Bhagya

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 9 Understanding Elementary Shapes Ex 9.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 9 Understanding Elementary Shapes Ex 9.5

1. Which of the following are polygons and there is no polygon. Give the reason:

Solution:
(i) It is not a closed figure. Therefore it is not a polygon.
(ii) It is made up of lines segment. Therefore it is polygon.
(iii) It is not a polygon, because it is not made of line segments.
(iv) It is not closed by line segment. Therefore, it is not a polygon.
(v) It is not polygon because line segments are intersecting each other.
(vi) It is made up of line segments, therefore it is a polygon.

2. Classify the following as concave or convex polygons:

Solution:
(i) Concave Polygon
(ii) Convex Polygon
(iii) Concave Polygon
(iv) Concave Polygon
(v) Convex Polygon
(vi) Convex Polygon.

3. Tick in the boxes, if the property holds true for a particular quadrilateral otherwise eroes out ‘x’:

Quadrilateral Properties	Rectangle	Parallelogram	Rhombus	Trapezium	Square
All sides are equal
Only opposite sides are equal
Diagonals are equal
Diagonals bisect each other
Diagonals are perpendicular to each other
Each angle is 90°

Solution:

Quadrilateral Properties	Rectangle	Parallelogram	Rhombus	Trapezium	Square
All sides are equal	×	×	√	×	√
Only opposite sides are equal	√	√	×	×	×
Diagonals are equal	√	×	×	×	√
Diagonals bisect each other	√	√	√	×	√
Diagonals are perpendicular to each other	×	×	√	×	√
Each angle is 90°	√	×	X	×	√

4. Fill in the blanks:

Question (i)
…………… is a quadrilateral with only one pair of opposite sides parallel.
Solution:
Trapezium

Question (ii)
…………….. is a quadrilateral with all sides equal and diagonals of equal length.
Solution:
Square

Question (iii)
A polygon with atleast one angle is reflex is called ……………….. .
Solution:
Concave polygon

Question (iv)
………….. is a regular quadrilateral.
Solution:
Square

Question (v)
…………… is a quadrilateral with opposite sides equal and diagonals of unequal length.
Solution:
Parallelogram.

5. State True or False:

Question (i)
A rectangle is always a rhombus.
Solution:
False

Question (ii)
The diagonals of a rectangle are perpendicular to each other.
Solution:
False

Question (iii)
A square is a parallelogram.
Solution:
True

Question (iv)
A trapezium is a parallelogram.
Solution:
False

Question (v)
Opposite sides of a parallelogram are parallel.
Solution:
True.

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 15 Visualising Solid Shapes Ex 15.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.1

1. Match the two dimensional figure with the names.

Answer:
(i) (e)
(ii) (d)
(iii) (a)
(iv) (b)
(v) (c)

2. Match the three dimension shapes with the names.

Answer:
(i) (d)
(ii) (e)
(iii) (a)
(iv) (c)
(v) (b)

3. Identify the nets which can be used to make cubes (cut out copies of the nets and try it).


Answer:
(i), (iv)

4. Draw the net for a square pyramid with base as square of sides 5 cm and slant edges 7 cm.
Answer:

5. Draw a net for the following cylinder.

Answer:

6. Draw the net of the solid given in figure.

Answer:

7. Dice are cubes with dots on each face opposite faces of a die always have a total of seven dots on them following are two nets to make dice (cuber) the number inserted in each square indicate the number of dots in that box insert suitable number in the blank squares.


Solution:

8. Which solid will be obtained by folding the following net ?

Solution:

9. Complete the following table.


Solution:
(i) Faces : 6.
(ii) Edges : 2. vertices : NIL,
(iii) Faces : 7. Edges : 15.
(iv) faces : 5, vertices : 5.

10. Multiple Choice questions :

Question (i).
Out of following which is 3-D figure ?
(a) Square
(b) Triangle
(c) Sphere
(d) Circle
Answer:
(c) Sphere

Question (ii).
Total number of faces a cylinder has :
(a) 0
(b) 2
(c) 1
(d) 3
Answer:
(d) 3

Question (iii).
How many edges are there in a square pyramid ?
(a) 5
(b) 8
(c) 1
(d) 4
Answer:
(b) 8

Question (iv).
Sum of number on the opposite faces of a die is :
(a) 8
(b) 7
(c) 9
(d) 6
Answer:
(b) 7

Question (v).
Which is not a solid figure ?
(a) Cuboid
(b) Sphere
(c) Quadrilateral
(d) Pyramid
Answer:
(c) Quadrilateral

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 9 Understanding Elementary Shapes Ex 9.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 9 Understanding Elementary Shapes Ex 9.4

1. Classify each of the following triangles as scalene, isosceles or equilateral:

Solution:
(i) Here, two sides of triangle are equal in length.
∴ It is an isosceles triangle.
(ii) Here, all the three sides of the triangle are equal in length.
∴ It is an equilateral triangle.
(iii) Here, no two sides are equal in length.
∴ It is scalene triangle.
(iv) Here, two sides of triangle are equal in length.
∴ It is an isosceles triangle.
(v) Here, no two sides are equal in length.
∴ It is scalene triangle.
(vi) Here, all the three sides of the triangle are equal in length.
∴ It is equilateral triangle.

2. Classify each of the following triangles as acute, obtuse or right triangle:

Solution:
(i) Here, one angle is 120°, which is obtuse angle.
∴ It is an obtuse-angled triangle.
(ii) Here, one angle is 90°, which is right angle.
∴ It is an right-angled triangle.
(iii) Here, each angle is acute angle.
∴ It is an acute-angled triangle.
(iv) Here, one angle is 90°, which is right angle.
∴ It is an right-angled triangle.
(v) Here, one angle is 120°, which is obtuse angle.
∴ It is an obtuse-angled triangle.
(vi) Here, each angle is 60°, which is actute angle.
∴ It is an actute angled triangle.

3. Which of the following triangles are possible with the given angles?

Question (i)
60°, 60°, 60°
Solution:
In a triangle sum of the three angles of a triangle is equal to 180°.
Here, sum of the three angles of triangle is:
60° + 60° + 60° = 180°
∴ This triangle is possible.

Question (ii)
110°, 50°, 30°
Solution:
Here, sum of the three angles of triangle is:
110° + 50° + 30°= 190° ≠ 180°
∴ This triangle is not possible.

Question (iii)
65°, 55°, 60°
Solution:
Here, sum of the three angles of triangle is:
65°+ 55°+ 60°= 180°
∴ This triangle is possible.

Question (iv)
90°, 40°, 50°
Solution:
Here, sum of the three angles of triangle is:
90°+ 40°+ 50°= 180°
∴ This triangle is possible.

Question (v)
48°, 62°, 50°
Solution:
Here, sum of the three angles of triangle is:
48°+ 62°+ 50°= 160° ≠ 180°
∴ This triangle is not possible.

