# RS Aggarwal Solutions Class 9 Maths Chapter 10 Quadrilaterals (Updated For 2021-22) RS Aggarwal Solutions Class 9 Maths Chapter 10 Quadrilaterals: You can start the preparation for your Class 9 Maths exam with the RS Aggarwal Solutions Class 9 Maths. All your doubts will be answered in the RS Aggarwal Solutions Class 9 Maths Chapter 10 Quadrilaterals. The solutions of RS Aggarwal Solutions Class 9 Maths Chapter 10 Quadrilaterals are designed by subject matter experts and are realy accurate and credible.

RS Aggarwal Solutions Class 9 Maths Chapter 10 Quadrilaterals

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#### Question 1:

Three angles of a quadrilateral are 75°, 90° and 75°. Find the measure of the fourth angle.

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

Let the fourth angle be x.

Using angle sum property of quadrilateral,

75°+90°+75°+x=360°240°+x=360°75°+90°+75°+x=360°⇒240°+x=360°

x=360°240°x=120°⇒x=360°-240°⇒x=120°

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

#### Question 2:

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

Let A = 2x​.
Then B = (4x)C = (5x) and 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
∴  A = 40B = 80C = 100D = 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.

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

Again ,  B + C = 180
⇒  C  = 180 − B
∴  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
EDC  =    DEC = ​DCE =  60.
To prove:  AE = BE and DAE = 15
Proof: In ∆ADE and ∆BCEwe have:
AD = BC              [Sides of a square]

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

DA = DC     [Sides of a square]
DC = DE      [Sides of an equilateral triangle]
So, DA = DE
ADE and BCE are isosceles triangles.
i.e., DAE = DEA = 12(180°  150°) = 30°2 =15°12180° – 150° = 30°2 =15°

#### Question 5:

In the adjoining figure, BM ⊥ AC and DN ⊥ AC. 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 ∆ABE and ∆ADEwe have:
∠BAE = ∠DAE​                               (Proven above)
AE is common.
∴ ∆ABE ≅  ∆ADE                          (SAS congruence rule)
⇒ BE = DE                                                 (By CPCT)

(iii)   ∆ABC ≅  ∆ADC                  (Proven above)
∴ ∠ABC = ∠AD​C                           (By CPCT)​

#### 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 = 12(180°  90°) = 90°2 = 4512180° – 90° = 90°2 = 45°

#### 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 ∆AOCOA + OC > AC

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

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#### 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 ∆ABCAB + BC > AC            …(1)

In ∆ACDCD + 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 ∆ABCAB + BC > AC
In ∆ACDCD + DA > AC
In ∆ BCDBC CD > BD
In ∆ ABDDA + AB > 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 ​= ∠and B ​= ∠D ​ ​
∴ ∠C = 72o
A and ∠B are adajcent angles.
i.e., ∠A ​+ ∠B​ = 180o
⇒ ∠B = 180o   ∠A
⇒ ∠B​ = 180o  72o = 108o
∴​ ∠B​ =​ ∠D = 108o
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, M is the midpoint of side BC of a parallelogram ABCD such that ∠BAM = ∠DAM. Prove that AD = 2CD.  Given: parallelogram ABCD, M is the midpoint of side BC and ∠BAM = ∠DAM.

Proof:

So, DAM=AMB∠DAM=∠AMB      (Alternate interior angles)

But, DAM=BAM∠DAM=∠BAM (Given)

Therefore, AMB=BAM∠AMB=∠BAM

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

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

2AB=2CD(AB+AB)=2CD⇒2AB=2CD⇒(AB+AB)=2CD

BM+MC=2CD⇒BM+MC=2CD      (AB BM and MC = BM)

#### Question 4:

In the adjoining figure, ABCD is a parallelogram in which ∠A = 60°. If the bisectors of ∠A and ∠B meet DC at P, prove that (i) ∠APB = 90°, (ii) AD = DP and PB = PC = BC, (iii) DC = 2AD. ABCD is a parallelogram.
∴ ​∠A = ​∠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 = 60°2=30°60°2=30° and ∠PBA = 120°2=60°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 andtherefore, PB = PC = BC.​

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

#### Question 5:

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

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

#### Question 6:

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

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 =  32 − 5 = 91o
Hence, x = 32o, ∠A = C = 89o and ∠B = D = 91o

#### Question 7:

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 = ∠and B = ∠D           (Opposite angles)
Let A = xo and ∠B = (4x5)4×5°
Now, ∠​A + ∠B = 180o                 (Adjacent angles are supplementary)
x + 4x5 = 180o9x5 = 180ox = 100oNow, A = 100o and B = (45)×100°

= 80o⇒x + 4×5 = 180o⇒9×5 = 180o⇒x = 100oNow, ∠A = 100o and ∠B = 45×100° = 80o
HenceA = C = 100o; B = ∠D = 80o

#### Question 8:

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
HenceA = C = 70o; ∠B = ∠D = 110o

