − ∠A
⇒ ∠B = 180o − 75o = 105o
Now, ∠B =​ ∠ABD + ∠CBD
⇒​​ 105o​ = ∠ABD + 60o
⇒ ∠ABD​ = 105o − 60o = 45o
⇒ ∠ABD = ​∠BDC​ = 45         (Alternate angles)

#### Page No 349: We know that diagonals of rhombus bisect each other at 90°.

Then, in ΔBOC,

90° + 50° + ∠OBC = 180°        (Angle sum property of triangle)

$⇒$ ∠OBC = 180° $-$ 140°

$⇒$ OBC = 40°

But OBC = ADB     (Alternate interior angles)

Thus,

Hence, the corerct option is (a).

#### Page No 349:

(d) Rectangle.

The diagonals of a rectangle are equal.

#### Page No 349:

(d) rhombus

The diagonals of a rhombus bisect each other at right angles.

#### Page No 350:

(a) 10 cm

Explanation: Let ABCD be the rhombus.
∴ AB = BC = CD = DA
Here, AC and BD are the diagonals of ABCD, where AC = 16 cm and BD = 12 cm.
Let the diagonals intersect each other at O.
We know that the diagonals of a rhombus are perpendicular bisectors of each other.
∴​ ∆​AOB is a right angle triangle, in which OA = AC /2 = 16/2 = 8 cm and OB = BD/2 = 12/2 = 6 cm.
Now, AB2 = OA2 + OB2              [Pythagoras theorem]
⇒ ​AB2 = (8)2 + (6)2
⇒​ AB2 =​ 64 + 36 = 100
⇒​ AB =​ 10 cm

Hence, the side of the rhombus is 10 cm.

#### Page No 350:

(b) 12 cm

Explanation: Let ABCD be the rhombus.
∴ AB = BC = CD = DA = 10 cm
Let AC and BD be the diagonals of the rhombus.
Let
AC be x and BD be 16 cm and O be the intersection point of the diagonals.
We know that the diagonals of a rhombus are perpendicular bisectors of each other.
∴​ AOB is a right angle triangle in which OA = AC $÷$2 = $÷$
÷ 2  and OB = BD $÷$÷2 = 16 $÷$÷ 2 = 8 cm.

Now,
AB2= OA2 + OB2              [Pythagoras theorem]

102 = (x2)2 + 82100  64 = x2436 ×4 =

#### Page No 350: Given: In rectangle ABCD, ∠OAD = 35
°.

$⇒$ ∠OAB = 90° $-$ 35° = 55°

In ΔOAB,

Since, OA = OB      (Diagonals of a rectangle are equal and bisect each other)

$⇒$ OAB = OBA = 55°     (Angles opposite to equal sides are equal)

Now, in ΔODA,

55° + 55° + ∠DOA = 180°       (Angle sum property of a triangle)

$⇒$ DOA = 180° $-$ 110°

$⇒$ DOA = 70°

Thus, the acute angle between the diagonals is 70°.

Hence, the correct option is (b).

#### Page No 350:

(c) Rectangle

Explanation:
A = ∠B
Then A + ∠B = 180o
2A = 180o
⇒ ∠A​ = 90o
⇒ ∠A​ =​ ∠B​ =​∠C​ =​​∠D = 90o
∴​ The parallelogram is a rectangle.

#### Page No 350:

(b) 50o

​​Explanation: C = 70o and ∠D = 30o
Then A + ∠B = 360o - (70 +30)o = 260o
$\frac{1}{2}$(∠A +B) =$\frac{1}{2}$ (260o) = 130o
In ∆​ AOB, we have:
AOB​ = 180o - [$\frac{1}{2}$(∠A +B)​]
⇒ ∠AOB​ = ​180 - 130 = 50o

#### Page No 350:

(d) 90°

Explanation:
Sum of two adjacent angles = 180o
Now, sum of angle bisectors of two adjacent angles =
∴ Intersection angle of bisectors of two adjacent angles =  180o − 90o =  90o

#### Page No 350:

(c) Rectangle

The bisectors of the angles of a parallelogram encloses a rectangle.