Question (vi)
90°, 95°, 30°.
Solution:
Here, sum of the three angles of triangle is:
90°+ 95°+ 30° =215° ≠ 180°
∴ This triangle is not possible.

4. Classify each of the following triangles as scalene, isosceles or equilateral triangle:

Question (i)
4 cm, 5 cm, 6 cm
Solution:
The sides of triangle are 4 cm, 5 cm, 6 cm
No, two sides of this triangle are equal.
∴ This is a scalene triangle.

Question (ii)
5 cm, 7 cm, 5 cm
Solution:
The sides of triangle are 5 cm, 7 cm, 5 cm
Here, two sides are equal each of 5 cm in length.
∴ This is an isosceles triangle.

Question (iii)
4.2 m, S3 m, 6.1 m
Solution:
The sides of triangle are 4.2 m, 5.3 m, 6.1 m
Here, all sides are of different length.
∴ This is a scalene triangle.

Question (iv)
3.5 cm, 3.5 cm, 33 cm
Solution:
The sides of triangle are 3.5 cm, 3.5 cm, 3.5 cm
All the sides of triangle are of equal length.
∴ This is an equilateal triangle.

Question (v)
8 cm, 4.2 cm, 4.2 cm
Solution:
The sides of triangle are 8 cm, 4.2 cm, 4.2 cm
Here, two sides of the triangle are of equal length.
∴ This is an isosceles triangle.

Question (vi)
2 cm, 3 cm, 4 cm.
Solution:
The sides of triangle are 2 cm, 3 cm, 4 cm
All the sides of the triangle are of different lengths
∴ This is a scalene triangle.

5. Name the following triangles in both ways: (Based on sides and angles)

Solution:
(i) Based on sides: In this triangle, no two sides of the triangle are equal.
∴ This is a scalene triangle.
Based on angles: All the three angles of the triangle are acute.
∴ This is an acute-angled triangle.

(ii) Based on sides: In this triangle, two sides are of equal length each is 4 cm.
∴ This is an isosceles triangle.
Based on angles: In this triangle, one angle is of 90° which is a right angle.
∴ This is a right-angled triangle.

(iii) Based on sides: In this triangle, two sides are of equal length.
∴ This is an isosceles triangle.
Based on angles: In this triangle one angle is of 110°, which is obtuse angle.
∴ This is an obtuse-angled triangle.

(iv) Based on sides: In this triangle, all the sides are of equal length i.e. each = 4 cm.
∴ This is an equilateral triangle.
Based on angles: In this triangle, all the angles are acute angles.
∴ This is an acute-angled triangle.

(v) Based on sides: In this triangle, all the three sides are of different lengths.
∴ This is a scalene triangle.
Based on angles: In this triangle, one angle is 105°, which is obtuse angle.
∴ This is an obtuse-angled triangle.

6. Fill in the blanks:

Question (i)
A triangle has …………. sides.
Solution:
3

Question (ii)
A triangle has …………. vertices.
Solution:
3

Question (iii)
A triangle has …………. angles.
Solution:
3

Question (iv)
A triangle has …………. parts.
Solution:
6

Question (v)
A triangle whose all sides are different is known as ………………. .
Solution:
Scalene triangle

Question (vi)
A triangle whose all angles are acute is known as ……………….. .
Solution:
Acute angled triangle

Question (vii)
A triangle whose two sides are equal is known as ……………….. .
Solution:
Isosceles triangle

Question (viii)
A triangle whose one angle is obtuse is known as ……………….. .
Solution:
obtuse-angled triangle

Question (ix)
A triangle whose all sides are equal is known as ……………….. .
Solution:
Equilateral triangle

Question (x)
A triangle whose one angle is right angle is known as ……………….. .
Solution:
Right-angled triangle

7. State True or False:

Question (i)
Each equilateral triangle is an isosceles triangle.
Solution:
True

Question (ii)
Each acute-angled triangle is a scalene triangle.
Solution:
False

Question (iii)
Each isosceles triangle is an equilateral triangle.
Solution:
False

Question (iv)
There are two obtuse angles in an obtuse triangle.
Solution:
False

Question (v)
In right triangle, there is only one right angle.
Solution:
True

Question (vi)
Right triangle can never be isosceles.
Solution:
False.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 7 Coordinate Geometry Ex 7.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.2

Q.uestion 1.
Find the co-ordinates of the point which divides the join (- 1, 7) and (4, – 3) in the ratio 2 : 3.
Solution:
Let required point be P (x, y) which divides the join of given points A (- 1, 7)
and B (4, – 3) in the ratio of 2 : 3.
(-1, 7) (x, y) (4, – 3)

∴ x = \(\frac{2 \times 4+3 \times-1}{2+3}=\frac{8-3}{5}=\frac{5}{5}=1\)

and y = \(\frac{2 \times-3+3 \times 7}{2+3}=\frac{-6+21}{5}=\frac{15}{5}=3\)
Hence, required point be (1, 3).

Question 2.
Find the co-ordinates of the points of trisection of the line segment joining (4, – 1) and (2, – 3).
Solution:
Let P (x₁, y₁) and Q (x₂, y₂) be the required points which trisect the line segment joining A (4, – 1)and B (- 2, – 3) i.e., P(x₁, y₁) divides AB in ratio 1: 2 and Q divides AB in ratio 2 : 1.

∴ x₁ = \(\frac{1 \times-2+2 \times 4}{1+2}=\frac{-2+8}{3}=\frac{6}{3}=2\)

and y₁ = \(\frac{1 \times-3+2 \times-1}{1+2}=\frac{-3-2}{3}=-\frac{5}{3}\)
∴ P(x₁, y₁) be (2, \(-\frac{5}{3}\))

Now, x₂ = \(\frac{2 \times-2+1 \times 4}{2+1}\)
= \(\frac{-4+4}{3}\) = 0

y₂ = \(\frac{2 \times-3+1 \times-1}{2+1}=\frac{-6-1}{3}=-\frac{7}{3}\)

∴ Q(x₂, y₂) be (0, \(-\frac{7}{3}\))
Hence, required points be (2, \(-\frac{5}{3}\)) and (0, \(-\frac{7}{3}\)).

Question 3.
To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1m from each other along AD, as shown in fig. Niharika runs \(\frac{1}{4}\) th the distance AD on the 2nd line and posts a green flag.

Preet runs \(\frac{1}{5}\) th the distance AD on the eighth line and posts a red flag. What is the distance betweenboth the flags? If Rashmi has to post a blue flag exactly half way between the line (segment) joining the two flags, where should she post her flag?

