## Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral

**Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral**

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**IMPORTANT POINTS**

**4.Quadrilateral:** A quadrilateral is a plane figure enclosed by four sides. It has four sides, four interior angles and four vertices.

In quadrilateral ABCD, shown alongside:

**(i)** four sides are : AB, BC, CD and DA.

**(ii)** four angles are : ∠ABC,∠BCD, ∠CDA and ∠DAB ; which are numbered∠1, ∠2, ∠3 and ∠4 respectively.

**(iii)** four vertices are : A, B, C and D.

**5. Diagonals of a Quadrilateral :** The line segments joining the opposite vertices of a quadrilateral are called its diagonals.

The given figure shows a quadrilateral PQRS with diagonals PR and QS.

**6. Types of Quadrilaterals :**

1. Trapezium: A trapezium is a quadrilateral in which one pair of opposite sides are parallel.

The figure, given alongside, shows a trapezium as its sides AB and DC are parallel i.e. AB || DC.

When the non-parallel sides of the trapezium are equal in length, it is called an isosceles trapezium.

The given figure shows a trapezium ABCD whose non-parallel sides AD and BC are equal in length i.e. AD = BC; therefore it is an isosceles trapezium.

Also, in an isosceles trapezium :

(i) base angles are equal:

i.e. ∠A = ∠B and ∠D =∠C

(ii) diagonals are equal

i.e. AC = BD.

**2.Parallelogram :** A parallelogram is a quadrilateral, in which both the pairs of opposite sides are parallel.

The quadrilateral ABCD, drawn alongside, is a parallelogram; since, AB is parallel to DC and AD is parallel to BC i.e.

AB || DC and AD || BC.

Also, in a parallelogram ABCD:

(i) opposite sides are equal:

i.e. AB = DC and AD = BC.

(ii) opposite angles are equal:

i.e. ∠ABC = ∠ADC and ∠BCD = ∠BAD

(iii) diagonals bisect each other :

i.e. OA = OC = \(\frac { 1 }{ 2 }\) AC and OB = OD = \(\frac { 1 }{ 2 }\) BD.

**7. Some special types of Parallelograms**

**(a) Rhombus :** A rhombus is a parallelogram in which all its sides are equal.

∴In a rhombus ABCD :

(i) opposite sides are parallel:

i.e. AB||DC and AD||BC.

(ii) all the sides are equal:

i.e. AB = BC = CD = DA.

(iii) opposite angles are equal:

i.e. ∠A = ∠C and ∠B = ∠D.

(iv) diagonals bisect each other at right angle :

i.e. OA= OC = \(\frac { 1 }{ 2 }\) AC ; OB = OD = \(\frac { 1 }{ 2 }\) BD.

and ∠AOB= ∠BOC = ∠COD = ∠AOD = 90°

(v) diagonals bisect the angles at the vertices :

i.e. ∠1 =∠2 ;∠3 = ∠4 ; ∠5 = ∠6 and ∠7 =∠8.

**(b) Rectangle :** A rectangle is a parallelogram whose any angle is 90°.

A rectangle is also defined as a quadrilateral whose each angle is 90°.

Note : If any angle of a parallelogram is 90° ; automatically its each angle is 90° ; the reason being that the opposite angles of a parallelogram are equal.

Also, in a rectangle:

(i) opposite sides are parallel.

(ii) opposite sides are equal.

(iii) each angle is 90°.

(iv) diagonals are equal.

(v) diagonals bisect each other.

**(c) Square :** A square is a parallelogram, whose all side are equal and each angle is 90°.

A square can also be defined as :

(i) a rhombus whose any angle is 90°.

(ii) a rectangle whose all sides are equal.

(iii) a quadrilateral whose all sides are equal and each angle is 90°.

∴ If ABCD is a square :

(i) all its sides are equal, i.e. AB = BC = CD = DA

(ii) each angle of it is 90°.

i.e. ∠A = ∠B = ∠C = ∠D = 90°.

Also,

(iii) diagonals are equal.

i.e. AC = BD.

(iv) diagonals bisect each other at 90°.

i.e. OA = OC =\(\frac { 1 }{ 2 }\) AC;OB = OD = \(\frac { 1 }{ 2 }\) BD

and ∠AOB = ∠BOC = ∠COD = ∠DOA = 90°.

Since, diagonals AC and BD are equal; therefore ; OA = OC = OB = OD.

(v) diagonals bisect the angles at the vertices

i.e. ∠1 = ∠2 = 45° [∵ ∠1 + ∠2 = 90°]

Similarly; ∠3 = ∠4 = 45° ;

∠5 – ∠6 = 45° and ∠7 = ∠8 = 45°.

