Understanding Quadrilaterals Class-8 ML Aggarwal ICSE Class-8 Maths

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Understanding Quadrilaterals Class-8 ML Aggarwal ICSE Mathematics Solutions Chapter-13. We provide step by step Solutions of Exercise / lesson-13 Understanding Quadrilaterals Class-8th ML Aggarwal ICSE Mathematics.

Our Solutions contain all type Questions with Exe-13.1 , Exe-13.2, Exe-13.3 , Objective Type Questions (including Mental Maths Multiple Choice Questions, HOTS,) and Check Your Progress to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-8 Mathematics .

Understanding Quadrilaterals Class-8 ML Aggarwal ICSE Mathematics Solutions Chapter-13


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Exe-13.1 Understanding Quadrilaterals Class-8 ML Aggarwal ICSE  Maths Solutions

Question 1.
Some figures are given below.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.1 1
Classify each of them on the basis of the following:
(a) Simple curve
(b) Simple closed curve
(c) Polygon
(d) Convex polygon
(e) Concave polygon

Answer

(a) (i), (ii), (iii), (v) and (vi) are simple curves.
(b) (iii), (v), (vi) are simple closed curves.
(c) (iii) and (vi) are polygons.
(d) (iii) is a convex polygon.
(e) (v) is a concave polygon.

Question 2.
How many diagonals does each of the following have?
(a) A convex quadrilateral
(b) A regular hexagon

Answer

(a) A convex quadrilateral: It has two diagonals.
(b) A regular hexagon: It has 9 diagonals as shown.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.1 2

Question 3.
Find the sum of measures of all interior angles of a polygon with number of sides:
(i) 8
(ii) 12

Answer

(i) Sum of measures of all interior angles of
8-sided polygon = (2n – 4) × 90°
= (2 × 8 – 4) × 90°
= 12 × 90° = 1080°
(ii) Sum of measures of all interior angles of
12-sided polygon = (2n – 4) × 90°
= (2 × 12 – 4) × 90°
= 18 × 90°= 1800°

Question 4.
Find the number of sides of a regular polygon whose each exterior angles has a measure of
(i) 24°
(ii) 60°
(iii) 72°

Answer

(i) Let number of sides of the polygon = n
Each exterior angle = 24°
∴ n = \frac{360^{\circ}}{24^{\circ}} = 15 sides
∴ Polygon is of 15 sides.
(ii) Each interior angle of the polygon = 60°
Let number of sides of the polygon = n
∴ n = \frac{360^{\circ}}{60^{\circ}} = 6
∴ Number of sides = 6
(iii) Each interior angle of the polygon = 72°
Let number of sides of the polygon = n
∴ n = \frac{360^{\circ}}{72^{\circ}} = 5
∴ Number of sides = 5

Question 5.
Find the number of sides of a regular polygon if each of its interior angles is
(i) 90°
(ii) 108°
(iii) 165°

Answer

(i) Each interior angle = 90°
Let number of sides of the regular polgyon = n
∴ 90° = \frac{2 n-4}{n} × 90°
⇒ \frac{2 n-4}{n}=\frac{90^{\circ}}{90^{\circ}} = 1
⇒ 2n – 4 = n
⇒ 2n – n = 4
⇒ n = 4
⇒ n = 4
∴ It is a square.
(ii) Each interior angle = 108°
Let number of sides of the regular polygon = n
∴ 108° = \frac{2 n-4}{n} × 90°
⇒ \frac{2 n-4}{n}=\frac{108^{\circ}}{90^{\circ}}=\frac{6}{5}
⇒ 10n – 20 = 6n ⇒ 10n – 6n = 20
⇒ 4n = 20
⇒ n = \frac{20}{4} = 5
∴ It is a pentagon.
(iii) Each interior angle = 165°
Let number of sides of the regular polygon = n
∴ 165° = \frac{2 n-4}{n} × 90°
⇒ \frac{2 n-4}{n}=\frac{165^{\circ}}{90^{\circ}}=\frac{11}{6}
⇒ 12n – 24 = 11n
⇒ 12n – 11n = 24
⇒ n = 24
∴ It is 24-sided polygon.

Question 6.
Find the number of sides in a polygon if the sum of its interior angles is:
(i) 1260°
(ii) 1980°
(iii) 3420°

Answer

We know that, sum of interior angles of polygon
is given by (2n – 4) at right angles.
(i) 1260°
∴ 1260 = (2n – 4) × 90
⇒ \frac{1260}{90} = 2n – 4
⇒ 14 = 2n – 4
⇒ n = 9
(ii) 1980°
∴ 1980 = (2n – 4) × 90
⇒ \frac{1980}{90} = 2n – 4
⇒ 22 = 2n – 4
⇒ n = 13.
(iii) 3420°
∴ 3420 = (272 – 4) × 90 3420
⇒ \frac{3420}{90} = 2n – 4
⇒ 38 = 2n – 4
⇒ n = 21

Question 7.
If the angles of a pentagon are in the ratio 7 : 8 : 11 : 13 : 15, find the angles.

Answer

Ratio in the angles of a polygon = 7 : 8 : 11 : 13 : 15
Sum of angles of a pentagon = (2n – 4) × 90°
= (2 × 5 – 4) × 90°
= 6 × 90° = 540°
Let the angles of the pentagon be
7x, 8x, 11x, 13x, 15x
∴ 7x + 8x + 11x + 13x + 15x = 540°
⇒ 54x = 540° ⇒ x = \frac{540^{\circ}}{54} = 10°
∴ Angles are 7 × 10° = 70°, 8 × 10° = 80°,
11 × 10° = 110°, 13 × 10° = 130° and 15 × 10°= 150°
∴ Angles are 70°, 80°, 110°, 130° and 150°

Question 8.
The angles of a pentagon are x°, (x – 10)°, (x + 20)°, (2x – 44)° and (2x – 70°) Calculate x.

Answer

Angles of a pentaon are x°, (x – 10)°, (x + 20)°,
(2x – 44)° and (2x – 70°)
But sum of angles of a pentagon
= (2n – 4) × 90°
= (2 × 5 – 4) × 90°
= 6 × 90° = 540°
∴ x + x – 10° + x + 20° + 2x – 44° + 2x – 70° = 540°
⇒ 7x – 104° = 540°
⇒ 7x = 540° + 104° = 644°
⇒ x = \frac{644^{\circ}}{7} = 92°
∴ x = 92°

Question 9.
The exterior angles of a pentagon are in ratio 1 : 2 : 3 : 4 : 5. Find all the interior angles of the pentagon.

Answer

Let the exterior angles of the pentagon are x, 2x, 3x, 4x and 5x.
We know that sum of exterior angles of polygon is 360°.
∴ x + 2x + 3x + 4x + 5x = 360°
⇒ 15x = 360°
⇒ x = \frac{360^{\circ}}{15}
⇒ x = 24°
∴ Exterior angles are 24°, 48°, 72°, 96°, 120°
Interior angles are 180° – 24°, 180° – 48°, 180° – 72°, 180° – 96°,
180° – 120° i.e. 156°, 132°, 108°, 84°, 60°.

