Bhagya

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

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 2 Linear Equations in One Variable Ex 2.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.4

Question 1.
Amina thinks of a number and subtracts \(\frac {5}{2}\) from it. She multiplies the result by 8. The result now obtained is 3 times the same number she thought of. What is the number?
Solution:
Let the number be x.
Subtracting \(\frac {5}{2}\) from it, we get, x – \(\frac {5}{2}\)
By multiplying this result by 8,
we get 8 (x – \(\frac {5}{2}\))
According to the condition,
8(x – \(\frac {5}{2}\)) = 3x
∴ 8x – 20 = 3x
∴ 8x = 3x + 20 [Transposing (-20) to RHS]
∴ 8x – 3x = 20 (Transposing 3x to LHS)
∴ 5x = 20
∴ \(\frac{5 x}{5}=\frac{20}{5}\) (Dividing both the sides by 5)
∴ x = 4
Thus, the number though of by Amina is 4.

Question 2.
A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?
Solution:
Let the smaller number be x
∴ The greater number is 5x
On adding 21 to both the numbers,
we get (5x + 21) and (x + 21)
According to the condition,
5x + 21 = 2 (x + 21)
∴ 5x + 21 = 2x + 42
∴ 5x = 2x + 42 – 21 (Transposing 21 to RHS)
∴ 5x = 2x + 21
∴ 5x – 2x = 21 (Transposing 2x to LHS)
∴ 3x = 21
∴ \(\frac{3 x}{3}=\frac{21}{3}\) (Dividing both the sides by 3)
∴ x = 7
The smaller number x = 7
The greater number = 5x = 5 × 7 = 35
Thus, the required numbers are 7 and 35.

Question 3.
Sum of the digits of a two-digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original number by 27. What is the two-digit number?
Solution:
Let the digit at the units place be x.
Then, the digit at the tens place = (9 – x)
The original number = 10(9 – x) + x
= 90 – 10x + x
= 90 – 9x
On interchanging the digits, the new number =10x + (9 – x)
= 10x + 9 – x
= 9x + 9
According to the condition,
New number = Original number + 27
∴ 9x + 9 = (90 – 9x) + 27
∴ 9x + 9 = 90 – 9x + 27
∴ 9x + 9 = 117 – 9x
∴ 9x = 117 – 9x – 9 (Transposing 9 to RHS)
∴ 9x = 108 – 9x
∴ 9x + 9x = 108 [Transposing (-9x) to LHS]
∴ 18x = 108
∴ \(\frac{18 x}{18}=\frac{108}{18}\) (Dividing both the sides by 18)
∴ x = 6
∴ Original number = 90 – 9x
= 90 – 9(6)
= 90 – 54 = 36
Thus, the original number is 36.

Question 4.
One of the two digits of a two-digit number is three times the other digit. I If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 88. What is the original number?
Solution:
Let the digit at units place be x and the digit at tens place be 3x.
The number = 10 (3x) + x
= 30x + x
= 31x
On interchanging the digits, the number – 10x + 3x = 13x
According to the condition,
31x + 13x = 88
44x = 88
∴ \(\frac{44 x}{44}=\frac{88}{44}\) (Dividing both the sides by 44)
∴x = 2
The number = 31x = 31 × 2 = 62
Thus, original number is either 62 or 26.

Question 5.
Saroj’s mother’s present age is six times ; Saroj’s present age. Saroj’s age five years from now will be one third of his mother’s present age. What are their present ages?
Solution:
Let Saroj’s present age be x years and mother’s present age be 6x years
After 5 years:
Saroj’s age will be x + 5 years
Mother’s age will be 6x + 5 years
According to the condition,
\(\frac {1}{3}\) (Mother’s present age) = Saroj’s age after 5 years
∴ \(\frac {1}{3}\)(6x) = x + 5
∴ 2x = x + 5
∴ 2x – x = 5 (Transposing x to LHS)
∴ x = 5
Saroj’s present age = x = 5 years
Mother’s present age = 6x = 6 × 5
= 30 years
Thus, present ages of Saroj and her mother are 5 years and 30 years respectively.