### Quadrilateral Exercise 27A – Selina Concise Mathematics Class 6 ICSE Solutions

**Question 1.**

**Two angles of a quadrilateral are 89° and 113°. If the other two angles are equal; find the equal angles.**

**Solution:**

Let the other angle = x°

According to given,

89° + 113° + x° + x° = 360°

2x° = 360° – 202°

2x° = 158°

x° = \(\frac { 158 }{ 2 }\) =79°

∴other two angles = 79° each

**Question 2.**

**Two angles of a quadrilateral are 68° and 76°. If the other two angles are in the ratio 5 : 7; find the measure of each of them.**

**Solution:**

Two angles are 68° and 76°

Let other two angles be 5x and 7x

∴ 68°+76°+5x + 7x = 360°

12x + 144° = 360°

12x = 360° – 144°

12x= 216°

x = 18°

angles are 5x and 7x

i.e. 5×18° and 7×18° i.e. 90° and 126°

**Question 3.**

**Angles of a quadrilateral are (4x)°, 5(x+2)°, (7x-20)° and 6(x+3)°. Find**

**(i) the value of x.**

**(ii) each angle of the quadrilateral.**

**Solution:**

Angles of quadrilateral are,

(4x)°, 5(x+2)°, (7x-20)° and 6(x+3)°.

4x+5(x+2)+(7x-20)+6(x+3) = 360°

4x+5x+10+7x-20+6x+18 = 360° 22x+8 = 360°

22x = 360°-8°

22x = 352°

x = 16°

Hence angles are,

(4x)° = (4×16)° = 64°,

5(x+2)° = 5(16+2)° = 90°,

(7x-20)° = (7×16-20)° = 92°

6(x+3)° = 6(16+3) = 114°

**Question 4.**

**Use the information given in the following figure to find :**

**(i) x**

**(ii) ∠B and ∠C**

**Solution:**

∵ ∠A = 90° (Given)

∠B = (2x+4°)

∠C = (3x-5°)

∠D = (8x – 15°)

∠A + ∠B + ∠C + ∠D = 360°

90° + (2x + 4°) + (3x – 5°) + (8x – 15°) = 360°

90° + 2x + 4° + 3x – 5° + 8x – 15° = 360°

⇒ 74° + 13x = 360°

⇒13x = 360° – 74°

⇒ 13x = 286°

⇒ x = 22°

∵ ∠B = 2x + 4 = 2. x 22° + 4 – 48°

∠C = 3x – 5 = 3×22° -5 = 61°

Hence (i) 22° (ii) ∠B = 48°, ∠C = 61°

**Question 5.**

**In quadrilateral ABCD, side AB is parallel to side DC. If ∠A : ∠D = 1 : 2 and ∠C : ∠B = 4:5**

**(i) Calculate each angle of the quadrilateral.**

**(ii) Assign a special name to quadrilateral ABCD.**

**Solution:**

∵∠A : ∠D = 1:2

Let ∠A = x and ∠B = 2x

∵∠C : ∠B = 4 : 5 Let ∠C = 4y and ∠B = 5y

∵AB || DC

∠A + ∠D = 180° x + 2x = 180°

3x = 180° x = 60°

∴A = 60°

∠D = 2x = 2 x 60 = 120° Again ∠B + ∠C = 180°

5y + 4y= 180°

9y = 180°

y = 20°

∴∠B = 5y- = 5 x 20 = 100°

∠C = 4y = 4 x 20 = 80°

Hence ∠A = 60° ; ∠B = 100° ; ∠C = 80° and ∠D = 120°

**Question 6.**

**From the following figure find ;**

**(i) x,**

**(ii) ∠ABC,**

**(iii) ∠ACD.**

**Solution:**

(i) In Quadrilateral ABCD,

x + 4x + 3x + 4x + 48° = 360°

12x = 360° – 48°

12x =312

(ii) ∠ABC = 4x

4 x 26 = 104°

(iii) ∠ACD = 180°-4x-48°

= 180°-4×26°-48°

= 180°-104°-48°

= 180°-152° = 28°

**Question 7.**

**Given : In quadrilateral ABCD ; ∠C = 64°, ∠D = ∠C – 8° ;**

**∠A = 5(a+2)° and ∠B = 2(2a+7)°.**

**Calculate ∠A.**

**Solution:**

∵∠C = 64° (Given)

∴∠D = ∠C- 8°

= 64°- 8°

= 56°

∠A = 5 (a + 2)°

∠B = 2(2a + 7)°

Now ∠A + ∠B + ∠C + ∠D = 360°

5(a + 2)° + 2(2a + 7)° + 64°+ 56° = 360°

5a + 10 + 4a + 14° + 64° + 56° = 360°

9a + 144° = 360°

9a = 360°-144°

9a = 216°

a = 24°

∴∠A = 5(a + 2)

