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#### Page No 287:

#### Question 1:

Three angles of a quadrilateral measure 56°, 115° and 84°. Find the measure of the fourth angle.

#### Answer:

Let the measure of the fourth angle be *x*^{o}.

Since the sum of the angles of a quadrilateral is 360^{o}, we have:

56 + 115 + 84 + *x* = 360

⇒ 255 + *x* = 360

⇒ *x* = 105

Hence, the measure of the fourth angle is 105^{o}.

#### Page No 287:

#### Question 2:

The angles of a quadrilateral are in the ratio 2: 4 : 5 : 7. Find the angles.

#### Answer:

Let $\angle $A = 2*x*^{∘}.

Then $\angle $B = (4*x*)^{∘}; $\angle $C = (5*x*)^{∘} and $\angle $D = (7*x*)^{∘}

Since the sum of the angles of a quadrilateral is 360^{o}, we have:

2*x* + 4*x* + 5*x* + 7*x* = 360^{∘}

⇒ 18 *x* = 360^{∘}

⇒ *x* = 20^{∘}

∴ $\angle $A = 40^{∘}; $\angle $B = 80^{∘}; $\angle $C = 100^{∘}; $\angle $D = 140^{∘}

#### Page No 287:

#### Question 3:

In the adjoining figure, *ABCD* is a trapezium in which *AB* || *DC*. If ∠*A* = 55° and ∠*B* = 70°, find ∠*C* and ∠*D*.

#### Answer:

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 288:

#### Question 4:

In the adjoining figure, *ABCD* is a square and ∆*EDC* is an equilateral triangle. Prove that (i) *AE* = *BE*, (ii) ∠*DAE* = 15°.

#### Answer:

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]

$\angle $ADE = $\angle $BCE = 90

^{∘}

^{ }+ 60

^{∘}= 150

^{∘}

*ADE*

*≅*∆

*BCE*

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 = $\frac{1}{2}\left(180\xb0-150\xb0\right)=\frac{30\xb0}{2}=15\xb0$

#### Page No 288:

#### Question 5:

In the adjoining figure, *BM* ⊥ *AC* and *DN* ⊥ *AC*. If *BM* = *DN*, prove that AC bisects *BD*.

#### Answer:

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 ∆ON*D** *and ∆OMB, we have:

∠OND = ∠OMB (90^{o} 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 288:

#### Question 6:

In the given figure, *ABCD* is a quadrilateral in which *AB* = *AD* and *BC* = *DC*. Prove that

(i) AC bisects ∠*A* and ∠*C*,

(ii) *BE* = *DE*,

(iii) ∠*ABC* = ∠*ADC*.

#### Answer:

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

(i)

In ∆ABC* *and ∆ADC*, *we have:

AB = AD (Given)

AC is common.

i.e., ∆ABC ≅ ∆ADC

*(SSS congruence rule)*

Thus, AC bisects ∠A and ∠ C.

(ii)

Now, in ∆ABE and ∆ADE

*,*we have:

AB = AD (Given)

∠BAE = ∠DAE (Proven above)

AE is common.

∴ ∆ABE ≅ ∆ADE

*(SAS congruence rule)*

⇒ BE = DE (By CPCT)

(iii) ∆ABC ≅ ∆ADC (Proven above)

∴ ∠ABC = ∠ADC (By CPCT)

#### Page No 288:

#### Question 7:

In the given figure, *ABCD* is a square and ∠*PQR* = 90°. If *PB* = *QC* = *DR*, prove that

(i) *QB* = *RC*,

(ii) *PQ* = *QR*,

(iii) *QPR* = 45°.

#### Answer:

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 (90^{o} each)

i.e., ∆RCQ ≅ ∆QBP (SAS congruence rule)

∴ QR = PQ (By CPCT)

(iii) ∆RCQ ≅ ∆QBP and QR = PQ (Proven above)

∴ In ∆RPQ, ∠QPR = ∠QRP = $\frac{1}{2}\left(180\xb0-90\xb0\right)=\frac{90\xb0}{2}={45}^{\xb0}$

#### Page No 288:

#### Question 8:

If *O* is a point within a quadrilateral *ABCD*, show that *OA* + *OB* + *OC* + *OD* > *AC* + *BD*.

#### Answer:

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*

*BOD*,

*OB + OD > BD*

*OA + OC*) + (

*OB + OD*) > (

*AC + BD*)

⇒

*OA + OB + OC + OD*>

*AC + BD*

#### Page No 288:

#### Question 9:

In the adjoining figure, *ABCD* is a quadrilateral and *AC* is one of its diagonals. Prove that:

(i) *AB* + *BC* + *CD* + *DA* > 2*AC*

(ii) *AB* + *BC* + *CD* > *DA*

(iii) *AB* + *BC* + *CD* + *DA* > *AC* + *BD*

#### Answer:

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)

*ACD*,

*CD*+

*DA*>

*AC*...(2)

*AB*+

*BC*+

*CD*+

*DA*> 2

*AC*

(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*

*ACD*,

*CD*+

*DA*>

*AC*

*BCD*,

*BC*+

*CD*>

*BD*

*ABD*,

*DA*+

*AB*> BD

Adding these inequalities, we get:

2(

*AB*+

*BC*+

*CD*+

*DA)*>

*2(AC*+

*BD*)

⇒ (

*AB*+

*BC*+

*CD*+

*DA)*>

*(AC*+

*BD*)

#### Page No 288:

#### Question 10:

Prove that the sum of all the angles of a quadrilateral is 360°.

#### Answer:

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* = 180^{o} ...(i)

In ∆*BCD*, we have:

∠3 + ∠4 + ∠*C* = 180^{o} ...(ii)

On adding (i) and (ii), we get:

^{o}

⇒ ∠

*A*+ ∠

*C*+ ∠

*B*+ ∠

*D*= 360

^{o}[ ∵ ∠1 + ∠3 = ∠

*B*; ∠4 + ∠2 = ∠

*D*]

∴ ∠

*A*+ ∠

*C*+ ∠

*B*+ ∠

*D*= 360

^{o}

#### Page No 300:

#### Question 1:

In the adjoining figure, *ABCD* is a parallelogram in which ∠*A* = 72°. Calculate ∠*B*, ∠*C* and ∠*D*.

#### Answer:

ABCD is parallelogram and ∠*A* = 72°.

We know that opposite angles of a parallelogram are equal.

∴∠*A* = ∠*C *and* *∠*B* = ∠*D*

∴ ∠C = 72^{o}

∠*A* and ∠B are adajcent angles.

i.e., ∠*A* + ∠B = 180^{o}

⇒ ∠B = 180^{o}^{ } *−* ∠*A*

**⇒ ∠B = 180^{o} *−* 72^{o} = 108^{o}

∴ ∠B = ∠*D** = *108^{o}

Hence, ∠B = ∠*D** = *108^{o} and ∠C = 72^{o}

#### Page No 300:

#### Question 2:

In the adjoining figure, *ABCD* is a parallelogram in which ∠*DAB* = 80° and ∠*DBC* = 60°. Calculate ∠*CDB* and ∠*ADB*.

#### Answer:

Given: ABCD is parallelogram and ∠*DAB* = 80° and ∠*DBC* = 60°

To find: Measure of ∠*CDB* and ∠*ADB*

In parallelogram *ABCD, AD || BC*

∴ ∠*DBC* = ∠ *ADB* = 60^{o} (Alternate interior angles) ...(i)

As ∠*DAB* and ∠*ADC** *are adajcent angles, ∠*DAB* + ∠*ADC* = 180^{o}

∴ ∠*ADC* = 180^{o}^{ } − ∠*DAB*

* *⇒∠*ADC* = 180^{o} − 80^{o} = 100^{o}

Also, ∠*ADC* = ∠*ADB* + ∠*CDB*

∴ ∠*ADC* =* *100^{o}

⇒ ∠*ADB* + ∠*CDB = *100^{o} ...(ii)

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

60^{o} + ∠*CDB = *100^{o}* *

⇒ ∠*CDB = *100^{o} − 60^{o} = 40^{o}

Hence, ∠*CDB* = 40^{o} and ∠*ADB* = 60^{o}

#### Page No 301:

#### Question 3:

In the adjoining figure, *ABCD* is a parallelogram in which ∠*A* = 60°. If the bisectors of ∠*A* and ∠*B* meet *DC* at *P*, prove that (i) ∠*APB* = 90°, (ii) *AD* = *DP* and *PB* = *PC* *= BC*, (iii) *DC* = 2*AD*.

#### Answer:

*ABCD* is a parallelogram.