Solution:
In the given figure, we take A as origin. Taking x-axis along AB and y-axis along AD.
Position of green flag = distance covered by Niharika
= Niharika runs \(\frac{1}{4}\)th distance AD on the 2nd line
= \(\frac{1}{4}\) × 100 = 25 m
∴ Co-ordinates of the green flag are (2, 25)
Now, position of red flag = distance covered by Preet = Preet runs \(\frac{1}{5}\)th the distance AD on the 8th line
= \(\frac{1}{5}\) × 100 = 20 m.
Co-ordinates of red flag are (8, 20)
∴ distance between Green and Red flags = \(\sqrt{(8-2)^{2}+(20-25)^{2}}\)
= \(\sqrt{36+25}=\sqrt{61}\) m.
Position of blue flag = mid point of green flag and red flag
= \(\left(\frac{2+8}{2}, \frac{25+20}{2}\right)\)
= (5, 22.5).
Hence, blue flag is in the 5th line and at a distance of 22.5 m along AD.

Question 4.
Find the ratio in which (he segment joining the points (- 3, 10) and (6, – 8) is divided by (- 1, 6).
Solution:
Let point P (- 1, 6) divides the line segment joining the points A (- 3, 10) and B (6, – 8) the ratio K : 1.

∴ -1 = \(\frac{6 \times \mathrm{K}-3 \times 1}{\mathrm{~K}+1}\)
or – K – 1 = 6K – 3
or – K – 6K = – 3 + 1
or – 7K = – 2
K : 1 = \(\frac{2}{7}\) : 1 = 2 : 7
Hence, required ratio is 2 : 7.

Question 5.
Find the ratio in which the line segment joining A (1, – 5) and B (- 4, 5) is divided by the x-axis. Also find the co
ordinates of the point of division.
Solution:
Let required point on x-axis is P (x, 0) which divides the line segment joining the points A (1, – 5) and B (- 4, 5) in the
ratio K : 1.

Consider, y co-ordinates of P ¡s:
0 = \(\frac{5 \times \mathrm{K}+(-5) \times 1}{\mathrm{~K}+1}\)

or 0 = \(\frac{5 \mathrm{~K}-5}{\mathrm{~K}+1}\)
or 5K – 5 = 0
or 5K = 5
or K = \(\frac{5}{5}\) = 1
∴ Required ratio is K : 1 = 1 : 1.
Now, x co-ordinate of P is:
x = \(\frac{-4 \times K+1 \times 1}{K+1}\)
Putting the value of K = 1, we get:
x = \(\frac{-4 \times 1+1 \times 1}{1+1}=\frac{-4+1}{2}\)
x = \(-\frac{3}{2}\)
Hence, required point be (\(-\frac{3}{2}\), 0).

Question 6.
If (1, 2); (4, y); (x, 6) and (3, 5)are the vertices of a parallelogram taken in order, find x and y.
Solution:
Let points of parallelogram ABCD are A (1, 2) (4, y) ; C (x, 6) and D (3, 5)
But diagonals of a || gm bisect each other.
Case I. When E is the mid point of A (1, 2) and C (x, 6)
∴ Co-ordinates of E are:
E = \(\left(\frac{x+1}{2}, \frac{6+2}{2}\right)\)
E = (\(\frac{x+1}{2}\), 4) …………..(1)

Case II. When E is the mid point B (4, y) and D (3, 5)
∴ Co-ordinates of E are:

E = \(\left(\frac{3+4}{2}, \frac{5+y}{2}\right)\)

E = \(\left(\frac{7}{2}, \frac{5+y}{2}\right)\) …………….(2)
But values of E in (1) and (2) are same, so comparing the coordinates, we get
\(\frac{x+1}{2}=\frac{7}{2}\)
or x + 1 = 7
or x = 6.

and 4 = \(\frac{5+y}{2}\)
or 8 = 5 + y
or y = 3
Hence, values of x and y are 6 and 3.

Question 7.
Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, – 3) and B is (1, 4).
Solution:
Let, coordinates of A be (x, y). But, centre is the’ niij ioint of the vertices of the diameter.

∴ O is the mid point of A(x, y) and B(1, 4)
∴ \(\left(\frac{x+1}{2}, \frac{y+4}{2}\right)\) = (2, -3)
On comparing, we get
\(\frac{x+1}{2}\)
or x + 1 = 4
or x = 3

and \(\frac{y+4}{2}\) = – 3
or y + 4 = – 6
or y = – 10
Hence, required point A be (3, – 10).

Question 8.
If A and B are (- 2, – 2) and (2, – 4) respectively, find the coordinates of P such that AP = AB and P lies ¡n the line segment AB.
Solution:
Let required point P be (x, y)
Also AP = \(\frac{3}{7}\) AB …(Given)
But, PB = AB – AP
= AB – \(\frac{3}{7}\) AB = \(\frac{7-3}{7}\) AB
PB = \(\frac{4}{7}\) AB
∴ \(\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\frac{3}{7} \mathrm{AB}}{\frac{4}{7} \mathrm{AB}}=\frac{3}{4}\).

∴ P divides given points A and B in ratio 3 : 4.
Now,
x = \(\frac{3 \times 2+4 \times-2}{3+4}\)

or x = \(\frac{6-8}{7}=-\frac{2}{7}\)

and y = \(\frac{3 \times-4+4 \times-2}{3+4}\)
= \(\frac{-12-8}{7}=-\frac{20}{7}\)

Hence, coordinates of P be (\(-\frac{2}{7}\), \(-\frac{20}{7}\)).

Question 9.
Find the coordinates of the points which divides the line segment joining A (- 2, 2) and B (2, 8) into four equal parts.
Solution:
Let required points are C, D and E which divide the line segment joming the points A (- 2, 2) and B (2, 8) into four equal parts. Then D is mid point of A and B ; C is the mid point of A and D ; E is the mid point of D and B such that
AC = CD = DE = EB

Now, mid point of A and B (i.e., Coordinates of D)
= \(\left(\frac{-2+2}{2}, \frac{2+8}{2}\right)\) = (0, 5)

Mid point of A and D (i.e., Coordinates of C)
= \(\left(\frac{-2+0}{2}, \frac{2+5}{2}\right)=\left(-1, \frac{7}{2}\right)\)

Mid point of D and B (i.e., Coordinates of E)
= \(\left(\frac{2+0}{2}, \frac{8+5}{2}\right)=\left(1, \frac{13}{2}\right)\)

Hence, requned points be (0, 5), (-1, \(\frac{7}{2}\)), (1, \(\frac{13}{2}\)).

Question 10.
Find the area of a rhombus if the vertices are (3, 0); (4, 5); (- 1, 4) and(- 2, – 1) taken in order.
[Hint: Areas of a rhombus = \(\frac{1}{2}\) (Product of its diagonals)]
Solution:
Let coordinates of rhombus ABCD are A (3, 0); B(4, 5); C(-1, 4) and D(- 2, – 1).
Diagonal, AC = \(\sqrt{(-1-3)^{2}+(4-0)^{2}}\)
= \(\sqrt{16+16}=\sqrt{32}=\sqrt{16 \times 2}\) = 4√2

and diagonal BD
BD = \(\sqrt{(-2-4)^{2}+(-1-5)^{2}}\)
= \(\sqrt{36+36}=\sqrt{72}=\sqrt{36 \times 2}\) = 6√2.