Question 10.
In a quadrilateral ABCD, AB || DC. If ∠A : ∠D = 2:3 and ∠B : ∠C = ∠7 : 8, find the measure of each angle.

Answer

As AB || CD
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.1 3
∠A + ∠D = 180° and ∠B + ∠C = 180°
⇒ 2x + 3x = 180° and 7y + 8y = 180°
5x = 180° and 15y = 180°
x = 36° and y = 12°
∴ ∠A = 2 × 36 = 72°
and ∠D = 3 × 36 = 108°
∠B = 7y = 7 × 12 = 84°
and ∠C = 8y = 8 × 12 = 96°

Question 11.
From the adjoining figure, find
(i) x
(ii) ∠DAB
(iii) ∠ADB
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.1 4

Answer

(i) ABCD is a quadrilateral
∴ ∠A + ∠B + ∠C + ∠D = 360°
⇒ (3x + 4) + (50 + x) + (5x + 8) + (3x + 10) = 360
⇒ 3x + 4 + 50 + x + 5x + 8 + 3x + 10 = 360°
⇒ 12x + 72 = 360°
⇒ 12x = 288
⇒ x = 24
(ii) ∠DAB = (3x + 4) = 3 × 24 + 4 = 76°
(iii) ∠ADB = 180°- (76° + 50°) = 54° (∵ ABD is a ∆)

Question 12.
Find the angle measure x in the following figures:
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.1 5

Answer

(i) In quadrilateral three angles are 40°, 140° and 100°
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.1 6
But sum of Four angles = 360°
⇒ 40° + 140°+ 100° + x = 360°
⇒ 280° + x = 360°
⇒ x = 360° – 280° = 80°
(ii) In the given figure, ABCDE is a pentagon.
Where side AB is produced to both sides0
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.1 7
∠1 + 60° = 180° (Linear pair)
∠1 = 180°- 60°= 120°
Similarly ∠2 + 80° = 180°
∴ ∠2 = 180°- 80°= 100°
Now, sum of angles of a pentagon = (2n – 4) × 90°
= (2 × 5 – 4) × 90° = 6 × 90° = 540°
∴ ∠A + ∠B + ∠C + ∠D + ∠E = 540°
⇒ 120° + 100° + x + 40° + x = 540°
⇒ 260° + 2x = 540°
⇒ 2x = 540° – 260° = 280°
⇒ x = \frac{280^{\circ}}{2} = 140°
(iii) In the given figure, ABCD is a quadrilateral
whose side AB is produced is both sides ∠A = 90°
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.1 8
But ∠A + ∠B + ∠C + ∠D = 360°
(Sum of angles of a quadrilateral)
⇒ 90°+ 60°+ 110° + x = 360°
⇒ 260° + x = 360°
⇒ x = 360° – 260° = 100°
∴ x = 100°
(iv) In the given figure, ABCD is a quadrilateral
whose side AB is produced to E.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.1 9
∠A = 90°, ∠C = 83°, ∠D = 110°
∠B + x = 180° (Lienar pair)
∠B = 180° – x
But ∠A + ∠B + ∠C + ∠D = 360°
⇒ 90° + (180° – x) + 83° + 110° = 360°
(Sum of angles of a quadrilateral)
⇒ 283°+ 180° – x = 360°
⇒ x = 283° + 180°- 360°
⇒ x = 463° – 360°= 103°

Question 13.
(i) In the given figure, find x + y + z.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.1 10
(ii) In the given figure, find x + y + z + w.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.1 11

Answer

(i) In ∆ABC, Sides AB, BC, CA are produce in order in one side.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.1 12
∠B = 70°, ∠C = 90°
∴ ∠A = 180° – (∠B + ∠C)
= 180°- (70°+ 90°)
= 180°- 160° = 20°
But x + 90° = 180° (Linear pair)
∴ x = 180° – 90° = 90°
Similarly, y = 180° – 70° = 110°
y = 180° – 20°= 160°
x + y + z = 90° + 110°+ 160° = 360°
(ii) In the given figure,
ABCD is a quadrilateral whose sides are produced in order.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.1 13
∠A = 130°, ∠B = 80°, ∠C = 70°
∴ ∠D = 360° – (∠A + ∠B + ∠C)
= 360° – (130° + 80° + 70°)
= 360° – 280° = 80°
Now, x + 130° = 180° (Linear pair)
∴ x = 180°- 130° = 50°
Similarly, y = 180° – 80° = 100°
z = 180° – 70°= 110°
w = 180°-80°= 100°
∴ x + y + z + w = 52° + 100° +110°+ 100°
= 360°

Question 14.
A heptagon has three equal angles each of 120° and four equal angles. Find the size of equal angles.

Answer

Sum of angles of a heptagon = (2 × n – 4) × 90°
= (2 × 7 – 4) × 90°
= 10 × 90° = 900°
Sum of three angles are each equal i.e. 120°
= 120° × 3 = 360°
Sum of remaining 4 equal angles
= 900° – 360° = 540°
∴ Each angle = \frac{540^{\circ}}{4} = 135°

Question 15.
The ratio between an exterior angle and the interior angle of a regular polygon is 1 : 5. Find
(i) the measure of each exterior angle
(ii) the measure of each interior angle
(iii) the number of sides in the polygon.

Answer

Ratio between an exterior and an interior angle = 1 : 5
Let exterior angle = x
Then interior angle = 5x
But sum of interior angle and exterior angle = 180°
∴ x + 5x = 180°
⇒ 6x = 180°
⇒ \frac{180^{\circ}}{6} = 30°
(i) Measure of exterior angle x = 30°
(ii) and measure of interior angle = 5x = 5 × 30° = 150°
(iii) Let number of sides = n, then
\frac{2 n-4}{n} × 90° = 150°
⇒ \frac{2 n-4}{n}=\frac{150^{\circ}}{90}=\frac{5}{3}
⇒ 6n – 12 = 5n
⇒ 6n – 5n = 12
⇒ n = 12
∴ Number of sides = 12

Question 16.
Each interior angle of a regular polygon is double of its exterior angle. Find the number of sides in the polygon.

Answer

In a polygon,
Let exterior angle = x
Then interior angle = 2x
But sum of interior angle and exterior angle = 180°
∴ 2x + x = 180°
⇒ 3x = 180°
⇒ x = \frac{180^{\circ}}{3} = 60°
∴ Interior angle = 2 × 60° = 120°
Let number of sides of the polygon = x
Then \frac{2 n-4}{n} × 90° = 120°
⇒ \frac{2 n-4}{n}=\frac{120^{\circ}}{90}=\frac{4}{3}
⇒ 6n – 12 = 4n
⇒ 6n – 4n = 12
⇒ 2n = 12
⇒ n = \frac{12}{2} = 6
∴ Number of sides = 6


Understanding Quadrilaterals Class-8 ML Aggarwal ICSE Mathematics Solutions  Ex 13.2

Question 1.

