RD Sharma Solutions Class 9 Maths Chapter 14 – Quadrilaterals (Updated for 2024)

RD Sharma Solutions Class 9 Maths Chapter 14 – Quadrilaterals: Quadrilateral: In this blog, we will be dealing with the important aspects of RD Sharma Solutions Class 9 Maths Chapter 14 which will guide you to bring changes in your methods of preparation. The parts that we will be concentrating on are the benefits of following Chapter 14 Solutions along with the exercise-wise explanations.

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RD Sharma Class 9 Solutions Chapter 14

Exercise-Wise RD Sharma Solutions Class 9 Maths Chapter 14 

RD Sharma Class 9 Chapter 14 Exercise 14A

RD Sharma Class 9 Chapter 14 Exercise 14B

RD Sharma Class 9 Chapter 14 Exercise 14C

RD Sharma Class 9 Chapter 14 Exercise 14D

Access answers of RD Sharma Solutions Class 9 Maths Chapter 14 

RD Sharma Class 9 Solution Chapter 14 Quadrilaterals Ex 14.1

Question 1: Three angles of a quadrilateral are respectively equal to 110⁰, 50⁰ and 40⁰. Find its fourth angle.

Solution:

Three angles of a quadrilateral are 110⁰, 50⁰ and 40⁰

Let the fourth angle be ‘x’

We know, sum of all angles of a quadrilateral = 360⁰

110⁰ + 50⁰ + 40⁰ + x⁰ = 360⁰

⇒ x = 360⁰ – 200⁰

⇒x = 160⁰

Therefore, the required fourth angle is 160⁰.

Question 2: In a quadrilateral ABCD, the angles A, B, C and D are in the ratio of 1:2:4:5. Find the measure of each angles of the quadrilateral.

Solution:

Let the angles of the quadrilaterals are A = x, B = 2x, C = 4x and D = 5x

We know, sum of all angles of a quadrilateral = 360⁰

A + B + C + D = 360⁰

x + 2x + 4x + 5x = 360⁰

12x = 360⁰

x = 360⁰/12 = 30⁰

Therefore,

A = x = 30⁰

B = 2x = 60⁰

C = 4x = 120⁰

D = 5x = 150⁰

Question 3: In a quadrilateral ABCD, CO and DO are the bisectors of ∠C and ∠D respectively. Prove that ∠COD = 1/2 (∠A + ∠B).

Solution:

In ΔDOC,

∠CDO + ∠COD + ∠DCO = 180⁰ [Angle sum property of a triangle]

or 1/2∠CDA + ∠COD + 1/2∠DCB = 180⁰

 

∠COD = 180⁰ – 1/2(∠CDA + ∠DCB) …..(i)

Also

We know, sum of all angles of a quadrilateral = 360⁰

∠CDA + ∠DCB = 360⁰ – (∠DAB + ∠CBA) ……(ii)

Substituting (ii) in (i)

∠COD = 180⁰ – 1/2{360⁰ – (∠DAB + ∠CBA) }

We can also write, ∠DAB = ∠A and ∠CBA = ∠B

∠COD = 180⁰ − 180⁰ + 1/2(∠A + ∠B))

∠COD = 1/2(∠A + ∠B)

Hence Proved.

Question 4: The angles of a quadrilateral are in the ratio 3:5:9:13. Find all the angles of the quadrilateral.

Solution:

The angles of a quadrilateral are 3x, 5x, 9x and 13x respectively.

We know, sum of all interior angles of a quadrilateral = 360⁰

Therefore, 3x + 5x + 9x + 13x = 360⁰

30x = 360⁰

or x = 12⁰

Hence, angles measures are

3x = 3(12) = 36⁰

5x = 5(12) = 60⁰

9x = 9(12) = 108⁰

13x = 13(12) = 156⁰

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Exercise 14.2

Question 1: Two opposite angles of a parallelogram are (3x – 2)⁰ and (50 – x) ⁰. Find the measure of each angle of the parallelogram.

Solution:

Given: Two opposite angles of a parallelogram are (3x – 2)⁰ and (50 – x) ⁰.

We know, opposite sides of a parallelogram are equal.