∴ Area of rhombus ABCD = \(\frac{1}{2}\) × AC × BD
ABCD = [\(\frac{1}{2}\) × 4√2 × 6√2] sq. units
(\(\frac{1}{2}\) × 24 × 2) sq. units
= 24 sq. units
Hence, area of rhombus is 24 sq. units.

Punjab State Board PSEB 8th Class English Book Solutions English Grammar Determiners Exercise Questions and Answers, Notes.

PSEB 8th Class English Grammar Determiners

Determiners वे शब्द है जो संज्ञा (noun) से पूर्व उसके अर्थ को किसी न किसी रूप से स्माष्ट करने या निरिचत करने के लिए प्रयुक्त होते हैं। यहां पर हम Determiners की एक सूची दे रहे हैं-
(i) Demonstratives. this, that, these, those.
(ii) Possessives. my, your, his, her, its, our, their
(iii) Cardinal numbers. one, two, three, etc.
(iv) Ordinal numbers. first, second, third, etc.
(v) Articles. a, an, the.
(vi) Miscellaneous. Some, any, both, certain, enough, few, every, least, less, little, more, most, much, next, other, own, plenty, several, such, many, another, each, no, a few, a large number of, a small number of, a great deal of, a good deal of, a large quantity of, a small quantity of, all, etc.

(i) Demonstratives (Paderat ya). This, that, these, those the procesu alert a Ridha करते हैं। अत: वे निर्देशक कहलाते हैं और इनसे Determiners की एक श्रेणी बनती है; जैसे:

• This book was purchased by me.
• That book was purchased by me.
• These girls are singing.
• Those girls are dancing.

Note. This और that एकवचन हैं जबकि these और those बहुवचन हैं।

(ii) Possessives (सम्बन्धवचक सब्द). My, her, your, his its, our, their सम्बन्धवचक सर्वनाम है ओर इनका प्रयोग एकवचन और बहुवचन दोनों प्रकार की संज्ञाओं के साथ किया जाता है; जैसे,
1. (a) I like my book. (Singular)
(b) I kept my books in order. (Plural)

2. (a) He told his story in tears. (Singular)
(b) People tell his stories in tears (Plural)

3. (a) Let me see your note-book. (Singular)
(b) Teacher will check your note-books. (Plural)

Note. सम्बन्धवाचक शब्द अगणनीय संज्ञाओं से पूर्व भी प्रयुक्त किए जा सकते हैं; जैसे-

• I took my bath. (uncountable)
• His courage failed him. (uncountable)
• We cannot doubt their patriotism. (uncountable)

(iii) and (iv) Cardinal and Ordinal Numbers.
Cardinal and Ordinal संख्याएं भी Determiners का एक भाग हैं।

• Two miles are left to be covered.
• I am reading the third chapter now.
• I need fifty rupees for this.

(v) The Articles. a, an तथा the को अंग्रेजी व्याकरण में articles के नाम से पुकारा जाता है। ये वास्तव में Adjective determiners हैं।

Filling the Blanks with Articles

1. Ram is ………. honest man. (an)
2. ………. book you gave me is lost. (The)
3. ………. cow is a useful animal. (The)
4. ………. Himalayas have many ranges. (The)
5. Gardening is ………. useful hobby. (a)
6. I have been waiting for you for ……. hour. (an)
7. I saw ……. Sri Harmandar Sahib. (the)
8. Who is ……… head of your family ? (the)
9. He is ……… boy who stole my pen. (the)
10. Our Principal is ……… intelligent man. (an)
11. ………. stitch in time saves nine. (A)
12. Do not make ………. noise (a)
13. My brother is ………. M.A., you are ………. B.A. (an, a)
14. New Delhi is ……… capital of India. (the)
15. This is ……… interesting story. (an)
16. He has ……… lot of money. (a)
17. He has ………. monthly income of Rs. 7000. (a)
18. ………. stars are a beautiful sight. (The)
19. Give me ……… apple or ………. orange. (an, an)
20. You should always show sympathy to ……… poor. (the)
21. Get me ………. nice cup of tea. (a)
22. This is ………. excellent watch. (an)
23. ………. rose smells sweet. (The)
24. Where is ………. book which I gave you? (the)
25. He is ……….. able student but not ………. ablest. (an, the)
26. Why are you in ………. hurry? (a)
27. I hate ………. dishonest servant. (a)
28. He is ……. idiot. (an)
29. ……… rich are not always happy. (The)
30. …….. moon has risen. (The)
31. There was ………. army of monkeys near …….. temple. (an, the)
32. This is quite …….. new idea. (a)
33. He is ………. hardworking boy. (a)
34. I need ………. pen and ………. inkpot. (a, an)
35. I read ……… Hindustan Times everyday. (the)
36. I have not seen ……… Taj as yet. (the)
37. ………. bad workman quarrels with his tools. (A)
38. ………. apple ………. day keeps ………. doctor away. (An, a, the)
39. …….. Ganga is ……… sacred river. (The, a)

More Exercises (Solved)

1. Fill up the blanks with the Articles (a, an, the):

1. He can read ………. Vedas.
2. She is ……… intelligent girl.
3. …….. sun shines brightly.
4. Life is not ………. bed of roses.
5. ……… higher you climb ……… colder it gets.
6. In ………. park I saw ………. one-eyed beggar.
7. Mumbai is ………. Manchester of India.
8. ………. Sikhs are a brave nation.
9. ………. owl cannot see during day time.
10. ………. Ganges is ………. sacred river.
11. We started late in ………. afternoon.
12. ………. rose smells sweet.
13. He reads ………. Bible everyday.
14. Kalidas was …….. Shakespeare of India.
15. Einstein was ………. Newton.
16. I read ………. Tribune daily.
17. I am in ……… hurry.
18. Help ……… poor.
19. Both ………. robbers were arrested.
20. Keep to ….. left.
21. ……… housemaid pulls up ……….. blind.
22. What ………. lovely girl !
23. What ……… ripe apple !
24. Not ……… word was said.
25. Help ……… poor, ……… needy and ………. miserable.
Hints:
1. the
2. an
3. The
4. a
5. The, the
6. the, a
7. the
8. The
9. An
10. The, a
11. the
12. The
13. the
14. the
15. the
16. the
17. a
18. the
19. the
20. the
21. The, the
22. a
23. a
24. a
25. the, the, the.

II. Correct the following sentences:

1. Only few. men are honest.
2. The man is mortal.
3. He acted like man.
4. Beas flows in Punjab.
5. You are in wrong but he is in right.
6. He is by far ablest boy.
7. Nobody likes a person with bad temper
8. The iron is useful metal.
9. Not word was said.
10. He has too high a opinion of you.
11. Learn this poem by the heart.
12. Never tell lie.
Hints:
1. Only a few men are honest.
2. Man is mortal.
3. He acted like a man.
4. The Beas flows in the Punjab.
5. You are in the wrong but he is in the right.
6. He is by far the ablest boy.
7. Nobody likes a person with a bad temper.
8. Iron is a useful metal.
9. Not a word was said.
10. He has too high an opinion of you.
11. Learn this poem by heart.
12. Never tell a lie.

