RS AGGARWAL CLASS 9 CHAPTER 10 QUADRILATERALS EXERCISE 10B

 EXERCISE 10B

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Question 1:

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

Answer 1:

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​

Question 2:

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

Answer 2:

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

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Question 3:

In the adjoining figure, M is the midpoint of side BC of a parallelogram ABCD such that ∠BAM = ∠DAM. Prove that AD = 2CD.

Answer 3:

Given: parallelogram ABCD, M is the midpoint of side BC and ∠BAM = ∠DAM.

To prove: AD = 2CD
 
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, $\angle AMB=\angle BAM$

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

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

$$\begin{gathered} \Rightarrow 2AB=2CD \\ \Rightarrow (AB+AB)=2CD \end{gathered}$$

$\Rightarrow BM+MC=2CD$      (AB = BM and MC = BM)

$$\begin{gathered} \Rightarrow BC=2CD \\ \therefore AD=2CD \left(AD=BC, \mathrm{Opposite} \mathrm{sides} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{are} \mathrm{equal}.\right) \end{gathered}$$

Question 4:

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 = 2AD.

Answer 4:

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 = $\frac{60°}{2}=30°$ and ∠PBA = $\frac{120°}{2}=60°$
   ∴​ ∠​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

Question 5:

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 5:

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

Question 6:

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

Answer 6:

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

Question 7:

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

Answer 7:

 Let ABCD be a parallelogram. 
∴ ∠​A = ∠C and ∠B = ∠D           (Opposite angles)
Let ∠A = x^o and ∠B = ${\left(\frac{4x}{5}\right)}^{°}$
Now, ∠​A + ∠B = 180^o                 (Adjacent angles are supplementary)
  $$\begin{gathered} \Rightarrow x + \frac{4x}{5} = {180}^{\mathrm{o}} \\ \Rightarrow \frac{9x}{5} = {180}^{\mathrm{o}} \\ \Rightarrow x = {100}^{\mathrm{o}} \\ \mathrm{Now}, \angle A = {100}^{\mathrm{o}} \mathrm{and} \angle B = \left(\frac{4}{5}\right)\times 100° = {80}^{\mathrm{o}} \end{gathered}$$
 Hence, ∠A = ∠C = 100^o; ∠B = ∠D = 80^o

Question 8:

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

Answer 8:

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

Question 9:

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 9:

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

Question 10:

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

Answer 10:

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 ∆​AB​D, 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

Question 11:

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 11:

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.

Question 12:

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 12:

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 = x ^$÷$ 2 and OB = BD ^$÷$2 = 16 ^$÷$ 2 = 8 cm.

Now, AB²= OA² + OB²              [Pythagoras theorem]
$$\begin{gathered} \Rightarrow {10}^2 = {\left(\frac{x}{2}\right)}^2 + 8^2 \\ \Rightarrow 100 - 64 = \frac{x^2}{4} \\ \Rightarrow 36 \times 4 = x \end{gathered}$$²

⇒x² =144
∴ x = 12 cm

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

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Question 13:

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

Answer 13:

(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 = 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 = ∠OBA = $\frac{1}{2}\times \left(180° - 110°\right) = {35}^{\mathrm{o}}$
∴​ 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​​

Question 14:

In a rhombus ABCD, the altitude from D to the side AB bisects AB. Find the angles of the rhombus.

Answer 14:

Given: ABCD is a rhombus, DE is altitude which bisects AB i.e. AE = EB

$\mathrm{In} ∆AED \mathrm{and} ∆BED,$

$DE=DE$                         (Common side)

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

$AE=EB$                          (Given)

$\therefore ∆AED\cong ∆BED$         (By SAS congruence Criteria)

$\Rightarrow AD=BD$                    (CPCT)

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

$\Rightarrow AD=AB=BD$

Thus, $∆ABD$is an equilateral triangle.

Therefore, $\angle A=60°$

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

$\mathrm{And}, \angle ABC+\angle BCD=180°$           (Adjacent angles of rhombus are supplementary.)

$$\begin{gathered} \Rightarrow \angle ABC+60°=180° \\ \Rightarrow \angle ABC=180°-60° \\ \Rightarrow \angle ABC=120° \\ \Rightarrow \angle ADC=\angle ABC=120° \end{gathered}$$

Hence, the angles of rhombus are $60°, 120°, 60° \mathrm{and} 120°$.

