HomeNCERT SOLUTIONS8th CLASSClass 8 Maths Chapter 3 Understanding Quadrilaterals Solutions

Class 8 Maths Chapter 3 Understanding Quadrilaterals Solutions

Quadrilaterals

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NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.1

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.1

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Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.2

NCERT Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.2 Solutions

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.2

Ex 3.2 Class 8 Maths Question 1.
Find x in the following figures.
NCERT Solutions for Class 8 Maths Understanding Quadrilaterals Ex 3.2 Q1
Solution:
(a) We know that the sum of all the exterior angles of a polygon = 360°
125° + 125° + x = 360°
⇒ 250° + x = 360°
NCERT Solutions for Class 8 Maths Understanding Quadrilaterals Ex 3.2 Q1.1
x = 360° – 250° = 110°
Hence x = 110°
(b) Here ∠y = 180° – 90° = 90°
NCERT Solutions for Class 8 Maths Understanding Quadrilaterals Ex 3.2 Q1.2
and ∠z = 90° (given)
x + y + 60° + z + 70° = 360° [∵ Sum of all the exterior angles of a polygon = 360°]
⇒ x + 90° + 60° + 90° + 70° = 360°
⇒ x + 310° = 360°
⇒ x = 360° – 310° = 50°
Hence x = 50°

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Ex 3.2 Class 8 Maths Question 2.
Find the measure of each exterior angle of a regular polygon of
(i) 9 sides
(ii) 15 sides
Solution:
(i) We know the sum of all the exterior angles of polygon = 360°
Measure of each angle of 9 sided regular polygon = \frac { 360 }{ 9 } = 40°
(ii) Sum of all the exterior angles of a polygon = 360°
Measure of each angle of 15 sided regular polygon = \frac { 360 }{ 15 } = 24°

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Ex 3.2 Class 8 Maths Question 3.
How many sides does a regular polygon have if the measure of an exterior angle is 24°?
Solution:
Sum of all exterior angles of a regular polygon = 360°
Number of sides
NCERT Solutions for Class 8 Maths Understanding Quadrilaterals Ex 3.2 Q3
Hence, the number of sides = 15

Ex 3.2 Class 8 Maths Question 4.
How many sides does a regular polygon have if each of its interior angles is 165°?
Solution:
Let re be the number of sides of a regular polygon.
Sum of all interior angles = (n – 2) × 180°
and, measure of its each angle
NCERT Solutions for Class 8 Maths Understanding Quadrilaterals Ex 3.2 Q4
Hence, the number of sides = 24

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Ex 3.2 Class 8 Maths Question 5.
(a) Is it possible to have a regular polygon with measure of each exterior angle a is 22°?
(b) Can it be an interior angle of a regular polygon? Why?
Solution:
(a) Since, the sum of all the exterior angles of a regular polygon = 360° which is not divisible by 22°.
It is not possible that a regular polygon must have its exterior angle 22°.
(b) Sum of all interior angles of a regular polygon of side n = (n – 2) × 180°
NCERT Solutions for Class 8 Maths Understanding Quadrilaterals Ex 3.2 Q5
not a whole number.
Since number of sides cannot be in fractions.
It is not possible for a regular polygon to have its interior angle = 22°.

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Ex 3.2 Class 8 Maths Question 6.
(a) What is the minimum interior angle possible for a regular polygon? Why?
(b) What is the maximum exterior angle possible for a regular polygon?
Solution:
(a) Sum of all interior angles of a regular polygon of side n = (n – 2) × 180°
The measure of each interior angle
NCERT Solutions for Class 8 Maths Understanding Quadrilaterals Ex 3.2 Q6
The minimum measure the angle of an equilateral triangle (n = 3) = 60°.
(b) From part (a) we can conclude that the maximum exterior angle of a regular polygon = 180° – 60° = 120°.