∴ ∠*A* = ∠*C *and* *∠*B* = ∠D (Opposite angles)

And ∠*A* + ∠*B* = 180^{o} (Adjacent angles are supplementary)

∴ ∠*B* = 180^{o} − ∠*A
⇒ *180

^{o}− 60

^{o}= 120

^{o}( ∵∠

*A*= 60

^{o})

∴ ∠

*A*= ∠

*C =*60

^{o}

*and*

*∠*

*B*= ∠D = 120

^{o}

(i) In

*∆ APB,*∠

*PAB*= $\frac{60\xb0}{2}=30\xb0$

^{ }and ∠

*PBA*= $\frac{120\xb0}{2}=60\xb0$

∴ ∠

*APB* = 180

^{o}− (30

^{o}+ 60

^{o}) = 90

^{o}

(ii) In

*∆ ADP,*∠PAD = 30

^{o}and ∠

*ADP*= 120

^{o}

∴ ∠

*APB*= 180

^{o}− (30

^{o}+ 120

^{o}) = 30

^{o}

Thus, ∠

*PAD*= ∠

*APB*= 30

^{o}

Hence,

*∆ADP*is an isosceles triangle and

*AD = DP*.

In

*∆ PBC,*∠ PBC = 60

^{o},

*∠*

*BPC*= 180

^{o}− (90

^{o}+30

^{o}) = 60

^{o}

^{ }and

^{ }∠ BCP = 60

^{o}(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]

⇒

*DC = AD + AD*

*DC = 2 AD*

#### Page No 301:

#### Question 4:

In the adjoining figure, *ABCD* is a parallelogram in which ∠*BAO* = 35°, ∠*DAO* = 40° and ∠*COD* = 105°. Calculate (i) ∠*ABO*, (ii) ∠*ODC*, (iii) ∠*ACB*, (iv) ∠*CBD*.

#### Answer:

ABCD is a parallelogram.

∴ AB ∣∣ DC and BC ∣∣ AD

(i) In ∆AOB, ∠*BAO* = 35°, ∠*AOB* = ∠*COD* = 105° (Vertically opposite angels)

∴ ∠*ABO* = 180^{o} − (35^{o} + 105^{o}) = 40^{o}

(ii)∠*ODC *and ∠*ABO *are alternate interior angles.

∴ ∠*ODC *= ∠*ABO = *40^{o}

(iii) ∠*ACB** =* ∠CAD = 40^{o}^{ }(Alternate interior angles)

(iv) ∠*CBD** = *∠*ABC** *−* *∠*ABD** ...*(i)

*ABC*

*=*180

^{o}− ∠

*BAD*

^{ }(Adjacent angles are supplementary)

**⇒∠

*ABC*

*=*180

^{o}− 75

^{o}= 105

^{o}

^{ }

⇒∠

*CBD*

*=*105

^{o}−

*∠*

*ABD*

*(∠*

*ABD*

*=*∠

*ABO*)

⇒∠

*CBD*

*=*105

^{o}

*−*

*40*

^{o}

*=*65

^{o}

#### Page No 301:

#### Question 5:

In a || gm *ABCD*, if ∠*A* = (2*x* + 25)° and ∠*B* = (3*x* − 5)°, find the value of *x* and the measure of each angle of the parallelogram.

#### Answer:

ABCD is a parallelogram.

i.e., ∠*A** = *∠*C** *and* *∠*B** = *∠D (Opposite angles)

Also, ∠*A** *+* *∠B = 180^{o} (Adjacent angles are supplementary)

∴ (2*x* + 25)° + (3*x* − 5)° = 180

⇒ 5*x* +20 = 180

⇒ 5*x = *160

⇒* x* = 32^{o}

∴∠*A = *2* ⨯* 32 + 25 = 89^{o} and ∠*B = *3 *⨯* 32 − 5 = 91^{o}

Hence, *x* = 32^{o}, ∠*A** = *∠*C** = *89^{o} and ∠*B** = *∠*D** = *91^{o}* *

#### Page No 301:

#### Question 6:

If an angle of a parallelogram is four-fifths of its adjacent angle, find the angles of the parallelogram.

#### Answer:

Let *ABCD* be a parallelogram.

∴ ∠*A* = ∠*C *and* *∠*B* = ∠*D** *(Opposite angles)

Let* *∠*A* = *x*^{o} and ∠*B* = ${\left(\frac{4x}{5}\right)}^{\xb0}$

Now, ∠*A* + ∠*B* = 180^{o} (Adjacent angles are supplementary)

$\Rightarrow x+\frac{4x}{5}={180}^{\mathrm{o}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{9x}{5}={180}^{\mathrm{o}}\phantom{\rule{0ex}{0ex}}\Rightarrow x={100}^{\mathrm{o}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\angle A={100}^{\mathrm{o}}\mathrm{and}\angle B=\left(\frac{4}{5}\right)\times 100\xb0={80}^{\mathrm{o}}\phantom{\rule{0ex}{0ex}}$

Hence*, *∠*A** = *∠*C* = 100^{o};* *∠*B* = ∠*D** = *80^{o}

#### Page No 301:

#### Question 7:

Find the measure of each angle of a parallelogram, if one of its angles is 30° less than twice the smallest angle.

#### Answer:

Let *ABCD* be a parallelogram.

∴ ∠*A* = ∠*C *and* *∠*B* = ∠*D** *(Opposite angles)

Let* *∠*A* be the smallest angle whose measure is *x*^{o}.

∴ ∠*B* = (2*x* − 30)^{o}

Now, ∠*A* + ∠*B* = 180^{o} (Adjacent angles are supplementary)

⇒ *x* + 2*x* − 30^{o} = 180^{o}

⇒ 3*x* = 210^{o}

⇒ *x* = 70^{o}

∴ ∠*B* = 2 ⨯ 70^{o} − 30^{o} = 110^{o}

Hence*, *∠*A** = *∠*C* = 70^{o}; ∠*B* = ∠*D** = *110^{o}

#### Page No 301:

#### Question 8:

*ABCD *is a parallelogram in which *AB* = 9.5 cm and its perimeter is 30 cm. Find the length of each side of the parallelogram.

#### Answer:

*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 + 2*x* = 30

⇒ 2*x* = 11

⇒ *x* = 5.5 cm

Hence, *AB = DC = *9.5 cm and *BC = DA* = 5.5 cm

#### Page No 301:

#### Question 9:

In each of the figures given below, *ABCD* is a rhombus. Find the value of *x* and *y* in each case.

#### Answer:

ABCD is a rhombus and a rhombus is also a parallelogram. A rhombus has four equal sides.

(i) In ∆*ABC*, ∠*BAC* = ∠*BCA* = $\frac{1}{2}\left(180-110\right)={35}^{\mathrm{o}}$

i.e., *x* = 35^{o}

Now, ∠*B* + ∠*C* = 180^{o} (Adjacent angles are supplementary)

But ∠C = *x* +* y* = 70^{o}*
*⇒

*y =*70

^{o }−

^{ }

*x*

⇒

*y*= 70

^{o}− 35

^{o}= 35

^{o}

Hence

*, x =*35

^{o};

*y =*35

^{o}

(ii) The diagonals of a rhombus are perpendicular bisectors of each other.

*So,*

*in ∆*

*AOB*, ∠

*OAB*= 40

^{o}, ∠

*AOB*= 90

^{o}and ∠

*ABO*= 180

^{o}− (40

^{o}+ 90

^{o}) = 50

^{o}

∴

*x*= 50

^{o}

In ∆

*ABD*,

*AB = AD*

So, ∠

*ABD*= ∠

*ADB*= 50

^{o}

Hence

*, x =*50

^{o};

*y =*50

^{o}

(iii) ∠

*BAC*= ∠

*DCA*(Alternate interior angles)

i.e.,

*x*= 62

^{o}

In ∆

*BOC*

*,*∠

*BCO*= 62

^{o}

^{ }[In ∆

*ABC*,

*AB*=

*BC*, so ∠

*BAC*= ∠

*ACB*]

Also, ∠

*BOC*= 90

^{o}

∴ ∠

*OBC*= 180

^{o}− (90

^{o}+ 62

^{o}) = 28

^{o}

Hence

*, x =*62

^{o};

*y =*28

^{o}

#### Page No 301:

#### Question 10:

The lengths of the diagonals of a rhombus are 24 cm and 18 cm respectively. Find the length of each side of the rhombus.

#### Answer:

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, *AB ^{2}= OA^{2} + OB^{2}* [Pythagoras theorem]

⇒

*AB*

^{2}= (12)

^{2}+ (9)

^{2}

⇒

*AB*

^{2}= 144 + 81 = 225

⇒

*AB*= 15 cm

Hence, the side of the rhombus is 15 cm

*.*

#### Page No 301:

#### Question 11:

Each side of a rhombus is 10 cm long and one of its diagonals measures 16 cm. Find the length of the other diagonal and hence find the area of the rhombus.

#### Answer:

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 ^{$\xf7$}* 2 =

*x*2 and

^{$\xf7$}*OB = BD*2 = 16

^{$\xf7$}^{$\xf7$ }2 = 8 cm.

