Bhagya

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 10 Circles Ex 10.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 10 Circles Ex 10.1

Question 1.
How many tangents can a circle have?
Solution:
Since at any point on a circle, there can be one and only one tangeni. But circle is a collection of infinite points, so we can draw infinite number of tangents to a circle.

Question 2.
Fill in the blanks:
(i) A tangent to a circle intersects it in ………………. point(s).
Solution:
one

(ii) A line intersecting a circle in two points is called a ………………..
Solution:
secant.

(iii) A circle can have ……………. parallel tangents at the most.
Solution:
two

(iv) The common point of a lingent h, circle and the circit is called ………………
Solution:
point of contact.

Question 3.
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is:
(A) 12 cm
(B) 13 cm
(C) 8.5 cm
(D) \(\sqrt{119}\) cm.
Solution:
According to given information we draw the figure such that,

OP = 5 cm and OQ = 12 cm
∵ PQ is a tangent and OP is the radius
∵ ∠OPQ = 90°
Now, In right angled ∆OPQ.
By Pythagoras Theorem,
OQ² = OP² + QP²
Or (12)² = (5)² + QP²
Or QP² = (12)² – (5)²
Or QP² = 144 – 25 = 119
Or QP = \(\sqrt{119}\) cm.
Hence, option (D) is correct.

Question 4.
Draw a circle and two lines parallel to given line such that one is a tangent and other a secant to the circle.
Solution:
According to thc given information we draw a circle having O as centre and l is the given line.

Now, m and n be two lines parallel to a given line l such that m is tangent as well as parallel to l and n is secant to the circle as well as parallel to l.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 9 Some Applications of Trigonometry Ex 9.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1

Question 1.
A circus artist is climbing 220 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, ¡f the angle made by the rope with the ground level is 30° (see fig.).

Solution:
Let AB be the heignt of pole;
AC = 20 m be the length of rope.
The angle of elevation in this situation is 30°.
Various arrangements are as shown in figure.

In right angled ∆ABC,
\(\frac{\mathrm{AB}}{\mathrm{AC}}\) = sin 30°

or \(\frac{\mathrm{AB}}{20}=\frac{1}{2}\)

or AB = \(\frac{1}{2}\) × 20 = 10
Hence, height of pole is 10 m.

Question 2.
A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 300 with it. The distance between the foot of the tree to the point where the top touches the ground is 8 rn Find the height of the tree.
Solution:
Let BD be length of tree before storm.
After storm AD = AC = length of broken part of tree.
The angle of elevation in this situation is 30°.
Various arrangements are as shown in the figure.

In right angled ∆ABC,

\(\frac{\mathrm{AB}}{\mathrm{AC}}\) = tan 30°

or \(\frac{h_{1}}{8}=\frac{1}{\sqrt{3}}\)
or h₁ = \(\frac{8}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{8}{3} \sqrt{3}\) m ……….(1)

\(\frac{\mathrm{BC}}{\mathrm{AC}}\) = cos 30°

or \(\frac{8}{h_{2}}=\frac{\sqrt{3}}{2}\)

or \(h_{2}=\frac{8 \times 2}{\sqrt{3}}=\frac{16}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\)

h₂ = \(\frac{16}{3}\) √3 …………..(2)

Total height of the tree = h₁ + h₂
= \(\frac{8}{3}\) √3 + \(\frac{16}{3}\) √3 [Using (1) & (2)]

= \(\left(\frac{8+16}{3}\right) \sqrt{3}=\frac{24}{3} \sqrt{3}\) = 8√3 m.
Hence, height of the tree is 8√3 m.

Question 3.
A contractor plants to install two slides for the children to play in a park. For the children below the age of 5 years, she
prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3 m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?
Solution:
Case I:
For children below 5 years.
Let AC = l₁ m denote the length of slide and BC = 1.5 m be the height of slide. The angle of elevation is 30°.
Various arrangements are shown in figure.

In right angled ∆ABC,

\(\frac{\mathrm{BC}}{\mathrm{AC}}\) = sin 30°

or \(\frac{1 \cdot 5}{l_{1}}=\frac{1}{2}\)

or l₁ = 1.5 × 2 = 3 m.

Case II:
For Elder children
Let AC = 12 m represent the length of slide and BC = 3 m be the height of slide. The angle of elevation is 60°. Various arrangements are as shown in figure.

In right angled ∆ABC,

\(\frac{\mathrm{BC}}{\mathrm{AC}}\) = sin 60°

or \(\frac{3}{l_{2}}=\frac{\sqrt{3}}{2}\)

or l₂ = \(\frac{3 \times 2}{\sqrt{3}}=\frac{6}{\sqrt{3}}\)

= \(\frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{6 \sqrt{3}}{3}\)

= 2√3 m.

Hence, length of slides for children below 5 years and elder children are 3 m and 2 m.

Question 4.
The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.
Solution:
Let BC = h m be the height of tower and AB = 30 m be the distance at ground level. Various arrangements are as shown in figure.

In right angled ∆ABC,

\(\frac{\mathrm{BC}}{\mathrm{AB}}\) = tan 30°

or \(\frac{h}{30}=\frac{1}{\sqrt{3}}\)

or h = \(\frac{30}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{30 \sqrt{3}}{3}\)

= 10√3 = 10 × 1.732
h = 17.32 (approx).
Hence, height of tower is 17.32 m.

Question 5.
A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.
Solution:
Let us suppose position of the kite is at point CAC = l m be length of string with which kite is attached. The angle of elevation for this situation be 60°. Various arrangements are as shown in the figure.

In right angled ∆ABC,

\(\frac{\mathrm{CB}}{\mathrm{AB}}\) = sin 60°

or \(\frac{60}{l}=\frac{\sqrt{3}}{2}\)

or l = \(\frac{60 \times 2}{\sqrt{3}}=\frac{120}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\)

= \(\frac{120 \sqrt{3}}{3}\) = 40√3 m.
Hence, length of the string be 40√3 m.

Question 6.
A 15 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.
Solution. Let ED = 30 m be the height of building and EC = l5 m be the height of boy.
The angle of elevation at different situation are 30° and 60° respectively.
Various arrangements are as shown in figure.

In right angled ∆ABC,

\(\frac{\mathrm{DC}}{\mathrm{AC}}\) = tan 30°

or \(\frac{28 \cdot 5}{x+y}=\frac{1}{\sqrt{3}}\)

or x + y = 28.5 × √3 m ………………(1)

Now, in right angled ∆BCD,

\(\frac{\mathrm{DC}}{\mathrm{BC}}\) = tan 60°

or \(\frac{28 \cdot 5}{y}=\sqrt{3}\)

or y = \(\frac{28 \cdot 5}{\sqrt{3}}\)

or y = \(\frac{28 \cdot 5}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{28 \cdot 5 \times \sqrt{3}}{3}\) ……….(2)

Distance covered towards building = x = (x + y) – y
= (28.5 × √3) – (\(\frac{28.5}{3}\) × √3) m [sing (1) and (2)]

= 28.5 (1 – \(\frac{1}{3}\)) √3 m

= 28.5 (\(\frac{3-1}{4}\)) √3 m

= [28.5 × \(\frac{2}{3}\)]√3 m = 19√3 m.

Hence, distance covered by boy towards the building is 19√3 m.

Question 7.
From a point on the ground, the angles of elevation of the bottom and top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.
Solution:
Let BC = 20 m be the height of building and DC = h m be the height of transmission tower. The angle of elevation of
the bottom and top of a transmission tower are 45° and 60° respectively.
Various arrangements are as shown in the figure.

In right angled ∆ABC,
\(\frac{A B}{B C}\) = cot 45°

or \(\frac{\mathrm{AB}}{20}\) = 1
or AB = 20 m ………………..(1)
Also, in right angled ∆ABD,
\(\frac{A B}{B C}\) = cot 60°

or \(\frac{\mathrm{AB}}{20+h}=\frac{1}{\sqrt{3}}\)

AB = \(\frac{(20+h)}{\sqrt{3}}\) ………….(2)

From (1) and (2), we get

20 = \(\frac{(20+h)}{\sqrt{3}}\)
or 20√3 = 20 + h
or h = 20√3 – 20
or h = 20 (√3 – 1) m
= 20 (1.732 – 1) m
= 20 × 0.732 = 14.64 m.

Hence, height of the tower is 14.64 m.

Question 8.
A statue 1.6 m tall stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.
Solution. Let BC = h m be the height of Pedestal and CD = 1.6 m be the height of statue.
The angle of elevation of top of statue and top of pedestal are 60° and 45° respectively. Various arrangements are as shown in the figure.

In right angled ∆ABC,
\(\frac{A B}{B C}\) = cot 45°

or \(\frac{A B}{h}\) = 1

or AB = h m ………….(1)

In right angled ∆ABC,
\(\frac{\mathrm{AB}}{\mathrm{BD}}\) = cot 60°

or \(\frac{\mathrm{AB}}{h+1.6}=\frac{1}{\sqrt{3}}\)

or AB = \(\frac{h+1.6}{\sqrt{3}}\) ……….(2)

From (1) and (2), we get
h = \(\frac{h+1.6}{\sqrt{3}}\)
or √3h = h + 1.6
or (√3 – 1) h = 1.6
or (1.732 – 1) h = 16
or (0.732) h = 1.6
or h = \(\frac{1.6}{0.732}\) = 2.1857923
= 2.20 m (approx.)
Hence, height of pedestal is 2.20 m.