Question 6.
There is a narrow rectangular plot, reserved for a school, in Mahuli village. The length and breadth of the plot are in the ratio 11 : 4. At the rate ₹ 100 per metre it will cost the village panchayat ₹ 75,000 to fence the plot. What are the dimensions of the plot ?
Solution :
Length : Breadth = 11 : 4
Let the length be 11x metres.
Then, the breadth = 4x metres.
Perimetre = 2 (length + breadth)
= 2 (11x + 4x)
= 2 (15x) = 30x
Cost of fencing = ₹ 100 × 30x
= ₹3000 x
But, the cost of fencing = ₹ 75,000 (Given)
∴ 3000 x = 75000
∴ \(\frac{3000 x}{3000}=\frac{75000}{3000}\) (Dividing both the sides by 3000)
∴ x = 25
Length = 11x
= 11 × 25
= 275 metres
Breadth = 4x
= 4 × 25
= 100 metres
Thus, the length and breadth of the rectangular plot are 275 metres and 100 metres respectively.

Question 7.
Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him ₹ 50 per metre and trouser material that costs him ₹ 90 per metre.
For every 3 metres of the shirt material he buys 2 metres of the trouser material. He sells the materials at 12% and 10 % profit respectively. His total sale is ₹ 36,600. How much trouser material did he buy?
Solution:
Let the length of cloth for trousers be 2x metres
Then, the length of cloth for shirts = 3x metres
Cost of trouser’s cloth = 2x × ₹ 90
= ₹ 180x
Cost of shirt’s cloth = 3x × ₹ 50
= ₹ 150x
10 % profit is made on trouser’s cloth.
If C.E of trouser’s cloth is ₹ 100, then S.E is ₹ 110.
S.E of trouser’s cloth at 10 % profit = ₹\(\frac {110}{100}\) × 180x
= ₹ 198x
12 % profit is made on shirt’s cloth. If C.P of shirt’s cloth is ₹ 100, then S.P is ₹ 112.
S.P of shirt’s cloth at 12 % profit = ₹\(\frac {112}{100}\) × 150x
= ₹ 168x
∴ Total S.P = ₹ 198x + ₹ 168x
= ₹ 366x
But the total S.P. = ₹ 36,600 (Given)
366x = 36600
∴ \(\frac {112}{100}\) (Dividing both the sides by 366)
∴ x = 100
Cloth for trousers = 2x = 2 × 100 = 200
Thus, he bought 200 metres of cloth for trousers.

Question 8.
Half of a herd of deer are grazing | in the field and three fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.
Solution:
Let the total number of deer be x.
Number of deer grazing in the field = \(\frac{x}{2}\)
Remaining deer = x – \(\frac{x}{2}\) = \(\frac{x}{2}\)
Deer playing near by = \(\frac {3}{2}\) × (Remaining no. of deer)
= \(\frac {3}{4}\) × \(\frac{x}{2}\)
= \(\frac{3 x}{8}\)
Number of deer drinking water = 9
∴ Total number of deer = \(\frac{x}{2}\) + \(\frac{3 x}{8}\) + 9
∴ \(\frac{x}{2}\) + \(\frac{3 x}{8}\) + 9 = x
\(\frac{x}{2}\) + \(\frac{3 x}{8}\) = x – 9 (Transposing 9 to RHS)
\(\frac{x}{2}\) + \(\frac{3 x}{8}\) -x = -9 (Transposing x to LHS)
\(\frac{4 x+3 x-8 x}{8}\) = -9 (LCM = 8)
∴ \(\frac{-x}{8}\) = -9
∴ \(\frac{-x}{8}\) × 8 = -9 × 8 (Multiplying both the sides by 8)
∴ -x = – 72
∴ x = 72 [∵ Multiplying both sides by (- 1)]
Thus, total number of deer in herd is 72.

Question 9.
A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.
Solution:
Let the present age of granddaughter be x years.
Grandfather’s age is 10x years.
Fresent age of daughter + 54 = Grandfather’s age
∴ x + 54 = 10x
∴ 10x = x + 54 (Interchanging the sides)
∴ 10x – x = 54 (Transposing x to LHS)
∴ 9x = 54
∴ \(\frac{9 x}{9}=\frac{54}{9}\) (Dividing both the sides by 9)
∴ x = 6
Granddaughter’s age = x = 6 years
Grandfather’s age = 10x = 10 × 6 = 60 years
Thus, granddaughter’s age is 6 years and grandfather’s age is 60 years.