= 5(24 + 2)

= 130°

**Question 8.**

**In the given figure :**

**∠b = 2a + 15**

**and ∠c = 3a+5; find the values of b and c.**

**Solution:**

∠b = 2a + 15

& ∠c = 3a + 5

∵Sum of angles of quadrilateral = 360°

70° + a + 2a + 15 + 3a + 5 – 360°

6a+90° = 360°

6a = 270°

a = 45°

∴ b = 2a+15= 2×45+15 = 105°

c = 3a+5 = 3×45+5 = 140°

105° and 140°

**Question 9.**

**Three angles of a quadrilateral are equal. If the fourth angle is 69°; find the measure of equal angles.**

**Solution:**

Let each equal angle be

x° x + x + x + 69° = 360°

3x = 360°-69 3x =291 x = 97°

Each equal angle = 97°

**Question 10.**

**In quadrilateral PQRS, ∠P : ∠Q : ∠R : ∠S = 3 : 4 : 6 : 7.**

**Calculate each angle of the quadrilateral and then prove that PQ and SR are parallel to each other. Is PS also parallel to QR ? **

**Solution:**

∵∠P : ∠Q : ∠R : ∠S = 3 : 4 : 6 : 7

Let ∠P = 3x

∠Q = 4x

∠R = 6x & ∠S = 7x

∴∠P+∠Q+∠R+∠S = 360°

3x + 4x + 6x + lx = 360°

20x = 360°

x = 18°

∴ ∠P = 3x = 3×18 = 54°

∠Q = 4x = 4×18 = 72°

∠R = 6x = 6×18 = 108°

∠S = 7x = 7×18 = 126°

∠Q + ∠R = 72°+108° = 180° or ∠P +∠S = 54°+126° = 180°

Hence PQ || RS

As ∠P + ∠Q = 72° +54° = 126°

Which is * 180°.

∴PS and QR are not parallel.

**Question 11.**

**Use the information given in the following figure to find the value of x.**

**Solution:**

Take A, B, C, D as the vertices of quadrilateral and BA is produced to E (say).

Since ∠EAD = 70°

∴∠DAB = 180° – 70° = 110° [∵ EAB is a straight line and AD stands on it]

∴∠EAD + ∠DAB = 180°

∴110° + 80° + 56° + 3x = 360°

[∵ sum of interior angles of a quadrilateral = 360°]

∴3x = 360° – 110° – 80° – 56° + 6°

3x = 360° – 240° = 120°

∴x = 40°

**Question 12.**

**The following figure shows a quadrilateral in which sides AB and DC are parallel.**

**If ∠A : ∠D = 4 : 5, ∠B = (3x – 15)° and ∠C = (4x + 20)°, find each angle of the quadrilateral ABCD.**

**Solution:**

Let ∠A = 4x

∠D = 5x

Since ∠A + ∠D = 180° [∵ AB || DC]

∴4x + 5x = 180°

⇒ 9x = 180° ⇒x = 20°

∴∠A = 4 (20) = 80°, ∠D = 5 (20) = 100° Again ∠B + ∠C = 180° [ ∵ AB || DC]

∴ 3x – 15° + 4x + 20° = 180°

7x = 180°-5°

⇒ 7x = 175° ⇒ x = 25°

∴∠B = 75°-15° = 60° and ∠C = 4(25) + 20 = 100° + 20° = 120°

### Quadrilateral Exercise 27B – Selina Concise Mathematics Class 6 ICSE Solutions

**Question 1.**

**In a trapezium ABCD, side AB is parallel to side DC. If ∠A = 78° and ∠C = 120°, find angles B and D.**

**Solution:**

∵ AB || DC and BC is transversal

∴∠B and ∠C, ∠A and ∠D are Cointerior angles with their sum = 180°

i.e. ∠B + ∠C = 180°

⇒ ∠B + 120° = 180°

⇒ ∠B = 180° – 120°

⇒ ∠B = 60°

Also ∠A + ∠D = 180°

⇒ 78° + ∠D = 180°

⇒ ∠D = 180° – 78°

∠D = 102°

**Question 2.**

**In a trapezium ABCD, side AB is parallel to side DC. If ∠A = x° and ∠D = (3x – 20)°; find the value of x.**

**Solution:**

∵AB || DC and BC is transversal

∴∠A and ∠B are Co-interior angles with their sum = 180°

i.e. ∠A + ∠D = 180°

⇒ x° + (3x – 20)° = 180°

⇒ x° + 3x° – 20° = 180°

⇒ 4x° = 180° + 20°

x° = \(\frac { 200 }{ 4 }\) = 50°

∴Value of x = 50°

**Question 3.**

**The angles A, B, C and D of a trapezium ABCD are in the ratio 3 : 4 : 5 : 6.**

**Le. ∠A : ∠B : ∠C : ∠D = 3:4: 5 : 6. Find all the angles of the trapezium. Also, name the two sides of this trapezium which are parallel to each other. Give reason for your answer**

**Solution:**

**Question 4.**

**In an isosceles trapezium one pair of opposite sides are ….. to each Other and the other pair of opposite sides are ….. to each other.**

**Solution:**

In an isosceles trapezium one pair of opposite sides are **parallel** to each other and the other pair of opposite sides are **equal** to each other.