In the given figure, ABCD is a parallelogram. Complete each statement along with the definition or property used.
(i) AD = ………..
(ii) DC = ………..
(iii) ∠DCB = ………..
(iv) ∠ADC = ………..
(v) ∠DAB = ………..
(vi) OC = ………..
(vii) OB = ………..
(viii) m∠DAB + m∠CDA = ………..
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.2 1

Answer

In paralleloram ABCD
(i) AD = 6 cm (Opposite sides of parallelogram)
(ii) DC = 9 cm (Opposite sides of parallelogram)
(iii) ∠DCB = 60° (∵ ∠DCB + ∠CBA = 180°)
(iv) ∠ADC = ∠ABC = 120°
(v) ∠DAB = ∠DCB = 60°
(vi) OC = AO = 7 cm
(vii) OB = OD = 5 cm
(viii) m∠DAB + m∠CDA = 180°

Question 2.
Consider the following parallelograms. Find the values of x, y, z in each.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.2 2

Answer

(i) ABCD is a parallelogram.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.2 3
Side BC is produced to E
∠DCE = 120°
But ∠DCE + ∠DCB = 180° (Linear pair)
⇒ 120° + ∠DCB = 180°
⇒ ∠DCB = 180° – 120° = 60°
But ∠A = ∠C
⇒ x = 60°
∠DCE = ∠ABC (Corresponding angles)
∴ y = 120°
But z = y (Opposite angle of a ||gm)
⇒ z = 120°
Hence x = 60°, y = 120°, z° = 120°

(ii) In parallelogram ABCD, diagonals bisect each other at O.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.2 4
∠DAC = 40°, ∠CAB = 30°, ∠DOC = 100°
∠ACB = ∠DAC = 40° (Alternate angles)
∴ z = 40°
∠ACD = ∠CAB (A ltemate angles)
⇒ ∠ACD = 30°
In ∆OCD,
∠DOC + ∠CDO + ∠OCD = 180° (Angles of a triangle)
⇒ 100° + x + 30° = 180°
⇒ x + 130° = 180°
⇒ x = 180°- 130° = 50°
Ext. ∠COD = y + z
100° = y + 40°
⇒ y = 100° – 40° = 60°
∴ x = 50°, y = 60°, z = 40°

(iii) In parallelogram ABCD, AC is its diagonal.
∠B = 120°, ∠DAC = 35°
∠DAB + ∠ABC = 180° (Co-interior angles)
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.2 5
35° + z + 120°= 180°
⇒ 155°+ z = 180°
⇒ z = 180° – 155° = 25°
But x = z (Alternate angles)
∴ x = 25°
y = ∠B (Opposite angles of a ||gm)
y = 120°
Hence x = 25°, y = 120°, z = 25°

(iv) In parallelogram ABCD
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.2 6
∠B = 70°, ∠DAC = 67°
∠D = ∠B (Opposite angles of a ||gm)
⇒ z = 70°
In ∆DAC
Ext. DCE = ∠D + ∠DAC
y = z + 67°
y = 70° + 67°= 137°
and ∠DCA + ∠DCE = 180° (Linear pair)
∠DCA + 137°= 180°
⇒ ∠DCA = 180° – 137° = 43°
But ∠CAB = ∠DCA (Alternate angles)
∴ x = 43°
∴ x = 43°, y = 137°, z = 70°

Question 3.
Two adjacent sides of a parallelogram are in the ratio 5 : 7. If the perimeter of parallelogram is 72 cm, find the length of its sides.

Answer

In ||gm ABCD
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.2 7
AD : AB = 5 : 7
Perimeter of ||gm = 72 cm
⇒ 2(DA + AB) = 72 cm
∴ DA + AB = \frac{72}{2} = 36 cm
Let DA = 5x and AB = 7x
5x + 7x = 36
⇒ 12x = 36
⇒ x = \frac{36}{12} = 3
∴ AB = 7x = 7 × 3 = 21 cm
AD = 5x = 5 × 3 = 15 cm

Question 4.
The measure of two adjacent angles of a parallelogram are in the ratio 4 : 5. Find the measure of each angle of the parallelogram.

Answer

In ||gm ABCD
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.2 8
∠A : ∠B = 4 : 5
Let ∠A = 4x, ∠B = 5x
But ∠A + ∠B = 180° (Cointerior angle)
∴ 4x + 5x = 180° ⇒ 9x= 180°
⇒ x = \frac{180^{\circ}}{9} = 20°
∴ ∠A = 4x = 4 × 20° = 80°
∠B = 5x = 5 × 20° = 100°
But ∠C = ∠A = 80° and ∠D = ∠B = 100°
(Opposite angles of a ||gm are equal)

Question 5.
Can a quadrilateral ABCD be a parallelogram, give reasons in support of your

Answer

(i) ∠A + ∠C= 180°?
(ii) AD = BC = 6 cm, AB = 5 cm, DC = 4.5 cm?
(iii) ∠B = 80°, ∠D = 70°?
(iv) ∠B + ∠C= 180°?

Answer

Quadrilateral ABCD can be a parallelogram of opposite sides
are equal and opposite angles are equal.
∴ ∠A = ∠C and ∠B = ∠D
and AB = DC, AD = BC
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.2 9
(i) ∠A + ∠C = 180°
It may be a parallelogram and may not be.
(ii) ∵ AD = BC = 6 cm, AB = 5 cm, DC = 4.5 cm
∵ AB ≠ DC
(iii) ∠B = 80°, ∠D = 70°
But there are opposite angles and ∠B ≠ ∠D
(iv) ∴ ∠B + ∠C = 180°
It may be or it may not be.

Question 6.
In the following figures HOPE and ROPE are parallelograms. Find the measures of angles x, y and z. State the properties you use to find them.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.2 10

Answer

(i) In parallelogram HOPE, HO is produced to D
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.2 11
∠AOP + ∠POD = 180° (Linear pair)
∴ ∠AOP + 70°= 180°
∠AOP = 180°- 70°= 110°
But ∠AOP = ∠HEP (Opposite angles of a ||gm)
∠HEP = 110°
⇒ x = 110°
∠HPO = ∠EHP (Alternate angles)
∴ y = 40°
In ∆HOP, Ext. ∠POD = y + z
⇒ 70° = y + z
⇒ 70° = 40° + z
⇒ z = 70° – 40° = 30°
∴ x= 110°, y = 40°, z = 30°

(ii) In ||gm ROPE, RO is produced to D
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.2 12
∠POD = 80°, ∠EOP = 60°
∠P = ∠POD (Alternate angles)
∴ y = 80°
∠ROE + ∠EOP + ∠POD = 180° (Angles on one side of a line)
x + 60° + 80° = 180° ⇒ x + 140° = 180°
∴ x = 180°- 140° = 40°
z = x (Alternate angles)
∴ z = 40°
Hence, x = 40°, y = 80°, z = 40°

Question 7.
In the given figure TURN and BURN are parallelograms. Find the measures of x and y (lengths are in cm).
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.2 13
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.2 14

Answer

(i) We know that opposite sides of a parallelogram are equal.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.2 15
∴ TU = RN
4x + 2 = 28 ⇒ 4x = 28 – 2
⇒ 4x = 26
⇒ x = \frac{26}{4} = 6.5 cm
and 5y – 1 = 24
⇒ 5y = 24 + 1
⇒ 5y = 25
⇒ y = \frac{25}{5} = 5
∴ x = 6.5 cm, y = 5 cm