Question 9.
The angle of elevation of the top of a building from the foot of the tower is 300 and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the
building.
Solution:
Let BC = 50 m be height of tower and AD = h m be height of building. The angle of elevation of the top of a building from the foot of tower and top of tower from foot of the building are 30° and 60° respectively. Various arrangement are as shown in figure.

In right angled ∆ABC,
\(\frac{A B}{B C}\) = cot 60°

or \(\frac{\mathrm{AB}}{50}=\frac{1}{\sqrt{3}}\)

or AB = \(\frac{50}{\sqrt{3}}\) …………(1)

Also, in right angled ∆DAB,
\(\frac{\mathrm{AB}}{\mathrm{DA}}\) = cot 30°

or \(\frac{A B}{h}\) = √3
or AB = h√3 ……………(2)

From (1) and (2), we get
\(\frac{50}{\sqrt{3}}\) = h√3

or \(\frac{50}{\sqrt{3}} \times \frac{1}{\sqrt{3}}\) = h

or h = \(\frac{50}{3}\) = 16.6666

or h = 16.70 m (approx).
Hence, height of building is 16.70 m.

Question 10.
Two poles of equal heights are tanding opposite each other on either side of he road, which is 80 m wide. From a point
between them on the road the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.
Solution:
Let BC = DE = h m he height of two equal poles and point A be the required position where the angle of elevations of top of two poles are 30° and 60° respectively. Various arrangement are as shown in the figure.

In right angled ∆ADE,

\(\frac{E D}{D A}\) = tan 30°

or \(\frac{h}{x}=\frac{1}{\sqrt{3}}\)

or h = \(\frac{x}{\sqrt{3}}\) ……………(1)

In right angled ∆ABC,

\(\frac{\mathrm{BC}}{\mathrm{AB}}\) = tan 60°

or \(\frac{h}{80-x}\) = √3

or h = (80 – x) √3 …………(2)

From (1) and (2), we get
\(\frac{x}{\sqrt{3}}\) = (80 – x)
or x = (80 – x) √3 × √3
or x = (80 – x) 3
or x = 240 – 3x
or 4x = 240
or x = \(\frac{240}{4}\) = 60
Substitute this value of x in (I), we get
h = \(\frac{60}{\sqrt{3}}=\frac{60^{\circ}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\)

= \(\frac{60 \sqrt{3}}{3}=20 \sqrt{3}\)

= (20 × 1.732) m = 34.64 m
DA = x = 60 m
and AB = 80 – x = (80 – 60) m = 20 m.
Hence, heigth of the poles are 3464 m and the distances of the point from the poles are 20 m and 60 m respectively.

Question 11.
A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From a point 20 m away from this point on the same bank, the angle of elevation of the top of the tower is 30° (see fig.). Find the height of the tower and the width of the canal.

Solution:
Let BC = x m be the width of canal and CD = h m be height of TV tower. The angles of elevation of top of tower at different position are 30° and 60° respectively. Various arrangements are as shown in figure.

In right angled ∆ABC,

\(\frac{\mathrm{AB}}{\mathrm{BC}}\) = tan 60°

or \(\frac{h}{x}\) = √3
or h = √3x …………..(1)

Also, in right angled ∆ABD,
\(\frac{\mathrm{AB}}{\mathrm{BD}}\) = tan 30°

or \(\frac{h}{20+x}=\frac{1}{\sqrt{3}}\)

or h = \(\frac{20+x}{\sqrt{3}}\) ……………….(2)

From (1) and (2), we get

√3x = \(\frac{20+x}{\sqrt{3}}\)
or √3(√3x) = 20 + x
or 3x = 20 + x
or 2x = 20
or x = \(\frac{20}{2}\) = 10

Substitute this value of x in (1), we get
h = 10(√3)
= 10 × 1.732
h = 17.32 m
Hence, height of TV tower is 17.32 m and. width of the canal is 10 m.

Question 12.
From the top of a 7m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.
Solution:
Let BD = hm be the height of cable tower and AE = 7 m be the height of building. The angle of elevation of the top of a cable tower and angle of depression of its foot from top of a building are 60° and 45° respectively.
Various arrangements are as shown in figure.

In right angled ∆ABC,

\(\frac{\mathrm{AB}}{\mathrm{AE}}\) = cot 45°

or \(\frac{\mathrm{AB}}{7}\) = 1

or AB = 7 m. ……………..(1)

Also, in right angled ∆DCE,

\(\) = cot 60°
or \(\frac{\mathrm{EC}}{h-7}=\frac{1}{\sqrt{3}}\)

or EC = \(\frac{h-7}{\sqrt{3}}\) ……………..(2)

But AB = EC ………….(Given)
7 = \(\frac{h-7}{\sqrt{3}}\) [Using (1) and (2)]
or 7√3 = h – 7
h = 7√3 + 7 = 7 (√3 + 1)
or h = 7 (1.732 + 1) = 7(2.732)
or h = 19.124
or h = 19.20 m (approx.)
Hence, height of the tower is 19.20 m.

Question 13.
As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Solution:

Let CD = 75 m be the height of light house and point D be top of light house from w’here angles of depression of two ships are 30° and 45° respectively. Various arrangements are as shown in the figure.

In right angled ∆BCD,
\(\frac{\mathrm{BC}}{\mathrm{CD}}\) = cot 45°

or \(\frac{y}{75}\) = 1
or y = 75 m ……………(1)

Also, in right angled ∆ACD
\(\frac{\mathrm{AC}}{\mathrm{CD}}\) = cot 30°

or \(\frac{x+y}{75}\) = √3
or x + y = 75√3
or x + 75 = 75√3 [using (1)]
or x = 75√3 – 75
= 75 (√3 – 1)
= 75( 1.732 – 1)
= 75 (0.732)
or x = 54.90
Hence, distance between the two ships is 54.90 m.

Question 14.
A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant ¡s 60°. After some time, the angle of elevation reduces to 30° (see fig.). Find the distance travelled by the balloon during the interval.

Solution:
Let ‘AB’ be the position of 1.2 m tall girl, at the point of the angles of elevation of balloon at
different distances are 30° and 60° respectively. Various arrangements are as shwon in th figure.
According to question,
FG = ED = CE – CD
= 88.2 m – 1.2 m
= 87 m

In right angled ∆AGF,
\(\frac{A G}{G F}\) = cot 60°

or \(\frac{x}{87}=\frac{1}{\sqrt{3}}\)

or x = \(\frac{87}{\sqrt{3}}\) m.

Also, in right angled ∆ADE,
\(\frac{A D}{E D}\) = cot 30°

or \(\frac{x+y}{87}\) = √3

or x + y = 87√3
or \(\frac{87}{\sqrt{3}}\) + y = 87√3
or y = 87√3 – \(\frac{87}{\sqrt{3}}\)

or y = 87√3 – \(\frac{1}{\sqrt{3}}\)

or y = 87 \(\frac{3-1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\)

or y = \(\frac{87 \times 2 \times \sqrt{3}}{3}\)

or y = 58√3
or y = 58(1.732) = 100.456
or y = 100.456 m.
Hence, distance travelled by the balloon during the interval is 100.46 m.

Question 15.
A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the further t(me taken by the car to reach the foot of the tower.
Solution:
Let CD = h m. be the tower of height.
Let A be initial position of the car and after six seconds the car be at 13. The angles of depression at A and B are 30° and 60° respectively. Various arrangements are as shown in figure.

Let speed of the car be υ metre per second using formula, Distance = Speed x Time
AB = Distance covered by car in 6 seconds
AB = 6υ metre
Also, time taken by car to reach the tower be ‘n’ seconds.
∴ BC = nυ metre
In right angled ∆ACD.
\(\frac{\mathrm{CD}}{\mathrm{AC}}\) = tan 30°

or \(\frac{h}{6 v+n v}=\frac{1}{\sqrt{3}}\)

or h = \(\frac{6 v+n v}{\sqrt{3}}\) ……………….(1)

Also, in right angled ∆BCD,
\(\frac{C D}{B C}\) = tan 60°

or \(\frac{h}{n v}\) = √3
h = nv (√3) ……….(2)

From (1) and (2), we get
\(\frac{6 v+n v}{\sqrt{3}}\) = nυ(√3)
or 6υ + nυ = nυ(√3)
or 6υ + nυ = 3nυ
or 6υ = 2nυ
or n = \(\frac{6 v}{2 v}\) = 3
Hence, time taken by car to reach the foot of tower is 3 seconds.

Question 16.
The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.
Solution:
Let CD = h m be the height of tower and B ; A be the required points which are at a distance of 4 m and 9 m from the tower respectively. Various arrangements are as shown in the figure.

In right angled ∆BCD
\(\frac{\mathrm{CD}}{\mathrm{BC}}\) = tan θ

or \(\frac{h}{4}\) = tan θ ………….(1)

Also, in right angled ∆ACD,
\(\frac{C D}{A C}\) = tan (90 – θ)

or \(\frac{h}{9}\) = cot θ

Multiplying (1) and (2), we get
\(\frac{h}{4} \times \frac{h}{9}\) = tan θ cot θ

or \(\frac{h^{2}}{36}=\tan \theta \times \frac{1}{\tan \theta}\)

or h² = 36 = (6)²
or h = 6
Hence, height of the tower is 6 m.

