Quadrilateral ICSE Class-6th Concise Selina Maths Solutions Ch-27

Quadrilateral ICSE Class-6th Concise Selina Mathematics Solutions Chapter-27. We provide step by step Solutions of Exercise / lesson-27 Quadrilateral for ICSE Class-6 Concise Selina Mathematics.Our Solutions contain all type Questions of Exe-27 A and Exe-27 B with Notes on Quadrilateral to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-6.

Quadrilateral ICSE Class-6th Concise Selina Mathematics Solutions Chapter-27

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Notes on Quadrilateral

Exe-27 A,

Exe-27 B,

A quadrilateral is a plane figure that has four sides or edges, so having four corners or vertices. Quadrilaterals will typically be of standard shapes with four sides like rectangle, square, trapezoid, and kite or irregular and uncharacterized

Types of Quadrilaterals

There are many types of quadrilaterals. As the word ‘Quad’ means four, all these types of a quadrilateral have four sides and the sum of angles of these shapes is 360 degrees.

Trapezium
Parallelogram
Squares
Rectangle
Rhombus
Kite

Another way to classify the types of quadrilaterals are:

Convex Quadrilaterals: Both the diagonals of a quadrilateral are completely contained within a figure.
Concave Quadrilaterals: At least one of the diagonals lies partly are entirely outside of the figure.
Intersecting Quadrilaterals: Intersecting quadrilaterals are not simple quadrilaterals in which the pair of non-adjacent sides intersect. This kind of quadrilaterals are known as self-intersecting or crossed quadrilaterals

Quadrilateral Properties

A quadrilateral is a 4-sided plane figure. Below are some important properties of quadrilaterals :

Every quadrilateral has 4 vertices, 4 angles, and 4 sides
The total of its interior angles = 360 degrees
A quadrilateral with four equal sides and four right angles is a square
A quadrilateral that has opposite sides equal and measure of every angle is 90 degrees is a rectangle
Parallelogram, trapezium, rhombus, and kite are other examples of quadrilaterals

Notes on Quadrilateral

A quadrilateral is a trapezoid or a trapezium if 2 of its sides parallel to each other.

A quadrilateral is a parallelogram if 2 pairs of sides parallel to each other.

Squares and Rectangles are special types of parallelograms. Below are some special properties.– All internal angles are of “right angle” (90 degrees).– Each figure contains 4 right angles.– Sides of a square are of the same length (all sides are congruent) – Opposite sides of a rectangle are same.– Opposite sides of a rectangle and square are parallel.

A quadrilateral is a rhombus, if All the sides are of equal length-Specified 2 pairs of sides are parallel to each other.

A kite is a special sort of quadrilateral, wherever 2 pairs of adjacent sides are equal to each other.

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

Question -1.

Two angles of a quadrilateral are 89° and 113°. If the other two angles are equal; find the equal angles.
Answer-1

Let the other angle = x°
According to given,
89° + 113° + x° + x° = 360°
2x° = 360° – 202°
2x° = 158°
x° = 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.
Answer-2

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.

Answer-3

(i) the value of x.

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°
(ii) each angle of the quadrilateral.

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

Answer-4

(i) x

∵ ∠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°
(ii) ∠B and ∠C

∵ ∠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.
Answer-5

(i) Calculate each angle of the quadrilateral.

∵∠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°

(ii) Assign a special name to quadrilateral ABCD.

Trapezium

Question- 6.

From the following figure find ;

(i) x,
(ii) ∠ABC,
(iii) ∠ACD.
Answer-6

(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.
Answer-7

∵∠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.

Answer-8

∠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.
Answer-9

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 ?
Answer-10

∵∠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.
Answer-11

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

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.

Answer-12

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°