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Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.3

NCERT Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.3 Solutions

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.3

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Ex 3.3 Class 8 Maths Question 1.
Given a parallelogram ABCD. Complete each statement along with the definition or property used.
(i) AD = …………
(ii) ∠DCB = ………
(iii) OC = ………
(iv) m∠DAB + m∠CDA = ……..
NCERT Solutions for Class 8 Maths Understanding Quadrilaterals Ex 3.3 Q1
Solution:
(i) AD = BC [Opposite sides of a parallelogram are equal]
(ii) ∠DCB = ∠DAB [Opposite angles of a parallelogram are equal]
(iii) OC = OA [Diagonals of a parallelogram bisect each other]
(iv) m∠DAB + m∠CDA = 180° [Adjacent angles of a parallelogram are supplementary]

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Ex 3.3 Class 8 Maths Question 2.
Consider the following parallelograms. Find the values of the unknowns x, y, z.
NCERT Solutions for Class 8 Maths Understanding Quadrilaterals Ex 3.3 Q2
Solution:
(i) ABCD is a parallelogram.
NCERT Solutions for Class 8 Maths Understanding Quadrilaterals Ex 3.3 Q2.1
∠B = ∠D [Opposite angles of a parallelogram are equal]
∠D = 100°
⇒ y = 100°
∠A + ∠B = 180° [Adjacent angles of a parallelogram are supplementary]
⇒ z + 100° = 180°
⇒ z = 180° – 100° = 80°
∠A = ∠C [Opposite angles of a ||gm]
x = 80°
Hence x = 80°, y = 100° and z = 80°
(ii) PQRS is a parallelogram.
NCERT Solutions for Class 8 Maths Understanding Quadrilaterals Ex 3.3 Q2.2
∠P + ∠S = 180° [Adjacent angles of parallelogram]
⇒ x + 50° = 180°
x = 180° – 50° = 130°
Now, ∠P = ∠R [Opposite angles are equal]
⇒ x = y
⇒ y = 130°
Also, y = z [Alternate angles]
z = 130°
Hence, x = 130°, y = 130° and z = 130°
(iii) ABCD is a rhombus.
[∵ Diagonals intersect at 90°]
NCERT Solutions for Class 8 Maths Understanding Quadrilaterals Ex 3.3 Q2.3
x = 90°
Now in ∆OCB,
x + y + 30° = 180° (Angle sum property)
⇒ 90° + y + 30° = 180°
⇒ y + 120° = 180°
⇒ y = 180° – 120° = 60°
y = z (Alternate angles)
⇒ z = 60°
Hence, x = 90°, y = 60° and z = 60°.
(iv) ABCD is a parallelogram
NCERT Solutions for Class 8 Maths Understanding Quadrilaterals Ex 3.3 Q2.4
∠A + ∠B = 180° (Adjacent angles of a parallelogram are supplementary)
⇒ x + 80° = 180°
⇒ x = 180° – 80° = 100°
Now, ∠D = ∠B [Opposite angles of a |jgm]
⇒ y = 80°
Also, z = ∠B = 80° (Alternate angles)
Hence x = 100°, y = 80° and z = 80°
(v) ABCD is a parallelogram.
NCERT Solutions for Class 8 Maths Understanding Quadrilaterals Ex 3.3 Q2.5
∠D = ∠B [Opposite angles of a ||gm]
y = 112°
x + y + 40° = 180° [Angle sum property]
⇒ x + 112° + 40° = 180°
⇒ x + 152° = 180°
⇒ x = 180° – 152 = 28°
z = x = 28° (Alternate angles)
Hence x = 28°, y = 112°, z = 28°.

Ex 3.3 Class 8 Maths Question 3.
Can a quadrilateral ABCD be a parallelogram if
(i) ∠D + ∠B = 180°?
(ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm?
(iii) ∠A = 70° and ∠C = 65°?
Solution:
(i) For ∠D + ∠B = 180, quadrilateral ABCD may be a parallelogram if following conditions are also fulfilled.
(a) The sum of measures of adjacent angles should be 180°.
(b) Opposite angles should also be of same measures. So, ABCD can be but need not be a parallelogram.
(ii) Given: AB = DC = 8 cm, AD = 4 cm, BC = 4.4 cm
In a parallelogram, opposite sides are equal.
Here AD ≠ BC
Thus, ABCD cannot be a parallelogram.
(iii) ∠A = 70° and ∠C = 65°
Since ∠A ≠ ∠C
Opposite angles of quadrilateral are not equal.
Hence, ABCD is not a parallelogram.