III. Fill in the blanks with suitable determiners:

1. I went to ………. window which commanded a large green garden.
2. I have heard so ……… about your school.
3. Look out of the window for ………. minute.
4. It is really something of ………. joke.
5. There you have …….. essential part of our system.
6. I asked her by way of ………. opening.
7. But I had ………. idea of all this.
8. Come down into ………. garden.
9. Having ………. arm tied up is troublesome.
10. It educates both ………. blind and the helpers.
Hints:
1. the
2. much
3. a
4. a
5. an
6. an
7. an
8. the
9. an
10. the.

IV. Insert the determiners this, that, these, those in the blanks:

1. ………. shirt is costly but ……… shirt is cheap.
2. Would you like to take … book or ………. one ?
3. ……… sum cannot be solved by ………. silly boys.
4. ……… flowers are beautiful but ………. flowers are ordinary ones.
5. He likes this pair of trousers but he does not like ……… one.
Hints :
1. This, that
2. this, that
3. This, those
4. These, those
5. that.

V. Insert ‘a few’ or ‘the few’ whichever is suitable:

1. ………. books she had were all lost.
2. It is question of spending ………. rupees.
3. ………. suggestions she gave were all carried out.
4. …….. hints on essay-writing are quite to the point.
5. ………. boys attended the class.
Hints:
1. The few
2. a few
3. The few
4. A few
5. A few.

VI. Insert ‘little’, a little, or ‘the little’ whichever is suitable

1. ………. knowledge is a dangerous thing.
2. There is ………. hope of his recovery.
3. He has ………. money with him.
4. ……… Strength he had in him proved useless.
5. He takes ……… interest in me.
Hints:
1. A little
2. little
3. a little
4. The little
5. little.

VII. Fill in the blanks with determiners ‘each, every, either or neither:

1. ………. pen costs ten rupees.
2. He comes here on ………. Sunday.
3. ………. of you may take this book.
4. ………. man wants to rise in the world.
5. ………. seat in the hall was occupied.
Hints:
1. Each
2. every
3. Either
4. Every
5. Every.

Punjab State Board PSEB 8th Class English Book Solutions English Grammar Parts of Speech Exercise Questions and Answers, Notes.

PSEB 8th Class English Grammar Parts of Speech

शब्दों को प्रयोग के अनुसार आठ भागों अथवा श्रेणियों में विभाजित किया जाता है, जिन्हें Parts of Speech (शब्द भेद) कहते हैं। ये आठ भाग हैं

1. Noun
2. Pronoun
3. Adjective
4. Verb
5. Adverb
6. Preposition
7. Conjunction
8. Interjection.

1. Noun- किसी व्यक्ति, स्थान, वस्तु अथवा गुण-दोष के नाम को Noun (संज्ञा) कहते हैं; जैसे Ram, Shyam, Mumbai, Pen, Pencil, Rose, Dog, Beauty, Honesty आदि।

2. Pronoun- जो शब्द किसी संज्ञा के स्थान पर प्रयोग होता है, उसे Pronoun (सर्वनाम) कहते हैं; जैसे – I, we, you, they, he, she, it आदि।

3. Adjective- जो शब्द किसी Noun अथवा Pronoun की विशेषता प्रकट करता है, उसे Adjective (विशेषण) कहते हैं; जैसे good, bad, thin, fat, tall, many, useful आदि। .

4. Verb- वह शब्द जिससे कार्य का होना अथवा करना प्रकट हो या जो शब्द किसी व्यक्ति, स्थान अथवा वस्तु के बारे में कुछ बतलाए, वह शब्द Verb (क्रिया) कहलाता है; जैसे-

• The girl reads a book.
• Ram is a good boy.
• Mohan goes to school daily.
• Seema feels sad.
• I helped the poor.

5. Adverb- जो शब्द किसी Verb, Adjective अथवा किसी अन्य Adverb की विशेषता प्रकट करे, तो उसे Adverb (क्रिया विशेषण) कहते हैं; जैसे-

• He ran fast. (Verb की विशेषता)
• Rose is very beautiful. (Adjective की विशेषता)
• He works too slowly. (Adverb की विशेषता)

6. Preposition- जो शब्द किसी Noun अथवा Pronoun से पहले प्रयोग किया जाता है और जो उनका सम्बन्ध वाक्य के अन्य शब्दों से जोड़ता है, उसे Preposition (सम्बन्धवाचक शब्द) कहते हैं; जैसे-

• Mohan is at the gate.
• Ram is proud of his teacher.
• He is fond of music.
• I am writing with a pen.

7. Conjunction-जो शब्द, शब्दों अथवा वाक्यों को जोड़ता है, उसे Conjunction (संयोजक) कहते हैं; जैसे-

• Ram and Shyam are friends.
• You will get through if you work hard.
• He was late because he missed the bus.
• Either he or his brother is at fault.

8. Interjection- अकस्मात् भावना को व्यक्त करने वाला शब्द Interjection (विस्मयादिबोधक अव्यय)
कहलाता है; जैसे-, Alas ! Hurrah! Oh! Lo ! आदि।

Exercises From Board’s Grammar (Solved)

I. Write in the space provided the name of the part of speech to which the underlined words belong in the following sentences:

1. Seema is a beautiful girl.
2. Alas! His dog is dead.
3. The sun sets in the west.
4. The lion is a ferocious animal.
5. Delhi is a very big city.
6. Rahim is poor but honest.
7. Honesty is the best policy.
8. The cat is under the table.
Hints:
1. Adjective
2. Interjection
3. Verb
4. Adjective
5. Adverb
6. Conjunction
7. Noun
8. Preposition.

II. Complete the following sentences with appropriate ‘Nouns’:

1. ………. is a good boy.
2. She goes to the ……….. everyday.
3. They go for a ……… daily.
4. ……… is the Capital of India.
5. ……….. is the best policy.
6. The ……….. rises in the …………….
7. Chandigarh is the ………. of Punjab and Haryana.
8. Shimla is a beautiful ………..
Hints:
1. Mohan
2. temple/Gurudwara
3. walk
4. Delhi
5. Honesty
6. sun, east
7. capital
8. place.

III. Fill in the blanks with suitable ‘Pronouns:

1. My son is playing with ………. toys.
2. ……….. father is working in Mumbai.
3. ………. has gone abroad for higher studies.
4. This is ……….. house.
5. The girls are doing ……….. homework.
6. ……… is the bread-winner of the family.
7. We should respect ……… parents.
8. He loves ……….. native place very much.
Hints:
1. his
2. My
3. He
4. our
5. their
6. She
7. our
8. his.