Question 10.
Aman’s age is three times his son’s age. Ten years ago he was five times his son’s age. Find their present ages.
Solution :
Let the present age of son be x years
Then, the present age of Aman = 3x years
Ten years ago their ages:
Son’s age was (x – 10) years
Aman’s age was (3x – 10) years
5 × (Son’s age 10 years ago) = Aman’s age 10 years ago
∴ 5 (x – 10) = (3x – 10)
∴ 5x – 50 = 3x – 10
∴ 5x = 3x – 10 + 50 [Transposing (-50) to RHS]
∴ 5x = 3x + 40
∴ 5x – 3x = 40 (Transposing 3x to LHS)
∴ 2x = 40
∴ \(\frac{2 x}{2}=\frac{40}{2}\)(Dividing both the sides by 2)
∴ x = 20
Son’s present age = x = 20 years
Aman’s present age = 3x = 3 × 20
= 60 years
Thus, present ages of Aman and his son are 60 years and 20 years respectively.

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

PSEB 8th Class English Voice Messages Writing

1.

Main Point of Voicemail Message

Message

Call from Ramesh Nagar Need two wheeler Not big budget Mobile No …………..

Manju : Hello, this is Manju. Thanks for calling. I’m not at home so please leave a message and I’ll call you back. Preeti : Hello, Manju this is Preeti calling from Ramesh Nagar. Your neighbourer Mr. Mbhan gave me your number. He said you are interested in selling your two wheeler. I was in need of second hand two wheeler. I can’t afford a big budget. Your two wheeler will serve my purpose. I am free on Sunday. Could I come to have a look of .the two wheeler. Thanks a lot. I look forward to hear from you. My Mobile No. is 98102-70 …….

2.

Main Point of Voicemail Message

Message

Business man or service man Location Number of rooms Minimum rent

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Calling from Hi-tech Washing powder, Washing soap etc. Brochure on table Live demo

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Calling from Dubey Builders Two rooms, three rooms or one room set Fully or semi furnished Visiting card

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Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 12 Perimeter and Area Ex 12.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 12 Perimeter and Area Ex 12.2

1. Find the approximate area of each of the following figures by countaing the number of squares – complete, more than half and exactly half.

Solution:
(i) Number of complete squares m = 7
Here we do not have any half square or more than half.
∴ n = 0, p = 0
∴ Area of plane figure = m + n + \(\frac {1}{2}\)p
= 7 + 0 + 0
= 7 sq. units

(ii) Number of complete squares m = 2
Number of more than half squares n = 4
Number of half squares p = 0
∴ Area of plane figure = m + n + \(\frac {1}{2}\)p
= 4 + 2 + 0
= 6 sq. units.

(iii) Number of complete squares m = 10
Number of more than half squares n = 0
Number of exactly half squares p = 2
∴ Area of plane figure = m + n + \(\frac {1}{2}\)p
= (10 + 0 + \(\frac {1}{2}\) × 2) sq. units
= (10 + 1) sq. units
= 11 sq. units

(iv) Number of complete squares m = 11
Number of more than half squares n = 0
Number of exactly half squares p = 2
∴ Area of plane figure = m + n + \(\frac {1}{2}\)p
= 11 + 0 + \(\frac {1}{2}\) × 2
= 11 + 1 = 12 sq. units

(v) Number of complete squares m = 10
Number of more than half squares n = 3
Number of exactly half squares p = 0
∴ Area of plane figure = m + n + \(\frac {1}{2}\)p
= 10 + 3 + 0
= 13 sq. units

2. Find the area of rectangle whose:

Question (i)
length = 12 cm, breadth = 16 cm
Solution:
Length of rectangle = 12 cm
and Breadth of rectangle = 16 cm.
∴ Area of rectangle = Length × Breadth
= 12 cm × 16 cm
= 192 sq. cm

Question (ii)
length = 25 m, breadth = 18 m
Solution:
Length of rectangle = 25 m
and Breadth of rectangle = 18 m
∴ Area of rectangle = Length × Breadth
= 25 m × 18 m
= 450 sq. m

Question (iii)
length = 2.7 m, breadth = 45 cm
Solution:
Length of rectangle = 2.7 m = 270 cm
and Breadth of rectangle = 45 cm
∴ Area of rectangle = Length × Breadth
= 270 cm × 45 cm
= 12150 sq. cm

Question (iv)
length 4.2 cm, breadth = 1.5 cm
Solution:
Length of rectangle = 4.2 cm
and Breadth of rectangle = 1.5 cm
∴ Area of rectangle = Length × Breadth
= 4.2 cm × 1.5 cm
= 6.3 sq. cm

Question (v)
length = 3.8 mm, breadth = 4 mm.
Solution:
Length of rectangle = 3.8 mm
and Breadth of rectangle = 4 mm
∴ Area of rectangle = Length × Breadth
= 3.8 mm × 4 mm
= 15.2 sq. mm

3. Find the area of the square with side:

Question (i)
19 cm
Solution:
Side of the square = 19 cm
∴ Area of the square = side × side
= 19 cm × 19 cm = 361 sq. cm.

