Rs Aggarwal 2017 Solutions for Class 9 Math Chapter 9 Quadrilaterals And Parallelograms are provided here with simple step-by-step explanations. These solutions for Quadrilaterals And Parallelograms are extremely popular among Class 9 students for Math Quadrilaterals And Parallelograms Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2017 Book of Class 9 Math Chapter 9 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2017 Solutions. All Rs Aggarwal 2017 Solutions for class Class 9 Math are prepared by experts and are 100% accurate.

#### Question 1:

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

Let the measure of the fourth angle be xo.
Since the sum of the angles of a quadrilateral is 360o, we have:
56 + 115 + 84 + x = 360
⇒ 255 + x = 360 ​
x = 105
Hence, the measure of the fourth angle is 105o.

#### Question 2:

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

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

#### Question 3:

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

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

#### Question 4:

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

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

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

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

#### Question 5:

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

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

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

#### 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,

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

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

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

(iii)

#### 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°.

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

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

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

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

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

#### Question 8:

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

Let ABCD be a quadrilateral whose diagonals are AC and BD and O is any point within the quadrilateral.
Join O with A, B, C, and D.
We know that the sum of any two sides of a triangle is greater than the third side.
So, in ∆AOC, OA + OC > AC

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

#### Question 9:

In the adjoining figure, ABCD is a quadrilateral and AC is one of its diagonals. Prove that:
(i) AB + BC + CD + DA > 2AC
(ii) AB + BC + CD > DA
(iii) AB + BC + CD + DA > AC + BD

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

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

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

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

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

#### Question 10:

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

Let ABCD be a quadrilateral and ∠1, ∠2, ∠3 and ∠4  are its four angles as shown in the figure.
Join BD which divides ABCD in two triangles, ∆ABD and ∆BCD.
In ∆ABD, we have:
∠1 + ∠2 + ​∠A = 180o      ...(i)
In ​∆BCD, we have:
∠3 + ∠4 + ∠C = 180o     ...(ii)
On adding (i) and (ii), we get:

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

#### Question 1:

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

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

#### Question 2:

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

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

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

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

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

#### 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 = 2AD.

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

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

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

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

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

#### 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.

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

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

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

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

(iv) ∠CBDABC  ABD            ...(i)

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

#### Question 5:

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.

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

#### Question 6:

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

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

Hence, AC = 100o; B = ∠D = 80o

#### Question 7:

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

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

#### 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.

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 9:

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

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

⇒​ y = 70o    x

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

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

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

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

#### 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.

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 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.

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

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

x2 =144
x = 12 cm

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

#### Question 12:

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

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

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

#### 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.

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

#### 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.

ABCD is a parallelogram.
(i) In ∆ALD and CMB, we have:
∠​ALD = ∠​CMB                         (90o each)
∠​ADL = ∠​CBM                         (Alternate interior angle)
∴ ∆ALD   ∆CMB

(ii) As ∆ALD ≅ ∆CMB               (Proved above)
∴​ AL = CM                               (By CPCT)

#### 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°.

ABCD is a parallelogram and ∠A and ∠B are adjacent angles.
∴ ​∠ A + ∠B = 180o.
⇒ ​
In ∆​ APB, we have:
∠​PAB = ∠A​ /2
∠​PBA = ∠B​ /2​
∴ ∠​APB = 180 − (∠​PAB​ + ∠​PBA​)           [ Angle sum property of triangle]
⇒ ​∠APB = 180 −
⇒ ∠APB = 180 − 90 = 90o
Hence, proved.

#### 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.

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

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

#### 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.

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 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.

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 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 = 2AB.

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 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 $BC=\frac{1}{2}QR$.

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 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.

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

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

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

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

#### 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.

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)

#### 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.

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

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

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

#### 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.

Given: AB || DC, AP = PD and BQ = CQ

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

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

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

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

#### Question 4:

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

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.

#### Question 5:

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

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

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

#### 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 = 2EB.

(i) In ∆​ DCG, we have:
DG || EB
DE = EC            (is the midpoint of DC)​
Also, GB = BC            (By midpoint theorem)
∴​ B is the midpoint of GC.