#### Page No 350: Given: In quadrilateral ABCDAS, BQ, Cand DS are angle bisectors of angles A, B, C and D, respectively.

QPS = APB        (Vertically opposite angles)          ...(1)

In $∆$APB,

APB + PAB + ABP180°        (Angle sum property of triangle.)

$⇒$ ∠APB + $\frac{1}{2}$A + $\frac{1}{2}$B = 180°

$⇒$ APB  = 180° – $\frac{1}{2}$(∠A + B)                  ...(2)

From (1) and (2), we get

QPS = 180° – $\frac{1}{2}$(∠+ ∠B)                          ...(3)

Similarly, QRS = 180° – $\frac{1}{2}$(∠C + D)        ...(4)

From (3) and (4), we get

QPS + QRS = 360° – $\frac{1}{2}$(∠A + ∠B + C + D)

= 360° – $\frac{1}{2}$(360°)

= 360° – 180°

= 180°

So, PQRS is a quadrilateral whose opposite angles are supplementary.

Hence, the correct option is (d).

#### Page No 350:

(d) parallelogram

The figure formed by joining the mid points of the adjacent sides of a quadrilateral is a parallelogram.

#### Page No 350:

(b) Square

The figure formed by joining the mid points of the adjacent sides of a square is a square.

#### Page No 350:

(d) parallelogram.
The figure made by joining the mid points of the adjacent sides of a parallelogram is  a parallelogram.

#### Page No 350:

(a) rhombus

The figure formed by joining the mid points of the adjacent sides of a rectangle is a rhombus.

#### Page No 351:

(c) Rectangle

The figure formed by joining the mid points of the adjacent sides of a rhombus is a rectangle.

#### Page No 351:

Since,

The quadrilateral formed by joining the mid-points of the sides of a parallelogram is parallelogram ,

The quadrilateral formed by joining the mid-points of the sides of a rectangle is rhombus,

​The quadrilateral formed by joining the mid-points of the sides of a quadrilateral with diagonals equal is rhombus, and

​The quadrilateral formed by joining the mid-points of the sides of a quadrilateral with diagonals perpendicular to each other is rectangle.

Hence, the correct option is (d).

#### Page No 351:

Given: The quadrilateral ABCD is a rhombus.

So, the sides AB, BC, CD and AD are equal.

Now, in $∆PQS,$ we have

$DC=\frac{1}{2}QS$     (Using mid-point theorem)          ...(1)

Similarly, in $∆PSR,$

$BC=\frac{1}{2}PR$                     ...(2)

As, BC = DC

[From (1) and (2)]

So, QS = PR

Thus, the diagonals of PQRS are equal.

Hence, the correct option is (c).

#### Page No 351:

Since,

The quadrilateral formed by joining the mid-points of the sides of a rhombus is rectangle,

The quadrilateral formed by joining the mid-points of the sides of a quadrilateral with diagonals equal is rhombus,

​The quadrilateral formed by joining the mid-points of the sides of a quadrilateral with diagonals perpendicular is rectangle, and

​The quadrilateral formed by joining the mid-points of the sides of a quadrilateral with diagonals equal and perpendicular is square.

Hence, the correct option is (d).

#### Page No 351:

(c) 72°​

Explanation:
Let ABCD be a parallelogram.
∴A = ∠C and ∠B = ∠D      (Opposite angles)
Let Ax and ∠​B$\frac{2}{3}x$(4x5)
∴ ∠​A + ∠B = 180o                (Adjacent angles are supplementary)
x $\frac{2}{3}$x 180o
$\frac{5}{3}x=180°$
⇒ x = 108o
∴ B = $\frac{2}{3}×$ (108o) = 72o
Hence
AC  = 108o  and B = D = 72o
9x5 = 180ox = 100oA = 100o an