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

PSEB 8th Class English Grammar Tenses

Tense शब्द लातीनी भाषा के शब्द tempus से बना है जिसका अर्थ है समय (Time) । अर्थात् Tense हमें किसी कार्य या घटना के समय या काल का ज्ञान कराता है। नीचे दिए गए वाक्य पढ़ें:

1. I play some game daily.
2. I played cricket yesterday.
3. I shall play hockey tomorrow.

पहले वाक्य के verb (play) से Present Time का बोध होता है। दूसरे वाक्य के verb (played) से Past. Time का बोध होता है। तीसरे वाक्य के verb (shall play) से Future Time का बोध होता है। इस तरह-
(1) जिस verb से Present Time का बोध होता है, वह Present Tense कहलाता है-
I take bath in the morning.
I go for a walk in the evening.

(2) जिस verb से Past Time का बोध होता है, वह Past Tense कहलाता है
I took bath in the,morning.
I went for a walk in the evening.

(3) जिस verb से Future Time का बोध होता है, वह Future Tense कहलाता है
I shall take bath in the morning.
I shall go for a walk in the evening.
इस प्रकार तीन मुख्य Tenses हैं-

1. The Present Tense
2. The Past Tense
3. The Future Tense.

अब यह जान लेना भी आवश्यक है कि प्रत्येक काल में कार्य अथवा क्रिया (verb) की स्थिति भिन्न-भिन्न होती है। हो सकता है कि काम चल रहा हो। यह भी सम्भव है कि कार्य पूरा हो चुका हो अथवा किसी अनिश्चित (indefinite) स्थिति में हो। इस तरह कार्य की मुख्य रूप से चार अवस्थाएं होती हैं। हम यूं भी कह सकते हैं कि प्रत्येक मुख्य Tense के चार रूप होते हैं और कुल मिला कर 12 Tenses होते हैं। यहां हम इन सभी Tenses का अलग-अलग अध्ययन करेंगे।

1. Simple Present Tense
OR
Present Indefinite Tense
Present Indefinite or Simple Present Tense का प्रयोग होता है-
(i) किसी आदत का वर्णन करने के लिए; जैसे,
He takes bath daily.
He goes for a walk everyday.

(ii) किसी सर्वमान्य सत्य को व्यक्त करने के लिए; जैसे,
The sun rises in the east.

(iii) किसी कहानी में किसी बीती हुई घटना को बताने में; जैसे,
The brave dog now kills the snake and waits for his master to come.

(iv) भविष्य में होने वाली किसी घटना को व्यक्त करने के लिए जो किसी योजना अथवा व्यवस्था का भाग हो; जैसे,
Our examination begins on Monday.

(v) Time तथा Condition की Clauses में साधारण Future Tense के स्थान पर; जैसे,
If it rains, we shall not go out for a walk.

(vi) कथनों को व्यक्त करने के लिए ; जैसे,
They say, “To err is human.

(vii) किसी ऐसी स्थिति को दर्शाने के लिए जो न बदलने वाली हो; जैसे,
Our house faces the east.

2. Present Continuous Tense
Present Continuous Tense का प्रयोग होता है-
(i) किसी ऐसे विशेष कार्य को प्रकट करने के लिए जो अभी पूरा न हुआ हो अथवा जारी हो; जैसे,
My brother is singing a song.

3. Present Perfect Tense
Present Perfect Tense का प्रयोग किया जाता है-
(i) ऐसे कार्य को प्रकट करने के लिए जो भूतकाल से अब तक जारी हो; जैसे,
I have never tasted tea. (I still do not drink it.)

(ii) किसी ऐसे पूर्ण कार्य अथवा घटना को प्रकट करने के लिए जो भूतकाल में विशेष समय को लेकर वर्तमान काल तक किया जाए; जैसे,
There have been two accidents on the road during 1984.

4. Present Perfect Continuous Tense.
यह Tense उस कार्य को व्यक्त करने के लिए प्रयोग होता है जो अतीत में किसी समय आरम्भ हुआ हो और अब भी चल रहा हो। समय को व्यक्त करने के लिए for (अनिश्चित समय) और since (निश्चित समय) का प्रयोग करते हैं; जैसे-
The match has been going on for an hour.
The man has been waiting for a reply.
You have been wasting your time since morning.

5. Past Indefinite Tense
OR
Simple Past Tense
Past Indefinite Tense का प्रयोग किया जाता है-
(i) भूतकाल की किसी आदत, लोकप्रिय कार्य अथवा सर्वमान्य तथ्य को प्रकट करने के लिए; जैसे,
People then believed that the sun moved round the earth.

(ii) ऐसे कार्य को व्यक्त करने के लिए जिस में भूतकाल में काफी समय लगा हो परन्तु जो अब समाप्त हो चुका हो; जैसे-
He lived in Delhi for ten years. (but he does not live there now)

(iii) भूतकाल में पूरे किए गए किसी कार्य को व्यक्त करने के लिए। इस प्रकार प्रायः बीते हुए समय को व्यक्त . करने के लिए किसी Adverb या Adverb phrase का प्रयोग किया जाता है: जैसे,
My father left for Delhi yesterday.

Note 1. कभी-कभी समय की अभिव्यक्ति Adverb की बजाये भाव (समय का) से भी हो सकती है; जैसे
I bought this watch in Mumbai.

Note 2. क्रमबद्ध घटनाओं में भी Time के Adverb की आवश्यकता नहीं पड़ती; जैसे,
He came. He saw. He conquered.

(iv) किसी प्रश्न का उत्तर देने में; जैसे,
How did he go to school ?
Answer:
He went on foot.

6. Past Continuous Tense
Past Continuous (Progressive) Tense का प्रयोग किया जाता है-
(i) किसी ऐसे कार्य को व्यक्त करने के लिए जो भूतकाल में किसी विशेष समय पर किया जा रहा हो, भले। ही कार्य करने का समय बताया गया हो, या न बताया गया हो; जैसे,-
At 7 a.m. this morning I was reading the newspaper.

7. Past Perfect Tense
Past Perfect Tense का प्रयोग किया जाता है-
(1) भूतकाल में एक साथ घटित होने वाले कार्यों में से पहले पूरा होने वाले कार्य के लिए; जैसे,
The patient had died before the doctor came.

8. Past Perfect Continuous Tense इस Tense का प्रयोग किसी ऐसे कार्य या घटना का वर्णन करने के लिए किया जाता है जो अतीत में किसी एक निश्चित समय (Point of Time) तक या किसी अवधि (Period of Time) में जारी रहे; जैसे,-
The phone had been ringing for a minute before Ram lifted it. वाक्य में स्पष्ट है कि जब राम ने फोन उठाया तब उससे पहले एक मिनट तक फोन की घण्टी बजती रही थी।

9. Future Indefinite Tense Simple Future Tense का प्रयोग ऐसे कार्य के लिए किया जाता है जो भविष्य में अभी किया जाना है; जैसे,
I shall finish the work tomorrow.
Tomorrow will be Monday.

10. Future Continuous Tense:
Future Continuous Tense का प्रयोग उस कार्य को करने के लिए किया जाता है जो भविष्य में किसी समय चल रहा हो; जैसे,
I shall be writing the letters then.
When I reach the station, the train will be moving.

11. Future Perfect Tense

• They will have heard the news by the time you reach.
• The teacher will have taken the roll-call before you enter the class.

12. Future Perfect Continuous Tense Future Perfect Continuous Tense का प्रयोग ऐसे कार्य का उल्लेख करने के लिए होता है जो भविष्य में किसी निश्चित समय के बाद भी जारी रहने का बोध कराता है; जैसे-

• The students will have been studying History since morning.
• He will have been studying Law for two years by next April.

Exercises From Board’s Grammar (Solved)

I. Fill in the blanks with the Simple Present or Present Continuous forms of the verbs given in the brackets:

1. The population of India ……………. very fast. (increase)
2. Water ……………. at 0° Celsius. (freeze)
3. The sun ……………. in the West. (not rise)
4. ……………. you ……………. Mr. Jain? (know)
5. ……………. he ever ……………… cricket? (play)
6. The Ganges …………… into the Bay of Bengal. (flow)
7. Why …………….. you ……………… this ? (eat)
8. She …………….. a bath. (have)
9. I …………… cricket everyday, but today I …………… tennis.. (play)
10. She usually ……………. a skirt but today she ……………… trousers. (wear)
Hints:
1. is increasing
2. freezes
3. does not rise
4. Do, know
5. Does, play
6. flows
7. are, eating
8. is having
9. play, am playing
10. wears, is wearing.

II. Fill in the blanks with the Present Perfect or Present Perfect Continuous forms of the verbs given in brackets:

1. Someone ……………. the window. (break)
2. Rita …………….. her pen. (lose)
3. The train ……………… just ………….. at the platform. (arrive)
4. We …………….. many medals. (win)
5. I ………. for a house for two months. (search)
6. …………. he ……………. a beard ? (grow)
7. ……………. you ……………. the Bible ? (read)
8. ……….. my uncle for months. (not visit)
9. She ……………… to China twice. (be)
10. We ……………. already ……………… our breakfast. (have)
Hints:
1. has broken
2. has lost
3. has, arrived
4. have won
5. have been searching
6. Has, grown
7. Have, read
8. have not visited
9. has been
10. have, had.

III. Fill in the blanks with the Simple Past Tense forms of the verbs given in the brackets:

Sher Singh smiled. He tossed his revolver in the air and ………… (catch) it by the handle. He ………… (take) careful aim at an empty sardine tin and …….3….. (fire) another six shots. The bullets ……4….. (go) through into the earth kicking up whiffs of dust. His Alsatian dog …….5….. (begin) to bark with excitement. He ………… (leap) up with a growl and ………… (run) down the canal embankment. He …….8….. (sniff) at the tin and …….9….. (take) it up in his mouth and …….10….. (run) back with it and ……. 11…. (lay) it at his master’s feet.
Hints:
1. caught
2. took
3. fired
4. went
5. began
6. leapt
7. ran
8. sniffed
9. took
10. ran
11. laid.

