RD Sharma Solutions for Class 9 Maths Chapter 14 Quadrilaterals

RD Sharma Solutions Class 9 Maths Chapter 14 – Free PDF Download

RD Sharma Solutions Class 9 Maths Chapter 14 Quadrilaterals consist of questions based on a polygon with 4 vertices or edges and 4 sides, which is termed as a quadrilateral. With 4 interior angles present, the sum of the interior angles of a quadrilateral is 360 degrees. Consider a quadrilateral PQRZ, with internal angles P, Q, R & Z. Then, ∠P + ∠Q + ∠R + ∠Z = 360^o. For a better grasp of the given concepts, students are suggested to practise RD Sharma Solutions on a daily basis.

Basically, quadrilaterals are classified on the basis of their intersecting nature. If they are not intersecting they are called simple quadrilaterals. Otherwise, if they happen to self-intersect, it is a complex quadrilateral. Simple quadrilaterals are further classified into concave and convex quadrilaterals based on the position of diagonals and their interior angles. The RD Sharma Class 9 Solutions explain all the concepts from the chapter Quadrilaterals based on CBSE 2023-24 syllabus.

RD Sharma Solutions for Class 9 Maths Chapter 14 Quadrilaterals

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Access Answers to Maths RD Sharma Solutions for Class 9 Chapter 14 Quadrilaterals

Exercise 14.1 Page No: 14.4

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

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

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 .

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

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

Therefore, ARPQ is a parallelogram.

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

AR = QP = 15 cm

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

RP = AQ = 10.5cm

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

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

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.

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 .

RD Sharma Solutions for Class 9 Maths Chapter 14 Quadrilaterals

In the 14th Chapter of Class 9, RD Sharma Solutions students will study important concepts. Some are listed below:

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