#### Page No 351:

(c)112°​

Explanation:
Let ABCD is a parallelogram.
∴ ∠​A = ∠C and ∠​B = ∠D          (Opposite angles)
Let A be the smallest angle whose measure is x.
∴​∠B  = (2x − 24)o
Now, ∠​A + ∠B = 180o                (Adjacent angles are supplementary)
x + 2x − 24o = 180o
3x =  204o
⇒ x = 68o
∴​∠​B = 2 ⨯ 68o − 24o = 112o
HenceA = C68o and B = D = 112o

#### Page No 351:

(c) Trapezium

Explanation:

Let the angles be (3x), (7x), (6x) and (4x)​.
Then 3x + 7x + 6x + 4x = 360o
x = 18o
​Thus, the angles are 3 ⨯18o = 54o, ​7 ⨯ 18o = 126o, ​6 ⨯ 18o = 108o and ​4 ⨯18o = 72o.
But 54o + 126o = 180o and 72o + 108o  = 180o
ABCD is a trapezium.

#### Page No 351:

(c) Opposite angles are bisected by the diagonals.

#### Page No 351:

(c) Rectangle

If APB and CQD are two parallel lines, then the bisectors of APQ, ∠BPQ, ∠CQP and ∠PQD enclose a rectangle.

#### Page No 352:

(c)​ 60°

Explanation:
∠BAD = ∠BCD = 75o       [Opposite angles are equal]
In ∆ BCD, ∠ C = 75o
∴ ​∠CBD = 180o (75o + 45o) = 60o

#### Page No 352:

(c) A < B

Explanation:
Let h be the height of parallelogram.
Then clearly, h < b
∴ ​A = a ⨯​ h < a ⨯ b = B
Hence, A < B

#### Page No 352:

(b) AF = 2 AB

Explanation:
I​n parallelogram ABCD, we have:
AB || DC
DCE = ∠​ EBF            (Alternate interior angles)
In ∆ DCE and ​ ∆ BFE, we have:
DCE = ∠ EBF              (Proved above)

DEC = ∠ BEF              (Vertically opposite angles)
BECE           ( Given)
i.e., ∆ DCE ≅​ ∆ BFE     (By ASA congruence rule)
∴  DC = BF         (CPCT)

But DC= AB, as ABCD is a parallelogram.
DC = AB =  BF                 ...(i)

Now, AF = AB + BF             ...(ii)
From (i), we get:
∴ AF = AB + AB = 2AB

#### Page No 352: Given: In ∆ABCMND and E are the mid-points of BP, CP, AB and AC, respectivley.

In ∆ABP,

$\because$ D and M are the mid-points of AB,and BP, respectively.      (Given)

$\therefore$ BM$\frac{1}{2}$AP and BM || AP          (Mid-point theorem)       ...(i)

Again, in ∆ACP,

$\because$ E and N are the mid-points of AC,and CP, respectively.      (Given)

$\therefore$ EN = $\frac{1}{2}$AP and EN || AP          (Mid-point theorem)       ...(ii)

From (i) and (ii), we get

BMEN and BM || EN

But this a pair of opposite sides of the quadrilateral DENM.

So, DENM is a parallelgram.

Hence, the correct option is (b).

#### Page No 352:

(b) $\frac{1}{2}\left(a+b\right)$

Explanation: Suppose ABCD is a trapezium.
Draw EF parallel to AB.
Join BD to cut EF at M.
Now, in ∆ DAB, E is the midpoint of AD and EM || AB.
∴ M is the mid point of BD and EM =
$\frac{1}{2}\left(a\right)$
Similarly, M is the mid point of BD and MF || DC.
i.e., F is the midpoint of BC and MF = $\frac{1}{2}\left(b\right)$

∴ EF =  EM + MF$\frac{1}{2}\left(a+b\right)$

#### Page No 352:

(d)

Explanation: Join CF and produce it to cut AB at G.
Then ∆CDF  ≅ GBF                [∵ DF = BF, ​DCF = ​BGF and ​CDF = ​GBF]
∴ CD = GB
Thus, in
∆​CAG, the points E and F are the mid points of AC and CG, respectively.
∴ EF$\frac{1}{2}\left(AG\right)$ =