Now,

*AB*[Pythagoras theorem]

^{2}= OA^{2}+ OB^{2}$\Rightarrow {10}^{2}={\left(\frac{x}{2}\right)}^{2}+{8}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 100-64=\frac{{x}^{2}}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow 36\times 4=x$

^{2}

*x*

^{2}

*x*

Hence, the other diagonal of the rhombus is 12 cm

*.*

∴ Area of the rhombus = $\frac{1}{2}\times \left(12\times 16\right)=96{\mathrm{cm}}^{2}$

#### Page No 301:

#### Question 12:

In each of the figures given below, *ABD* is a rectangle. Find the values of *x* and *y* in each case.

#### Answer:

(i) *ABCD* is a rectangle.

The diagonals of a rectangle are congruent and bisect each other. Therefore, in ∆ AOB, we have:

*OA = OB*

∴ ∠*OAB = *∠O*BA = *35^{o}

∴* x* = 90^{o} − 35^{o} = 55^{o}

And ∠*AOB* = 180^{o} − (35^{o} + 35^{o}) = 110^{o}

∴ *y* = ∠*AOB* = 110^{o} [Vertically opposite angles]

Hence, *x* = 55^{o} and *y* = 110^{o}

(ii) In ∆AOB, we have:

^{ } *OA = OB*

Now, ∠*OAB = *∠O*BA** = $\frac{1}{2}\times \left(180\xb0-110\xb0\right)={35}^{\mathrm{o}}$*

∴ *y* = ∠BAC *= *35^{o} [Interior alternate angles]

Also, *x* = 90^{o} − *y* [ ∵∠C = 90^{o} = *x* + *y* ]

⇒ *x* = 90^{o} − 35^{o} = 55^{o}

Hence, *x* = 55^{o} and *y* = 35^{o}

#### Page No 302:

#### Question 13:

In the adjoining figure, *ABCD* is a square. *A* line segment *CX* cuts *AB* at *X* and the diagonal *BD* at *O* such that ∠*COD* = 80° and ∠*OXA* = *x*°. Find the value of *x*.

#### Answer:

The angles of a square are bisected by the diagonals.

∴ ∠*OBX* = 45^{o}^{ }[∵∠*ABC* = 90^{o} and *BD* bisects ∠*ABC*]

And ∠*BOX* = ∠*COD* = 80^{o} [Vertically opposite angles]

∴ In ∆*BOX*, we have:

*∠AXO = ∠OBX *+* ∠BOX* [Exterior angle of ∆*BOX*]

⇒ ∠*AXO* = 45^{o} + 80^{o} = 125^{o}

∴ *x* =125^{o}

#### Page No 302:

#### Question 14:

In the adjoining figure, *AL* and *CM* are perpendiculars to the diagonal *BD* of a || gm *ABCD*. Prove that (i) ∆*ALD* ≅ ∆*CMB*, (ii) *AL* = *CM*.

#### Answer:

ABCD is a parallelogram.

∴ AD || BC and AD = BC

(i) In ∆*ALD** *and* *∆*CMB**,* we have:

AD = BC

∠*ALD = *∠*CMB *(90^{o} each)

∠*ADL = *∠*CBM *(Alternate interior angle)

∴ ∆*ALD** *≅* * ∆*CMB*

(ii) As ∆*ALD** *≅ ∆*CMB** *(Proved above)

∴ *AL = CM* (By CPCT)

#### Page No 302:

#### Question 15:

In the adjoining figure, *ABCD* is a parallelogram in which the bisectors of ∠*A* and ∠*B* intersect at a point *P*. Prove that ∠*APB* = 90°.

#### Answer:

*ABCD* is a parallelogram and ∠*A* and ∠*B* are adjacent angles.

∴ ∠ *A* + ∠*B* = 180^{o}.

⇒ $\frac{\angle A}{2}+\frac{\angle B}{2}=\frac{180}{2}={90}^{o}....................\left(i\right)$

In ∆ *APB*, we have:

∠*PAB* = ∠*A* /2

∠*PBA* = ∠*B* /2

∴ ∠*APB = *180 −* (*∠*PAB* + ∠*PBA*)^{ }[ Angle sum property of triangle] ^{ }

⇒ ∠*APB* = 180 − $\left(\frac{\angle A}{2}+\frac{\angle B}{2}\right)$

⇒ ∠*APB* = 180 − 90 = 90^{o}

Hence, proved.

#### Page No 302:

#### Question 16:

In the adjoining figure, *ABCD* is a parallelogram. If *P* and *Q* are points on *AD* and *BC* respectively such that $AP=\frac{1}{3}AD$ and $CQ=\frac{1}{3}BC$, prove that *AQCP* is a parallelogram.

#### Answer:

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 $AP=\frac{1}{3}AD\mathrm{and}CQ=\frac{1}{3}BC$.

∴ *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 302:

#### Question 17:

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:

In ∆*ODF* and ∆*OBE*, we have:

*OD = OB* (Diagonals bisects each other)

∠*DOF* = ∠*BOE** *(Vertically opposite angles)

∠F*DO* = ∠*OBE** * (Alternate interior angles)

i.e., ∆*ODF* ≅ ∆*OBE*

∴ *OF = OE *(CPCT)

Hence, proved.

#### Page No 302:

#### Question 18:

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:

*ODC*and ∆

*OEB*, we have:

*DC = BE*(∵

*DC = AB*)

∠C

*OD*= ∠

*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 302:

#### Question 19:

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

#### Answer:

**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)

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 303:

#### Question 20:

A ∆*ABC* is given. If lines are drawn through *A*, *B*, *C*, parallel respectively to the sides *BC*, *CA* and *AB*, forming ∆*PQR*, as shown in the adjoining figure, show that $BC=\frac{1}{2}QR$.

#### Answer:

* 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* = 2*BC*

∴ *BC* = $\frac{1}{2}QR$

#### Page No 303:

#### Question 21:

In the adjoining figure, ∆*ABC* is a triangle and through *A*, *B*, *C*, lines are drawn, parallel respectively to *BC*, *CA* and *AB*, intersecting at *P*, *Q* and *R*. Prove that the perimeter of ∆*PQR* is double the perimeter of ∆*ABC*.

#### Answer:

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* = 2*BC*

∴ *BC* = $\frac{1}{2}QR$

Similarly, *CA* = $\frac{1}{2}PQ$ and *AB* = $\frac{1}{2}PR$

*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 314:

#### Question 1:

In the adjoining figure, *ABCD* is a trapezium in which *AB* || *DC* and *E* is the midpoint of *AD*. A line segment *EF* || *AB* meets *BC* at *F*. Show that *F* is the midpoint of *BC*.

#### Answer:

Join *BD* to cut *EF* at *M*.

Now, in ∆*DAB*, *E* is the mid point of *AD* and *EM || AB.*

∴ *M* is the midpoint of *BD*. (By converse of mid point theorem)

Again, in ∆*BDC*, *M* is the mid point of *BD* and *MF || DC.*

∴ *F* is the midpoint of *BC.* (By converse of mid point theorem)

#### Page No 314:

#### Question 2:

In the adjoining figure, *ABCD* is a || gm in which *E* and *F* are the midpoints of *AB* and *CD* respectively. If *GH* is a line segment that cuts *AD*, *EF* and *BC* at *G*, *P* and *H* respectively, prove that *GP* = *PH*.

#### Answer:

In parallelogram *ABCD*, we have:

*AD || BC *and* AB || DC*

*AD = BC*and

*+*

*AB = DC*

AB = AEAB = AE

*and*

*BE**+*

*DC = DF*

*FC**AE = BE*=

*DF = FC*

*,*

*DF = AE*and DF || AE.

*AEFD*is a parallelogram.

**∴

*AD || EF*

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 314:

#### Question 3:

In the adjoining figure, *ABCD* is a trapezium in which *AB* || *DC* and *P*, *Q* are the midpoints of *AD* and *BC* respectively. *DQ* and *AB* when produced meet at *E*. Also, *AC* and *PQ* intersect at *R*. Prove that (i) *DQ* = *QE*, (ii) *PR* || *AB*, (iii) *AR* = *RC*.

#### Answer:

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

⇒ AB || PR || DC

(iii)

*PQ*

*, AB*and

*DC*are the three lines cut by transversal

*AD*at

*P*such that

*AP = PD*.

These lines

*PQ*

*, AB,*

*DC*are also cut by transversal

*BC*at

*Q*such that

*BQ = QC*.

Similarly, lines

*PQ*

*,*

*AB*and

*DC*are also cut by

*AC*at

*R*.

∴

*AR = RC*(By intercept theorem)

#### Page No 314:

#### Question 4:

In the adjoining figure, *AD* is a median of ∆*ABC* and *DE* || *BA*. Show that *BE* is also a median of ∆*ABC*.