(ii) We know that the diagonal of a parallelogram bisect each other.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.2 16
∴ BO = OR
⇒ x + y = 20 ………..(i)
and UO = ON
⇒ x + 3 = 18
⇒ x = 18 – 3 = 15 From (i)
15 + y = 20
⇒ y = 20 – 15 = 5
∴ x = 15, y = 5

Question 8.
In the following figure both ABCD and PQRS are parallelograms. Find the value of x.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.2 17

Answer

Two parallelograms ABCD and PQRS in which
∠A = 120° and ∠R = 50°
∠A + ∠B = 180° (Co-interior angles)
120° + ∠B = 180°
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.2 18
⇒ ∠B = 180°- 120° = 60°
∠P = ∠R (Opposite angles of a ||gm)
∠P = 50°
Now in ∆OPB,
∠POB + ∠P + ∠B = 180° (Angles of a triangle)
x + 50° + 60° = 180°
x + 110° = 180° ⇒ x = 180°- 110° = 70°
∴ x = 70°

Question 9.
In the given figure, ABCD, is a parallelogram and diagonals intersect at O. Find :
(i) ∠CAD
(ii) ∠ACD
(iii) ∠ADC
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.2 19

Answer

(i) ∠DBC = ∠BDA = 46° (alternate angles)
In ∆ AOD,
46° + 68° + ∠CAD = 180° (∵ ∠CAD = ∠OAD)
∠CAD = 180°- 114° = 66°
(ii) ∠AOD + ∠COD = 180° (straight angle)
∴ ∠COD= 180°- 68°= 112°
In ∆COD, 112° + 30° + ∠ACD = 180° (∵ ∠ACD = ∠OCD)
∠ACD = 180° – 112° – 30° = 38°
(iii) ∠ADC = 30° + 46° = 76° (∵ ∠ADC = ∠ADO + ∠ODC)

Question 10.
In the given figure, ABCD is a parallelogram. Perpendiculars DN and BP are drawn on diagonal AC. Prove that:
(i) ∆DCN ≅ ∆BAP
(ii) AN = CP
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.2 20

Answer

In the given figure,
ABCD is a parallelogram AC is it’s one diagonal.
BP and DN are perpendiculars on AC.
To prove :
(i) ∆DCN ≅ ∆BAP
(ii) AN = CP
Proof: In ∆DCN and ∆BAP
DC=AB (Opposite sides of a ||gm)
∠N = ∠P (Each 90°)
∠DCN = ∠PAB (Alternate angle)
∴ ∆DCN ≅ ∆BAP (AAS axiom)
∴ NC = AP (c.p.c.t.)
Subtracting NP from both sides.
NC – NP = AP – NP
∴ AN = CP

Question 11.
In the given figure, ABC is a triangle. Through A, B and C lines are drawn parallel to BC, CA and AB respectively, which forms a ∆PQR. Show that
2(AB + BC + CA) = PQ + QR + RP.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.2 21

Answer

In the given figure, ABC is a triangle.
Through A, B and C lines are drawn parallel to
BC, CA and AB respectively which forms ∆PQR.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.2 22
To prove:
2(AB + BC + CA) = PQ + QR + RP
∵ BC || PR, AC || RQ
∴ ARBC is a ||gm
∴ AR = CB ….(i)
Similarly ABCP is a ||gm
∴ AP = BC …(ii)
From (i) and (ii),
AR = AP or PR = 2BC …(iii)
Similarly we can prove that
RQ = 2A and PQ = 2AB
Now perimeter of ∆PQR = PQ + QR + RP
= 2AB + 2AC + 2BC
= 2(AB + BC + CA)
Hence PQ + QR + RP = 2(AB + BC + CA)


Understanding Quadrilaterals Exe-13.3 Class-8 ML Aggarwal ICSE Mathematics Solutions

Question 1.

Identify all the quadrilaterals that have
(i) four sides of equal length
(ii) four right angles.

Answer

(i) Any quadrilateral whose four sides are equal
in length is a square or rhombus.
(ii) A quadrilateral having four right angles
is a square or a rectangle.

Question 2.
Explain how a square is
(i) a quadrilateral
(ii) a parallelogram
(iii) a rhombus
(iv) a rectangle.

Answer

(i) A square is a quadrilateral which has four sides
and four angles whose sum is 360°.
(ii) A square is a parallelogram whose opposite sides are parallel.
(iii) A square is a parallelogram whose sides are equal
and so, it is a rhombus.
(iv) A square is a parallelogram whose each angle is 90°.
So, it is a rectangle.

Question 3.
Name the quadrilaterals whose diagonals
(i) bisect each other
(ii) are perpendicular bisectors of each other
(iii) are equal.

Answer

(i) Rectangle, square, rhombus, parallelogram.
(ii) Square, rhombus.
(iii) Square, rectangle.

Question 4.
One of the diagonals of a rhombus and its sides are equal. Find the angles of the rhombus.

Answer

In a rhombus, side and one diagonal are equal.
∴ Angles will be 60° and 120°

Question 5.
In the given figure, ABCD is a rhombus, find the values of x, y and z.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.3 2

Answer

In rhombus ABCD.
∵ The diagonals of rhombus bisect each other at right angles.
∴ AO = OC and BO = OD
AO = x, OC = 8 cm, BO =y and OD = 6 cm
∴ x = 8 cm and y = 6 cm
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.3 3
In ∆AOB,
AB2 = AO2 + BO2
AB2 = 82 + 62
AB2 = 64 + 36
AB2 = 100 = (10)2
AB = 10 cm

Question 6.
In the given figure, ABCD is a trapezium. If ∠A : ∠D = 5 : 7, ∠B = (3x + 11)° and ZC = (5x – 31)°, then find all the angles of the trapezium.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.3 4

Answer

In the given figure,
ABCD is a trapezium in which DC || AB
∠A : ∠D = 5 : 7
∠B = (3x + 11)° and
∠C = (5x – 31)°
∵ ABCD is a trapezium
∴ ∠B + ∠C = 180° (Cointerior angle)
3x + 11° + 5x – 31° = 180°
8x – 20° = 180° ⇒ 8x = 180° + 20° = 200°
⇒ x = x=\frac{200^{\circ}}{8}=25^{\circ}
2x = 180°- 118°
2x = 62° ⇒ x = 31°
∴ ∠ABO = 31°

(ii) Also, ∠AOB + ∠AOD = 180° (Linear pair)
∴ ∠AOD = 180°- 118° = 62°
Now, In ∆AOD, AO = DO = y°
∴ 62 + 2y= 180°
2y = 180° – 62°
2y = 180°- 62°
2y = 118° ⇒ y = 59°

(iii) ∠OCB = ∠OAD = 59° (Alternate angles)

Question 7.
In the given figure, ABCD is a rectangle. If ∠CEB : ∠ECB = 3 : 2 find
(i) ∠CEB,
(ii) ∠DCF
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.3 5