IV. Fill in the blanks with the Simple Past or Past Perfect forms of the verbs given in the brackets:

1. The plane ……………. When we reached the airport. (leave)
2. Ramesh …………… home when I phoned him. (return)
3. …………… he ……………. his old car before he bought a new one ? (sell)
4. The children ……………. before I came home. (sleep)
5. The film had already begun when we …………….. the theatre. (reach)
6. The teacher ……………. the book before the examination began. (finish)
7. The robber had run away before the police. (come)
8. Tom …………. sleepy after having a good lunch. (feel)
9. I ……………. the message before you came. (receive)
10. He ……………… for India last year. (play)
Hints:
1. had left
2. had returned
3. Had, sold
4. had slept
5. reached
6. had finished
7. came
8. felt
9. had received
10. played.

V. Correct the following sentences:

1. The rain has stopped yesterday.
2. He had been born in 1950.
3. He is suffering from fever since last night.
4. Stephenson has invented the steam engine.
5. He will reach home before the storm will come.
6. I left Bihar before the earthquake occurred.
7. She will reach the station before the train will go.
8. The great reformer had died in 1977.
9. I waited at home for her since 9 o’clock.
10. She finished her dinner when I saw her.
Hints:
1. The rain stopped yesterday.
2. He was born in 1950.
3. He has been suffering from fever since last night.
4. Stephenson invented the steam engine.
5. He will have reached home before the storm comes.
6. I had left Bihar before the earthquake occurred.
7. She will have reached the station before the train goes.
8. The great reformer died in 1977.
9. I had been waiting at home for her since 9 o’clock.
10. She had finished her dinner when I saw her.

Errors in the Use of Tenses

The Simple Past is often used wrongly for the Present Perfect Tense; as,

Incorrect : He did not write the letter yet.
Correct : He has not written the letter yet.
Incorrect : We did not hear from him for a week.
Correct : We have not heard from him for a week.
Incorrect : I lived in Ambala since 1990.
Correct : I have lived in Ambala since 1990.

The Present Perfect is often used wrongly for the Simple Past; as,

Incorrect – Columbus has discovered America.
Correct – Columbus discovered America.
Incorrect – Babar has won the First Battle of Panipat.
Correct – Babar won the First Battle of Panipat.
Incorrect – The servant has not answered when called.
Correct – The servant did not answer when called.

The Present Perfect Tense में Past Time को व्यक्त करने वाला adverb या कोई अन्य शब्द प्रयोग नहीं किया जा सकता जैसे-

Incorrect – I have made a call to him yesterday.
Correct – I made a call to him yesterday.
Incorrect – A new bookshop has been opened last Monday.
Correct – A new bookshop was opened last Monday.
Incorrect – I have finished my work last night.
Correct – I finished my work last night.

The Past Perfect is often used wrongly for the Simple Past; as,

Incorrect – I had visited her yesterday.
Correct – I visited her yesterday.
Incorrect – He had gone to Kolkata last year.
Correct – He went to Kolkata last year.
Incorrect – We had gone for a picnic last Sunday.
Correct – We went for a picnic last Sunday.
Incorrect – Nehru had died in 1964.
Correct – Nehru died in 1964.

The Simple Past is often used wrongly for the Past Perfect; as,

Incorrect – The patient died before the doctor came.
Correct – The patient had died before the doctor came.
Incorrect – The train left before we bought the tickets.
Correct – The train had left before we bought the tickets.
Incorrect – I finished my work before my father came.
Correct – I had finished my work before my father came.

The Past Perfect or Perfect Continuous, and not the Simple Past or Past Continuous, is used to express something that continued up to a past time after beginning at a still earlier time; as,

Incorrect – He told me that he was ill for four days.
Correct – He told me that he had been ill for four days.
Incorrect – She was writing a novel for six weeks when I visited her.
Correct – She had been writing a novel for six weeks when I visited her.

The Simple Future is often used wrongly for the Future Perfect; as;

Incorrect – We shall reach home before the sun will set.
Correct – We shall have reached home before the sun sets.
Incorrect – I shall leave for Ludhiana by the time he will come.
Correct – I shall have left for Ludhiana by the time he comes.

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

PSEB 8th Class English Notice Writing

नोटिस (Notice) किसी घटित होने वाली या घटित हो चुकी घटना के बारे में सूचना होती है। इसके द्वारा किसी आदेश, प्रार्थना अथवा चेतावनी पर अमल करने की सूचना दी जाती है।

याद रखने योग्य बातें-

1. नोटिस की भाषा उद्देश्यपूरक होनी चाहिए। इसमें निजीपन नहीं होना चाहिए।
2. दी जाने वाली सूचना पूर्ण होनी चाहिए।
3. इसका प्रस्तुतीकरण संक्षिप्त और स्पष्ट होना चाहिए। नोटिस का उद्देश्य भी स्पष्ट हो।
4. जहां तक हो सके नोटिस में “I’ तथा ‘You’ के प्रयोग से बचें।
5. नोटिस लिखे जाने की तिथि का उल्लेख अवश्य करें।
6. विषय-वस्तु से सम्बन्धित स्थान, समय तथा कार्यक्रम आदि की स्पष्ट जानकारी दें।
7. नोटिस जारी करने वाले व्यक्ति के पद का नाम तथा उसके हस्ताक्षर भी नोटिस में होने चाहिएं।
8. नोटिस दी गई शब्द-सीमा में ही लिखना चाहिए।

Format of a Notice

Notice

Date

Heading / Title
Content

Signatory

Important Notices

1. Notice About Something Found
You have found a purse lying in one of the lawns of your school. Write a notice asking the owner of the purse to contact you.

NOTICE

March 7, 20…..
FOUND ! FOUND ! FOUND!

This is to inform all the students that a purse has been found lying in one of the i school lawns. It is a black leather purse containing some money. The owner should contact the undersigned.

Gulshan Rai
Roll No. 10, VIII A

2. Notice About a Tour
Your school is organising a tour to Delhi and Agra. You are the secretary of Tour Organising Committee. Draft a notice asking the students to give you their names.

NOTICE
JOINING A HISTORICAL TOUR

March 8, 20……

Our school is organising a historical tour to Delhi and Agra. The duration of the tour is six days starting on March 12. Those interested should give their names to the undersigned latest by March 10.

Mohan Lai
(Secretary Tour Organising Committee)

3. Notice About Paper Reading Contest
You are the incharge of Junior Humanities Forum of your school. The Forum is organising a Paper Reading Contest. Draft a notice inviting the participants to give you their names.

NOTICE
PAPER READING CONTEST

March 9, 20….

The Junior Humanities Forum of the school is organising a Paper Reading Contest on March 16 in the school hall. Students of class VI to VIII are eligible to participate. Those interested should give their names to the undersigned before March 12.

Raman
Secretary
(Junior Humanities Forum)

4. Notice for a School Function
You are Shashi Mehta, the Sports Secretary of your school. Your school is organising the Annual Sports meet next week. Write a notice in about 50 words to be put on the school notice board to this effect.

NOTICE
ANNUAL SPORTS MEET

January 15, 20….

The Annual Sports meet of the school is going to be held on 28th and 29th of this month. Students who wish to take part in any event should give their names to the undersigned by 24th of this month positively.

Shashi Mehta
(Sports Secretary)

5. Notice for a Lost Wrist Watch
You have lost a wrist watch in your school. Write a notice about the loss giving the particulars of the watch. Also announce a reward for the finder.

NOTICE
LOST ! LOST ! LOST !

April 10, 20….

This is to inform all students that I lost my wrist watch yesterday in the school during the recess period. It is an H.M.T. watch with a golden case and a golden chain. He/She who happens to find it should contact the undersigned immediately. He (She) will be suitably rewarded.

Aastha Jindal.
R. No. 2, VIII D

6. Notice about a Lost Pen
You have lost your pen somewhere in your school. Write a notice about it.

NOTICE
LOST ! LOST ! LOST !

12 March, 20 …….
A new gel pen has been lost somewhere in the school ground. The pen is of Reynold make with blue colour. The finder is requested to return it to the undersigned or deposit it with the school office.

Kulbir Singh
Roll No. 2
VIII B

7. Books for Sale
Ashok Mathur of Class VIII A has just passed his annual examination. Two of his books are in fairly good condition and he wants to sell them at reduced prices. He puts up a notice on the school notice-board giving all the necessary details. Write this notice in the space below, using not more than 50 words.

NOTICE
BOOKS FOR SALE

March 22, 20

Two VIII Class books fairly in good condition are for sale. The books and their reduced prices are given below :
1. History of India Rs. 50/- only
2. General Science Rs. 60/- only
Those who are interested should contact Ashok Mathur, Class IX A

8. Taking Part in Debate
K.C. Sharma, Senior English teacher, invites applications from the students of VIII class who want to take part in debate to be held at Ludhiana. March 15 is the last date for the applications to reach the undersigned. In this connection, he puts up a notice on the school notice-board. Write this notice in about 40 words.

NOTICE
TAKING PART IN DEBATE

March 10, 20 ………..

Applications are invited from the students of class IX who want to take part in the debate to be held at Ludhiana. The students should apply before 15th March, giving details of their proficiency.
Sd/
K.C. Sharma
Senior English Teacher

9. Shoes From School Red Cross Fund
Anil Kumar Sharma, the school Red Cross teacher, invites applications from the poor students who want to take shoes from the Red Cross. In this connection, he puts up a notice on, the school notice-board. Write this notice in not more than 30 words.