Question (ii)
24 mm
Solution:
Side of square = 24 mm
∴ Area of square = side × side
= 24 mm × 24 mm
= 576 sq. mm

Question (iii)
3.5 cm
Solution:
Side of the square = 3.5 cm
Area of square = side × side
= 3.5 cm × 3.5 cm
= 12.25 sq. cm

Question (iv)
2.6 cm
Solution:
Side of the square = 2.6 cm
Area of square = side × side
= 2.6 cm × 2.6 cm
= 6.76 sq. cm

Question (v)
8.2 cm.
Solution:
Side of the square = 8.2 cm
Area of the square = side × side
= 8.2 cm × 8.2 cm
= 67.24 sq. cm.

4. The area of a rectangle is 216 sq. cm and its length is 12 cm. Find its breadth.
Solution:
The area of the rectangle = 216 sq. cm.
Length of the rectangle = 12 cm
Breadth of the rectangle = \(\frac{\text { Area }}{\text { Length }}\)
= \(\frac {216}{12}\) cm
= 18 cm.

5. The area of a rectangle is 225 sq. m and its breadth is 9 in. Find its length.
Solution:
The area of the rectangle = 225 sq. m
and Breadth of the rectangle = 9 m
∴ Length of the rectangle = \(\frac{\text { Area }}{\text { Breadth }}\)
= \(\frac {225}{9}\) m
= 25 m

6. The length and breadth of a ground are 32 m and 24 m. Find the cost of levelling the ground at the rate of ₹ 3 per sq. m.
Solution:
Length of ground = 32 m
and Breadth of ground = 24 m
Area of the ground = Length × Breadth
= 32 m × 24 m
= 768 sq. m.
Levelling cost of 1 sq. m = ₹ 3
Levelling cost of 768 sq. m = ₹ 3 × 768
= ₹ 2304

7. Find the perimeter of a rectangle whose area is 324 sq. cm and its one side is 36 cm.
Solution:
Area of rectangle = 324 sq. cm
One side of rectangle = 36 cm
∴ Other side of rectangle = \(\frac{\text { Area }}{\text { One side }}\)
= \(\frac {324}{36}\) = 9 cm
Perimeter of rectangle
= 2 × (1st side + 2nd side)
= 2 × (36 + 9) cm
= 2 × 45 cm
= 90 cm

8. The perimeter of a square field is 100 m. Find its area.
Solution:
The perimeter of square field = 100 m
∴ 4 × side of square = 100 m
∴ Side of a square = \(\frac {100}{4}\) = 25 m
Hence area of square field = side × side
= 25 m × 25 m
= 625 sq. m.

9. Area of a rectangle of length 20 cm is 340 sq. cm. Find its perimeter.
Solution:
Area of rectangle = 340 sq. cm
and Length of rectangle = 20 cm
Breadth of rectangle = \(\frac{\text { Area }}{\text { Length }}\)
= \(\frac {340}{20}\) cm = 17 cm
Perimeter of rectangle
= 2 × (Length + Breadth)
= 2 × (20 + 17) cm
= 2 × 37 cm
= 74 cm

10. A marble tile measure 15 cm × 20 cm. How many tiles will be required to cover a wall of size 4 m × 6 m?
Solution:
Area of the wall = 4 m × 6 m
= 400 cm × 600 cm
= 240000 sq. cm
Area of a marble tile = 15 cm × 20 cm
= 300 sq. cm
Number of tiles required to cover wall
= \(\frac{\text { Area of wall }}{\text { Area of tile }}=\frac{240000}{300}\) = 800
Hence number of tiles required to cover wall = 800

11. Find the cost of levelling the square field of side 75 m at rate of ₹ 5 per square metre.
Solution:
Side of the square field = 75 m
Area of square field = side × side
= 75 m × 75 m
= 5625 sq. m
Cost of levelling 1 sq. m = ₹ 5
Cost of levelling 5625 sq. m = ₹ 5 × 5625
= ₹ 28125

12. How many stamps of size 2 cm × 1.5 cm can be pasted on a sheet of paper of size 6 cm × 12 cm?
Solution:
Area of the sheet of paper
= 6 cm × 12 cm = 72 sq. cm
Area of one stamp = 2 cm × 1.5 cm = 3 sq cm
Number of stamps pasted on sheet of paper
= \(=\frac{\text { Area of paper sheet }}{\text { Area of stamp }}=\frac{72}{3}\) = 24.