Question 15:

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 15:

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

Question 16:

In a rhombus ABCD show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.

Answer 16:

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 $∆ABC$,

$AB = BC$       (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.)

$$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 $\angle C \mathrm{and} \angle A$

In $∆DAB$,

$AD = AB$                 (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.)

$$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 $\angle D \mathrm{and} \angle B.$

Question 17:

In a parallelogram ABCD, points M and N have been taken on opposite sides AB and CD respectively such that AM = CN. Show that AC and MN bisect each other.

Answer 17:

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.

$$\begin{gathered} \Rightarrow AB\parallel DC \\ \Rightarrow AM\parallel NC \end{gathered}$$

Also, AM = CN           (Given)

Thus, AMCN is a parallelogram.

Since, diagonals of a parallelogram bisect each other.

Hence, AC and MN bisect each other.

 

Question 18:

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 18:

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.

Question 19:

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 19:

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.

Question 20:

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

Answer 20:

​Given: In parallelogram ABCD, DP⊥ AB, AQ ⊥ BC and ∠PDQ = 60°

In quadrilateral DPBQ, by angle sum property, we have

$$\begin{gathered} \angle PDQ+\angle DPB+\angle B+\angle BQD=360° \\ \Rightarrow 60°+90°+\angle B+90°=360° \\ \Rightarrow \angle B=360°-240° \\ \Rightarrow \angle B=120° \end{gathered}$$

Therefore, $\angle B = 120°$

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.)

$$\begin{gathered} \Rightarrow \angle A+120°=180° \\ \Rightarrow \angle A=180°-120° \\ \Rightarrow \angle A=60° \end{gathered}$$

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

So, the angles of a parallelogram are $60°, 120°, 60° \mathrm{and} 120°.$

Question 21:

ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that (i) ABCD is a square, (ii) diagonal BD bisects ∠B as well as ∠D.

Answer 21:

Given: In rectangle ABCD, AC 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, $\angle 2 = \angle 4$

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.

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Question 22:

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 22:

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.
 

Question 23:

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

Answer 23:

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. 

Question 24:

Two parallel lines l and m are intersected by a transversal t. Show that the quadrilateral formed by the bisectors of interior angles is a rectangle.

Answer 24:

Given: l || m and the bisectors of interior angles intersect at B and D.

To prove: ABCD is a rectangle.

Proof:

Since, l || m                       (Given)

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

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

$\Rightarrow \angle BAC=\angle ACD$

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

$\Rightarrow AB\parallel DC$

Similarly, $BC\parallel AD$

So, ABCD is a parallelogram.

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

$$\begin{gathered} \Rightarrow \frac{1}{2}\angle PAC+\frac{1}{2}\angle CAS=90° \\ \Rightarrow \angle BAC+\angle CAD=90° \\ \Rightarrow \angle BAD=90° \end{gathered}$$

But, this an angle of the parallleogram ABCD.

Hence, ABCD is a rectangle.

Question 25:

K, L, M and N are points on the sides AB, BC, CD and DA respectively of a square ABCD such that AK = BL = CM = DN. Prove that KLMN is a square.

Answer 25:

​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

$\Rightarrow$ 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,

$∆NAK\cong ∆KBL$

$\Rightarrow NK=KL$     (CPCT)           ...(2)

Similarly,

$$\begin{gathered} ∆MDN\cong ∆NAK \\ ∆DNM\cong CML \\ ∆MCL\cong LBK \end{gathered}$$

$\Rightarrow$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 $∆NAK$,

$\angle NAK = 90°$

Let $\angle AKN = x°$

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

$$\begin{gathered} \Rightarrow \angle DNM+\angle MNK=90°+x° \\ \Rightarrow x°+\angle MNK=90°+x° \\ \Rightarrow \angle MNK=90° \end{gathered}$$

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.

Question 26:

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 26:

 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 = $\frac{1}{2}QR$

Question 27:

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 27:

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  = $\frac{1}{2}QR$ + $\frac{1}{2}PQ$ + $\frac{1}{2}PR$

                              = $\frac{1}{2}\left(PR + QR + PQ\right)$
 
i.e., Perimeter of ∆​ABC  = $\frac{1}{2}$ (Perimeter of ∆​PQR)
∴ Perimeter of ∆PQR = 2 ⨯ Perimeter of ∆ABC