NOTICE
SHOES FROM SCHOOL RED-CROSS FUND

March 2, 20….

Applications are invited from the poor students who want to take shoes from the Red Cross. All applications should reach the undersigned by 15th March.

Anil Kumar Sharma
Red Cross Teacher

10. Selection to the Volleyball Team
Mohan Singh has been selected the captain of the School Volleyball Team. He is to invite applications from the students of class IX for selection to the Volleyball team. So he puts up a notice on the school notice-board in this connection. Write this notice in about 40 words.

NOTICE
SELECTION TO THE VOLLEYBALL TEAM

Applications are invited from the students of class IX for selection to the Volley-ball team. Apply before 15th September, giving details of your proficiency in the game.

Mohan Singh
Captain Volleyball Team

11. Free Yoga Classes
You are the PTI of your school. Write a notice asking the students to enrol for free yoga classes.

NOTICE
FREE YOGA CLASSES

2 April, 20

Attention !
Students interested in attending free yoga classes from 10th April every morning from 6 a.m. to 7 a.m. should contact the undersigned before 7th April.

B.S. Bedi
PTI

12. Notice for returning Library Books
You are the librarian of your school. Write a notice asking the students to return borrowed books before the school closes down for the summer vacation.

NOTICE

May 10, 20……..

ATTENTION !
The school breaks up for the summer vacation next week. All the students who have borrowed any book from the library must return it before the school breaks up for the vacation.

Raman Kumar
Librarian

13. Notice for a Lost Item
You are Kulbir Singh of Class VIII. You have lost your new water bottle. Write the notice that you would like to put up on the school notice-board.

NOTICE
LOST ! LOST ! LOST !

12 March, 20
A new water botde has been lost somewhere in the school garden. The bottle is of Eagle make with blue colour. The finder is requested to return it to the undersigned or deposit it with the school office.

Kulbir Singh
Roll No. 2
VIII B

14. Notice for Blood Donation
You are the Sarpanch of your village. Write a notice inviting adults to donate blood at the blood donation camp to be held at the community centre.

NOTICE
BLOOD DONATION CAMP

The village Panchayat is going to organise a blood donation camp in the community centre on 8th April from 9 a.m. All the adults of the village should come forward and donate blood to save lives of accident victims.

Balwant Singh
Village Sarpanch

15. Notice for a Misplaced Library Book
You have misplaced a library book ‘Panchtantra Tales’. Write a notice that you would like to put up in the classroom.

NOTICE
BOOK MISPLACED

March 10, 20…. .

The book ‘Panchtantra Tales’ has been misplaced somewhere in the classroom. It was borrowed from the library yesterday. The finder is requested to return it to the undersigned or hand it over to the class teacher.

Rajni
Roll No. 5
VIIIA

16. Notice for Sports Participants
You are the sports captain of your school. Write a notice to all participants to submit their names and event in which they are taking part.

NOTICE
ATTENTION ! SPORTS PARTICIPANTS

March 10, 20….

All the participants who are interested to take part in the school sports are requested to give their names to the undersigned before Sunday, the 6th May. They must mention the event they are taking part in.

Raman Kumar
(Sports Captain)

17. Help for Tsunami Victims
You are the Head Boy/Girl of your school. Draft a notice requesting the students to come forward to help the tsunami victims.

NOTICE
HELPING TSUNAMI VICTIMS

March 18, 20
This is to inform all the students of the school that our school is raising a fund to help the tsunami victims. All should come forward with open heart and contribute to the fund as much as possible. Deposit the money with the respective class teachers.

Sham/Suman
Head Boy/Girl

Punjab State Board PSEB 8th Class English Book Solutions English Role Play Writing Exercise Questions and Answers, Notes.

PSEB 8th Class English Role Play Writing

1. Imagine your father is ill. You went to the hospital to get medicine for him and you got late for school. The headmaster asked you the reason for being late. You will play the role of the student and your friend of the headmaster. The beginning of the conversation is given. You will start with the given conversation.

1. Headmaster : Why are you late today ?
Student : I had gone to the hospital, sir.
2. Headmaster : ………………….
Student : ………………….
3. Headmaster : ………………….
Student : ………………….
4. Headmaster : ………………….
Student : ………………….
5. Headmaster : ………………….
Student : ………………….
6. Headmaster : ………………….
Student : ………………….
Answer:
2. Headmaster : Why did you go to the hospital ?
Student : I had gone to get medicine for my father.
3. Headmaster : How long had you to wait for your turn ?
Student : I had to wait there for one hour for my turn.
4. Headmaster : How is your father now ?
Student : He is improving, sir.
5. Headmaster : Who is looking after him now ?
Student : My mother is looking after him now.
6. Headmaster : How do you come to school ?
Student : I come to school on foot.

2. Imagine you are a football player. Your team won a football match yesterday. Your father wants to know about the match. You will play the role of son and the father. The beginning of the conversation is given.

1. Father : Did you play the match yesterday ?
Son : Yes, I played the match yesterday.
2. Father : ………………….
Son : ………………….
3. Father : ………………….
Son : ………………….
4. Father : ………………….
Son : ………………….
5. Father : ………………….
Son : ………………….
6. Father : ………………….
Son : ………………….
Answer:
2. Father : Where was the match played ?
Son : It was played on our school ground.
3. Father : Which team was stronger ?
Son : Both the teams were equally strong.
4. Father : Which team scored the first goal ?
Son : Our team scored the first goal.
5. Father : How many goals were scored ?
Son : Two goals were scored.
6. Father : Did you score any goal ?
Son : Yes, I scored one goal.

3. Suppose a friend (Ramesh) of yours wants to know about your school library. You will play the role of you and Ramesh. The beginning of the conversation is given. You will start with the given conversation.

1. Ramesh : Has your school a big library ?
You : Yes, our school has a big library.
2. Ramesh : ………………….
You : ………………….
3. Ramesh : ………………….
You : ………………….
4. Ramesh : ………………….
You : ………………….
5. Ramesh : ………………….
You : ………………….
6. Ramesh : ………………….
You : ………………….
7. Ramesh : ………………….
You : ………………….
8. Ramesh : ………………….
You : ………………….
9. Ramesh : ………………….
You : ………………….
10. Ramesh : ………………….
You : ………………….
Answer:
2. Ramesh : How many sections has it ?
You : It has three sections—Punjabi, Hindi and English.
3. Ramesh : Who is its incharge ?
You : Mr. Hari Singh is its incharge.
4. Ramesh : How many almirahs are there in the library ?
You : There are fifteen big steel almirahs in the library.
5. Ramesh : How many books are there in the library ?
You : There are more than six thousand books in the library.
6. Ramesh : Are there novels and story books in it ? ‘
You : Yes, there are many novels and story books in it.
7. Ramesh : Has your library any books on general knowledge ?
You : Yes, our library has general knowledge books too.
8. Ramesh : Are there chairs and tables in the library ?
You : Yes, there are chairs and tables in our library.
9. Ramesh : How often does your class go to the library ?
You : Our class goes twice a week to the library.
10. Ramesh : How many books can you take out at a time ?
You : We can take out only one book at a time.

4. You gave some clothes for dry cleaning. Your suit was spoiled. You went to the dry cleaner and made the complaint. Now you will play the role of yourself and dry cleaner. The beginning of the conversation is given. You will start with the given conversation.

1. Dry Cleaner : Do you want to get your suit dry cleaned ?
You : No, I have come to complain about my suit.
2. Dry Cleaner : ………………….
You : ………………….
3. Dry Cleaner : ………………….
You : ………………….
4. Dry Cleaner : ………………….
You : ………………….
5. Dry Cleaner : ………………….
You : ………………….
6. Dry Cleaner : ………………….
You : ………………….
7. Dry Cleaner : ………………….
You : ………………….
8. Dry Cleaner : ………………….
You : ………………….
9. Dry Cleaner : ………………….
You : ………………….
10. Dry Cleaner : ………………….
You : ………………….
Answer:
2. Dry Cleaner : Whom did you give your suit for dry cleaning ?
You : I gave my suit to your servant for dry cleaning.
3. Dry Cleaner : What has gone wrong with your suit ?
You : It has been completely spoiled.
4. Dry Cleaner : When did you buy it ?
You : I bought it last month.
5. Dry Cleaner : How much did you spend on it ?
You : I spent one thousand rupees on it.
6. Dry Cleaner : What can I do to make up your loss ?
You : You should pay me seven hundred rupees.
7. Dry Cleaner : Who is responsible for this damage ?
You : You alone are responsible for this damage.
8. Dry Cleaner : Why should I pay you for this damage ?
You : ‘Because you are the owner of this shop.
9. Dry Cleaner : Can you wait for a few days ?
You : No, I want it immediately.
10. Dry Cleaner : What will you do if I do not pay the money ?
You : If you do not pay the money, I will go to the police station.

5. Imagine you have returned from Delhi after visiting your uncle. Your father asks you a few questions in connection with your visit. Now you will play the role of you and your father. Start you conversation as:

1. Father : Why did you not come back yesterday ?
You : Uncle did not allow me to come yesterday.
2. Father : ………………….
You : ………………….
3. Father : ………………….
You : ………………….
4. Father : ………………….
You : ………………….
5. Father : ………………….
You : ………………….
6. Father : ………………….
You : ………………….
7. Father : ………………….
You : ………………….
Answer:
2. Father : Why did he not allow you ?
You : He is not keeping good health.
3. Father : What is wrong with him ?
You : He is having high blood pressure.
4. Father : Did you request him to visit us ?
You : Yes, I requested him to visit us some time.
5. Father : Did he promise to come to Chandigarh ?
You : Yes, he promised to come to Chandigarh in June.
6. Father : Did you visit any important place in Delhi ?
You : Yes, I visited many places such as Qutab Minar, Red Fort, Appu Ghar, etc.
7. Father : Did you enjoy your visit ?
You : Yes, I enjoyed it very much.