Answer

In ∆ BCE, ∠B = 90° (∵ ABCD is a rectangle)
∴ ∠CEB + ∠ECB = 90°
3x + 2x = 90°
⇒ x= 18°
∴ ∠CEB = 3x = 3 × 18° = 54°
Now, ∠CEB = ∠ECD = 54° (Alternate angles)
Also ∠ECD + ∠DCF = 180° (Linear pair)
⇒ ∠DCF = 180 – 54= 126°

Question 8.
In the given figure, ABCD is a rectangle and diagonals intersect at O. If ∠AOB = 118°, find
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.3 6
(i) ∠ABO
(ii) ∠ADO
(iii) ∠OCB

Answer

In ∆ ABO, OA = OB
(∵ diagonals of a rectangle bisect each other)
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.3 7
∴ ∠OAB = ∠OBA = x °
⇒ 118° + x + x = 180°
2x= 180°- 118°
2x = 62° ⇒ x = 31°
∴ ∠ABO = 31°

(ii) Also, ∠AOB + ∠AOD = 180° (Linear pair)
∴ ∠AOD = 180° – 118° = 62°
Now, In ∆ AOD, AO = DO = y°
∴ 62 + 2y = 180°
2y = 180°- 62°
2y = 118° ⇒ y = 59°

(iii) ∠OCB = ∠OAD = 59° (Alternate angles)

Question 9.
In the given figure, ABCD is a rhombus and ∠ABD = 50°. Find :
(i) ∠CAB
(ii) ∠BCD
(iii) ∠ADC
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.3 8

Answer

(i) We know that diagonals of a rhombus
are ⊥ to each other.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.3 9
∴ ∠BOA = 90°
In ∆ AOB,
∠OAB + ∠BOA + ∠ABO = 180°
∠OAB + 90° + 50° = 180°
∠OAB = 180 – 140 = 40°
∴ ∠CAB = ∠OAB = 40°

(ii) ∠BCD = 2 ∠ACD = 2 × 40° = 80°
(∵ ∠CAB = ∠ACD alternate angles)
(iii) ∠ADC = 2 ∠BDC = 2 × 50° = 100°
(∵ ∠ABD = ∠BDC alternate angles)

Question 10.
In the given isosceles trapezium ABCD, ∠C = 102°. Find all the remaining angles of the trapezium.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.3 10

Answer

AB || CD
∠B + ∠C = 180°
(∵ adjacent angles on the same side of
transversal are supplementary)
⇒ ∠B + 102° = 180°
∠B = 180°- 102° = 78°
As AD = BC (Given)
∴ ∠A = ∠B = 78°
∠A + ∠B + ∠C + ∠D = 360°
78° + 78° + 102° + ∠D = 360°
∠D + 258° = 360°
∠D = 102°

Question 11.
In the given figure, PQRS is a kite. Find the values of x and y.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.3 11

Answer

In the figure, PQRS is a kite
∠Q = 120° and ∠R = 50°
∴ ∠Q = ∠S
∴ x = 120°
∠P + ∠R = 360° – (120° + 120°)
∠P + ∠R = 360° – 240° = 120°
But ∠R = 50°
∴ ∠P = y = 120°- 50° = 70°
Hence, x = 120°, y = 70°


Objective Type Questions, Chapter-13 Understanding Quadrilaterals Class-8 ML Aggarwal ICSE Mathematics Solutions

Mental Maths

Question 1.
Fill in the blanks:
(i) The curves which have different beginning and end points are called …………
(ii) A curve which does not cross itself at any point is called a …………
(iii) A simple closed curve made up entirely of line segments is called a …………
(iv) A polygon in which each interior angle is less than 180° is called a …………
(v) 10 sided polygon is called …..
(vi) A polygon which has all its sides of equal length and all its angles of equal measure is called a …………
(vii) Sum of all exterior angles of a polygon is …………
(vi) A polygon which has all its sides of equal length and all its angles of equal measure is called a …………
(vii) Sum of all exterior angles of a polygon is …………
(viii) Sum of all interior angles of a n-sided polygon is …………
(ix) The adjacent angles of a parallelogram are …………
(x) If an angle of a parallelogram is a right angle, then it is called a …………
(xi) If two adjacent sides of a parallelogram are equal, then it is called a …………
(xii) It two adjacent sides of a rectangle are equal, then it is called a …………
(xiii) The diagonal of a rhombus bisect each other at …………
(xiv) A quadrilateral in which one pair of opposite sides is parallel is called a …………
(xv) A quadrilateral in which two pairs of adjacent sides are equal is called a …………
(xvi) If two non-parallel sides of a trapezium are equal then it is called …………

Answer

(i) The curves which have different beginning
and end points are called open curves.
(ii) A curve which does not cross itself
at any point is called a simple curve.
(iii) A simple closed curve made up entirely
of line segments is called a polygon.
(iv) A polygon in which each interior angle
is less than 180° is called a convex polygon.
(v) 10 sided polygon is called decagon.
(vi) A polygon which has all its sides of equal length
and all its angles of equal measure is called a regular polygon.
(vii) Sum of all exterior angles of a polygon is 360°.
(viii) Sum of all interior angles of a n-sided polygon is
(n – 2) × 180° or (2n – 4) × 90°.
(ix) The adjacent angles of a parallelogram are supplementary.
(x) If an angle of a parallelogram is a right angle,
then it is called a rectangle.
(xi) If two adjacent sides of a parallelogram are equal,
then it is called a rhombus.
(xii) It two adjacent sides of a rectangle are equal,
then it is called a square.
(xiii) The diagonal of a rhombus bisect each other at right angles.
(xiv) A quadrilateral in which one pair of
opposite sides is parallel is called a trapezium.
(xv) A quadrilateral in which two pairs of
adjacent sides are equal is called a kite.
(xvi) If two non-parallel sides of a trapezium
are equal then it is called an isosceles trapezium.

Question 2.

State whether the following statements are true (T) or false (F):
(i) The curves which have same beginning and end points are called open curves.
(ii) The region of the plane that lies inside the curve is called interior of curve.
(iii) A polygon in which atleast one interior angle is greater than 180° is called convex polygon.
(iv) 6 sided polygon is called hexagon.
(v) Sum of all interior angles of a quadrilateral is 180°.
(vi) Each interior angle of a n-sided regular polygon is \frac{(2 n-4) \times 90^{\circ}}{n}.
(vii) The diagonals of a parallelogram bisect each other at right angles.
(viii) The opposite angles of a parallelogram are of equal measure.
(ix) The diagonals of a rhombus bisect the angles of rhombus.
(x) The diagonals of a square are not equal.
(xi) Co-interior angles of a parallelogram are supplementary.
(xii) The diagonals of a kite bisect at right angles.
(xiii) All rectangles are squares.
(xiv) All rhombuses are parallelograms.
(xv) All squares are rhombuses and also rectangles.
(xvi) All squares are not parallelograms.
(xvii) All kites are rhombuses.
(xviii) All rhombuses are kites.
(xix) All parallelograms are trapeziums.
(xx) All squares are trapeziums.