IV. Fill in the blanks with suitable ‘Adjectives:

1. She is my ……….. friend.
2. Pudding is my ……….. dish.
3. The scenery of Mussoorie is ………..
4. She likes to wear ……….. dresses.
5. The Bible is a ……….. book.
6. Cricket is a ……….. game.
7. John is an ……….. teacher.
8. Mango is a ……….. fruit.
Hints:
1. best
2. favourite
3. beautiful
4. white
5. holy
6. good
7. English
8 juicy.

V. Fill in the blanks with suitable ‘Verbs’:

1. She ……….. ice-cream.
2. They ……….. a lot.
3. My mother ……….. food.
4. John ………. in a factory.
5. They ………. football…
6. We should ………. a bath everyday.
7. The school peon ……….. the bell.
8. Seema ……… for a walk daily.
Hints:
1. is eating
2. worked
3. cooks
4. works
5. play
6. take
7. rings
8. goes.

VI. Fill up the blanks with suitable ‘Adverbs?:

1. Sohan walks ………….
2. Everyone should work ……..
3. She sings …………
4. His dad is a ……… respectable man.
5. Our teacher speaks ………. politely.
6. My sister sleeps ………
7. John is a ………. hardworking boy.
8. Girls sang …………..
Hints:
1. slowly
2. hard
3. sweetly
4. very
5. very
6. early
7. very
8. nicely.

VII. Complete the following sentences with suitable ‘Prepositions:

1. My grandfather is hard ………. hearing.
2. My mom is fond ……….. music.
3. A burglar broke ……….. our house last night.
4. The students should listen ……… their teachers attentively.
5. Yesterday we went ………. the Rose Garden.
6. Ayushi is playing ……….. the piano.
7. Children have been playing ……….. morning.
8. My uncle has been living ………. Canada ………. fifteen years.
9. I will play …….. finishing my homework.
10. I am standing ………. Anshu. Anshu is in front of me.
Hints:
1. of
2. of
3. into
4. to
5. to
6. on
7. since
8. in, for
9. after
10. behind.

VIII. Fill in the blanks with suitable Conjunctions:

1. My younger brother is both intelligent ……….. hardworking.
2. He says ………. he is a doctor.
3. ……… Seema ………. Rita is at fault.
4. Our servant is poor ………. honest.
5. He ………. his nephew manage the shop.
6. Ram went on leave ………. he was injured.
7. You will get the tickets ……….. you reach there before 6 o’clock.
8. ……….. he was late, yet he was able to catch the bus.
Hints:
1. and
2. that
3. Either, or
4. but
5. and
6. because
7. if
8. Although

IX. Fill in the blanks with suitable ‘Interjections:

1. ……….. We have won the game
2. ………… The man is dead.
3. ………….. They have come.
4. ………….. Is this the place?
5. ……………. Well done.
6. ……………. He has lost all his money in gambling.
7. ……….. You are hurt.
8. ………….. He has failed again.

Hints:
1. Hurrah !
2. Alas!
3. Lo !
4. Oh !
5. Bravo!
6. Alas!
7. Oh !
8. Alas!

Punjab State Board PSEB 8th Class English Book Solutions English Vocabulary Synonyms Exercise Questions and Answers, Notes.

PSEB 8th Class English Vocabulary Synonyms

A Synonym is a word with the same meaning (एक सामान अये) as another word; as

Word – Synonym
abode – home
actual – real
allow – permit
add – plus
annual – yearly
arrive – reach
beauty – loveliness
beautiful – pretty
begin – start
big – large
brave – bold
briefa – short
calm – peaceful
centre – middle
clever – intelligent

close – shut
costly – expensive
corn – grain
damp – wet
daily – everyday
definite – certain
difficult – tough
enemy – foe
example – instance
excellent – wonderful
fear – terror/horror
foolish – stupid/silly
fun – enjoyment
grief – sorrow
happy – glad
hollow – empty
hot – warm
kind – generous
loving – affectiodate
perfect – ideal
quiet – silent
reply – answer
right – correct
small – tiny
smell – scent
soft – tender
taste – flavour
timeless – unending
unitet – cooperate
vacant – empty
wealthy – rich
wide – broad

Punjab State Board PSEB 8th Class English Book Solutions English Vocabulary One Word Substitution Exercise Questions and Answers, Notes.

PSEB 8th Class English Vocabulary One Word Substitution

Group of Words	Words
1. one who has become dependent on something	Addict
2. one who produces works of art	Artist
3. a number of sheep	Flock
4. one who writes books	Author
5. one who cuts hair	Barber
6. a building in which aircraft housed	Hanger
7. a building where an audience	auditorium
8. one who plays a note in a play or movie	sits
9. one who treats all teeth problems	Actor
10. one who grows crops	Dentist
11. one who works in gold	Farmers
12. a period of ten years	Goldsmith
13. a performance given by a number of musicians	decade
14. one who works in iron	Concert
15. one who sells meat	Blacksmith
16. one who works with wood	Butcher
17. a study of animals	Carpenter
18. one who lives at the same time	Zoology
19. one who sells medicines	Contemporary
20 one who mends shoes	Chemist
21. one who settles in another country	Cobbler
22. one who treats the sick	immigrant
23. person working in the same place	Doctor
24. one who sells flowers	Colleague
25. a doctor who treats mental illness	Florist

Punjab State Board PSEB 8th Class English Book Solutions English Vocabulary Words as per Dictionary Order Exercise Questions and Answers, Notes.

PSEB 8th Class English Vocabulary Words as per Dictionary Order

यदि English भाषा के शब्दों को अंग्रेज़ी Albhabet (A, B, C ….. Y, Z) के क्रम में लिखा जाए तो शब्दों का Dictionary order या Alphabetical Order कहा जाता है। Dictionary में कोई शब्द ढूंढ़ने के लिए हमारे लिए इस क्रम को जानना बहुत ज़रूरी है। वैसे भी शब्दों के किसी समूह को जब Dictionary Order में लिखा जाता है, तो उसमें से किसी भी शब्द को ढूंढ़ना आसान हो जाता है।

Dictionary Order में लिखने का तरीका-किसी दिए गए शब्द समूह को Dictionary Order में लिखने के लिए निम्नलिखित विधि अपनाएं:
1. सबसे पहले शब्दों के पहले letter (अक्षर) पर ध्यान दें। Alphabet में पहले आने वाले अक्षर से शुरू होने वाला शब्द सबसे पहले आयेगा। जैसे-पहले A, फिर B …..
2. यदि दो या दो से अधिक शब्दों का letter एक जैसा हो तो शब्द के दूसरे letter को follow करें।
3. हो सकता है किन्हीं दो शब्दों के पहले दो अक्षर एक जैसे हों। ऐसे में तीसरे अक्षर (letter) के अनुसार शब्दों को क्रम-बद्ध किया जाएगा।

इसी विधि का अनुसरण करते हुए आगे बढ़ते जाएं, जब तक कि पूरा शब्द-समूह क्रमबद्ध न हो जाए।

Worked Out Examples

Write the given words as per dictionary order:
Question 1.
Apple, Orange, Banana, Mango, Grapes.
Answer:
Apple, Banana, Grapes, Mango, Orange.