6. Suppose you learn that your friend (Arun) has been involved in an accident and is in the hospital. You go to the hospital and meet his father. Now play the role of yourself and his father. The beginning of the conversation is given. You will start with the given conversation.

1. You : Where is Arun, my friend ?
His Father : He is in the operation theatre.
2. You : ………………….
His Father : ………………….
3. You : ………………….
His Father : ………………….
4. You : ………………….
His Father : ………………….
5. You : ………………….
His Father : ………………….
6. You : ………………….
His Father : ………………….
7. You : ………………….
His Father : ………………….
Answer:
You : What has happened to him ?
His Father : He has a fracture in his left arm.
You : How did all this happen ?
His Father : His scooter was hit by a bus.
You : Was he alone ?
His Father : No, his sister was with him.
You : Is the girl safe ?
His Father : Thank God, she did not receive any injury.
You : Where did the accident occur ?
His Father : It occurred near the Rose Garden.
You : Do you need any kind of help from me ?
His Father : No, thank you. You should only pray for his early recovery.

7. Suppose somebody has stolen your hundred rupees. You go to the Principal of your school to complain. You will play the role of student and the Principal. The beginning of the conversation is given. You will start with the given conversation.

1. Principal : What brings you here ?
Student : Sir, I have come with a problem.
2. Principal : ………………….
Student : ………………….
3. Principal : ………………….
Student : ………………….
4. Principal : ………………….
Student : ………………….
5. Principal : ………………….
Yudent : ………………….
6. Principal : ………………….
Student : ………………….
7. Principal : ………………….
Student : ………………….
Answer:
2. Principal : What is your problem ?
Student : Sir, somebody has stolen my hundred rupees.
3. Principal : Where had you kept it ?
Student : I had kept it in my bag in a book, sir.
4. Principal : Where had you put your bag ?
Student : It was lying in the classroom.
5. Principal : Where had the boys of your class gone ?
Student : All the boys had gone to the ground for P.T.
6. Principal : Did you ask anybody about it ?
Student : Yes sir, I asked a number of my class-fellows.
7. Principal : Have you notified it on the notice-board ?
Student : Not yet sir, but I will put it lip on the notice-board.

8. Suppose you visit your cousin and ask him about his hobby. You will play the role of yourself and cousin. The beginning of the conversation is given. You will start with the given conversation.

1. You : What is your hobby ?
Cousin : Gardening is my hobby.
2. You : ………………….
Cousin : ………………….
3. You : ………………….
Cousin : ………………….
4. You : ………………….
Cousin : ………………….
5. You : ………………….
Cousin : ………………….
6. You : ………………….
Cousin : ………………….
7. You : ………………….
Cousin : ………………….
8. You : ………………….
Cousin : ………………….
9. You : ………………….
Cousin : ………………….
10. You : ………………….
Cousin : ………………….
Answer:
2. You : When do you work in your garden ?
Cousin : I work in my garden after the school time.
3. You : Do you apply manure to the plants ?
Cousin : Yes, I apply manure to the plants.
4. You : What have you grown there ?
Cousin : I have grown roses, pansy and petunia flowers there.
3. You : Do you have any fruit trees there ?
Cousin : Yes, I have some fruit trees there.
6. You : When do you cut the grass of the lawn ?
Cousin : I cut the grass of the lawn on Saturday.
7. You : Where do you bring the plants from ?
Cousin : I bring the plants from Green Nursery.
8. You : Do you have any vine there ?
Cousin : Yes, I have a vine of grapes there.
9. You : When do you take part in a Flower Show ?
Cousin : I take part in a Flower Show in the month of March.
10. You : When can I come to see your garden ?
Cousin : You can come to see my garden on any holiday.

9. Suppose you went to Rose Garden for a picnic. Your younger brother who could not go with you asks you about it. Two boys shall play the role of elder brother and younger brother.The beginning of the conversation is given. You will start with the given conversation.

1. Younger Brother : Where did you go for a picnic ?
Elder Brother : We went to the Rose Garden.
2. Younger Brother : ………………….
Elder Brother : ………………….
3. Younger Brother : ………………….
Elder Brother : ………………….
4. Younger Brother : ………………….
Elder Brother : ………………….
5. Younger Brother : …………………..
Elder Brother : ………………….
6. Younger Brother : ………………….
Elder Brother : ………………….
7. Younger Brother : ………………….
Elder Brother : ………………….
8. Younger Brother : ………………….
Elder Brother : ………………….
9. Younger Brother : ………………….
Elder Brother : ………………….
10. Younger Brother : ………………….
Elder Brother : ………………….
Answer:
2. Younger Brother : How many friends were you ?
Elder Brother : We were a party of ten friends.
3. Younger Brother : How did you go there ?
Elder Brother : We went there on bicycles.
4. Younger Brother : What did you take with you ?
Elder Brother : We took with us a camera, a transistor, a pack of cards, a stove, some milk, tea and sugar.
5. Younger Brother : YHow much time did you take to reach Rose Garden ?
Elder Brother : It took us an hour to reach Rose Garden.
6. Younger Brother : Where did you see children playing merrily ?
Elder Brother : We saw children playing merrily in the garden.
7. Younger Brother : How did you feel there ?
Elder Brother : We felt very happy there.
8. Younger Brother : Where did you sit ?
Elder Brother : We sat on a mat under a shady tree.
9. Younger Brother : What did you do there ?
Elder Brother : We played cards, took snaps, listened to songs and took tea there.
10. Younger Brother : Did you enjoy yourselves ?
Elder Brother : Yes, we really enjoyed ourselves.

10. Suppose Ram has just come out of the Examination Hall after finishing his paper. Another examinee named Sham has also come out. You will play the role of Ram and Sham. Now start your conversation.

1. Ram : How have you done your paper ?
Sham : I have done my paper well.
2. Ram : ………………….
Sham : ………………….
3. Ram : ………………….
Sham : ………………….
4. Ram : ………………….
Sham : ………………….
3. Ram : ………………….
Sham : ………………….
6. Ram : ………………….
Sham : ………………….
7. Ram : ………………….
Sham : ………………….
8. Ram : ………………….
Sham : ………………….
9. Ram : ………………….
Sham : ………………….
10. Ram : ………………….
Sham : ………………….
Answer:
2. Ram : Is the paper easy ?
Sham : It is not so easy.
3. Ram : Which question is difficult ? ‘
Sham : Question No. IV is rather difficult.
4. Ram : What do you think about the answer to this question ?
Sham : I think my answer to this question is also to the point.
5. Ram : From which chapter is the question set ?
Sham : The question is set from Chapter II of the book.
6. Ram : How do you find the question on ‘Fill in the blanks’ ?
Sham : The question on ‘Fill in the blanks’ is confusing.
7. Ram : What do you think about these two questions ?
Sham : These two questions are out of the course, prescribed.
8. Ram : How did you answer these questions ?
Sham : I gave rather poor answers to these questions.
9. Ram : How do you feel on the whole ?
Sham : On the whole I am satisfied with the answers I gave.
10. Ram : How many marks do you hope to get in this paper ?
Sham : I hope to get only 40 per cent marks in this paper.

11. Your friend bought a mobile phone. The mobile phone is not working properly. Now play the role of friend and shop owner. Write the conversation in a dialogue format.

1. Shop owner : Good morning, Sir.
Friend : Good’morning Sir.
2. Shop owner : ………………….
Friend : ………………….
3. Shop owner : ………………….
Friend : ………………….
4. Shop owner : ………………….
Friend : ………………….
5. Shop owner : ………………….
Friend : ………………….
Answer:
2. Shop owner : How can I help you ?
Friend : Sir, the mobile phone I bought yesterday from your shop is not working correctly.
3. Shop owner : Oh ! Please explain your problem in detail.
Friend : It is giving a lot of trouble. There is no ring tones on it. The sound/music/ audio quality is not good either.
4. Shop owner-: OK. Let me check it. Oh ! you are right. Tell me, what to do ?
Friend : It is of no use to me.
5. Shop owner : I understand.
Friend : Please take it back and refund my money.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 2 Linear Equations in One Variable Ex 2.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.6

Solve the following equations.