Answer

(i) The curves which have same beginning
and end points are called open curves. False
Correct:
It is called a closed curve.
(ii) The region of the plane that lies inside
the curve is called interior of curve. True
(iii) A polygon in which atleast one interior angle is
greater than 180° is called convex polygon. False
Correct:
It is called a concave polygon.
(iv) 6 sided polygon is called hexagon. True
(v) Sum of all interior angles of a quadrilateral is 180°. False
Correct:
The sum is 360°.
(vi) Each interior angle of a n-sided regular polygon is \frac{(2 n-4) \times 90^{\circ}}{n}. True
(vii) The diagonals of a parallelogram bisect each other at right angles. False
(viii)The opposite angles of a parallelogram are of equal measure. True
(ix) The diagonals of a rhombus bisect the angles of rhombus. True
(x) The diagonals of a square are not equal.
False
Correct:
Diagonals are equal.
(xi) Co-interior angles of a parallelogram are supplementary. True
(xii) The diagonals of a kite bisect at right angles. False
(xiii) All rectangles are squares. False
Correct:
Some rectangle whose sides are equal are squares.
(xiv) All rhombuses are parallelograms. True
(xv) All squares are rhombuses and also rectangles. True
(xvi) All squares are not parallelograms. False
Correct:
All squares are parallelograms.
(xvii) All kites are rhombuses. False
Correct : A kite with all sides equal is a rhombus.
(xviii) All rhombuses are kites. True
(xix) All parallelograms are trapeziums. True
(xx) All squares are trapeziums. True


MCQs 

Multiple Choice Questions

Choose the correct answer from the given four options (3 to 19):
Question 3.
Sum of all interior angles of a 11-sided polygon is
(a) 1620°
(b) 1440°
(c) 1260
(d) none of these

Answer

Sum of all interior angles of an 11-sided polygon is
= (2 × n – 4 ) × 90°
= (2 × 11 – 4) × 90°
= 18 × 90 = 1620 (a

Question 4.
If each interior angle of a regular polygon is 144°, then number of sides of polygon is
(a) 8
(b) 9
(c) 10
(d) 11

Answer

Each interior angle of a regular polygon is 144°
Then \frac{2 n-4}{n} × 90° = 144°
⇒ \frac{2 n-4}{n}=\frac{144^{\circ}}{90^{\circ}}
⇒ 10n – 20° = 8n
⇒ 10n – 8n = 20°
⇒ 2n = 20°
⇒ n = 10
∴ It is 10-sided polygon. (c)

Question 5.
If the sum of all interior angles of a polygon is 1260°, then number of sides of polygon is
(a) 6
(b) 7
(c) 8
(d) 9

Answer

Sum of all interior angles of a polygon = 1260°
∴ (2n – 4) × 90° = 1260°
⇒ 2n – 4 = \frac{1260^{\circ}}{90}
⇒ 2n = 14 + 4 = 18
⇒ n = \frac{18}{2} = 9
∴ Polygon is 9-sided. (d)

Question 6.
The sum of all exterior angles of a pentagon is
(a) 590°
(b) 360°
(c) 180°
(d) none of these

Answer

Sum of exterior angles of a pentagon = 360° (b)

Question 7.
If the ratio between an exterior and interior angle of a regular polygon is 1 : 5, then the number of sides of the polygon is
(a) 11
(b) 12
(c) 13
(d) 14

Answer

Ratio between exterior angle and interior angle
of a regular polygon = 1 : 5
But sum of angles = 180°
∴ Exterior angle = \frac{180^{\circ}}{1+5} × 1
\frac{180^{\circ}}{6} = 30°
∴ Number of sides = \frac{360^{\circ}}{30} =12 (b)
(Sum of exterior angles = 360°)

Question 8.
In the given figure, the value of x is
(a) 140°
(b) 50°
(c) 130°
(d) 40°
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Objective Type Questions 1

Answer

In the given figure,
Sum of exterior angles of a triangle = 360°
∴ 140° + x + 90° = 360°
⇒ x + 230° = 360°
∴ x = 360° – 230° = 130° (c)

Question 9.

In the given figure, the value of x is
(a) 120°
(b) 130°
(c) 140°
(d) 150°
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Objective Type Questions 2

Answer

In the given figure,
Sum of angles of a quadrilateral = 360°
∴ 60° + (180° – 120°) + 110° + x = 360°
⇒ 60° + 60° + 110° + x = 360°
230° + x = 360°
∴ x = 360°- 230° = 130° (b)

Question 10.
In the given figure, the value of x + y + z + w is
(a) 180°
(b) 270°
(c) 300°
(d) 360°
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Objective Type Questions 3

Answer

In the given figure,
Sum of exterior angles of a quadrilateral = 360°
∴ x + y + z + w = 360° (d)

Question 11.
In the given figure, the value of x + y is
(a) 180°
(b) 190°
(c) 170°
(d) 160°
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Objective Type Questions 4

Answer

In the given figure,
Sum of interior angles of a quadrilateral = 360°
60° + y + 110° + x = 360°
⇒ x + y + 170° = 360°
⇒ x + y = 360° – 170°
x + y = 190° (b)

Question 12.
The lengths of two adjacent sides of a parallelogram are in the ratio 1 : 2. If the perimeter of parallelogram is 60 cm, then length of its sides are
(a) 6 cm, 12 cm
(b) 8 cm, 16 cm
(c) 9 cm, 18 cm
(d) 10 cm, 20 cm

Answer

Ratio in the length of two adjacent sides of a parallelogram = 1 : 2
Perimeter = 60 cm
∴ Sum of two adjacent sides = \frac{60}{2} = 30 cm
Let first side = x, then second side = 2x
∴ x + 2x = 30 ⇒ 3x = 30
x = \frac{30}{2} = 10 cm
First side = 10 cm
and second side = 10 × 2 = 20 cm (d)

Question 13.
In the given figure, ABCD is a parallelogram, the values of x and y respectively are
(a) 1 cm, 1 cm
(b) 2 cm, 1 cm
(c) 1 cm, 2 cm
(d) 2 cm, 2 cm
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Objective Type Questions 5

Answer

In the given figure, ABCD is a parallelogram
∵ Diagonals of a parallelogram bisect each other
∴ AO = OC and BO = OD
∴ 6 = 5x + 1
⇒ 5x = 6 – 1 = 5
⇒ x = \frac{5}{5}
and y + 3 = 4
⇒ y = 4 – 3 = 1
∴ x = 1, y = 4 (a)

Question 14.
In the given figure, ABCD is a parallelogram, the values of x, y and z respectively are
(a) 60°, 60°, 70°
(b) 60°, 70°, 60°
(c) 70°, 60°, 60°
(d) none of these
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Objective Type Questions 6

Answer

In the given figure,
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Objective Type Questions 7
ABCD is a parallelogram, BD is its one diagonal
∠ABD + ∠DBC +∠CBE = 180°
(Angles on one side of a line)
⇒ 50° + x + 70° = 180°
x + 120° = 180°
∴ x = 180° – 120° = 60°
But y = x (Alternate angles)
∴ y = 60°
z = 70° (Alternate angles)
∴ x = 60°, y = 60°, z = 70° (a)

Question 15.
In the given figure, ABCD is a rectangle, the value of angle x is
(a) 60°
(b) 90°
(c) 120°
(d) none of these
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Objective Type Questions 8