#### Answer:

*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 315:

#### Question 5:

In the adjoining figure, *AD* and *BE* are the medians of ∆*ABC* and *DF* || *BE*. Show that $CF=\frac{1}{4}AC$.

#### Answer:

In * *∆*ABC**,* we have:

* AC = AE + EC *...(i)

*AE = EC ...*(ii) [*BE* is the median of ∆*ABC*]

∴ *AC = 2EC ...*(iii)

In ∆*BEC*, *DF* || *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 315:

#### Question 6:

In the adjoining figure, *ABCD* is a parallelogram. *E* is the midpoint of *DC* and through *D*, a line segment is drawn parallel to *EB* to meet *CB* produced at *G* and it cuts *AB* at *F*. Prove that (i) $AD=\frac{1}{2}GC$, (ii) *DG* = 2*EB*.

#### Answer:

(i) In ∆ *DCG, *we have:

* DG* || *EB*

* DE = EC *(*E *is the midpoint of* DC*)

Also, *GB = BC* (By midpoint theorem)

∴ *B* is the midpoint of *GC*.

Now, *GC = GB + BC*

⇒ *GC* = 2*BC*

⇒* GC** = *2 ⨯* AD *[*AD = BC*]

∴ *AD = *$\frac{1}{2}GC$

(ii) In ∆ *DCG, **DG* || *EB *and *E* is the midpoint of *DC* and *B* is the midpoint of *GC*.

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

i.e., *EB* = $\frac{1}{2}DG$

∴ *DG* = 2 ⨯ *EB*

#### Page No 315:

#### Question 7:

Prove that the line segments joining the middle points of the sides of a triangle divide it into four congruent triangles.

#### Answer:

∆*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 315:

#### Question 8:

In the adjoining figure, *D*, *E*, *F* are the midpoints of the sides *BC*, *CA* and *AB* respectively, of ∆*ABC*. Show that ∠*EDF* = ∠*A*, ∠*DEF* = ∠*B* and ∠*DEF* = ∠*C*.

Figure

#### Answer:

∆ *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 315:

#### Question 9:

Show that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a rectangle is a rhombus.

#### Answer:

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 ∆ D*AC**, *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*

*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*= 90

^{o}

*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 315:

#### Question 10:

Show that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a rhombus is a rectangle.

#### Answer:

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 ∆D*AC**, *the 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*

*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*= 90

^{o}

Now,

*RQ*∣∣

*DB*

⇒

*RE*∣∣

*FO*

Also,

*SR*∣∣

*AC*

⇒

*FR*∣∣

*OE*

∴

*OERF*is a parallelogram.

So, ∠

*FRE*= ∠

*EOF*= 90

^{o} (Opposite angles are equal)

Thus,

*PQRS*is a parallelogram with ∠

*R*= 90

^{o}.

∴

*PQRS*is a rectangle.

#### Page No 315:

#### Question 11:

Show that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a square is a square.

#### Answer:

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 ∆D*AC**, *the 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*

*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*= 90

^{o}

*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*= 90

^{o}

Now,

*RQ*∣∣

*DB*

⇒

*RE*∣∣

*FO*

Also,

*SR*∣∣

*AC*

⇒

*FR*∣∣

*OE*

∴

*OERF*is a parallelogram.

So, ∠

*FRE*= ∠

*EOF*= 90

^{o} (Opposite angles are equal)

Thus,

*PQRS*is a parallelogram with ∠

*R*= 90

^{o}and

*PQ = PS = SR = RQ.*

∴

*PQRS*is a square.

#### Page No 315:

#### Question 12:

Prove that the line segments joining the midpoints of opposite sides of a quadrilateral bisect each other.

#### Answer:

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 Δ*ABD*, *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 315:

#### Question 13:

In the given figure, *ABCD* is a quadrilateral whose diagonals intersect at right angles. Show that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides is a rectangle.

#### Answer:

*ABCD* is a quadrilateral and *P*,* Q*,* R* and *S* be the midpoints of *AB, BC, CD *and* DA,* respectively.

*AC *and* BD* are the diagonals which intersect each other at O. *RQ* intersects *AC* at *E* and *SR* intersects *BD* at *F*.

In ∆ *ABC, *we have:

∴ *PQ* ∣∣ *AC* and *PQ* = $\frac{1}{2}$*AC* [By midpoint theorem]

Again, in ∆D*AC**, *the 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*

*PQ = SR*[Each equal to $\frac{1}{2}$

*AC*] ...(i)

So,

*PQRS*is a parallelogram.

We know that the diagonals of the given quadrilateral bisect each other at right angles.

∴ ∠

*EOF*= 90

^{o}

Now,

*RQ*∣∣

*DB*

⇒

*RE*∣∣

*FO*

Also,

*SR*∣∣

*AC*

⇒

*FR*∣∣

*OE*

∴

*OERF*is a parallelogram.

So, ∠

*FRE*= ∠

*EOF*= 90

^{o} (Opposite angles are equal)

Thus,

*PQRS*is a parallelogram with ∠

*R*= 90

^{o}.

∴

*PQRS*is a rectangle.

#### Page No 318:

#### Question 1:

Three angles of a quadrilateral are 80°, 95° and 112°. Its fourth angle is

(a) 78°

(b) 73°

(c) 85°

(d) 100°

#### Answer:

(b) 73°

Explanation:

Let the measure of the fourth angle be *x*^{o}.

Since the sum of the angles of a quadrilateral is 360^{o}, we have:

80^{o} + 95^{o} + 112^{o} + *x* = 360^{o}

⇒ 287^{o} + *x* = 360^{o}

⇒ *x* = 73^{o}

Hence, the measure of the fourth angle is 73^{o}.

#### Page No 318:

#### Question 2:

The angles of a quadrilateral are in the ratio 3 : 4 : 5 : 6. The smallest of these angles is

(a) 45°

(b) 60°

(c) 36°

(d) 48°

#### Answer:

(b) 60°

Explanation:

Let ∠A = 3*x*, ∠B = 4*x*, ∠C = 5*x* and ∠D = 6*x.*

Since the sum of the angles of a quadrilateral is 360^{o}, we have:

3*x* + 4*x* + 5*x* + 6*x* = 360^{o}

⇒ 18*x* = 360^{o}

⇒ *x* = 20^{o}

∴ ∠A = 60^{o}, ∠B = 80^{o}, ∠C = 100^{o} and ∠D = 120^{o}

^{}^{Hence, the smallest angle is }^{60$\xb0$.}

#### Page No 318:

#### Question 3:

In the given figure, *ABCD* is a parallelogram in which ∠*BAD* = 75° and ∠*CBD* = 60°. Then, ∠*BDC* = ?

(a) 60°

(b) 75°

(c) 45°

(d) 50°

#### Answer:

(c) 45°

Explanation:

∠B = 180^{o} − ∠A

⇒ ∠B = 180^{o} − 75^{o} = 105^{o}

^{}Now, ∠B = ∠ABD + ∠CBD

⇒ 105^{o} = ∠ABD + 60^{o}

⇒ ∠ABD = 105^{o} − 60^{o} = 45^{o}

⇒ ∠ABD = ∠BDC = 45^{o } (Alternate angles)

#### Page No 318:

#### Question 4:

In which of the following figures are the diagonals equal?

(a) Parallelogram

(b) Rhombus

(c) Trapezium

(d) Rectangle

#### Answer:

(d) Rectangle.

The diagonals of a rectangle are equal.

#### Page No 318:

#### Question 5:

If the diagonals of a quadrilateral bisect each other at right angles, then the figure is a

(a) trapezium

(b) parallelogram

(c) rectangle

(d) rhombus

#### Answer:

(d) rhombus

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

#### Page No 318:

#### Question 6:

The lengths of the diagonals of a rhombus are 16 cm and 12 cm. The length of each side of the rhombus is

(a) 10 cm

(b) 12 cm

(c) 9 cm

(d) 8 cm

#### Answer:

(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, *AB ^{2}^{ }= OA^{2} + OB^{2}* [Pythagoras theorem]

⇒

*AB*

^{2}= (8)

^{2}+ (6)

^{2}

⇒

*AB*

^{2 }= 64 + 36 = 100

⇒

*AB*= 10 cm

Hence, the side of the rhombus is 10 cm.

#### Page No 318:

#### Question 7:

The length of each side of a rhombus is 10 cm and one if its diagonals is of length 16 cm. The length of the other diagonal is

(a) 13 cm

(b) 12 cm

(c) $2\sqrt{39}\mathrm{cm}$

(d) 6 cm

#### Answer:

(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 *^{$\xf7$}2 = *x ^{$\xf7$}*

*2 and*

^{÷}*OB = BD*

^{$\xf7$}*2 = 16*

^{÷}^{$\xf7$}

^{÷}

^{ }2 = 8 cm.