Question 1.
\(\frac{8 x-3}{3 x}=2\)
Solution:
\(\frac{8 x-3}{3 x}=2\)
∴ 3x\(\left(\frac{8 x-3}{3 x}\right)\) = 3x (2) (Multiplying both the sides by 3x)
∴ 8x – 3 = 6x
∴ 8x – 6x = 3 [Transposing 6x to LHS and (-3) to RHS]
∴ 2x = 3
∴ \(\frac{2 x}{2}=\frac{3}{2}\) (Dividing both the sides by 2)
∴ x = \(\frac {3}{2}\)

Question 2.
\(\frac{9 x}{7-6 x}=15\)
Solution:
\(\frac{9 x}{7-6 x}=15\)
∴ 9x = 15 (7 – 6x) (Cross multiplication)
∴ 9x = 105 – 90x
∴ 9x + 90x = 105 [Transposing (- 90x) to LHS]
∴ 99x = 105
∴ \(\frac{99 x}{99}=\frac{105}{99}\)(Dividing both the sides by 99)
∴ x = \(\frac {35}{33}\)

Question 3.
\(\frac{z}{z+15}=\frac{4}{9}\)
Solution:
\(\frac{z}{z+15}=\frac{4}{9}\)
∴ z(9) = 4(z + 15) (Cross multiplication)
∴ 9z = 4z + 60
∴ 9z – 4z = 60 (Transposing 4z to LHS)
∴ 5z = 60
∴ \(\frac{5 z}{5}=\frac{60}{5}\) (Dividing both the sides by 5)
∴ z = 12

Question 4.
\(\frac{3 y+4}{2-6 y}=\frac{-2}{5}\)
Solution:
\(\frac{3 y+4}{2-6 y}=\frac{-2}{5}\)
∴ 5(3y + 4) = -2(2 – 6y) (Cross multiplication)
∴ 15y + 20 = -4 + 12y
∴ 15y – 12y = – 4 – 20 (Transposing 12y to LHS and 20 to RHS)
∴ 3y = -24
∴ \(\frac{3 y}{3}=\frac{-24}{3}\) (Dividing both the sides by 3)
∴ y = (-8)

Question 5.
\(\frac{7 y+4}{y+2}=\frac{-4}{3}\)
Solution:
\(\frac{7 y+4}{y+2}=\frac{-4}{3}\)
∴ 3(7y + 4) = -4(y + 2) (Cross multiplication)
∴ 21y + 12 = – 4y – 8
∴ 21y + 4y = – 8 – 12 (Transposing -4y to LHS and 12 to RHS)
∴ 25y = -20
∴ \(\frac{25 y}{25}=\frac{-20}{25}\) (Dividing both the sides by 25)
∴ y = \(\frac {-4}{5}\)

Question 6.
The ages of Hari and Harry are in the ratio 5:7. Four years from now the ratio of their ages will be 3:4. Find their present ages.
Solution:
Age of Hari : Age of Harry
= 5 : 7
Let the present age of Hari be 5x years.
Then, the present age of Harry = 7x years.
After 4 years their ages :
Hari = (5x + 4) years
Harry = (7x + 4) years
∴ (5x + 4) : (7x + 4) = 3 : 4
∴ \(\frac{5 x+4}{7 x+4}=\frac{3}{4}\)
∴ 4(5x + 4) = 3(7x + 4) (Cross multiplication)
∴ 20x + 16 = 21x + 12
∴ 20x – 21x = 12 – 16 (Transposing 21x to LHS and 16 to RHS)
∴ -x = – 4
∴ x = 4 [Multiplying both the sides by (- 1)]
∴ Hari’s present age = 5x = 5 × 4
= 20 years
∴ Harry’s present age = 7x = 7 × 4
= 28 years
Thus, Hari’s present age is 20 years and Harry’s present age is 28 years.

Question 7.
The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is \(\frac {3}{2}\). Find the rational number.
Solution:
Let the numerator be x.
Denominator (8 more than numerator) = x + 8
New numerator = x + 17
(After adding 17)
New denominator = x + 8 – 1
= x + 7
(After decreasing 1)
But new number = \(\frac{x+17}{x+7}\)
But this rational number is \(\frac {3}{2}\)
\(\frac{x+17}{x+7}=\frac{3}{2}\)
∴ 2(x + 17) = 3(x + 7) (Cross multiplication)
∴ 2x + 34 = 3x + 21
∴ 2x – 3x = 21 – 34 (Transposing 3x to LHS and 34 to RHS)
∴ -x = – 13
∴ x = 13 [Multiplying both the sides by (-1)]
∴ Numerator = x = 13
Denominator = x + 8
= 13 + 8
= 21
The rational number = \(\frac {13}{21}\)
Thus, the rational number is \(\frac {13}{21}\).

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 13 Symmetry Ex 13.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 13 Symmetry Ex 13.2

1. Draw the reflection of following figures along the dotted line :

Solution:

2. Write ‘yes’ for right reflection and ‘no’ for wrong reflection:

Solution:
(a) Yes
(b) Yes
(c) Yes
(d) Yes.

3. Trace the figures on the graph paper and draw the reflections. The dotted line is the line of symmetry:

Solution:

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 13 Symmetry Ex 13.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 13 Symmetry Ex 13.1

1. Classify the figure as symmetrical or non-symmetrical. Also draw the line/ lines of symmetry (if any).

Solution:

2. Which Capital letter of English alphabet have:

Question (i)
No line of symmetry.
Solution:
Capital letters of English alphabet having no line of symmetry.
F, G, J, L, N, P, Q, R, S, Z.

Question (ii)
1 line of symmetry.
Solution:
Capital letters of English alphabet having 1 line of symmetry.
A, B, C, D, E, K, M, T, U, V, W, Y.

Question (iii)
2 lines of symmetry.
Solution:
2 lines of symmetry.
O, X, H, I.

3. Find file numbers of line/lines of symmetry for the following:

Solution:

4. Draw the line (s) of symmetry the following figures:

Question (a)
Rhombus
Solution:
Rhombus: A rhombus has two lines of symmetry.

Question (b)
Scalene Triangle
Solution:
Scalene Triangle: A scalene triangle has no line of symmetry.

Question (c)
Parallelogram
Solution:
Parallelogram: A parallelogram has no line of symmetry.

Question (d)
Rectangle
Solution:
Rectangle: A rectangle has two lines of symmetry.

Question (e)
Square
Solution:
Square: A square has four lines of symmetry.

Question (f)
Regular Pentagon.
Solution:
Regular Pentagon. A regular pentagon has five lines of symmetry.

5. Complete each of the figure using both lines of symmetry:

Solution:

6. Draw a triangle which has:

Question (i)
No line of symmetry
Solution:
No line of symmetry: Scalene triangle has no line of symmetry.

Question (ii)
Exactly one line of symmetry
Solution:
Exactly one line of symmetry: Isosceles triangle has exactly one line of symmetry.

Question (iii)
Exactly three lines of symmetry.
Solution:
Exactly three lines of symmetry. Equilateral triangle has exactly three lines of symmetry.

7. List any three symmetrical objects from your day-to-day life.
Solution:
Glass, Lock Pencil are three symmetrical objects.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 2 Linear Equations in One Variable Ex 2.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5

Solve the following linear equations.

Question 1.
\(\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}\)
Solution:
\(\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}\)
∴ \(\frac {1}{2}\) [Transposing \(\frac{x}{3}\) to LHS and \(\frac {-1}{2}\) to RHS]
∴ \(\frac{3 x-2 x}{6}=\frac{1 \times 5+1 \times 4}{20}\) [LCM = 6, LCM = 20]
∴ \(\frac{x}{6}=\frac{5+4}{20}\)
∴ \(\frac{x}{6}=\frac{9}{20}\)
∴ \(\frac{x}{6} \times 6=\frac{9}{20} \times 6\) (Multiplying both the sides by 6)
∴ \(\frac {27}{10}\)
∴ x = 2.7

Question 2.
\(\frac{n}{2}-\frac{3 n}{4}+\frac{5 n}{6}\) = 21
Solution:
\(\frac{n}{2}-\frac{3 n}{4}+\frac{5 n}{6}\) = 21
∴ \(\frac{n \times 6}{2 \times 6}-\frac{3 n \times 3}{4 \times 3}+\frac{5 n \times 2}{6 \times 2}\) = 21
∴ \(\frac{6 n-9 n+10 n}{12}\) = 21
∴ \(\frac {7 n}{12}\) = 21
∴ 7n = 21 × 12 (Multiplying both the sides by 12)
∴ n = \(\frac{21 \times 12}{7}\) (Dividing both the sides by 7)
∴ n = 36

Question 3.
\(x+7-\frac{8 x}{3}=\frac{17}{6}-\frac{5 x}{2}\)
Solution:
\(x+7-\frac{8 x}{3}=\frac{17}{6}-\frac{5 x}{2}\)
(Multipljdng both the sides by 6, the LCM of 3, 6 and 2.)
(6 × x) + (6 × 7) – \(\left(\frac{6 \times 8 x}{3}\right)\) = \(\left(6 \times \frac{17}{6}\right)-\left(\frac{6 \times 5 x}{2}\right)\)
∴ 6x + 42 – 16x = 17- 15x
∴ – 10x + 42 = 17 – 15x
∴ – 10x + 15x = 17 – 42 [Transposing (-15x) to LHS and 42 to RHS]
∴ 5x = – 25
∴ x = – 5 (Dividing both the sides by 5)

Question 4.
\(\frac{x-5}{3}=\frac{x-3}{5}\)
Solution:
\(\frac{x-5}{3}=\frac{x-3}{5}\)
∴ 5(x – 5) = 3 (x – 3) (Cross multiplication)
∴ 5x – 25 = 3x – 9
∴ 5x – 3x = 25 – 9 [Transposing 3x to LHS and (-25) to RHS]
∴ 2x = 16
∴ \(\frac{2 x}{2}=\frac{16}{2}\) (Dividing both the sides by 2)
∴ x = 8

Question 5.
\(\frac{3 t-2}{4}-\frac{2 t+3}{3}=\frac{2}{3}-t\)
Solution:
\(\frac{3 t-2}{4}-\frac{2 t+3}{3}=\frac{2}{3}-t\)
(Multiplying both the sides by 12, the LCM of 4 and 3)
12\(\left(\frac{3 t-2}{4}\right)\) – 12\(\left(\frac{2 t+3}{3}\right)\) = 12 × \(\frac {1}{2}\) – 12t
∴ 3(3t – 2) -4 (2t + 3) = 8 – 12t
∴ 9t – 6 – 8t – 12 = 8 – 12t
∴ t – 18 = 8 – 12t
∴ t + 12t = 8 + 18 [Transposing (-12t) to LHS and (-18) to RHS]
∴ 13t = 26
∴ \(\frac{13 t}{13}=\frac{26}{13}\) (Dividing both the sides by 13)
∴ t = 2

Question 6.
\(m-\frac{m-1}{2}=1-\frac{m-2}{3}\)
Solution:
\(m-\frac{m-1}{2}=1-\frac{m-2}{3}\)
(Multiplying both the sides by 6, the LCM of 2 and 3)
6m – 6 \(\left(\frac{m-1}{2}\right)\) = 1 × 6 – 6\(\left(\frac{m-2}{3}\right)\)
∴ 6m – 3(m- 1) = 6 – 2(m-2)
∴ 6m – 3m + 3 = 6 – 2m + 4
∴ 3m + 3 = 10 – 2m
∴ 3m + 2m = 10 – 3 [Transposing (-2 m) to LHS and 3 to RHS]
∴ 5m = 7
∴ \(\frac{5 m}{5}=\frac{7}{5}\) (Dividing both the sides by 5)
∴ m = \(\frac {7}{5}\)

Simplify and solve the following linear equations.