(3x – 2)⁰ = (50 – x) ⁰

3x + x = 50 + 2

4x = 52

x = 13

Angle x is 13⁰

Therefore,

(3x-2) ⁰ = (3(13) – 2) = 37⁰

(50-x) ⁰ = (50 – 13) = 37⁰

Adjacent angles of a parallelogram are supplementary.

x + 37 = 180⁰

x = 180⁰ − 37⁰ = 143⁰

Therefore, required angles are : 37⁰, 143⁰, 37⁰ and 143⁰.

Question 2: If an angle of a parallelogram is two-third of its adjacent angle, find the angles of the parallelogram.

Solution:

Let the measure of the angle be x. Therefore, measure of the adjacent angle is 2x/3.

We know, adjacent angle of a parallelogram is supplementary.

x + 2x/3 = 180⁰

3x + 2x = 540⁰

5x = 540⁰

or x = 108⁰

Measure of second angle is 2x/3 = 2(108⁰)/3 = 72⁰

Similarly measure of 3^rd and 4^th angles are 108⁰ and 72⁰

Hence, four angles are 108⁰, 72⁰, 108⁰, 72⁰

Question 3: Find the measure of all the angles of a parallelogram, if one angle is 24⁰ less than twice the smallest angle.

Solution:

Given: One angle of a parallelogram is 24⁰ less than twice the smallest angle.

Let x be the smallest angle, then

x + 2x – 24⁰ = 180⁰

3x – 24⁰ = 180⁰

3x = 108⁰ + 24⁰

3x = 204⁰

x = 204⁰/3 = 68⁰

So, x = 68⁰

Another angle = 2x – 24⁰ = 2(68⁰) – 24⁰ = 112⁰

Hence, four angles are 68⁰, 112⁰, 68⁰, 112⁰.

Question 4: The perimeter of a parallelogram is 22cm. If the longer side measures 6.5cm what is the measure of the shorter side?

Solution:

Let x be the shorter side of a parallelogram.

Perimeter = 22 cm

Longer side = 6.5 cm

Perimeter = Sum of all sides = x + 6.5 + 6.5 + x

22 = 2 ( x + 6.5 )

11 = x + 6.5

or x = 11 – 6.5 = 4.5

Therefore, shorter side of a parallelogram is 4.5 cm

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Exercise 14.3

Question 1: In a parallelogram ABCD, determine the sum of angles ∠C and ∠D.

Solution:

In a parallelogram ABCD , ∠C and ∠D are consecutive interior angles on the same side of the transversal CD.

So, ∠C + ∠D = 180⁰

Question 2: In a parallelogram ABCD, if ∠B = 135⁰, determine the measures of its other angles.

Solution:

Given: In a parallelogram ABCD, if ∠B = 135⁰

Here, ∠A = ∠C, ∠B = ∠D and ∠A + ∠B = 180⁰

∠A + 135⁰ = 180⁰

∠A = 45⁰

Answer:

∠A = ∠C = 45⁰

∠B = ∠D = 135⁰

Question 3: ABCD is a square. AC and BD intersect at O. State the measure of ∠AOB.

Solution:

We know, diagonals of a square bisect each other at right angle.

So, ∠AOB = 90⁰

Question 4: ABCD is a rectangle with ∠ABD = 40⁰. Determine ∠DBC.

Solution:

Each angle of a rectangle = 90^o

So, ∠ABC = 90⁰

∠ABD = 40⁰ (given)

Now, ∠ABD + ∠DBC = 90⁰

40⁰ + ∠DBC = 90⁰

or ∠DBC = 50⁰ .

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Exercise 14.4

Question 1: In a ΔABC, D, E and F are, respectively, the mid points of BC, CA and AB. If the lengths of sides AB, BC and CA are 7 cm, 8 cm and 9 cm, respectively, find the perimeter of ΔDEF.

Solution:

Given: AB = 7 cm, BC = 8 cm, AC = 9 cm

In ∆ABC,

In a ΔABC, D, E and F are, respectively, the mid points of BC, CA and AB.

According to Midpoint Theorem:

EF = 1/2BC, DF = 1/2 AC and DE = 1/2 AB

Now, Perimeter of ∆DEF = DE + EF + DF

= 1/2 (AB + BC + AC)

= 1/2 (7 + 8 + 9)

= 12

Perimeter of ΔDEF = 12cm

Question 2: In a ΔABC, ∠A = 50⁰, ∠B = 60⁰ and ∠C = 70⁰. Find the measures of the angles of the triangle formed by joining the mid-points of the sides of this triangle.