Question 2.
Tomato, Potato, Carrot, Radish, Onion.
Answer:
Carrot, Onion, Potato, Radish, Tomato.

Question 3.
Active, Absent, Addition, Aeroplane, Almond.
Answer: Absent, Active, Addition, Aeroplane, Almond.

Question 4.
January, June, July, April, August.
Answer:
April, August, January, July, June.

Question 5.
Black, Brown, Birthday, Blood, Beautiful.
Answer:
Beautiful, Birthday, Black, Blood, Brown.

Question 6.
Earth, Essay, Father, Farmer, Brother.
Answer:
Brother, Earth, Essay, Father, Farmer.

Question 7.
Watch, Which, Small, Smell, Bitch.
Answer:
Bitch, Small, Smell, Watch, Which.

Question 8.
School, Office, Post-office, Soldier, Knife.
Answer:
Knife, Office, Post-office, School, Soldier.

Question 9.
Barber, Author, Cobbler, Actor, Dancer.
Answer:
Actor, Author, Barber, Cobbler, Dancer.

Question 10.
Bear, Horse, Tiger, Lion, Lamb.
Answer:
Bear, Horse, Lamb, Lion, Tiger.

Exercise for Practice

Arrange the following group of words in Alphabetical/Dictionary Order:
1. Teacher, Artist, Student, Tailor, Gardener.
2. Bird, Book, Cycle, Circle, Circus.
3. Music, Monitor, May, Marker, March.
4. Agra, Delhi, Dehradun, Capital, Child.
5. Table, Toy, Chair, Friend, Tree.
6. Mouse, Magician, Rabbit, Manager, Prince.
7. King, Kite, Zebra, Zoo, Shirt
8. Grass, Green, Leaves, Pitcher, Picture.
9. Morning, Evening, Night, Afternoon, Dawn.
10. Coffee, Indian, Inkpot, Insect, Mountain.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 7 Coordinate Geometry Ex 7.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1

Question 1.
Find the distance between the following pairs of points:
(i) (2, 3); (4, 1)
(ii)(-5, 7); (-1, 3)
(iii) (a, b); (-a, -b).
Solution:
(i) Given points are: (2, 3); (4, 1)
Required distance = \(\sqrt{(4-2)^{2}+(1-3)^{2}}\)
\(\sqrt{4+4}=\sqrt{8}=\sqrt{4 \times 2}\)
= 2√2.

(ii) Given points are: (-5, 7); (-1, 3)
Required distance = \(\sqrt{(-1+5)^{2}+(3-7)^{2}}\)
\(\sqrt{16+16}=\sqrt{32}\)
= \(\sqrt{16 \times 2}\)
= 4√2.

(iii) Given points are: (a, b); (-a, -b)
Required distance = \(\sqrt{(-a-a)^{2}+(-b-b)^{2}}\)
= \(\sqrt{(-2 a)^{2}+(-2 b)^{2}}\)
= \(\sqrt{4 a^{2}+4 b^{2}}\)
= √4 \(\sqrt{a^{2}+b^{2}}\)
= \(2 \sqrt{a^{2}+b^{2}}\)

Question 2.
Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B
discussed in section 7.2.
Solution:
Given points are: A (0, 0) and B (36, 15)
Distance, AB = \(\sqrt{(0-36)^{2}+(0-15)^{2}}\)
\(\sqrt{1296+225}=\sqrt{1521}\) = 39.
According to Section 7.2
Draw the distinct points A (0, 0) and B (36, 15) as shown in figure.

Draw BC ⊥ on X-axis.
Now, In rt. ∠d ∆ACB,
AB = \(\sqrt{\mathrm{AC}^{2}+\mathrm{BC}^{2}}\)
= \(\sqrt{(36)^{2}+(15)^{2}}\)
= \(\sqrt{1296+225}=\sqrt{1521}\)
= 39.
Hence, required distance between points is 39.

Question 3.
Determine if the points (1, 5), (2, 3) and (- 2, – 11) are collinear.
Solution:
Given point are : A (1. 5); B (2.3) and C (- 2, – 11).
AB = \(\sqrt{(2-1)^{2}+(3-5)^{2}}\)
= \(\sqrt{1+4}=\sqrt{5}\)

BC = \(\sqrt{(-2-2)^{2}+(-11-3)^{2}}\)
= \(\sqrt{16+196}=\sqrt{212}\)

CA = \(\sqrt{(1+2)^{2}+(5+11)^{2}}\)
= \(\sqrt{9+256}=\sqrt{265}\)
From above distances, it is clear that sum of any two is not equal to third one.
Hence, given points are not collinear

Question 4.
Check whether (5, – 2); (6, 4) and (7, – 2) are the Vertices of an isosceles triangle.
Solution:
Given points be A (5, – 2); B (6, 4) and C (7, – 2).
AB = \(\sqrt{(5-6)^{2}+(-2-4)^{2}}\)
= \(\sqrt{1+36}=\sqrt{37}\)

BC = \(\sqrt{(6-7)^{2}+(4+2)^{2}}\)
= \(\sqrt{1+36}=\sqrt{37}\)

CA = \(\sqrt{(7-5)^{2}+(-2+2)^{2}}\)
= \(\sqrt{4+0}=2\)
From above discussion, it is clear that AB = BC = √37.
Given points are vertices of an isosceles triangle.

Question 5.
In a classroom, 4 friends are seated at the points A, B, C and D as shown in fig. Champa and Charnel walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square”? Chameli disagrees. Using distance formula, find which of them is correct, and why?

Solution:
In the given diagram, the vertices of given points are : A (3, 4); B (6, 7); C (9, 4) and D (6, 1).
Now,
AB = \(\sqrt{(6-3)^{2}+(7-4)^{2}}\)
= \(\sqrt{9+9}=\sqrt{18}\)

BC = \(\sqrt{(9-6)^{2}+(4-7)^{2}}\)
= \(\sqrt{9+9}=\sqrt{18}\)

CD = \(\sqrt{(6-9)^{2}+(1-4)^{2}}\)
= \(\sqrt{9+9}=\sqrt{18}\)

DA=\(\sqrt{(3-6)^{2}+(4-1)^{2}}\)
= \(\sqrt{9+9}=\sqrt{18}\)

AC = \(\sqrt{(9-3)^{2}+(4-4)^{2}}\)
= \(\sqrt{36+0}=6\)

BD = \(\sqrt{(6-6)^{2}+(1-7)^{2}}\)
= \(\sqrt{0+36}\) = 6
From above discussion, it is clear that
AB = BC = CD = DA = √18 and AC = BD = 6.
ABCD formed a square and Champa is correct about her thinking.