Now,

*AB*[Pythagoras theorem]

^{2}= OA^{2}+ OB^{2}$\Rightarrow {10}^{2}={\left(\frac{x}{2}\right)}^{2}+{8}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(\frac{x}{2}\right)}^{2}=36={6}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow x=2\times 6=12\mathrm{cm}$

#### Page No 319:

#### Question 8:

If *ABCD* is a parallelogram with two adjacent angles ∠*A* = ∠*B*, then the parallelogram is a

(a) rhombus

(b) trapezium

(c) rectangle

(d) none of these

#### Answer:

(c) Rectangle

Explanation:

∠*A* = ∠*B*

**Then* *∠*A* + ∠*B** = *180^{o}

⇒ *2*∠*A** = *180^{o}

⇒ ∠*A* = 90^{o}

⇒ ∠*A* = ∠*B* =∠*C* =∠*D = *90^{o}

∴ The parallelogram is a rectangle.

#### Page No 319:

#### Question 9:

In a quadrilateral *ABCD*, if *AO* and *BO* are the bisectors of ∠*A* and ∠*B* respectively, ∠*C* = 70° and ∠*D* = 30°. Then, ∠*AOB* = ?

(a) 40°

(b) 50°

(c) 80°

(d) 100°

#### Answer:

(b) 50^{o}

Explanation:

∠*C* = 70^{o} and ∠*D *=* *30^{o}

**Then* *∠*A* + ∠*B** = *360^{o }- (70 +30)^{o} = 260^{o}

∴* $\frac{1}{2}$*(∠*A +*∠*B*)* =$\frac{1}{2}$ *(260^{o}) = 130^{o}

In ∆ *AOB*, we have:

∠*AOB* = 180^{o} - [*$\frac{1}{2}$*(∠*A +*∠*B*)]

⇒ ∠*AOB* = 180 - 130 = 50^{o}

#### Page No 319:

#### Question 10:

The bisectors of any two adjacent angles of a parallelogram intersect at

(a) 40°

(b) 45°

(c) 60°

(d) 90°

#### Answer:

(d) 90°

Explanation:

Sum of two adjacent angles = 180^{o}

Now, sum of angle bisectors of two adjacent angles = $\frac{1}{2}\times \left({180}^{\mathrm{o}}\right)={90}^{\mathrm{o}}$

∴ Intersection angle of bisectors of two adjacent angles = 180^{o} − 90^{o} = 90^{o}

#### Page No 319:

#### Question 11:

The bisectors of the angles of a parallelogram enclose a

(a) rhombus

(b) square

(c) rectangle

(d) parallelogram

#### Answer:

(c) Rectangle

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

#### Page No 319:

#### Question 12:

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

(a) rhombus

(b) square

(c) rectangle

(d) parallelogram

#### Answer:

(d) parallelogram

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

#### Page No 319:

#### Question 13:

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

(a) rhombus

(b) square

(c) rectangle

(d) parallelogram

#### Answer:

(b) Square

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

#### Page No 319:

#### Question 14:

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

(a) rhombus

(b) square

(c) rectangle

(d) parallelogram

#### Answer:

(d) parallelogram.

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

#### Page No 319:

#### Question 15:

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

(a) rhombus

(b) square

(c) rectangle

(d) parallelogram

#### Answer:

(a) rhombus

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

#### Page No 319:

#### Question 16:

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

(a) rhombus

(b) square

(c) rectangle

(d) parallelogram

#### Answer:

(c) Rectangle

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

#### Page No 319:

#### Question 17:

If an angle of a parallelogram is two-third of its adjacent angle, the smallest angle of the parallelogram is

(a) 108°

(b) 54°

(c) 72°

(d) 81°

#### Answer:

(c) 72°

Explanation:

Let *ABCD* be a parallelogram.

∴ ∠*A* = ∠*C* and ∠*B* = ∠*D* (Opposite angles)

Let* *∠*A* = *x* and ∠*B* = $\frac{2}{3}x$

∴ ∠*A* + ∠*B* = 180^{o} (Adjacent angles are supplementary)

*⇒* *x*^{ }+ $\frac{2}{3}$*x*^{ }= 180^{o}

*⇒ $\frac{5}{3}x=180\xb0$*

*⇒ x* = 108^{o}

∴ ∠*B* = $\frac{2}{3}\times $ (108^{o}) = 72^{o}

Hence*, *∠*A** = *∠*C* = 108^{o}* *and ∠*B* = ∠*D** = *72^{o}

#### Page No 319:

#### Question 18:

If one angle of a parallelogram is 24° less than twice the smallest angle, then the largest angle of the parallelogram is

(a) 68°

(b) 102°

(c) 112°

(d) 136°

#### Answer:

(c)112°

Explanation:

Let *ABCD* is a parallelogram.

∴ *∠** A = ∠C *and

*(Opposite angles)*

*∠**B*= ∠*D*Let

*∠*

*A*be the smallest angle whose measure is

*x*.

∴∠

*B*= (2

*x*− 24)

^{o}

Now, ∠

*A*+ ∠

*B*= 180

^{o}(Adjacent angles are supplementary)

⇒

*x*+ 2

*x*− 24

^{o}= 180

^{o}

⇒ 3

*x*= 204

^{o}

⇒

*x*= 68

^{o}

∴∠

*B*= 2 ⨯ 68

^{o}− 24

^{o}= 112

^{o}

Hence

*,*∠

*A*

*=*∠

*C*= 68

^{o}

*and*

*∠*

*B*= ∠

*D*

*=*112

^{o}

#### Page No 320:

#### Question 19:

In the given figure, *ABCD* is a parallelogram in which ∠*BDC* = 45° and ∠*BAD* = 75°. Then, ∠*CBD* = ?

(a) 45°

(b) 55°

(c) 60°

(d) 75°

#### Answer:

(c) 60°

Explanation:

∠BAD = ∠BCD = 75^{o} [Opposite angles are equal]

In ∆ BCD, ∠ C = 75^{o}

∴ ∠CBD = 180^{o} − (75^{o} + 45^{o}) = 60^{o}

#### Page No 320:

#### Question 20:

If area of a || gm with sides *a* and *b* is *A* and that of a rectangle with sides *a* and *b* is *B*, then

(a) *A* > *B*

(b) *A* = *B*

(c) *A* < *B*

(d) *A* ≥ *B*

#### Answer:

(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 320:

#### Question 21:

In the given figure, *ABCD* is a || gm and *E* is the mid-point of *BC*. Also, *DE* and *AB* when produced meet at *F*. Then,

(a) $AF=\frac{3}{2}AB$

(b) *AF* = 2*AB*

(c) *AF* = 3*AB*

(d) *AF*^{2} = 2*AB*^{2}

#### Answer:

(b) *AF* = 2 *AB*

Explanation:

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)

*BE*=

*CE*( 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 =*2

*AB*

#### Page No 320:

#### Question 22:

The parallel sides of a trapezium are *a* and *b* respectively. The line joining the mid-points of its non-parallel sides will be

(a) $\frac{1}{2}(a-b)$

(b) $\frac{1}{2}(a+b)$

(c) $\frac{2ab}{(a+b)}$

(d) $\sqrt{ab}$

#### Answer:

(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 320:

#### Question 23:

In a trapezium *ABCD*, if *E* and *F* be the mid-point of the diagonals *AC* and *BD* respectively. Then, *EF* = ?

(a) $\frac{1}{2}AB$

(b) $\frac{1}{2}CD$

(c) $\frac{1}{2}(AB+CD)$

(d) $\frac{1}{2}(AB-CD)$

#### Answer:

(d) $\frac{1}{2}\left(AB-CD\right)$

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* = $\frac{1}{2}\left(AG\right)$ = $\frac{1}{2}\left(AB-GB\right)=\frac{1}{2}\left(AB-CD\right)$

#### Page No 320:

#### Question 24:

In the given figure, *ABCD* is a parallelogram, *M* is the mid-point of *BD* and *BD* bisects ∠*B* as well as ∠*D*. Then, ∠*AMB* = ?

(a) 45°

(b) 60°

(c) 90°

(d) 30°

#### Answer:

(c) 90°

Explanation:

∠B = ∠D

⇒ $\frac{1}{2}$∠B = $\frac{1}{2}$∠D

⇒ ∠ADB = ∠ABD

∴ ∆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 321:

#### Question 25:

In the given figure, *ABCD* is a rhombus. Then,

(a) *AC*^{2} + *BD*^{2} = *AB*^{2}

(b) *AC*^{2} + *BD*^{2} = 2*AB*^{2}

(c) *AC*^{2} + *BD*^{2} = 4*AB*^{2}

(d) 2(*AC*^{2} + *BD*^{2}) = 3*AB*^{2}

#### Answer:

(c) *AC*^{2} + *BD*^{2} = 4*AB*^{2}

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, *AB*^{2}= *OA*^{2} + *OB*^{2} = $\frac{1}{4}$(*AC*)^{2} + $\frac{1}{4}$(*BD*)^{2}

∴ 4*AB*^{2} = (*AC*^{2} + *BD*^{2})

#### Page No 321:

#### Question 26:

In a trapezium *ABCD*, if *AB* || *CD*, then (*AC*^{2} + *BD*^{2}) = ?