Solution:

In ΔABC,

D, E and F are mid points of AB,BC and AC respectively.

In a Quadrilateral DECF:

By Mid-point theorem,

DE ∥ AC ⇒ DE = AC/2

And CF = AC/2

⇒ DE = CF

Therefore, DECF is a parallelogram.

∠C = ∠D = 70⁰

[Opposite sides of a parallelogram]

Similarly,

ADEF is a parallelogram, ∠A = ∠E = 50⁰

BEFD is a parallelogram, ∠B = ∠F = 60⁰

Hence, Angles of ΔDEF are: ∠D = 70⁰, ∠E = 50⁰, ∠F = 60⁰.

Question 3: In a triangle, P, Q and R are the mid points of sides BC, CA and AB respectively. If AC = 21 cm, BC = 29 cm and AB = 30 cm, find the perimeter of the quadrilateral ARPQ.

Solution:

In ΔABC,

R and P are mid points of AB and BC

By Mid-point Theorem

RP ∥ AC ⇒ RP = AC/2

In a quadrilateral, ARPQ

RP ∥ AQ ⇒ RP = AQ

[A pair of side is parallel and equal]

Therefore, ARPQ is a parallelogram.

Now, AR = AB/2 = 30/2 = 15 cm

[AB = 30 cm (Given)]

AR = QP = 15 cm

[ Opposite sides are equal ]

And RP = AC/2 = 21/2 = 10.5 cm

[AC = 21 cm (Given)]

RP = AQ = 10.5cm

[ Opposite sides are equal ]

Now,

Perimeter of ARPQ = AR + QP + RP +AQ

= 15 +15 +10.5 +10.5

= 51

Perimeter of quadrilateral ARPQ is 51 cm.

Question 4: In a ΔABC median AD is produced to X such that AD = DX. Prove that ABXC is a parallelogram.

Solution:

In a quadrilateral ABXC,

AD = DX [Given]

BD = DC [Given]

From figure, Diagonals AX and BC bisect each other.

ABXC is a parallelogram.

Hence Proved.

Question 5: In a ΔABC, E and F are the mid-points of AC and AB respectively. The altitude AP to BC intersects FE at Q. Prove that AQ = QP.

Solution:

In a ΔABC

E and F are mid points of AC and AB (Given)

EF ∥ BC ⇒ EF = BC/2 and

[By mid-point theorem]

In ΔABP

F is the mid-point of AB, again by mid-point theorem

FQ ∥ BP

Q is the mid-point of AP

AQ = QP

Hence Proved.

Question 6: In a ΔABC, BM and CN are perpendiculars from B and C respectively on any line passing through A. If L is the mid-point of BC, prove that ML = NL.

Solution:

Given that,

In ΔBLM and ΔCLN

∠BML = ∠CNL = 90⁰

BL = CL [L is the mid-point of BC]

∠MLB = ∠NLC [Vertically opposite angle]

By ASA criterion:

ΔBLM ≅ ΔCLN

So, LM = LN [By CPCT]

Question 7: In figure, triangle ABC is a right-angled triangle at B. Given that AB = 9 cm, AC = 15 cm and D, E are the mid-points of the sides AB and AC respectively, calculate

(i) The length of BC (ii) The area of ΔADE.

Solution:

In ΔABC,

∠B=90⁰ (Given)

AB = 9 cm, AC = 15 cm (Given)

By using Pythagoras theorem

AC² = AB² + BC²

⇒15² = 9² + BC²

⇒BC² = 225 – 81 = 144

or BC = 12

Again,

AD = DB = AB/2 = 9/2 = 4.5 cm [D is the mid−point of AB

D and E are mid-points of AB and AC

DE ∥ BC ⇒ DE = BC/2 [By mid−point theorem]

Now,

Area of ΔADE = 1/2 x AD x DE

= 1/2 x 4.5 x 6

=13.5

Area of ΔADE is 13.5 cm²

Question 8: In figure, M, N and P are mid-points of AB, AC and BC respectively. If MN = 3 cm, NP = 3.5 cm and MP = 2.5 cm, calculate BC, AB and AC.

Solution:

Given: MN = 3 cm, NP = 3.5 cm and MP = 2.5 cm.