Question 6.
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) ( 1,- 2), (1, 0),(- 1, 2), (- 3, 0)
(ii) ( 3, 5), (3, 1), (0, 3), (- 1, – 4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2).
Solution:
(i) Given points be A (- 1, – 2); B(1, 0); C(- 1, 2) and D(- 3, 0).
AB = \(\sqrt{(1+1)^{2}+(0+2)^{2}}\)
= \(\sqrt{4+4}=\sqrt{8}\)

BC = \(\sqrt{(-1-1)^{2}+(2-0)^{2}}\)
= \(\sqrt{4+4}=\sqrt{8}\)

CD = \(\sqrt{(-3+1)^{2}+(0-2)^{2}}\)
= \(\sqrt{4+4}=\sqrt{8}\)

DA = \(\sqrt{(-1+3)^{2}+(-2+0)^{2}}\)
= \(\sqrt{4+4}=\sqrt{8}\)

AC = \(\sqrt{(-1+1)^{2}+(2+2)^{2}}\)
= \(\sqrt{0+16}=4\)

BD = \(\sqrt{(-3-1)^{2}+(0-0)^{2}}\)
= \(\sqrt{16+0}=4\)

From above discussion, it is clear that
AB = BC = CD = DA = √8 and AC = BD = 4.
Hence, given quadrilateral ABCD is a square.

(ii) Given points be A (- 3, 5); B (3, 1); C (0, 3) and D (- 1,- 4)
AB = \(\sqrt{(-3-3)^{2}+(5-1)^{2}}\)
= \(\sqrt{36+16}=\sqrt{52}=\sqrt{4 \times 13}\)
= 2√13

BC = \(\sqrt{(3-0)^{2}+(1-3)^{2}}\)
= \(\sqrt{9+4}=\sqrt{13}\)

CA = \(\sqrt{(0+3)^{2}+(3-5)^{2}}\)
= \(\sqrt{9+4}=\sqrt{13}\)
Now, BC + CA = \(\sqrt{13}+\sqrt{13}\) = 2√13 = AB
∴A, B and C are collinear then A, B, C and D do not form any quadrilateral.

(iii) Given points are A (4, 5); B (7, 6); C (4, 3) and D (1, 2)
AB = \(\sqrt{(7-4)^{2}+(6-5)^{2}}\)
= \(\sqrt{9+1}=\sqrt{10}\)

BC = \(\sqrt{(4-7)^{2}+(3-6)^{2}}\)
= \(\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}\)

CD = \(\sqrt{(1-4)^{2}+(2-3)^{2}}\)
= \(\sqrt{9+1}=\sqrt{10}\)

DA = \(\sqrt{(4-1)^{2}+(5-2)^{2}}\)
= \(\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}\)

AC = \(\sqrt{(4-4)^{2}+(3-5)^{2}}\)
= \(\sqrt{0+4}\) = 2

BD = \(\sqrt{(1-7)^{2}+(2-6)^{2}}\)
= \(\)

From above discussion, it is clear that AB = CD and BC = DA. and AC ≠ BD.
i.e., opposite sides are equal but their diagonals are not equal.
Hence, given quadrilateral ABCD is a parallelogram.

Question 7.
Find the points on the x-axis which is equidistant from (2, – 5) and (- 2, 9).
Solution:
Let required point be P (x, 0) and given points be A (2, – 5) and B (- 2, 9).
According to question,
PA = PB
(PA)² = (PB)²
or (2 – x)² + (- 5- 0)² = (- 2 – x)² + (9 – 0)²
or 4 + x² – 4x + 25 = 4 + x²+ 4x + 81
-8x = 56
x = \(\frac{4}{4}\) = – 7
Hence, required point be (- 7, 0).

Question 8.
Find the values of y for which the distance between the points P (2, – 3) and Q (10, y) is 10 units.
Solution:
Given points are P (2, – 3) and Q (10, y)
PQ = \(\sqrt{(10-2)^{2}+(y+3)^{2}}\)
= \(\sqrt{64+y^{2}+9+6 y}\)
= \(\sqrt{y^{2}+6 y+73}\)
According to question,
PQ = 10
or \(\sqrt{y^{2}+6 y+73}\) = 10
Squaring
or y² + 6y + 73 = 100
or y² + 6y – 27 = 0
or y² + 9y – 3y – 27 = 0
S = 6 P = – 27
or y (y + 9) – 3 (y + 9) = 0
or (y + 9) (y – 3) = 0
Either y + 9 = 0 or y – 3 = 0
y = – 9 or y = 3
Hence, y = – 9 and 3.

Question 9.
If Q (0, 1) is equidistant from P (5, – 3) and R (x, 6), find the values of x. Also find the distances QR and PR.
Solution:
Given points Q (0, 1); P (5, – 3) and R (x, 6)
QP = \(\sqrt{(5-0)^{2}+(-3-1)^{2}}\)
= \(\sqrt{25+16}=\sqrt{41}\)

and QR = \(\sqrt{(x-0)^{2}+(6-1)^{2}}\)
= \(\sqrt{x^{2}+25}\)

According to question,
QP = QR
or \(\sqrt{41}=\sqrt{x^{2}+25}\)
Squaring
or 41 = x² + 25
or x² = 16
or x = ± √16 = ± √4.

When x = 4 then R (4, 6).
QR = \(\sqrt{(4-0)^{2}+(6-1)^{2}}\)
= \(\sqrt{16+25}=\sqrt{41}\)

PR = \(\sqrt{(4-5)^{2}+(6+3)^{2}}\)
= \(\sqrt{1+81}=\sqrt{82}\)

When x = – 4 then R (- 4, 6).
QR = \(\sqrt{(-4-0)^{2}+(6-1)^{2}}\)
= \(\sqrt{16+25}=\sqrt{41}\)

PR = \(\sqrt{(-4-5)^{2}+(6+3)^{2}}\)
= \(\sqrt{81+81}=\sqrt{162}\).

Question 10.
Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (- 3, 4).
Solution:
Let required points be P (x, y) and given points are A (3, 6) and B (- 3, 4)
PA = \(\sqrt{(3-x)^{2}+(6-y)^{2}}\)
= \(\sqrt{9+x^{2}-6 x+36+y^{2}-12 y}\)
= \(\sqrt{x^{2}+y^{2}-6 x-12 y+45}\)

and PB = \(\sqrt{(-3-x)^{2}+(4-y)^{2}}\)
= \(\sqrt{9+x^{2}+6 x+16+y^{2}-8 y}\)
= \(\sqrt{x^{2}+y^{2}+6 x-8 y+25}\)

According to question,
PA = PB
\(\sqrt{x^{2}+y^{2}-6 x-12 y+45}\) = \(\sqrt{x^{2}+y^{2}+6 x-8 y+25}\)
sq,. both sides, we have,
or x² + y² – 6x – 12y + 45 = x² + y² + 6x – 8y – 25
or -12x – 4y + 20 = 0
or 3x + y – 5 = 0 is the required relation.