Now, GC = GB + BC
GC = 2BC
∴ 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

#### Question 7:

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

∆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.

#### 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

∆ 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)

#### Question 9:

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

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

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

#### Question 10:

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

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

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

#### Question 11:

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

Let ABCD be a square and P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively.
Join the diagonals AC and BD. Let BD cut SR at F and AC cut RQ at E. Let O be the intersection point of AC and BD.

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

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

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

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

∴​ PQRS is a square.

#### Question 12:

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

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 QRBD ... (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.

#### 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.

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

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

We know that the diagonals of the given quadrilateral bisect each other at right angles.
∴ ∠​ EOF = 90o
​Now, RQ ∣∣ DB
RE ∣∣ FO
Also, SR ∣∣ AC
⇒ FR ∣∣ OE
∴ OERF is a parallelogram.

So, ∠FRE = ∠EOF = 90o​         (Opposite angles are equal)
Thus, PQRS is a parallelogram with ∠R = 90o.
∴​ PQRS is a rectangle.

#### Question 1:

Three angles of a quadrilateral are 80°, 95° and 112°. Its fourth angle is
(a) 78°
(b) 73°
(c) 85°
(d) 100°

(b) 73°​

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

#### 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°

(b) 60°​

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

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

#### 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°

(c) 45°

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

#### Question 4:

In which of the following figures are the diagonals equal?
(a) Parallelogram
(b) Rhombus
(c) Trapezium
(d) Rectangle

(d) Rectangle.

The diagonals of a rectangle are equal.

#### 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

(d) rhombus

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

#### 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

(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.

#### 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)
(d) 6 cm

(b) 12 cm

Explanation:

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

Now,
AB2= OA2 + OB2              [Pythagoras theorem]

102 = (x2)2 + 82100  64 = x2436 ×4 =

#### 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

(c) Rectangle

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

#### 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°

(b) 50o

​​Explanation:

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

#### Question 10:

The bisectors of any two adjacent angles of a parallelogram intersect at
(a) 40°
(b) 45°
(c) 60°
(d) 90°

(d) 90°

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

#### Question 11:

The bisectors of the angles of a parallelogram enclose a
(a) rhombus
(b) square
(c) rectangle
(d) parallelogram

(c) Rectangle

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

#### 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

(d) parallelogram

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

#### 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

(b) Square

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

#### 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

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

#### 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

(a) rhombus

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

#### 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

(c) Rectangle

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

#### 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°

(c) 72°​

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

#### 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°

(c)112°​

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

#### 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°

(c)​ 60°

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

#### 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) AB

(c) A < B

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

#### 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 = 2AB
(c) AF = 3AB
(d) AF2 = 2AB2

(b) AF = 2 AB

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

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

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

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

#### 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) $\frac{1}{2}\left(a-b\right)$
(b) $\frac{1}{2}\left(a+b\right)$
(c) $\frac{2ab}{\left(a+b\right)}$
(d) $\sqrt{ab}$

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

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

(d)

Explanation:

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

#### 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°

(c) ​90°

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

#### Question 25:

In the given figure, ABCD is a rhombus. Then,
(a) AC2 + BD2 = AB2
(b) AC2 + BD2 = 2AB2
(c) AC2 + BD2 = 4AB2
(d) 2(AC2 + BD2) = 3AB2

(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)

#### Question 26:

In a trapezium ABCD, if AB || CD, then (AC2 + BD2) = ?
(b) AB2 + CD2 + 2ABCD
(c) AB2 + CD2 + 2ADBC
(d) BC2 + AD2 + 2ABCD

(c) BC2 + AD2 + 2AB.CD

Explanation:

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

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

#### 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

(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.

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

(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

#### 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

(c) Trapezium

Explanation:

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

#### 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.

(c) Opposite angles are bisected by the diagonals.

#### Question 31:

If APB and CQD are two parallel lines, then the bisectors of APQ, ∠BPQ, ∠CQP and ∠PQD enclose a
(a) square
(b) rhombus
(c) rectangle
(d) kite

(c) Rectangle

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

#### 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°

(a) 40°

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

#### 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

(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.

#### 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

(b) II and III

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

#### 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.

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

Hence, the correct answer is (a).