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Ex 3.3 Class 8 Maths Question 4.
Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure.
Solution:
ABCD is a rough figure of a quadrilateral in which m∠A = m∠C but it is not a parallelogram. It is a kite.
NCERT Solutions for Class 8 Maths Understanding Quadrilaterals Ex 3.3 Q4

Ex 3.3 Class 8 Maths Question 5.
The measures of two adjacent angles of a parallelogram are in the ratio 3 : 2. Find the measure of each of the angles of the parallelogram.
Solution:
Let ABCD is parallelogram such that
m∠B : m∠C = 3 : 2
NCERT Solutions for Class 8 Maths Understanding Quadrilaterals Ex 3.3 Q5
Let m∠B = 3x° and m∠C = 2x°
m∠B + m∠C = 180° (Sum of adjacent angles = 180°)
3x + 2x = 180°
⇒ 5x = 180°
⇒ x = 36°
Thus, ∠B = 3 × 36 = 108°
∠C = 2 × 36° = 72°
∠B = ∠D = 108°
and ∠A = ∠C = 72°
Hence, the measures of the angles of the parallelogram are 108°, 72°, 108° and 72°.

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Ex 3.3 Class 8 Maths Question 6.
Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram.
Solution:
Let ABCD be a parallelogram in which
∠A = ∠B
NCERT Solutions for Class 8 Maths Understanding Quadrilaterals Ex 3.3 Q6
We know ∠A + ∠B = 180° [Sum of adjacent angles = 180°]
∠A + ∠A = 180°
⇒ 2∠A = 180°
⇒ ∠A = 90°
Thus, ∠A = ∠C = 90° and ∠B = ∠D = 90°
[Opposite angles of a parallelogram are equal]

Ex 3.3 Class 8 Maths Question 7.
The adjacent figure HOPE is a parallelogram. Find the angle measures x, y and z. State the properties you use to find them.
NCERT Solutions for Class 8 Maths Understanding Quadrilaterals Ex 3.3 Q7
Solution:
∠y = 40° (Alternate angles)
∠z + 40° = 70° (Exterior angle property)
⇒ ∠z = 70° – 40° = 30°
z = ∠EPH (Alternate angle)
In ∆EPH
∠x + 40° + ∠z = 180° (Adjacent angles)
⇒ ∠x + 40° + 30° = 180°
⇒ ∠x + 70° = 180°
⇒ ∠x = 180° – 70° = 110°
Hence x = 110°, y = 40° and z = 30°.

Ex 3.3 Class 8 Maths Question 8.
The following figures GUNS and RUNS are parallelograms. Find x and y. (Lengths are in cm)
NCERT Solutions for Class 8 Maths Understanding Quadrilaterals Ex 3.3 Q8
Solution:
(i) GU = SN (Opposite sides of a parallelogram)
NCERT Solutions for Class 8 Maths Understanding Quadrilaterals Ex 3.3 Q8.1
⇒ y + 7 = 20
⇒ y = 20 – 7 = 13
Also, ON = OR
⇒ x + y = 16
⇒ x + 13 = 16
x = 16 – 13 = 3
Hence, x = 3 cm and y = 13 cm.

Ex 3.3 Class 8 Maths Question 9.
NCERT Solutions for Class 8 Maths Understanding Quadrilaterals Ex 3.3 Q9
In the above figure both RISK and CLUE are parallelograms. Find the value of x.
Solution:
Here RISK and CLUE are two parallelograms.
NCERT Solutions for Class 8 Maths Understanding Quadrilaterals Ex 3.3 Q9.1
∠1 = ∠L = 70° (Opposite angles of a parallelogram)
∠K + ∠2 = 180°
Sum of adjacent angles is 180°
120° + ∠2 = 180°
∠2 = 180° – 120° = 60°
In ∆OES,
∠x + ∠1 + ∠2 = 180° (Angle sum property)
⇒ ∠x + 70° + 60° = 180°
⇒ ∠x + 130° = 180°
⇒ ∠x = 180° – 130° = 50°
Hence x = 50°

Ex 3.3 Class 8 Maths Question 10.
Explain how this figure is a trapezium. Which of its two sides are parallel?
NCERT Solutions for Class 8 Maths Understanding Quadrilaterals Ex 3.3 Q10
Solution:
∠M + ∠L = 100° + 80° = 180°
∠M and ∠L are the adjacent angles, and sum of adjacent interior angles is 180°
KL is parallel to NM
Hence KLMN is a trapezium.