(a) *BC*^{2} + *AD*^{2} + 2*BC* ⋅ *AD*

(b) *AB*^{2} + *C**D*^{2}^{ }+ 2*AB* ⋅ *CD*

(c) *AB*^{2} + *C**D*^{2} + 2*AD* ⋅ *BC*

(d) *BC*^{2} + *AD*^{2} + 2*AB* ⋅ *CD*

#### Answer:

(c) *BC*^{2} + *AD*^{2} + 2*AB**.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.

∴ *AC*^{2}= *BC*^{2} + *AB*^{2} - 2*AB.AE*

In ∆*ABD*, ∠*A* is acute.

∴ *BD*^{2 }= *AD*^{2} + *AB*^{2} - 2*AB.AF*

∴ *AC*^{2} + *BD*^{2} = (*BC*^{2} + *AD*^{2}) + (*AB*^{2}^{ }+^{ }*AB*^{2} ) - 2*AB*(*AE* + *BF*)

= (*BC*^{2} + *AD*^{2}) + 2*AB*(AB - *AE* - *BF*) [∵ *AB *=* AE + EF + FB and AB - AE = BE*]

= (*BC*^{2} + *AD*^{2}) + 2*AB*(*BE* - *BF*)

= (*BC*^{2} + *AD*^{2}) + 2*AB**.EF *

*∴ **AC*^{2} + *BD*^{2} = (*BC*^{2} + *AD*^{2}) + 2*AB**.CD *

#### Page No 321:

#### Question 27:

Two parallelograms stand on equal bases and between the same parallels. The ratio of their areas is

(a) 1 : 2

(b) 2 : 1

(c) 1 : 3

(d) 1 : 1

#### Answer:

(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 321:

#### Question 28:

In the given figure, *AD* is a median of ∆*ABC* and *E* is the mid-point of *AD*. If *BE* is joined and produced to meet *AC* in *F*, then *AF* = ?

(a) $\frac{1}{2}AC$

(b) $\frac{1}{3}AC$

(3) $\frac{2}{3}AC$

(4) $\frac{3}{4}AC$

#### Answer:

(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*

#### Page No 321:

#### Question 29:

If ∠*A*, ∠*B*, ∠*C* and ∠*D* of a quadrilateral *ABCD* taken in order, are in the ratio 3 : 7 : 6 : 4, then *ABCD* is a

(a) rhombus

(b) kite

(c) trapezium

(d) parallelogram

#### Answer:

(c) Trapezium

Explanation:

Let the angles be (3*x*), (7*x*), (6*x*) and (4*x*).

Then 3*x* + 7*x* + 6*x* + 4*x* = 360^{o}

∴ *x* = 18^{o}

Thus, the angles are 3 ⨯18^{o} = 54^{o}, 7 ⨯ 18^{o} = 126^{o}, 6 ⨯ 18^{o} = 108^{o} and 4 ⨯18^{o} = 72^{o}.

But 54^{o} + 126^{o} = 180^{o} and 72^{o} + 108^{o} = 180^{o}

∴ *ABCD* is a trapezium.

#### Page No 321:

#### Question 30:

Which of the following is not true for a parallelogram?

(a) Opposite sides are equal.

(b) Opposite angles are equal.

(c) Opposite angles are bisected by the diagonals.

(d) Diagonals bisect each other.

#### Answer:

(c) Opposite angles are bisected by the diagonals.

#### Page No 321:

#### Question 31:

If *APB* and *CQD* are two parallel lines, then the bisectors of ∠*APQ*, ∠*BPQ*, ∠*CQP* and ∠*P**QD* enclose a

(a) square

(b) rhombus

(c) rectangle

(d) kite

#### Answer:

(c) Rectangle

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

#### Page No 322:

#### Question 32:

The diagonals *AC* and *BD* of a parallelogram *ABCD* intersect each other at the point *O* such that ∠*DAC* = 30° and ∠*AOB* = 70°. Then, ∠*DBC* = ?

(a) 40°

(b) 35°

(c) 45°

(d) 50°

#### Answer:

(a) 40°

Explanation:

∠*OAD* = ∠*OCB* = 30^{o} (Alternate interior angles)

∠*AOB* + ∠*BOC* = 180^{o} (Linear pair of angles)

∴ ∠*BOC* = 180^{o} − 70^{o} = 110^{o} (∠ *AOB* = 70^{o})

In ∆*BOC*, we have:

∠*OBC* = 180^{o} − (110^{o} + 30^{o}) = 40^{o}

∴ ∠*DBC* = 40^{o}

#### Page No 322:

#### Question 33:

Three statements are given below:

I. In a || gm, the angle bisectors of two adjacent angles enclose a right angle.

II. The angle bisectors of a || gm form a rectangle.

III. The triangle formed by joining the mid-points of the sides of an isosceles triangle is not necessarily an isosceles triangle.

Which is true?

(a) I only

(b) II only

(c) I and II

(d) II and III

#### Answer:

(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 322:

#### Question 34:

Three statements are given below:

I. In a rectangle *ABCD*, the diagonal *AC* bisects ∠*A* as well as ∠*C*.

II. In a square *ABCD*, the diagonal *AC* bisects ∠*A* as well as ∠*C*.

III. In a rhombus *ABCD*, the diagonal *AC* bisects ∠*A* as well as ∠*C*.

(a) I only

(b) II and III

(c) I and III

(d) I and II

#### Answer:

(b) II and III

*A*and ∠

*C*

*(∴ adjacent sides are not equal).*

#### Page No 322:

#### Question 35:

Is quadrilateral *ABCD* a || gm?

I. Diagonals *AC* and *BD* bisect each other.

II. Diagonals *AC* and *BD* are equal.

(a) if the question can be answered by one of the given statements alone and not by the other;

(b) if the question can be answered by either statement alone;

(c) if the question can be answered by both the statements together but not by any one of the two;

(d) if the question cannot be answered by using both the statements together.

#### Answer:

We know that if the diagonals of a quadrilateral bisects each other, then it is a parallelogram.

∴ I gives the answer.

If the diagonals of a quadrilateral are equal, then it is not necessarily a parallelogram.

∴ II does not give the answer.

#### Page No 323:

#### Question 36:

Is quadrilateral *ABCD* a rhombus?

I. Quad. *ABCD* is a || gm.

II. Diagonals *AC* and *BD* are perpendicular to each other.

(a) if the question can be answered by one of the given statements alone and not by the other;

(b) if the question can be answered by either statement alone;

(c) if the question can be answered by both the statements together but not by any one of the two;

(d) if the question cannot be answered by using both the statements together.

#### Answer:

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 323:

#### Question 37:

Is || gm *ABCD* a square?

I. Diagonals of || gm *ABCD* are equal.

II. Diagonals of || gm *ABCD* intersect at right angles.

(a) if the question can be answered by one of the given statements alone and not by the other;

(b) if the question can be answered by either statement alone;

(c) if the question can be answered by both the statements together but not by any one of the two;

(d) if the question cannot be answered by using both the statements together.

#### Answer:

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 323:

#### Question 38:

Is quad. *ABCD* a parallelogram?

I. Its opposite sides are equal.

II. Its opposite angles are equal.

(a) if the question can be answered by one of the given statements alone and not by the other;

(b) if the question can be answered by either statement alone;

(c) if the question can be answered by both the statements together but not by any one of the two;

(d) if the question cannot be answered by using both the statements together.

#### Answer:

#### Page No 323:

#### Question 39:

**Assertion:** If three angles of a quadrilateral are 130°, 70° and 60°, then the fourth angles is 100°.

**Reason:** The sum of all the angle of a quadrilateral is 360°.

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

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

(c) Assertion is true and Reason is false.

(d) Assertion is false and Reason is true.

#### Answer:

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

Explanation:

Fourth angle = 360^{o} − (130^{o} + 70^{o} + 60^{o}) = 100^{o}

Clearly, reason (R) and assertion (A) are both true and the reason gives the assertion.

Hence, the correct answer is (a).

#### Page No 323:

#### Question 40:

**Assertion:** *ABCD* is a quadrilateral in which *P*, *Q*, *R* and *S* are the mid-points of *AB*, *BC*, *CD* and *DA* respectively. Then, *PQRS* is a parallelogram.