#### Question 36:

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.

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).

#### 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.

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).

#### Question 38:

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.

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

#### 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.

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

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

#### 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.

(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).

#### 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.

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

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

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

#### 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.

(d) Assertion is false and Reason is true.

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

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

#### 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.

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

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

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

#### 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
(a) .....,
(b) .....,
(c) .....,
(d) .....,

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

#### 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
(a) ......,
(b) ......,
(c) ......,
(d) ......,

​(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

#### 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.

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

#### Question 2:

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

(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 ⨯ and base BD = DC]
Now, let h1 be the height of ∆BPD and PDC.
∴ ar(∆BPD) = ar(∆PDC)​

We know that ar(∆ABD) = ar (∆BPD ) + ​ar (∆ABP )
and ar(∆ADC) = ar(∆PDC) + ​ar(∆ACP)
∴ ar(∆ABP) = ar(∆ACP)​

#### Question 3:

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

Let the angles of the parallelogram be x, 3x, 5x and 6x.
∴ x + 3x + 5x + 6x =  360o
⇒​ 15x = 360o
⇒x = 24o
Thus, the angles are 24o, 24o ⨯ 3 = 72o, 24o ⨯ 5 = 120o and 24o ⨯ 6 = 144o.
Hence, the greatest angle is 144o.

#### 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.

In  ABCD 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

#### Question 5:

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

In  ABCD 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.

#### 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 = ?

Lines l ||​ m ||​ n. Also, p and q are transversal lines.
AB : BC = 5 : 15 = 1 : 3
DE : EF = 1 : 3         (By intercept theorem)

#### Question 7:

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

Let ABCD be a rectangle.
Here, AB = CD and BC = DA and AB = CD = 90o
Now, we have:
∠​ABD = ∠​DBC        [ ∵BD bisects ∠B ]
​∠​ADB = ​∠​DBC        [ Alternate interior angles]
As the two opposite angles of ∆ABD are equal, the opposite sides must also be equal.
i.e., AB = DA
∴ AB = CDDA = BC
So, when all the sides are equal and all the angles are of 90o, the quadrilateral is a square.
Hence, ABCD is a square.

#### 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°

(c)​ 65°

Explanation:
BOC = 50° = ​∠AOD       (Vertically opposite angles)
In ∆ AOD, OA = OD          (Diagonals are equal, so bisectors are also equal)
∴ ​∠OAD = ​∠ODA = $\frac{1}{2}\left(180°-50°\right)=65°$

#### 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
(a) ......,
(b) ......,
(c) ......,
(d) ......,

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

DISCLAIMER: There is an error in the book;
90o  should come in place of 45o in part (q),
because the diagonals of a square
bisect each other at an angle of 90o.

#### 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°

(b) ​40°

Explanation:
BDC = ∠ABD = 50o     (Alternate interior angles)
In ∆AOB, ∠ABD = 50o and ∠AOB = 90o
∴ ∠OAB180o − (90o + 50o) = 40o

#### Question 11:

ABCD is a trapezium in which AB || CD and AD = BC, then A = ∠B is
(a) true
(b) false

(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:
DF = CE              (Perpendicular distance between two parallel lines)
∠AFD∠BEC    (90o each)
∴​ ∠A​ = ∠B

#### 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

(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 = 90o
i.e., ∆ABE ≅​ ∆ ACF              ( By AAS congruence rule)
AB = AC      (CPCT)
i.e., BC = AB      ( CPCT)
AB = AC = BC

Hence, ∆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)
So,  ∆ABD ≅ ​​∆ADC  (By AAS conruency)
BD = DC
D is the mid point of BC or AD bisects BC.

#### 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).

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

#### Question 14:

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

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)​​

#### 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).

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)

#### Question 16:

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

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(∆ABC) ​

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

#### Question 17:

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

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)

#### 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.

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.

#### 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 = 2AB.

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)
CED = BEF                        (Vertically opposite angles)
ECD = 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

#### 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) $PR=\frac{1}{2}AC$.

∴​ PR ∣∣​ DB and PR$\frac{1}{2}$DB$\frac{1}{2}$AC  [∵​ AC = BD]