M and N are mid-points of AB and AC

By mid−point theorem, we have

MN∥BC ⇒ MN = BC/2

or BC = 2MN

BC = 6 cm

[MN = 3 cm given)

Similarly,

AC = 2MP = 2 (2.5) = 5 cm

AB = 2 NP = 2 (3.5) = 7 cm

Question 9: ABC is a triangle and through A, B, C lines are drawn parallel to BC, CA and AB respectively intersecting at P, Q and R. Prove that the perimeter of ΔPQR is double the perimeter of ΔABC.

Solution:

ABCQ and ARBC are parallelograms.

Therefore, BC = AQ and BC = AR

⇒AQ = AR

⇒A is the mid-point of QR

Similarly B and C are the mid points of PR and PQ respectively.

By mid−point theorem, we have

AB = PQ/2, BC = QR/2 and CA = PR/2

or PQ = 2AB, QR = 2BC and PR = 2CA

⇒PQ + QR + RP = 2 (AB + BC + CA)

⇒ Perimeter of ΔPQR = 2 (Perimeter of ΔABC)

Hence proved.

Question 10: In figure, BE ⊥ AC, AD is any line from A to BC intersecting BE in H. P, Q and R are respectively the mid-points of AH, AB and BC. Prove that ∠PQR = 90⁰.

Solution:

BE⊥AC and P, Q and R are respectively mid-point of AH, AB and BC. (Given)

In ΔABC, Q and R are mid-points of AB and BC respectively.

By Mid-point theorem:

QR ∥ AC …..(i)

In ΔABH, Q and P are the mid-points of AB and AH respectively

QP ∥ BH ….(ii)

But, BE⊥AC

From (i) and (ii) we have,

QP⊥QR

⇒∠PQR = 90⁰

Hence Proved.

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Exercise VSAQs

Question 1: In a parallelogram ABCD, write the sum of angles A and B.

Solution:

In parallelogram ABCD, Adjacent angles of a parallelogram are supplementary.

Therefore, ∠A + ∠B = 180⁰

Question 2: In a parallelogram ABCD, if ∠D = 115⁰, then write the measure of ∠A.

Solution:

In a parallelogram ABCD,

∠D = 115⁰ (Given)

Since, ∠A and ∠D are adjacent angles of parallelogram.

We know, Adjacent angles of a parallelogram are supplementary.

∠A + ∠D = 180⁰

∠A = 180⁰ – 115⁰ = 65⁰

Measure of ∠A is 65⁰.

Question 3: PQRS is a square such that PR and SQ intersect at O. State the measure of ∠POQ.

Solution:

PQRS is a square such that PR and SQ intersect at O. (Given)

We know, diagonals of a square bisects each other at 90 degrees.

So, ∠POQ = 90⁰

Question 4: In a quadrilateral ABCD, bisectors of angles A and B intersect at O such that ∠AOB = 75°, then write the value of ∠C + ∠D.

Solution:

∠AOB = 75^o (given)

In a quadrilateral ABCD, bisectors of angles A and B intersect at O, then

∠AOB = 1/2 (∠ADC + ∠ABC)

or ∠AOB = 1/2 (∠D + ∠C)

By substituting given values, we get

75 ^o = 1/2 (∠D + ∠C)

or ∠C + ∠D = 150 ^o

Question 5: The diagonals of a rectangle ABCD meet at O. If ∠BOC = 44^o, find ∠OAD.

Solution:

ABCD is a rectangle and ∠BOC = 44^o (given)

∠AOD = ∠BOC (vertically opposite angles)

∠AOD = ∠BOC = 44^o

∠OAD = ∠ODA (Angles facing same side)

and OD = OA

Since sum of all the angles of a triangle is 180 ^o, then

So, ∠OAD = 1/2 (180 ^o – 44 ^o) = 68 ^o

Question 6: If PQRS is a square, then write the measure of ∠SRP.

Solution:

PQRS is a square.

⇒ All side are equal, and each angle is 90^o degrees and diagonals bisect the angles.

So, ∠SRP = 1/2 (90 ^o) = 45^o

Question 7: If ABCD is a rectangle with ∠BAC = 32^o, find the measure of ∠DBC.

Solution:

ABCD is a rectangle and ∠BAC=32 ^o (given)

We know, diagonals of a rectangle bisects each other.

AO = BO

∠DBA = ∠BAC = 32 ^o (Angles facing same side)

Each angle of a rectangle = 90 degrees

So, ∠DBC + ∠DBA = 90 ^o

or ∠DBC + 32 ^o = 90 ^o

or ∠DBC = 58 ^o

Question 8: If ABCD is a rhombus with ∠ABC = 56^o, find the measure of ∠ACD.