Ex 3.3 Class 8 Maths Question 11.
Find m∠C in below figure if \bar { AB } || \bar { DC }
Solution:
Given that \bar { AB } || \bar { DC }
m∠B + m∠C = 180° (Sum of adjacent angles of a parallelogram is 180°)
NCERT Solutions for Class 8 Maths Understanding Quadrilaterals Ex 3.3 Q11
120° + m∠C = 180°
m∠C = 180° – 120° = 60°
Hence m∠C = 60°

Ex 3.3 Class 8 Maths Question 12.
Find the measure of ∠P and ∠S if \bar { SP } || \bar { RQ } in figure, is there any other method to find m∠P?)
NCERT Solutions for Class 8 Maths Understanding Quadrilaterals Ex 3.3 Q12
Solution:
Given that ∠Q = 130° and ∠R = 90°
\bar { SP } || \bar { RQ } (given)
∠P + ∠Q = 180° (Adjacent angles)
⇒ ∠P + 130° = 180°
⇒ ∠P = 180° – 130° = 50°
and, ∠S + ∠R = 180° (Adjacent angles)
⇒ ∠S + 90° = 180°
⇒ ∠S = 180° – 90° = 90°
Alternate Method:
∠Q = 130°, ∠R = 90° and ∠S = 90°
We know that
∠P + ∠Q + ∠R + ∠Q = 360° (Angle sum property of quadrilateral)
⇒ ∠P + 130° + 90° + 90° = 360°
⇒ ∠P + 310° = 360°
⇒ ∠P = 360° – 310° = 50°
Hence m∠P = 50°

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Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.4

NCERT Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.4 Solutions

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.4

Ex 3.4 Class 8 Maths Question 1.
State whether True or False.
(a) All rectangles are squares.
(b) All rhombuses are parallelograms.
(c) All squares are rhombuses and also rectangles.
(d) All squares are not parallelograms.
(e) All kites are rhombuses.
(f) All rhombuses are kites.
(g) All parallelograms are trapeziums.
(h) All squares are trapeziums.
Solution:
(a) False
(b) True
(c) True
(d) False
(e) False
(f) True
(g) True
(h) True

Ex 3.4 Class 8 Maths Question 2.
Identify all the quadrilaterals that have
(a) four sides of equal length
(b) four right angles
Solution:
(a) Squares and rhombuses.
(b) Rectangles and squares.

Ex 3.4 Class 8 Maths Question 3.
Explain how a square is
(i) a quadrilateral
(ii) a parallelogram
(iii) a rhombus
(iv) a rectangle
Solution:
(i) Square is a quadrilateral because it is closed with four line segments.
(ii) Square is a parallelogram due to the following properties:
(a) Opposite sides are equal and parallel.
(b) Opposite angles are equal.
(iii) Square is a rhombus because its all sides are equal and opposite sides are parallel.
(iv) Square is a rectangle because its opposite sides are equal and has equal diagonal.

Ex 3.4 Class 8 Maths Question 4.
Name the quadrilaterals whose diagonals
(i) bisect each other
(ii) are perpendicular bisectors of each other
(iii) are equal
Solution:
(i) Parallelogram, rectangle, square and rhombus
(ii) Square and rhombus
(iii) Rectangle and square

Ex 3.4 Class 8 Maths Question 5.
Explain why a rectangle is a convex quadrilateral.
Solution:
In a rectangle, both of its diagonal lie in its interior. Hence, it is a convex quadrilateral.

Ex 3.4 Class 8 Maths Question 6.
ABC is a right-angled triangle and O is the midpoint of the side opposite to the right angle. Explain why O is equidistant from A, B and C. (The dotted lines are drawn additionally to help you).
NCERT Solutions for Class 8 Maths Understanding Quadrilaterals Ex 3.4 Q6
Solution:
Since the right-angled triangle ABC makes a rectangle ABCD by the dotted lines.
Therefore OA = OB = OC = OD [Diagonals of a rectangle are equal and bisect each other]
Hence, O is equidistant from A, B and C.
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Extra Questions

Extra Questions Maths Chapter 3

Extra Questions for Class 8 Maths Chapter 3 Understanding Quadrilaterals

Understanding Quadrilaterals Class 8 Extra Questions Very Short Answer Type

Question 1.
In the given figure, ABCD is a parallelogram. Find x.
Understanding Quadrilaterals NCERT Extra Questions for Class 8 Maths Q1
Solution:
AB = DC [Opposite sides of a parallelogram]
3x + 5 = 5x – 1
⇒ 3x – 5x = -1 – 5
⇒ -2x = -6
⇒ x = 3

Question 2.
In the given figure find x + y + z.
Understanding Quadrilaterals NCERT Extra Questions for Class 8 Maths Q2
Solution:
We know that the sum of all the exterior angles of a polygon = 360°
x + y + z = 360°

