Rs Aggarwal 2021 2022 for Class 9 Maths Chapter 10 - Quadrilaterals

Answer:

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

Let the fourth angle be x.

Using angle sum property of quadrilateral,

$$\begin{gathered} 75°+90°+75°+x=360° \\ \Rightarrow 240°+x=360° \end{gathered}$$

$$\begin{gathered} \Rightarrow x=360°-240° \\ \Rightarrow x=120° \end{gathered}$$

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

Answer:

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 360^o, we have:
2x + 4x + 5x + 7x = 360^∘   
⇒ 18 x = 360^∘ ​
⇒ x = 20^∘
∴  $\angle$A = 40^∘; $\angle$B = 80^∘; $\angle$C = 100^∘; $\angle$D = 140^∘

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^∘
 

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]

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 ∆OND 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.

Answer:

​Given:  ABCD is a quadrilateral in which AB = AD and BC = DC  
(i)
In ∆ABC and ∆ADC, we have:
AB = AD                                                  (Given)

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 = $\frac{1}{2}\left(180° - 90°\right) = \frac{90°}{2} = {45}^{°}$

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

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)

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:

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​

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 + ∠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​
​

Answer:

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)

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

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 = $\frac{60°}{2}=30°$ 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

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)

Answer:

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 

Answer:

 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

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 = (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

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 + 2x = 30
⇒ 2x = 11
⇒ x = 5.5 cm
Hence, AB = DC = 9.5 cm and BC = DA = 5.5 cm

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 ∆​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

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, 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.

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 ^$÷$ 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 =  $\frac{1}{2}\times \left(12\times 16\right) = 96 {\mathrm{cm}}^2$

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 = ∠​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​​

Answer:

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 $60°, 120°, 60° \mathrm{and} 120°$.

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

Answer:

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 $\angle D \mathrm{and} \angle B.$

Answer:

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.

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.

Answer:

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.

Answer:

​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 $60°, 120°, 60° \mathrm{and} 120°.$

Answer:

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.

Answer:

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)

Answer:

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.

Answer:

​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.

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 = 2BC​
∴ BC = $\frac{1}{2}QR$

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 = 2BC​
∴ BC = $\frac{1}{2}QR$
Similarly, CA = ​$\frac{1}{2}PQ$ and AB = ​$\frac{1}{2}PR$

Answer:

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)

$\Rightarrow AC\parallel PQ \mathrm{and} PQ=\frac{1}{2}AC$            (Using mid-point theorem.)

(ii)
In $∆ADC$,

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

$\Rightarrow SR\parallel AC \mathrm{and} SR=\frac{1}{2}AC$           (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.

Answer:

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)

Answer:

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

Answer:

 
Given: A parallelogram ABCD
 
To prove: MN is bisected at O

Proof:

In  $∆$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,

$∆OAM \cong ∆OCN$

$\Rightarrow OM = ON$(CPCT)

Hence, MN is bisected at O.

Answer:

 
Given: In trapezium PQRS, PQ || SR, M 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)

Answer:

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

⇒ 2x = x + ∠MTR

⇒ ∠MTR = x

⇒ △TPQ is an isosceles triangle.

⇒ TQ = PQ = 12 cm

Now,

RT = TQ − QR

      = TQ − PS

      = 12 − 9

      = 3 cm

Answer:

Given: AB || DC, AP = PD and BQ = CQ

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

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.

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 =  $\frac{1}{4}AC$

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.

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)

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 ∆ 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

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 ∆DAC, 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

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 ∆DAC, 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

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 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.

Answer:

Answer:

 
Given: In quadrilateral ABCD, AC $\perp$ BD. P, Q, R and S are the mid-points of AB, BC, CD and AD, respectively.

To prove: PQRS is a rectangle.

Proof:

In ΔABC, P 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 ΔBCD, Q 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.

Answer:

Given: In quadrilateral ABCD, AC = BD and AC ⊥ BD. P, 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)

$\Rightarrow$ $\frac{1}{2}$AC = $\frac{1}{2}$BD

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

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

Hence, PQRS is a square.

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.

Answer:

(b) 60°​

Explanation:
Let ∠A = 3x​, ∠B = 4x, ∠C = 5x and ∠D = 6x.
Since the sum of the angles of a quadrilateral is 360^o, we have:
3x + 4x + 5x + 6x = 360^o   
⇒ 18x = 360^o ​
⇒ x = 20^o
∴ ∠A = 60^o​, ∠B = 80^o, ∠C = 100^o and ∠D = 120^{o
​Hence, the smallest angle is 60$°$.}

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)

Answer:

We know that diagonals of rhombus bisect each other at 90°.