**Question 5.**

**Two diagonals of an isosceles trapezium are x cm and (3x – 8) cm. Find the value of x.**

**Solution:**

∵The diagonals of an isosceles trapezium are of equal length

∴3x – 8 = x

⇒ 3x – x = 8 cm

⇒ 2x = 8 cm

⇒ x = 4 cm

∴ The value of x is 4 cm

**Question 6.**

**Angle A of an isosceles trapezium is 115° ; find the angles B, C and D.**

**Solution:**

**Question 7.**

**Two opposite angles of a parallelogram are 100° each. Find each of the other two opposite angles.**

**Solution:**

**Given :** Two opposite angles of a parallelogram are 100° each

∵ Adjacent angles of a parallelogram are supplementary,

∴∠A + ∠B = 180°

⇒ 100° + ∠B = 180°

⇒ ∠B = 180° – 100°

⇒ ∠B = 80°

Also, opposite angles of a parallelogram are equal

∴∠D = ∠B = 80°

∴∠B = ∠D = 80°

**Question 8.**

**Two adjacent angles of a parallelogram are 70° and 110° respectively. Find the other two angles of it.**

**Solution:**

Given two adjacent angles of a parallelogram are 70° and 110° respectively.

Since, we know that opposite angles of a parallelogram are equal

∴∠C = ∠A = 70° and ∠D = ∠B = 110°

**Question 9.**

**The angles A, B, C and D of a quadrilateral are in the ratio 2:3: 2 : 3. Show this quadrilateral is a parallelogram.**

**Solution:**

**Question 10.**

**In a parallelogram ABCD, its diagonals AC and BD intersect each other at point O.**

**If AC = 12 cm and BD = 9 cm ; find; lengths of OA and OD.**

**Solution:**

**Question 11.**

**In parallelogram ABCD, its diagonals intersect at point O. If OA = 6 cm and OB = 7.5 cm, find the length of AC and BD.**

**Solution:**

**Question 12.**

**In parallelogram ABCD, ∠A = 90°**

**(i) What is the measure of angle B.**

**(ii) Write the special name of the paralleogram**.

**Solution:**

**Question 13.**

**One diagnol of a rectangle is 18 cm. What is the length of its other diagnol?**

**Solution:**

∵ In a rectangle, diagnols are equal

⇒ AC = BD

Given, one diagnol of a rectangle = 18cm

∴ Other diagnol of a rectangle will be = 18cm

i.e. AC = BD = 18cm.

**Question 14.**

**Each angle of a quadrilateral is x + 5°. Find :**

**(i) the value of x**

**(ii) each angle of the quadrilateral.**

**Give the special name of the quadrilateral taken.**

**Solution:**

**Question 15.**

**If three angles of a quadrilateral are 90° each, show that the given quadrilateral is a rectangle.**

**Solution:**

The given quadrilateral ABCD will be a rectangle, if its each angle is 90°

Since, the sum of interior angles of a quadrilateral is 360°.

∴∠A +∠B + ∠C + ∠D = 360°

⇒ 90° + 90° + 90° + ∠D = 360°

⇒ 270° + ∠D = 360°

⇒ ∠D = 360° – 270°

⇒ ∠D = 90°

Since, each angle of the quadrilateral is 90°.

∴The given quadrilateral is a rectangle.

**Question 16.**

**The diagnols of a rhombus are 6 .cm and 8 cm. State the angle at which these diagnols intersect.**

**Solution:**

The diagnols of a Rhombus always intersect at 90°.

**Question 17.**

**Write, giving reason, the name of the figure drawn alongside. Under what condition will this figure be a square.**

**Solution:**

Since, all the sides of the given figure are equal.

i.e. AB = BC = CD = DA = 6 cm

∴ The given figure is a rhombus.

This figure shall be considered as a square, if any angle is 90°.

**Question 18.**

**Write two conditions that will make the adjoining figure a square.**

**Solution:**

The conditions that will make the ad-joining figure a square are :

(i) All the sides must be equal.

(ii) Any angle is 90°.