Question 3.
In the given figure, find x.
Understanding Quadrilaterals NCERT Extra Questions for Class 8 Maths Q3
Solution:
∠A + ∠B + ∠C = 180° [Angle sum property]
(x + 10)° + (3x + 5)° + (2x + 15)° = 180°
⇒ x + 10 + 3x + 5 + 2x + 15 = 180
⇒ 6x + 30 = 180
⇒ 6x = 180 – 30
⇒ 6x = 150
⇒ x = 25

Question 4.
The angles of a quadrilateral are in the ratio of 2 : 3 : 5 : 8. Find the measure of each angle.
Solution:
Sum of all interior angles of a quadrilateral = 360°
Let the angles of the quadrilateral be 2x°, 3x°, 5x° and 8x°.
2x + 3x + 5x + 8x = 360°
⇒ 18x = 360°
⇒ x = 20°
Hence the angles are
2 × 20 = 40°,
3 × 20 = 60°,
5 × 20 = 100°
and 8 × 20 = 160°.

Question 5.
Find the measure of an interior angle of a regular polygon of 9 sides.
Solution:
Measure of an interior angle of a regular polygon
Understanding Quadrilaterals NCERT Extra Questions for Class 8 Maths Q5

Question 6.
Length and breadth of a rectangular wire are 9 cm and 7 cm respectively. If the wire is bent into a square, find the length of its side.
Solution:
Perimeter of the rectangle = 2 [length + breadth]
= 2[9 + 7] = 2 × 16 = 32 cm.
Understanding Quadrilaterals NCERT Extra Questions for Class 8 Maths Q6
Now perimeter of the square = Perimeter of rectangle = 32 cm.
Side of the square = \frac { 32 }{ 4 } = 8 cm.
Hence, the length of the side of square = 8 cm.

Question 7.
In the given figure ABCD, find the value of x.
Understanding Quadrilaterals NCERT Extra Questions for Class 8 Maths Q7
Solution:
Sum of all the exterior angles of a polygon = 360°
x + 70° + 80° + 70° = 360°
⇒ x + 220° = 360°
⇒ x = 360° – 220° = 140°

Question 8.
In the parallelogram given alongside if m∠Q = 110°, find all the other angles.
Understanding Quadrilaterals NCERT Extra Questions for Class 8 Maths Q8
Solution:
Given m∠Q = 110°
Then m∠S = 110° (Opposite angles are equal)
Since ∠P and ∠Q are supplementary.
Then m∠P + m∠Q = 180°
⇒ m∠P + 110° = 180°
⇒ m∠P = 180° – 110° = 70°
⇒ m∠P = m∠R = 70° (Opposite angles)
Hence m∠P = 70, m∠R = 70°
and m∠S = 110°

Question 9.
In the given figure, ABCD is a rhombus. Find the values of x, y and z.
Understanding Quadrilaterals NCERT Extra Questions for Class 8 Maths Q9
Solution:
AB = BC (Sides of a rhombus)
x = 13 cm.
Since the diagonals of a rhombus bisect each other
z = 5 and y = 12
Hence, x = 13 cm, y = 12 cm and z = 5 cm.

Question 10.
In the given figure, ABCD is a parallelogram. Find x, y and z.
Understanding Quadrilaterals NCERT Extra Questions for Class 8 Maths Q10
Solution:
∠A + ∠D = 180° (Adjacent angles)
⇒ 125° + ∠D = 180°
⇒ ∠D = 180° – 125°
x = 55°
∠A = ∠C [Opposite angles of a parallelogram]
⇒ 125° = y + 56°
⇒ y = 125° – 56°
⇒ y = 69°
∠z + ∠y = 180° (Adjacent angles)
⇒ ∠z + 69° = 180°
⇒ ∠z = 180° – 69° = 111°
Hence the angles x = 55°, y = 69° and z = 111°

Question 11.
Find x in the following figure. (NCERT Exemplar)
Understanding Quadrilaterals NCERT Extra Questions for Class 8 Maths Q11
Solution:
In the given figure ∠1 + 90° = 180° (linear pair)
∠1 = 90°
Now, sum of exterior angles of a polygon is 360°, therefore,
x + 60° + 90° + 90° + 40° = 360°
⇒ x + 280° = 360°
⇒ x = 80°

Understanding Quadrilaterals Class 8 Extra Questions Short Answer Type

Question 12.
In the given parallelogram ABCD, find the value of x andy.
Understanding Quadrilaterals NCERT Extra Questions for Class 8 Maths Q12
Solution:
∠A + ∠B = 180°
3y + 2y – 5 = 180°
⇒ 5y – 5 = 180°
⇒ 5y = 180 + 5°
⇒ 5y = 185°
⇒ y = 37°
Now ∠A = ∠C [Opposite angles of a parallelogram]
3y = 3x + 3
⇒ 3 × 37 = 3x + 3
⇒ 111 = 3x + 3
⇒ 111 – 3 = 3x
⇒ 108 = 3x
⇒ x = 36°
Hence, x = 36° and y – 37°.