Solution:

In a rhombus ABCD,

<ABC = 56 ^o

So, <BCD = 2 (<ACD) (Diagonals of a rhombus bisect the interior angles)

or <ACD = 1/2 (<BCD) …..(1)

We know, consecutive angles of a rhombus are supplementary.

∠BCD + ∠ABC = 180 ^o

∠BCD = 180 ^o – 56 ^o = 124 ^o

Equation (1) ⇒ <ACD = 1/2 x 124 ^o = 62 ^o

Question 9: The perimeter of a parallelogram is 22 cm. If the longer side measure 6.5 cm, what is the measure of shorter side?

Solution:

Perimeter of a parallelogram = 22 cm. (Given)

Longer side = 6.5 cm

Let x be the shorter side.

Perimeter = 2x + 2×6.5

22 = 2x + 13

2x = 22 – 13 = 9

or x = 4.5

Measure of shorter side is 4.5 cm.

Question 10: If the angles of a quadrilateral are in the ratio 3:5:9:13, then find the measure of the smallest angle.

Solution:

Angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13 (Given)

Let the sides are 3x, 5x, 9x, 13x

We know, sum of all the angles of a quadrilateral = 360^o

3x + 5x + 9x + 13x = 360 ^o

30 x = 360 ^o

x = 12 ^o

Measure of smallest angle = 3x = 3(12) = 36^o .

Detailed Exercise-wise Explanation with Important Topics in the Exercise

RD Sharma Class 9 Chapter 14 Exercise 14A

In Chapter 14A RD Sharma Class 9 the students will be able to examine their performances by attempting objective-type questions that are associated with Quadrilaterals. After going through the solutions of RD Sharma Class 9 Chapter 14A the students will be able to understand the applications of quadrilaterals angles, the angle sum property of a quadrilateral, and many more.

The preparation style of the students will undergo a lot of changes which will be beneficial for their performances in the final Maths Exam. Hence, you must buy a copy for yourself right now. 

RD Sharma Class 9 Chapter 14 Exercise 14B

In the RD Sharma Class 9 Chapter 14B the students will learn about the techniques to handle problems based on Quadrilaterals. The main areas of RD Sharma Solutions Class 9 Maths Chapter 14B which you need to be careful about are various types of quadrilaterals like rectangles, squares, rhombus, and many more. The problems which the students will find over there are all connected with the application of the basic concept.

By looking at the step-wise explanations the students can attempt all kinds of questions from this chapter with ease. Practicing RD Sharma Solutions Class 9 Maths Chapter 14 is something you cannot ignore to strengthen your preparation style.

RD Sharma Class 9 Chapter 14 Exercise 14C

 In the RD Sharma Exercise 14C Class, 9 Chapter 14 students will get new inputs from the in-depth explanations. The main areas the students need to focus on are the conditions for a quadrilateral to be a parallelogram and some important theorems results.

If the students concentrate on each part of Chapter 14C RD Sharma Solutions carefully then they will be able to grasp the important concepts of Quadrilaterals.

RD Sharma Class 9 Chapter 14 Exercise 14D

The RD Sharma Class 9 Chapter 14D consists of the elements which are related to Quadrilaterals. The students are expected to practice all the problems which are given in RD Sharma Solutions Class 9 Chapter 14 Exercise 14B regularly.

The experts have presented the answers to the complex problems in RD Sharma Solutions Class 9 Chapter 14D to make things easier for the students. Solving RD Sharma Class 9 Chapter 14D is the key to success and to get good results you must make proper use of the key.  The essential areas are the triangle and theorems results of the triangle

In the last exercise, here students can check the progress of their performances by practicing objective-type questions in connection with Quadrilaterals. When you follow the important parts of the Chapter 14 solutions your confidence gets a boost and you will be automatically prepared to face more challenges, You can test yourself when you reach this part to know where you stand.

Important concepts from RD Sharma Solutions Class 9 Maths

• Quadrilaterals Introduction.
• Quadrilaterals angles.
• Angle sum property.
• Various types of Quadrilaterals.
• Conditions for a Quadrilaterals to be a Parallelogram.

FAQs on RD Sharma Solutions Class 9 Maths Chapter 14