Answer

In the given figure, ABCD is a rectangle
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Objective Type Questions 9
∴ ∠OBA = ∠OAB = 30°
In ∆AOB,
∠AOB = ∠COD (Vertically opposite angles)
∠AOB = x
∠AOB + ∠OBA + ∠OAB = 180° (Angles of a triangle)
⇒ x + 30° + 30° = 180°
⇒ x = 180°- 30°- 30° = 120°
∴ x = 120° (c)

Question 16.
In a rhombus ABCD, the diagonals AC and BD are respectively 8 cm and 6 cm. The length of each side of the rhombus is
(a) 7 cm
(b) 5 cm
(c) 6 cm
(d) 8 cm

Answer

In rhombus ABCD
Diagonals AC and BD are 8 cm and 6 cm
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Objective Type Questions 10
∴ AC = 8 cm and BD = 6 cm
∵ Diagonals of a rhombus bisect each other at right angles
AO = OC = \frac{8}{2} = 4 cm,
BO = OD = \frac{6}{2} = 3 cm
∴ In right ∆AOB
AB = \sqrt{\mathrm{AO}^{2}+\mathrm{BO}^{2}}=\sqrt{4^{2}+3^{2}}
\sqrt{16+9}=\sqrt{25} = 5 cm
Each side of rhombus = 5 cm (b)

Question 17.
In the given figure, ABCD is a square, the value of angle x is
(a) 30°
(b) 45°
(c) 60°
(d) not possible to find
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Objective Type Questions 11

Answer

In the given figure,
ABCD is a square whose diagonals AC and BD
bisect each other at O.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Objective Type Questions 12
∵ Diagonals of a square bisect the opposite angles.
∴ x = \frac{1}{2} × ∠B = \frac{1}{2} × 90° = 45° (b)

Question 18.
In the given figure, ABCD is a kite, the value of angle x is
(a) 86°
(b) 100°
(c) 104°
(d) none of these
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Objective Type Questions 13

Answer

In the given figure, ABCD is a kite whose
diagonals AC and BD intersect at O at right angles.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Objective Type Questions 14
In ∆OAB, ∠O = 90°
∴ ∠OAB + ∠ABO = 90°
⇒ ∠OAB + 36° = 90°
⇒ ∠OAB = 90° – 36° = 54°
But ∠OAD = ∠OCD = 50°
x = ∠DAO + ∠AOB
⇒ x = 50° + 54°= 104° (c)

Question 19.
In the given figure, ABCD is an isosceles trapezium. The values of x, y and z respectively are
(a) 110°, 110°, 70°
(b) 110°, 70°, 110°
(c) 70°, 110°, 110°
(d) none of these
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Objective Type Questions 15

Answer

In isosceles trapezium ∠A = 70°
But ∠B = ∠A = 70° ⇒ z = 70°
But x + 70° = 180°
⇒ x = 180°-70°= 110°
But y = x = 110°
∴ 110°, 110°, 70° (a)


HOTS

Higher Order Thinking Skills 

Question 1.
In the given figure, ABCDEF is a regular hexagon. Prove that quadrialterals ABDE and ACDF are parallelograms. Also prove that quadrilateral AGDH is a parallelogram.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Objective Type Questions 16

Answer

In the given figure, ABCDEF is a regular hexagon.
AC, AE, DF, DB are joined.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Objective Type Questions 17
To prove: ABDE, ACDF and AGDH are ||gm
Proof: In ∆BCD,
BC = CD
∴ ∠CBD = ∠CDB = 30° (∵ ∠C = 120° angle of hexagon)
But ∠B = ∠D (Angles of a regular hexagon)
∴ ∠B – ∠CBD = ∠D – ∠CDB
⇒ ∠ABD = ∠BDE = 90°
Similarly, ∠FAB = ∠AED = 90°
∴ ∠ABD + ∠BDE = 90° + 90° = 180°
But they are cointerior angle.
∴ AB || DE
∴ ABDE is a ||gm
Similarly we can prove that ACDF is a parallelogram.
∵ AC || DF and BD || AC
∴ AGDH is a parallelogram.

Question 2.
In the given figure, ABCD is a parallelogram and M, N are the mid-points of sides BC, AD respectively. Prove that EA = AB = BF.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Objective Type Questions 18

Answer

In the given figure,
ABCD is a parallelogram M and N are
midpoints of the sides BC and AD respectively.
To prove : EA = AB = BF.
Proof: In ∆AFD,
M is the midpoint of BC and BC || AD
∴ B is mid-point of AF
∴ AB = BF …(i)
Similarly in ∆EBC,
N is the midpoint of AD and AD || BC
∴ A is the midpoint of EB
∴ EA = AB …(ii)
From (i) and (ii),
EA = AB = BF

Question 3.
Prove that the quadrilateral formed by joining the mid-points of the adjacent sides of a rectangle is a parallelogram.

Answer

Given : ABCD is a rectangle.
P, Q, R and S are the mid points of the sides
AB, BC, CD and DA respectively.
PQ, QR, RS and SP are joined.
To prove : PQRS is a parallelogram.
Construction : Join AC and BD.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Objective Type Questions 19
Proof: In ∆ABC,
P and Q are mid points of AB and BC respectively
∴ PQ || AC and PQ = \frac{1}{2} AC …(i)
Similarly in ∆ADC
SR mid points of AD and CD respectively
∴ SR || AC and SR = \frac{1}{2} AC …(ii)
From (i) and (ii),
SR || PQ and SR = PQ
∴ PQRS is a parallelogram.


Check Your Progress, Understanding Quadrilaterals Class-8 ML Aggarwal ICSE Mathematics Solutions Chapter-13

Question 1.

From the given diagram, find the value of x.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Check Your Progress 1

Answer

Reflex angle B = 360 – 140 = 220°
Also, ∠ADC = 180 – 122 = 58° (Linear pair)
Now, ABCD is a quadrilateral
∴ ∠A + ∠B + ∠C + ∠D = 360°
⇒ x + 220 + 53 + 58 = 360
or x = 360 – 331 =29°

Question 2.
If two angles of a quadrilateral are 77° and 51°, and out of the remaining two angles, one angle is 10° smaller than the other, find these angles.

Answer

Two angles of a quadrilateral are 77° and 51°
Sum of angles of a quadrilateral = 360°
∴ Sum of other two angles = 360° – (77° + 51°)
= 360° – 134° = 226°
Let one angle among there two angles = x
Then other angle = x – 10°
∴ x + x – 10 = 226°
⇒ 2x = 226° + 10° = 236°
⇒ x = \frac{236^{\circ}}{2} = 118°
∴ One angle = 118°
and other angle =118 – 10 = 108°

Question 3.
In the given figure, AB || DC, ∠A = 74° and ∠B : ∠C = 4 : 5. Find
(i) ∠D
(ii) ∠B
(iii) ∠C
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Check Your Progress 2

Answer

∠A = 74° ∠B = 4x, ∠C = 5x.
As AB || CD
∴ 4x + 5x= 180°, 9x= 180°
⇒ x = 20°
∴ ∠B = 80° and ∠C = 100°
ABCD is a quadrilateral
∴ ∠A + ∠B + ∠C + ∠D = 360°
⇒ ∠D = 360° – (74° + 80° + 100°)
= 360°- 254 = 106°
Therefore, ∠D = 106° ∠B = 80°, ∠C = 100°

Question 4.
In quadrilateral ABCD, ∠A : ∠B : ∠C : ∠D = 3 : 4 : 6 : 7. Find all the angles of the quadrilateral. Hence, prove that AB and DC are parallel. Is BC also parallel to AD?