Question 13.
ABCD is a rhombus with ∠ABC = 126°, find the measure of ∠ACD.
Understanding Quadrilaterals NCERT Extra Questions for Class 8 Maths Q13
Solution:
∠ABC = ∠ADC (Opposite angles of a rhombus)
∠ADC = 126°
∠ODC = \frac { 1 }{ 2 } × ∠ADC (Diagonal of rhombus bisects the respective angles)
⇒ ∠ODC = \frac { 1 }{ 2 } × 126° = 63°
⇒ ∠DOC = 90° (Diagonals of a rhombus bisect each other at 90°)
In ΔOCD,
∠OCD + ∠ODC + ∠DOC = 180° (Angle sum property)
⇒ ∠OCD + 63° + 90° = 180°
⇒ ∠OCD + 153° = 180°
⇒ ∠OCD = 180° – 153° = 27°
Hence ∠OCD or ∠ACD = 27°

Question 14.
Find the values of x and y in the following parallelogram.
Understanding Quadrilaterals NCERT Extra Questions for Class 8 Maths Q14
Solution:
Since, the diagonals of a parallelogram bisect each other.
OA = OC
x + 8 = 16 – x
⇒ x + x = 16 – 8
⇒ 2x = 8
x = 4
Similarly, OB = OD
5y + 4 = 2y + 13
⇒ 3y = 9
⇒ y = 3
Hence, x = 4 and y = 3

Question 15.
Write true and false against each of the given statements.
(a) Diagonals of a rhombus are equal.
(b) Diagonals of rectangles are equal.
(c) Kite is a parallelogram.
(d) Sum of the interior angles of a triangle is 180°.
(e) A trapezium is a parallelogram.
(f) Sum of all the exterior angles of a polygon is 360°.
(g) Diagonals of a rectangle are perpendicular to each other.
(h) Triangle is possible with angles 60°, 80° and 100°.
(i) In a parallelogram, the opposite sides are equal.
Solution:
(a) False
(b) True
(c) False
(d) True
(e) False
(f) True
(g) False
(h) False
(i) True

Question 16.
The sides AB and CD of a quadrilateral ABCD are extended to points P and Q respectively. Is ∠ADQ + ∠CBP = ∠A + ∠C? Give reason.
(NCERT Exemplar)
Solution:
Join AC, then
∠CBP = ∠BCA + ∠BAC and ∠ADQ = ∠ACD + ∠DAC (Exterior angles of triangles)
Understanding Quadrilaterals NCERT Extra Questions for Class 8 Maths Q16
Therefore,
∠CBP + ∠ADQ = ∠BCA + ∠BAC + ∠ACD + ∠DAC
= (∠BCA + ∠ACD) + (∠BAC + ∠DAC)
= ∠C + ∠A

Understanding Quadrilaterals Class 8 Extra Questions Higher Order Thinking Skills (HOTS)

Question 17.
The diagonal of a rectangle is thrice its smaller side. Find the ratio of its sides.
Understanding Quadrilaterals NCERT Extra Questions for Class 8 Maths Q17
Solution:
Let AD = x cm
diagonal BD = 3x cm
In right-angled triangle DAB,
AD2 + AB2 = BD2 (Using Pythagoras Theorem)
x2 + AB2 = (3x)2
⇒ x2 + AB2 = 9x2
⇒ AB2 = 9x2 – x2
⇒ AB2 = 8x2
⇒ AB = √8x = 2√2x
Required ratio of AB : AD = 2√2x : x = 2√2 : 1

Question 18.
If AM and CN are perpendiculars on the diagonal BD of a parallelogram ABCD, Is ∆AMD = ∆CNB? Give reason. (NCERT Exemplar)
Solution:
Understanding Quadrilaterals NCERT Extra Questions for Class 8 Maths Q18
In triangles AMD and CNB,
AD = BC (opposite sides of parallelogram)
∠AMB = ∠CNB = 90°
∠ADM = ∠NBC (AD || BC and BD is transversal.)
So, ∆AMD = ∆CNB (AAS)

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NCERT Solutions for Class 8 Maths

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