#### Page No 352:

(c) ​90°

Explanation:
∠B = ∠D
$\frac{1}{2}$∠B = $\frac{1}{2}$∠D​
∴ ∆ABD is an isosceles triangle and M is the midpoint of BD. We can also say that M is the median of ∆ABD.
∴ AM ⊥ BD and, hence, ​∠AMB =​​ 90°

#### Page No 352:

(c) ​AC2 + BD2 = 4AB2

Explanation:
We know that the diagonals of a rhombus bisect each other at right angles.
Here, OA = $\frac{1}{2}$AC, OB = $\frac{1}{2}$BD and ∠​AOB ​= 90°
Now, AB2= OA2 + OB2$\frac{1}{4}$(AC)2$\frac{1}{4}$(BD)2
∴ 4AB2 = (AC2 + BD2)

#### Page No 353:

(c) BC2 + AD2 + 2AB.CD

Explanation: Draw perpendicular from D and C on AB which meets AB at E and F, respectively.
∴​ DEFC is a parallelogram and EF = CD.

In ∆ABC, ∠B is acute.
∴ AC2BC2 + AB2 - 2AB.AE
In ∆ABD, ∠A is acute.​
∴ ​BDAD2 + AB2 - 2AB.AF
∴ ​AC2BD2 = (BC2AD2) + (AB2 + AB2 ) - 2AB(AE + BF)
= (BC2 + AD2) + 2AB(AB - AE - BF)                [∵ AB = AE + EF + FB and AB - AE =  BE]
= (BC2 + AD2) + 2AB(BE - BF)
= (BC2 + AD2) + 2AB.EF
AC2 + BD2 = ​(BC2 + AD2) + 2AB.CD

#### Page No 353:

(d) 1:1

Area of a parallelogram = base ⨯ height
If both parallelograms stands on the same base and between the same parallels, then their heights are the same.
So, their areas will also be the same.

#### Page No 353:

(b) ⅓ AC

Explanation: Let G be the mid point of FC. Join DG.
​In BCF, D is the mid point of BC and G is the mid point of FC.
∴ DG || BF
⇒ DG || EF

​In ∆ ADG, E is the mid point of AD and EF || DG.
i.e.,
F is the mid point of AG.
Now,
AF = FG = GC       [∵ G is the mid point of FC]
∴ AF =⅓ AC
4AC

#### Page No 353:

(a) 40°

Explanation:
OAD = ​∠OCB = 30o              (Alternate interior angles)
AOB + ∠BOC = 180o              (Linear pair of angles)
∴ ∠BOC = 180o − 70o = 110o       (∠ AOB = 70o)
In ∆BOC, we have:
OBC = 180o − (110o + 30o) = 40o
∴ ​∠DBC = 40o

#### Page No 353:

(c) I and II

Statements I and II are true. Statement III is false, as the triangle formed by joining the midpoints of the sides of an isosceles triangle is an isosceles triangle.

#### Page No 353:

(b) II and III

Clearly, statements II and III are true. Statement I is false, as diagonal of a rectangle does not bisect ∠A and ∠C (∴ adjacent sides are not equal).

#### Page No 353:

Since, the opposite angles of the quadrilateral PQRS are equal.

$⇒$ Quadrilateral PQRS is a parallelogram.

(Opposite sides of a parallelogram are equal.)

#### Page No 354:

No, the statement is incorrect because the diagonals of a parallelogram bisect each other.

#### Page No 354:

Since, in quadrilateral PQRS,P + ∠S = 180°.

i.e. the sum of adjacent angles is 180$°$.

So, PQRS is a parallelogram.

#### Page No 354:

No, the statement is false because if all the four angles of a quadrilateral are less than 90$°$, then the sum of all four angles will be less than 360$°$.

#### Page No 354:

Yes, the statement is true because all the angles of a quadrilateral such as rectangle and square are right angles.