Answer

Let, four angles of quadrilateral be 3x, 4x, 6x, 7x.
∴ 3x + 4x + 6x + 7x = 360
20x = 360 ⇒ x = 18
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Check Your Progress 3
∴ ∠A = 3x = 3 × 18 = 54°
∠B = 4x = 4 × 18 = 72°
∠C = 6x = 6 × 18= 108°
∠D = 7x = 7 × 18= 126°
As, ∠A + ∠D = 54 + 126 = 180°
∴ AB || CD
(If two angles on the same side of transversal are supplementary,
then lines are parallel)
But ∠A + ∠B (54 + 72) ≠ 180
∴ BC is not parallel to AD.

Question 5.
One angle of a parallelogram is two-third of the other. Find the angles of the parallelogram.

Answer

In a parallelogram, one angle is \frac{2}{3} of the other.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Check Your Progress 4
Let one angle = x
Then second = \frac{2}{3}x
But x + \frac{2}{3}x = 180°
⇒ \frac{5}{3} = 180°
⇒ x =\frac{180^{\circ} \times 3}{5} = 108°
∴ ∠A = x = 108°
∠B = \frac{2}{3}x = 108° × \frac{2}{3} =72°
But ∠C = ∠A and ∠D = ∠B
(Opposite angles of a parallelogram)
∴∠C = ∠A = 108° and ∠D = ∠B = 72°
Hence, ∠A = 108°, ∠B = 72°, ∠C = 108°, ∠D = 72°

Question 6.
In the given figure, ABCD is a kite. If ∠BCD = 52° and ∠ADB = 42°, find the values of x, y, and z.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Check Your Progress 5

Answer

Join BD.
In ∆ ABD,
AB = AD (Given)
∴ ∠ABD = ∠ADB
(angles opposite to equal sides are equal)
⇒ x = 42°
In ∆ BCD,
BC = CD (given)
∴ ∠BDC = ∠DBC = z
∴ z + z + 52 = 180° ⇒ 2z = 128
⇒ z = 64°
ABCD is quadrilateral
∴ ∠A + ∠B + ∠C + ∠D = 360°
⇒ y + (x + z) + 52 + (42 + z) = 360°
⇒ y + 106 + 52 + 106 = 360
⇒ y = 360 – 264 = 96°

Question 7.
In the given figure, ABCD is a rectangle. Prove that AC = BD.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Check Your Progress 6

Answer

In ∆ ABC and ∆ABD
BC = AD (opposite sides of rectangle)
∠B = ∠A (each 90°)
AB = AB (common)
∴ ∆ABC = ∆ABD (S.A.S.)
∴ AC ≅ BD (c.p.c.t.)

Question 8.
In the given figure, ABCD is a rhombus and EDC is an equilateral triangle. If ∠DAB = 48°, find
(i) ∠BEC
(ii) ∠DEB
(iii) ∠BFC
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Check Your Progress 7

Answer

ABCD is a rhombus
∴ AB = BC = CD = DA
Also, EDC is an equilateral ∆
∴ ED = DC = CE …(ii)
From (i) and (ii)
We get, BC = CE.
In ∆ BCE, ∠BCE = 60 + 48 = 108
Also, BC = EC
∴ ∠BEC = ∠EBC = x
⇒ x + x + ∠BCE = 180°
⇒ 2x = 180 – 108 = 72°
⇒ x = 36°
∴ ∠BEC = x = 36°.

(ii) ∠DEB = ∠DEC – ∠BEC
= 60° – 36° = 24°

(iii) In ∆ DEF,
∠D = 60°, ∠DEF = 24°, ∠DFE = y.
60 + 24 + 7 = 180°
y = 180 – 84 = 96°
∠BFC = ∠DFE = 96° (Vertically opposite angles)

Question 9.
Find the number of sides of a regular polygon if each of its interior angle is 168°.

Answer

Each interior angle of a regular polygon = 168°
Let number of sides = n, then
\frac{2 n-4}{n} × 90° = 168°
\frac{2 n-4}{n}=\frac{168^{\circ}}{90^{\circ}}=\frac{28}{15}
∴ 30n – 60 = 28n
⇒ 30n – 28n = 60
⇒ 2n = 60
⇒ n = \frac{60}{2} = 30
∴ Number of sides of the polygon = 30

Question 10.
If the sum of interior angles of polygon is 3780° find the number of sides.

Answer

Sum of interior angles of polygon = (2n – 4) × 90
∴ 3780 = (2n – 4) × 90 ⇒ (2n – 4) = \frac{3780}{90}
⇒ 2n – 4 = 42
⇒ 2n = 46
⇒ n = 23

Question 11.
The angles of a hexagon are (2x + 5)°, (3x – 5)°, (x + 40)°, (2x + 20)°, (2x + 25)° and (2x + 35)°. Find the value of x.

Answer

Number of sides in hexagon = 6.
Sum of interior angles = (2 × 6 – 4) × 90 = 720°
∴ (2x + 5) + (3x – 5) + (x + 40) + (2x + 20)
+ (2x + 25) + (2x + 35) = 720
⇒ 12x + 120 = 720
⇒ 12x = 720 – 120
⇒ 12x = 600
⇒ x = 50.

Question 12.
Two angles of a polygon are right angles and every other angle is 120°. Find the number of sides of the polygon.

Answer

Let the number of sides = n
∴ 2 × 90 + (n – 2) × 120 = (2n – 4) × 90
⇒ 180 + 120n – 240 = 180n – 360
⇒ 120n – 60 = 180n – 360
⇒ 60n = 300
⇒ n = 5

Question 13.
The sum of interior angles of a regular polygon is twice the sum of its exterior angles. Find the number of sides of the polygon.

Answer

Sum of interior angles = (2n – 4) × 90
Sum of exterior angles = 360
According to the condition,
(2n – 4) × 90 = 2 × 360
⇒ 2n – 4 = 8
∴ n = 6

Question 14.
An exterior angle of a regular polygon is one- fourth of its interior angle. Find the number of sides in the polygon.

Answer

Let measure of interior angle = x°
Then exterior angle = \frac{1}{4}
∴ x + \frac{1}{4}x = 180° ⇒ \frac{5}{4} x = 180
⇒ x = 180 × \frac{4}{5} ⇒ x = 144°
Therefore, each interior angle is 144°
144=\frac{(2 n-4)}{n} \times 90
⇒ 144n= 180n – 360
⇒ 180n – 144n = 360
⇒ 36n = 360
⇒ n = 10

 

-:  End of  Understanding Quadrilaterals Class-8 ML Aggarwal Solutions :–

 

Return to –  ML Aggarwal Maths Solutions for ICSE Class -8


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