Question 7.
3(t – 3) = 5(2t + 1)
Solution:
3(t – 3) = 5(2t + 1)
∴ 3t – 9 = 10t + 5
∴ 3t – 10t = 5 + 9 [Transposing 10t to LHS and (-9) to RHS]
∴ – 7t = 14
∴ 7t = – 14 [Multiplying both the sides by (-1)]
∴ \(\frac{7 t}{7}=\frac{-14}{7}\) (Dividing both the sides by 7)
∴ t = -2

Question 8.
15 (y – 4) – 2 (y – 9) + 5 (y + 6) = 0
Solution:
15 (y – 4) – 2 (y – 9) + 5 (y + 6) = 0
∴ 15y – 60 – 2y + 18 + 5y + 30 = 0
∴ 15y – 2y + 5y – 60 + 18 + 30 = 0
∴ 15y + 5y – 2y + 18 + 30 – 60 = 0 (Arranging the terms)
∴ 18y – 12 = 0
∴ 18y = 12 [Transposing (- 12) to RHS]
∴ \(\frac{18 y}{18}=\frac{12}{18}\) (Dividing both the sides by 18)
∴ y = \(\frac {2}{3}\)

Question 9.
3(5z – 7) -2 (9z – 11) = 4(8z – 13) – 17
Solution:
3(5z – 7) -2 (9z – 11) = 4(8z – 13) – 17
∴ 15z – 21 – 18z + 22 = 32z – 52 – 17
∴ 15z – 18z – 21 + 22 = 32z + (-52 – 17)
∴ – 3z + 1 = 32z – 69
∴ – 3z – 32z = – 69 – 1 (Transposing 1 to RHS and 32z to LHS)
∴ – 35z = – 70
∴ 35z = 70[Multiplying both the sides by (-1)]
∴ \(\frac{35 z}{35}=\frac{70}{35}\) (Dividing both the sides by 35)
∴ z = 2.

Question 10.
0.25 (4f – 3) = 0.05(10f – 9)
Solution:
0.25 (4f – 3) = 0.05(10f – 9)
0.25 × 4f – 0.25 × 3
= 0.05 × 10f – 0.05 × 9

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 12 Perimeter and Area MCQ Questions with Answers.

PSEB 6th Class Maths Chapter 12 Perimeter and Area MCQ Questions

Multiple Choice Questions

Question 1.
The outer boundary of a closed figure is called ………….. .
(a) Perimeter
(b) Region
(c) Area
(d) Curve.
Answer:
(a) Perimeter

Question 2.
Find the perimeter of the given figures:

(a) 30 cm
(b) 31 cm
(c) 32 cm
(d) 33 cm.
Answer:
(b) 31 cm

Question 3.
Perimeter of an equilateral triangle = ………………… .
(a) 3 + Side
(b) Side × Side
(c) Side + Side
(d) 3 × Side.
Answer:
(d) 3 × Side.

Question 4.
Perimeter of Rectangle = …………………… .
(a) 2l + b
(b) 2 (l + b)
(c) l + 2b
(d) l × b.
Answer:
(b) 2 (l + b)

Question 5.
If side of an equilateral triangle is 4 cm then perimeter = ……………….. .
(a) 8 cm
(b) 7 cm
(c) 12 cm
(d) 16 cm.
Answer:
(c) 12 cm

Question 6.
If length and breath of a rectangle are 2.4 cm and 1.9 cm then its perimeter is ………………. .
(a) 4.3 cm
(b) 8.2 cm
(c) 4.2 cm
(d) 8.6 cm.
Answer:
(d) 8.6 cm.

Question 7.
The perimeter of square is 16 cm then its side is ……………………… .
(a) 4 cm
(b) 64 cm
(c) 24 cm
(d) 32 cm.
Answer:
(a) 4 cm

Question 8.
The perimeter of a rectangle is 50 cm and its length is 12 cm then breadth is ……………………….. .
(a) 38 cm
(b) 13 cm
(c) 62cm
(d) 18cm.
Answer:
(b) 13 cm

Question 9.
Two sides of a triangle are 4.8 cm and 3.9 cm. The perimeter of the triangle is 12 cm. Find the third side.
(a) 3.3 cm
(b) 4.3 cm
(c) 20.7 cm
(d) 3.7 cm.
Answer:
(b) 4.3 cm

Question 10.
Samandeep takes 3 rounds of square park side 125 m. Find the distance covered by her.
(a) 1.5 km
(b) 1500 km
(c) 500 m
(d) 375 m.
Answer:
(a) 1.5 km

Question 11.
The measurement of the region enclosed by a closed plane figure is called its ………….. .
(a) Circumference
(b) Curve
(c) Perimeter
(d) Area.
Answer:
(d) Area

Question 12.
If the length of a rectangle is x units and breadth is 5 units then its perimeter is ………….. .
(a) 5x
(b) 2 (x + 5)
(c) 10x
(d) 10 + x
Answer:
(b) 2 (x + 5)

Question 13.
Find the area of the given rectangle whose length is 16 m and breadth is 8 m.
(a) 42 sq.m
(b) 128 sq. m
(c) 72 sq. m
(d) 21 sq. m.
Answer:
(b) 128 sq. m

Question 14.
The area of a rectangle is 144 m . If its breadth is 9 m then find its length.
(a) 16 sq. m
(b) 12 m
(c) 16 m
(d) 18 m.
Answer:
(c) 16 m

Question 15.
1 sq. m = ……………. sq. cm.
(a) 100
(b) 10000
(c) 1000
(d) 1.
Answer:
(b) 10000

Question 16.
Find the area of a square having side 3.6 cm.
(a) 14.4 cm
(b) 12.96 cm
(c) 1.29 sq. cm
(d) 12.96 sq. cm.
Answer:
(d) 12.96 sq. cm.

Question 17.
The perimeter of a square is 68 m. Find its area.
(a) 289 sq.m
(b) 329 sq. m
(c) 279 sq. m
(d) 249 sq.m.
Answer:
(a) 289 sq.m

Question 18.
A marble tile is of side 25 cm by 25 cm. How many tiles will be required to cover a floor of 4 m by 3 m?
(a) 216
(b) 192
(c) 188
(d) 196.
Answer:
(b) 192

Question 19.
What will happen to the area of a square, if side is doubled?
(a) Double
(b) Half
(c) Four times
(d) Nochange.
Answer:
(c) Four times

Question 20.
Find the perimeter of a rectangle whose area is 234 sq. cm and its one side is 13 cm.
(a) 31 cm
(b) 62 cm
(c) 18 cm
(d) 24 cm.
Answer:
(b) 62 cm

Question 21.
How many cm² are in 1 m²?
(a) 1000
(b) 100
(c) 10000
(d) 10.
Answer:
(c) 10000

Question 22.
The distance covered along the boundary forming a closed figure when you go round the figure once is called its:
(a) Length
(b) Perimeter
(c) Breadth
(d) Area.
Answer:
(b) Perimeter

Question 23.
The amount of surface enclosed by a closed figure is called its:
(a) Perimeter
(b) Volume
(c) Area
(d) None of these.
Answer:
(c) Area

Question 24.
The formula to find perimeter of a rectangle is:
(a) Length + Breadth
(b) 2 (Length + Breadth)
(c) Length – Breadth
(d) Length × Breadth.
Answer:
(b) 2 (Length + Breadth)

Question 25.
The formula to find perimeter of a square is :
(a) 4 × side
(b) 3 × side
(c) 2 × side
(d) 5 × side.
Answer:
(a) 4 × side

Fill in the blanks:

Question (i)
The formula to find area of equilateral triangle is …………….. .
Answer:
3 × side

Question (ii)
The formula to find area of a rectangle is ……………… .
Answer:
Length × Breadth

Question (iii)
The formula to find area of a square is ………………. .
Answer:
Perimeter

Question (iv)
The sum of lengths of all sides of a polygon is called ………………… .
Answer:
Area

Write True/False:

Question (i)
Perimeter of square = 4 × side. (True/False)
Answer:
True

Question (ii)
1 sq. m = 1000 sq. cm. (True/False)
Answer:
False

Question (iii)
The outer boundary of a closed figure is called area. (True/False)
Answer:
False

Question (iv)
Perimeter of a triangle = 3 × side. (True/False)
Answer:
True

Question (v)
Area of rectangle = Length × Breadth. (True/False)
Answer:
True