Then, in ΔBOC,

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

$\Rightarrow$ ∠OBC = 180° $-$ 140°

$\Rightarrow$ ∠OBC = 40°

But ∠OBC = ∠ADB     (Alternate interior angles)

Thus, ∠ADB = 40°

Hence, the corerct option is (a).

Answer:

(d) Rectangle.

The diagonals of a rectangle are equal.

Answer:

(d) rhombus

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

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, 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.

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

Now, AB²= OA² + OB²              [Pythagoras theorem]
$$\begin{gathered} \Rightarrow {10}^2 = {\left(\frac{x}{2}\right)}^2 + 8^2 \\ \Rightarrow {\left(\frac{x}{2}\right)}^2 = 36 = 6^2 \\ \Rightarrow x = 2\times 6 =12 \mathrm{cm} \end{gathered}$$
⇒102 = (x2)2 + 82⇒100 − 64 = x24⇒36 ×4 = 

Answer:

 
Given: In rectangle ABCD, ∠OAD = 35°.

Since, ∠BAD = 90°

$\Rightarrow$ ∠OAB = 90° $-$ 35° = 55°

In ΔOAB,

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

$\Rightarrow$ ∠OAB = ∠OBA = 55°     (Angles opposite to equal sides are equal)

Now, in ΔODA,
 
55° + 55° + ∠DOA = 180°       (Angle sum property of a triangle)

$\Rightarrow$ ∠DOA = 180° $-$ 110°

$\Rightarrow$ ∠DOA = 70°

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

Hence, the correct option is (b).

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.

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

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

Answer:

(c) Rectangle

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

Answer:

Given: In quadrilateral ABCD, AS, BQ, CQ and DS are angle bisectors of angles A, B, C and D, respectively.

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

In $∆$APB,

∠APB + ∠PAB + ∠ABP = 180°        (Angle sum property of triangle.)

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

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

From (1) and (2), we get

∠QPS = 180° – $\frac{1}{2}$(∠A + ∠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).

Answer:

(d) parallelogram

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

Answer:

(b) Square

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

Answer:

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

Answer:

(a) rhombus

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

Answer:

(c) Rectangle

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

Answer:

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).

Answer:

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

$\Rightarrow \frac{1}{2}QS = \frac{1}{2}PR$        [From (1) and (2)]

So, QS = PR

Thus, the diagonals of PQRS are equal.

Hence, the correct option is (c).

Answer:

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).

Answer:

(c) 72°​

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

Answer:

(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 = 180^o                (Adjacent angles are supplementary) 
   ⇒ x + 2x − 24^o = 180^o
   ⇒ 3x =  204^o
  ⇒ x = 68^o
∴​∠​B = 2 ⨯ 68^o − 24^o = 112^o
 Hence, ∠A = ∠C = 68^o and ∠B = ∠D = 112^o

Answer:

(c) Trapezium

Explanation:

Let the angles be (3x), (7x), (6x) and (4x)​.
Then 3x + 7x + 6x + 4x = 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.

Answer:

(c) Opposite angles are bisected by the diagonals.

Answer:

(c) Rectangle

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

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

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

Answer:

(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)

Answer:

Given: In ∆ABC, M, N, D 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

BM = EN 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).

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 = $\frac{1}{2}\left(b\right)$

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

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°

Answer:

(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)

Answer:

(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.
∴ AC²= BC² + AB² - 2AB.AE
In ∆ABD, ∠A is acute.​
∴ ​BD² = AD² + AB² - 2AB.AF
∴ ​AC² + BD² = (BC² + AD²) + (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          

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.

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​
4AC

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

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.

Answer:

(b) II and III

Answer:

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

$\Rightarrow$ Quadrilateral PQRS is a parallelogram.

$\Rightarrow SR = PQ$  (Opposite sides of a parallelogram are equal.)

$\therefore SR = PQ = 2\mathit{ }\mathrm{cm}$

Answer:

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

Answer:

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

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

So, PQRS is a parallelogram.

Answer:

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$°$.

Answer:

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

Answer:

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$°$.

Answer:

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

So, we cannot form a quadrilateral with these angles.

Answer:

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

Let each angle of the quadrilateral be x.

x + x + x + x = 360°

⇒ 4x = 360°

⇒ x = 90°

⇒ All angles of the quadrilateral are 90°.

Hence, given quadrilateral is a rectangle.

Answer:

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.

Answer:

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.)

$$\begin{gathered} \Rightarrow 56°+\angle R=180° \\ \Rightarrow \angle R=180°-56° \\ \Rightarrow \angle R=124° \end{gathered}$$

Answer:

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
 

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.

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).