#### Page No 354:

No, the statement is false because if all angles are greater than 90$°$, then the sum of four obtuse angles will be greater than 360$°$.

#### Page No 354:

Since, the sum of all angles (i.e. 70° + 115° + 60° + 120° = 365$°$).

So, we cannot form a quadrilateral with these angles.

#### Page No 354:

We know that, sum of all angles in a quadrilateral is 360°.

Let each angle of the quadrilateral be x.

x + x = 360°

⇒ 4x = 360°

⇒ x = 90°

⇒ All angles of the quadrilateral are 90°.

Hence, given quadrilateral is a rectangle.

#### Page No 354:

In $∆$ABC,

Since, D and E are respectively the mid-points of sides AB and BC    (Given)

So, DE = $\frac{1}{2}$AC       (Uing mid-point theorem)

But AC = 3.6 cm      (Given)

DE $\frac{1}{2}$(3.6)

or, DE = 1.8 cm

Hence, the length of DE is 1.8 cm.

#### Page No 354:

Since, the diagonals PR and QS of quadrilateral PQRS bisect each other.        (Given)

So, PQRS is a parallelogram.

Now, $\angle Q+\angle R=180°$(Adjacent angles are supplementary.)

$⇒56°+\angle R=180°\phantom{\rule{0ex}{0ex}}⇒\angle R=180°-56°\phantom{\rule{0ex}{0ex}}⇒\angle R=124°$

#### Page No 354: Given: Parallelograms BDEF and AFDE.

Since, BF = DE     (Opposite sides of parallelogram BDEF)       ...(i)

And, AF = DE     (Opposite sides of parallelogram AFDE)       ...(ii)

From (i) and (ii), we get

AF = FB

#### Page No 354:

We know that if the diagonals of a ​quadrilateral bisects each other, then it is a parallelogram.
If the diagonals of a quadrilateral are equal, then it is not necessarily a ​parallelogram.
∴ II does not give the answer.

Hence, the correct answer is (a).

#### Page No 355:

Clearly, I alone is not sufficient to answer the given question.
Also, II alone is not sufficient to answer the given question.​
However, both I and II together will give the answer.
∴ Hence, the correct answer is (c).

#### Page No 355:

When the diagonals of a parallelogram are equal, it is either a rectangle or a square.
Also, if the diagonals intersects at a right angle, then it is a square.
∴ Both I and II together will give the answer.
Hence, the correct answer is (c).

#### Page No 355:

We know that a quadrilateral is a parallelogram when either I or II holds true.
​Hence, the correct answer is (b).

#### Page No 355:

(a)  Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

Explanation:
Fourth angle = 360o − (130o + 70o + 60o) = 100o
Clearly, reason (R) and assertion (A) are both true and the reason gives the assertion.
Hence, the correct answer is (a).

#### Page No 355:

(a)  Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

​Explanation:
Clearly, reason (R) and assertion (A) are both true and reason (R) gives assertion (A).
Hence, the correct answer is (a).

#### Page No 355:

(b)  Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.

​Explanation:
Clearly, reason (R) and assertion (A) are both true.

But reason (R) does not give assertion (A).
Hence, the correct answer is (b).

#### Page No 356:

(d) Assertion is false and Reason is true.

​Explanation:
We can easily prove reason (R). So, reason (R) is true.

Clearly,  assertion (A) is false (as every parallelogram is not necessarily a rectangle).
Hence, the correct answer is (d).

#### Page No 356:

​(b)  Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.

​Explanation:
Clearly, assertion (A) is true.
We can easily prove reason (R). So, (R) is also true.

But, reason (R) does not give assertion (A).
Hence, the correct answer is (b)

#### Page No 356:

(a) - (q), (b) - (r), (c) - (s), (d) - (p)

#### Page No 356:

(a) PQ$\frac{1}{2}$(AB+ CD) = $\frac{1}{2}$(17) = 8.5 cm
(b) OR$\frac{1}{2}$(PR) = ​$\frac{1}{2}$(13) = 6.5 cm