**Reason:** 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.

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

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

(c) Assertion is true and Reason is false.

(d) Assertion is false and Reason is true.

#### Answer:

(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 323:

#### Question 41:

**Assertion:*** *In a rhombus *ABCD*, the diagonal *AC* bisects ∠*A* as well as ∠*C*.

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

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

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

(c) Assertion is true and Reason is false.

(d) Assertion is false and Reason is true.

#### Answer:

(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.

#### Page No 324:

#### Question 42:

**Assertion:*** *Every parallelogram is a rectangle.

**Reason:** The angle bisectors of a parallelogram form a rectangle.

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

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

(c) Assertion is true and Reason is false.

(d) Assertion is false and Reason is true.

#### Answer:

(d) Assertion is false and Reason is true.

Explanation:

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

#### Page No 324:

#### Question 43:

**Assertion:*** *The diagonals of a || gm bisect each other.

**Reason:** If the diagonals of a || gm are equal and intersect at right angles, then the parallelogram is a square.

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

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

(c) Assertion is true and Reason is false.

(d) Assertion is false and Reason is true.

#### Answer:

(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.

#### Page No 324:

#### Question 44:

Match the following columns:

Column I |
Column II |

(a) Angle bisectors of a parallelogram form a | (p) parallelogram |

(b) The quadrilateral formed by joining the mid-points of the pairs of adjacent sides of a square is a | (q) rectangle |

(c) The quadrilateral formed by joining the mid-points of the pairs of adjacent sides of a rectangle is a | (r) square |

(d) The figure formed by joining the mid-points of the pairs of adjacent sides of a quadrilateral is a | (s) rhombus |

(b) .....,

(c) .....,

(d) .....,

#### Answer:

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

#### Page No 324:

#### Question 45:

Match the following columns:

Column I |
Column II |

(a) In the given figure, ABCD is a trapezium in which AB = 10 cm and CD = 7 cm. If P and Q are the mid-points of AD and BC respectively, then PQ = |
(p) equal |

(b) In the given figure, PQRS is a || gm whose diagonals intersect at O. If PR = 13 cm, then OR = |
(q) at right angles |

(c) The diagonals of a square are | (r) 8.5 cm |

(d) The diagonals of a rhombus bisect each other | (s) 6.5 cm |

(b) ......,

(c) ......,

(d) ......,

#### Answer:

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

Explanation:

(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

#### Page No 330:

#### Question 1:

Which is false?

(a) In a || gm, the diagonals are equal.

(b) In a || gm, the diagonals bisect each other.

(c) If a pair of opposite sides of a quadrilateral are equal, then it is a || gm.

(d) If the diagonals of a || gm are perpendicular to each other, then it is a rhombus.

#### Answer:

(a) In a || gm, the diagonals are equal.

#### Page No 330:

#### Question 2:

If *P* is a point on the median *AD* of a ∆*ABC*, then ar(∆*ABP*) = ar(∆*ACP*).

(a) True

(b) False

#### Answer:

(a) True

Explanation:

In ∆*ABC**, AD *is the median*. *

∴ *BD = DC*

Let *h* be the height of ∆*ABC**.*

∴ ar(∆*ABD*)* = *ar(∆*ADC*) [∵ Area = $\frac{1}{2}$base ⨯ *h *and base BD = DC]

Now, let *h*_{1} be the height of ∆*BPD** *and* *∆*PDC**.*

∴ ar(∆*BPD*)* = *ar(∆*PDC*)

We know that ar(∆*ABD) = ar *(∆*BPD* ) + *ar *(∆A*BP* )

and ar(∆*ADC) = ar*(∆*PDC*) + *ar*(∆A*CP*)

∴ ar(∆A*BP*)* = *ar(∆*ACP*)

#### Page No 330:

#### Question 3:

The angles of a quadrilateral are in the ratio 1 : 3 : 5 : 6. Find its greatest angle.

#### Answer:

Let the angles of the parallelogram be *x*, 3*x*, 5*x* and 6*x*.

∴ *x *+ 3*x** *+ 5*x** *+* *6*x** *= 360^{o}

⇒ 15*x* = 360^{o}

⇒*x* = 24^{o}

Thus, the angles are 24^{o}, 24^{o} ⨯ 3 = 72^{o}, 24^{o} ⨯ 5 = 120^{o} and 24^{o} ⨯ 6 = 144^{o}.

Hence, the greatest angle is 144^{o}.

#### Page No 330:

#### Question 4:

In a ∆*ABC*, *D* and *E* are the mid-points of *AB* and *AC* respectively and *DE* = 5.6 cm. Find the length of *BC*.

#### Answer:

In ∆*ABC*, *D* and *E* are the midpoints of *AB* and *AC,* respectively.

By mid point theorem, DE = $\frac{1}{2}$ (BC)

∴ BC = 2 ⨯ DE

⇒ BC = 2 ⨯ 5.6 = 11.2 cm (DE = 5.6 cm)

Hence, BC = 11.2 cm

#### Page No 330:

#### Question 5:

In the given figure, *AD* is the median and *DE* || *AB*. Prove that *BE* is the median.

#### Answer:

In ∆*ABC*, *D* is the mid-point of *BC* and *DE* ||* AB*.

∴ *AE* = *EC* and *DE* = $\frac{1}{2}$(*AB*) (By mid point theorem)

So,* E* is the mid point of *AC*.

∴ *BE* is the median.

#### Page No 330:

#### Question 6:

In the given figure, lines *l*, *m* and *n* are parallel lines and the lines *p* and *q* are transversals. If *AB* = 5 cm, *BC* = 15 cm, then *DE* : *EF* = ?

#### Answer:

Lines *l* || *m* || *n*. Also, *p *and *q* are transversal lines.

⇒ *AB* : *BC* = 5 : 15 = 1 : 3

∴ *DE* : *EF* = 1 : 3 (By intercept theorem)

#### Page No 330:

#### Question 7:

*ABCD* is a rectangle in which diagonal *BD* bisects ∠*B*. Show that *ABCD* is a square.

#### Answer:

Let *ABCD* be a rectangle.

Here, *AB* = *CD* and *BC* = *DA and *∠*A** = *∠*B** = *∠*C**= *∠*D** = *90^{o}

Now, we have:

∠*ABD* = ∠*DBC* [ ∵*BD* bisects ∠*B** *]

∠ADB = ∠*DBC* [ Alternate interior angles]

∠*ABD* = ∠*ADB*

As the two opposite angles of ∆*ABD** *are equal, the opposite sides must also be equal.

i.e., *AB* = *DA*

∴ *AB* = *CD* = *DA = BC*

So, when all the sides are equal and all the angles are of 90^{o}, the quadrilateral is a square.

Hence, *ABCD* is a square.

#### Page No 331:

#### Question 8:

The diagonals of a rectangle *ABCD* intersect at the point *O*. If ∠*BOC* = 50°, then ∠*OAD* = ?

(a) 50°

(b) 55°

(c) 65°

(d) 75°

#### Answer:

(c) 65°

Explanation:

∠*BOC* = 50° = ∠*AOD* (Vertically opposite angles)

In ∆ AOD, OA = OD (Diagonals are equal, so bisectors are also equal)

∴ ∠OAD = ∠ODA = $\frac{1}{2}\left(180\xb0-50\xb0\right)=65\xb0$

#### Page No 331:

#### Question 9:

Match the following column:

Column I |
Column II |

(a) Sum of all the angles of a quadrilateral is | (p) right angles |

(b) In a || gm, the angle bisectors of two adjacent angles intersect at | (q) rectangle |

(c) Angle bisectors of a || gm form a | (r) 45° |

(d) The diagonals of a square are equal and bisect each other at an angle of | (s) 4 right angles |

(b) ......,

(c) ......,

(d) ......,

#### Answer:

(a)-(s)

(b)-(p)

(c)-(q)

(d)-(r)

DISCLAIMER: There is an error in the book;

90^{o} should come in place of 45^{o} in part (q),

because the diagonals of a square

bisect each other at an angle of 90^{o}.

#### Page No 331:

#### Question 10:

The diagonals of a rhombus *ABCD* intersect at the point *O*. If ∠*BDC* = 50°, then ∠*OAB* = ?

(a) 50°

(b) 40°

(c) 25°

(d) 20°

#### Answer:

(b) 40°

Explanation:

∠*BDC *=* *∠A*BD *= 50^{o}* *(Alternate interior angles)

In ∆*AOB*, ∠A*BD *=* *50^{o} and ∠*AOB *=* *90^{o}

∴ ∠*OAB** = *180^{o} − (90^{o} + 50^{o}) = 40^{o}

#### Page No 331:

#### Question 11:

*ABCD* is a trapezium in which *AB* || *CD* and *AD* = *BC*, then ∠*A* = ∠*B* is

(a) true

(b) false

#### Answer:

(a) True

Explanation:

Draw perpendiculars from *D* and *C* to *AB* which cuts *AB* at *F* and *E*, respectively.