#### Question 9:

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

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

#### Question 10:

In each of the figures given below, ABCD is a rhombus. Find the value of x and y in each case. ABCD is a rhombus and a rhombus is also a parallelogram. A rhombus has four equal sides.
(i)​ In ∆ABC, ∠​BAC = ∠BCA = 12(180  110) = 35o12180 – 110 = 35o
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 ∆​ ABCAB = BC, so ∠​BAC = ∠​ACB]
Also, ∠​BOC = 90o
∴ ∠​OBC = 180o − (90o + 62o) = 28o
Hence, x = 62oy = 28o

#### Question 11:

The lengths of the diagonals of a rhombus are 24 cm and 18 cm respectively. Find the length of each side of the rhombus. Let ABCD be a rhombus.
∴ AB = BC = CD = DA
Here, AC and BD are the diagonals of ABCD, where AC = 24 cm and BD = 18 cm.
Let the diagonals intersect each other at O.
We know that the diagonals of a rhombus are perpendicular bisectors of each other.
∴​ ∆AOB is a right angle triangle in which OA = AC/2 = 24/2 = 12 cm and OB = BD/2 = 18/2 = 9 cm.
Now, AB2= OA2 + OB2              [Pythagoras theorem]
⇒​ AB2=​ (12)2 + (9)2
⇒​ AB2=​ 144 + 81 = 225
⇒​ AB=​ 15 cm

Hence, the side of the rhombus is 15 cm.

#### Question 12:

Each side of a rhombus is 10 cm long and one of its diagonals measures 16 cm. Find the length of the other diagonal and hence find the area of the rhombus. 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]
102 = (x2)2 + 82100  64 = x2436 ×4 = x⇒102 = x22 + 82⇒100 – 64 = x24⇒36 ×4 = x2

x2 =144
∴ x = 12 cm

Hence, the other diagonal of the rhombus is 12 cm.
∴ Area of the rhombus =  12×(12×16) = 96 cm212×12×16 = 96 cm2

Page No 330:

#### Question 13:

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 = 12×(180°  110°) = 35o12×180° – 110° = 35o
∴​ y = ∠BAC 35o                 [Interior alternate angles]
Also, x = 90o − y                          [ ​∵∠C = 90o =  x + y ]
⇒​ x = 90o − 35o = 55o
Hence, x = 55o and y = 35o​​

#### Question 14:

In a rhombus ABCD, the altitude from D to the side AB bisects AB. Find the angles of the rhombus. Given: ABCD is a rhombus, DE is altitude which bisects AB i.e. AE = EB

In ΔAED and ΔBED,In ∆AED and ∆BED,

DE=DEDE=DE                         (Common side)

DEA=DEB=90°∠DEA=∠DEB=90°     (Given)

AE=EBAE=EB                          (Given)

ΔAEDΔBED∴ ∆AED≅∆BED         (By SAS congruence Criteria)

Thus, ΔABD ∆ABD is an equilateral triangle.

Therefore, A=60°∠A=60°

C=A=60°⇒∠C=∠A=60°                         (Opposite angles of rhombus are equal)

And, ABC+BCD=180°And, ∠ABC+∠BCD=180°           (Adjacent angles of rhombus are supplementary.)

Hence, the angles of rhombus are 60°, 120°, 60° and 120°60°, 120°, 60° and 120°.

#### Question 15:

In the adjoining figure, ABCD is a square. A line segment CX cuts AB at X and the diagonal BD at O such that ∠COD = 80° and ∠OXA = x°. Find the value of x. 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 16:

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

AB = BC  AB = BC         (Sides of rhombus are equal.)

4=2∠4=∠2          (Angles opposite to equal sides are equal.)     …(1)

Now,

AC is transversal.

So, 1=4∠1=∠4        (Alternate interior angles)          …(2)

From (1) and (2), we get

1=2∠1=∠2

Thus, AC bisects A∠A.

Similarly,

Since, ABDCAB∥DC and AC is transversal.

So, 2=3∠2=∠3     (Alternate interior angles)     …(3)

From (1) and (3), we get

4=3∠4=∠3

Thus, AC bisects ∠C.

Hence, AC bisects C and A∠C and ∠A

In ΔDAB∆DAB,

Now,

DCABDC∥AB                        (Opposite sides of rhombus are parallel.)

BD is transversal.

So, CDB=DBA∠CDB=∠DBA        (Alternate interior angles)          …(5)

From (4) and (5), we get

Thus, DB bisects D∠D.

Similarly,

From (4) and (6), we get

CBD=ABD∠CBD=∠ABD

Thus, BD bisects ∠B.

Hence, BD bisects D and B.∠D and ∠B.

#### Question 17:

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

To prove: AC and MN bisect each other.

Construction: Join AN and MC.

Proof:
Since, ABCD is a parallelogram.

ABDCAMNC⇒AB∥DC⇒AM∥NC

Also, AM = CN           (Given)

Thus, AMCN is a parallelogram.

Since, diagonals of a parallelogram bisect each other.

Hence, AC and MN bisect each other.

This is the complete blog on the RS Aggarwal Solutions Class 9 Maths Chapter 10 Quadrilaterals. To know more about the CBSE Class 9 Maths exam, ask in the comments.