In ∆*ADF* and ∆*BCE*, we have:

*AD = BC* (Given)

*DF = CE* (Perpendicular distance between two parallel lines)

∠A*FD** = *∠BEC (90^{o} each)

i.e., ∆*ADF* ≅ ∆*BCE*

∴ ∠*A* = ∠*B*

#### Page No 331:

#### Question 12:

Look at the statements given below:

I. If *AD*, *BE* and *CF* be the altitudes of a ∆*ABC* such that *AD* = *BE* = *CF*, then ∆*ABC* is an equilateral triangle.

II. IF *D* is the mid-point of hypotenuse *AC* of a right ∆*ABC*, then *BD* = *AC*.

III. In an isosceles ∆*ABC* in which *AB* = *AC*, the altitude *AD* bisects *BC*.

Which is true?

(a) I only

(b) II only

(c) I and III

(d) II and III

#### Answer:

(c) I and III

Explanation:

Statements I and III are true.

For I:

In ∆*ABC*, the altitudes *AD =* *BE* = *CF *.

In ∆*ABE** and *∆*ACF**, *we have:

*BE = CF * (Given)

∠A is common.

∠*AEB* = ∠*AFC* = 90^{o}

i.e., ∆*ABE *≅* *∆ *ACF *( By AAS congruence rule)

∴ *AB = AC* (CPCT)

Similarly, ∆*BCF** *≅* *∆*BAD*

i.e., *BC *=* AB * ( CPCT)

∴*AB** *=* AC *=* BC*

*ABC*is an equilateral triangle.

For III:

Let ∆

*ABC*be an isosceles triangle.

*AD*is the altitude of ∆

*ABC*.

In ∆

*ABD*and ∆

*ADC*, we have:

*AB = AC*(Given)

∠

*B*= ∠

*C*(Angles opposite to equal sides are equal)

∠

*ADB*= ∠

*ADC*= 90

^{o}

So, ∆

*ABD*≅ ∆

*ADC*(By AAS conruency)

⇒

*BD*=

*DC*

∴

*D*is the mid point of

*BC*or

*AD*bisects

*BC*.

#### Page No 332:

#### Question 13:

In the given figure, *D* and *E* are two points on side *BC* of ∆*ABC* such that *BD* = *DE* = *EC*. Prove that ar(∆*ABD*) = ar(∆*ADE*) = ar(∆*AEC*).

#### Answer:

We know that area of a triangle = $\frac{1}{2}$(base $\times $ height)

Draw AL ⊥ BC.

Let* h *be the height of ∆*ABC*, AL.

∴ Height of ∆*ABD** *= height of ∆*ADE * = height of ∆*AEC*

The* *bases of* *∆*ABD, *∆*ADE** *and* *∆*AEC *are BD, DE and EC, respectively.

Given: *BD* = *DE* = *EC*

∴ ar(∆*ABD*) = ar(∆*ADE*) = ar(∆*AEC**)*

#### Page No 332:

#### Question 14:

In the given figure *ABCD*, *DCEF* and *ABFE* are parallelograms. Show that ar(∆*ADE*) = ar(∆*BCF*).

#### Answer:

In ∆*ADE** *and* *∆*BCF**, *we have:

*AD = BC * (Opposite sides of parallelogram *ABCD*)

*AE = BF* (Opposite sides of parallelogram *ABFE*)

*DE = CF* (Opposite sides of parallelogram *DCEF*)

i.e., ∆*ADE *≅* *∆*BCF*

∴ ar(∆*ADE*) = ar(∆*BCF*)

#### Page No 332:

#### Question 15:

In the given figure, *ABCD* is a trapezium in which *AB* || *DC* and diagonals *AC* and *BD* intersect at *O*. Prove that ar(∆*AOD*) = ar(∆*BOC*).

#### Answer:

In trapezium* ABCD*, *AB* || *DC* and *AC* and *BD* are the diagonals which intersect at *O*.

As ∆*ACD *and* *∆*BCD* lie on the same base and between the same parallel lines.

∴ ar(∆*ACD*) = ar(∆*BCD*)

⇒ ar(∆*ACD*) − ar(∆*COD*) = ar(∆*BCD*) − ar(∆*COD*)

Hence, ar(∆*AOD*) = ar(∆*BOC*)

#### Page No 332:

#### Question 16:

Show that a diagonal divides a parallelogram into two triangles of equal area.

#### Answer:

Let *ABCD *be the* *parallelogram* *and *AC* be one of its diagonal.

In ∆*CDA *and* *∆*ABC, *we have:

DA = BC (Opposite sides of parallelogram)

AC is common

CD = AB

∴ ∆*CDA** *≅ ∆*ABC* (SSS congruency)

i.e., ar(∆*CDA*) = ar(∆A*BC*)

Hence, the diagonal of a parallelogram divides it into two triangles of equal area.

#### Page No 332:

#### Question 17:

In the given figure, *AC* is a diagonal of quad. *ABCD* in which *BL* ⊥ *AC* and *DM* ⊥ *AC*. Prove that or (quad. *ABCD*) $=\frac{1}{2}\times AC\times (BL+DM)$.]

#### Answer:

*AC* is one of the diagonal of quadrilateral *ABCD. *Also, *BL *⊥ *AC* and *DM* ⊥ *AC.*

Then area of *ABCD = *ar*(*∆*ADC**) + *ar*(*∆*ABC**)*

Now, ar*(*∆*ADC**) *= $\frac{1}{2}$ ⨯ *AC* ⨯ *BL* (*BL *⊥ *AC)*

Also, ar*(*∆*ABC**) *= $\frac{1}{2}$ ⨯ *AC* ⨯ *DM* (*DM *⊥ *AC)*

∴ Area of *ABCD = *($\frac{1}{2}$ ⨯ *AC* ⨯ *BL*)* *+ ($\frac{1}{2}$ ⨯ *AC* ⨯ *DM*)

= $\frac{1}{2}$ ⨯ *AC *⨯ (*BL* + *DM*)

#### Page No 332:

#### Question 18:

|| gm *ABCD* and rectangle *ABEF* have the same base *AB* and are equal in areas. Show that the perimeter of the || gm is greater than that of the rectangle.

#### Answer:

We have:

A parallelogram *ABCD *and a rectangle *ABEF* are on the same base *AB* and in between the same parallel lines.

*AB = CD* and *AB = FE *

i.e., *CD = FE*

∴ *AB + CD = AB + FE* ...(i)

Now, we know that *AD > AF* and *BC > BE.* (∵ Hypotenuse is the longest side of a triangle)

∴ *AD + BC > AF + BE ..*.(ii)

We know that:

Perimeter of *ABCD = **AB + BC + CD + AD*

Perimeter of *ABEF = AB + BE+ FE+ AF*

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

∴ *AB+ CD + AD + BC > AB + FE+ AF + BE*

Hence, the perimeter of *ABCD* is greater than that of* **ABEF.*

#### Page No 333:

#### Question 19:

In the adjoining figure, *ABCD* is a || gm and *E* is the mid-point of side *BC*. If *DE* and *AB* when produced meet at *F*, prove that *AF* = 2*AB*.

#### Answer:

*ABCD* is a parallelogram.

i.e., *AB* ∣∣ *DC*

∴ *DC *∣∣* BF*

In ∆*DEC* and ∆*FEB*, we have:

* EC = EB* (*E* is the mid point of *BC*)

∠*C**ED* = ∠*BEF** *(Vertically opposite angles)

∠*E**CD* = ∠*EBF** * (Alternate interior angles)

∴ ∆*DEC* ≅ ∆*FEB*

⇒ *CD = BF *(CPCT)

Also, *CD = AB *(Opposite sides of a parallelogram)

Now, *AF = AB + BF
⇒ *

*AF = AB + AB*

∴

*AF = 2AB*

#### Page No 333:

#### Question 20:

In the adjoining figure, *ABCD* and *PQRC* are rectangles, where *Q* is the mid-point of *AC*. Prove that (i) *DP* = *PC* (ii) $PR=\frac{1}{2}AC$.

#### Answer:

(i)∠*CRQ* = ∠*CBA* = 90^{o} ⇒ *QR* ∣∣ *AB*

In ∆*ABC**, Q* is the mid point of *AC* and *QR* ∣∣ *AB*.

∴ R is the mid point of BC.

Similarly, *P* is the midpoint of *DC*.

∴ *DP* = *PC*

(ii) Join *BD*. In ∆*CDB*, *P* is the mid point of *DC* and *R* is the mid point of *BC*.

∴ *PR* ∣∣ *DB* and *PR* = $\frac{1}{2}$*DB* = $\frac{1}{2}$*AC* [∵ *AC* = *B*D]

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