13.

Question (i)
What will happen to the area of a square if its side is trebled (tripled)?
Solution:
Let side of the square = x cm
∴ Area of the square = (x × x) sq. m
Now, if the side is trebled, then side of new square = 3x cm
∴ Area of the new square
= [(3x) × (3x)] sq. m
= (3 × 3 × x × x) sq. m
= 9 (x × x) sq. cm
= 9 × (Area of original square)
∴ If side is trebled, then area becomes 9 times of original area.

Question (ii)
What will happen to the area of a rectangle if its length is halved and breadth is doubled?
Solution:
l cm and b cm be the length and breadth of the rectangle respectively.
∴ Area of rectangle = l × b
Now, If its length is halved and breadth is doubled.
∴ New length = \(\frac {1}{2}\) l
and new breadth = 2b
Thus area of new rectangle = length × breadth
= \(\frac {1}{2}\) × l × 2b
= \(\frac {1}{2}\) × 2 × (l × b)
= (l × b) = original area
Area will remain the same.

Question (iii)
What will happen to the area of a square if its side is halved?
Solution:
Let side of the square = x cm
∴ Area of the square = (x × x) sq. cm
Now, if side is halved, then
side of new square = \(\frac {1}{2}\)x cm
∴ Area of the new square sq.cm


Hence, if side is halved, then the area becomes one-fourth times original area.

14. Find the area of the following figures by splitting it into rectangles and squares:

Question (i)

Solution:
The given figure can be divided into 2 parts

Rectangle A of size 8 cm × 3 cm
Rectangle B of size 4 cm × 2 cm
∴ Area of rectangle A
= 8 cm × 3 cm = 24 sq. cm
and Area of rectangle B
= 4 cm × 2 cm
= 8 sq. cm
⇒ Total area of the figure
= 24 sq. cm + 8 sq. cm
= 32 sq. cm

Question (ii)

Solution:
The given figure can be divided into 2 parts

Rectangle A of size 10 cm × 2 cm
Rectangle B of size 8 cm × 2 cm
∴ Area of rectangle A
= 10 cm × 2 cm = 20 sq. cm
and Area of rectangle B
= 8 cm × 2 cm = 16 sq. cm
Total Area of the figure
= (20 + 16) sq. cm = 36 sq. cm

Question (iii)

Solution:
The given figure can be divided into 3 parts

Rectangle A of size 12 cm × 3 cm
Rectangle B of size 3 cm × 2 cm
Rectangle of size 12 cm × 3 cm
Area of rectangle A
= 12 cm × 3 cm = 36 sq. Cm
Area of rectangle B
= 3 cm × 2 cm = 6 sq. cm
Area of rectangle C
= 12 cm × 3 cm = 36 sq. cm
Total area of the figures
= (36 + 6 + 36) sq. cm
= 78 sq. cm

Question (iv)

Solution:
The given figure can be divided into 3 parts

Rectangle A of size 13 × 3 units
Rectangle B of size 5 × 3 units
Rectangle C of size 5 × 3
Area of rectangle A
= 13 × 3 = 39 sq. units
Area of rectangle B
= 5 × 3 = 15 sq. units
Area of rectangle C
= 5 × 3 = 15 sq. units
Hence total area of the figure
= (39 + 15 + 15) sq. units
= 69 sq. units Ans.

15. Fill in the blanks:

Question (i)
1 square metre = ……….. sq. cm.
Solution:
10000

Question (ii)
1 square cm = ………. sq. mm.
Solution:
100

Question (iii)
Area of Rectangle = …………. × ……………
Solution:
length, breadth

Question (iv)
Length = ………….. ÷ breadth.
Solution:
Area

Question (v)
Area of square = …………. × ……………
Solution:
side × side.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 12 Perimeter and Area Ex 12.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 12 Perimeter and Area Ex 12.1

1. Find the perimeter of the following shapes:

Question (i)

Solution:
Perimeter = AB + BC + CD + DA
= 8 cm + 7 cm + 12 cm + 9 cm
= 36 cm

Question (ii)

Solution:
Perimeter = XY + YZ + ZX
= 10m + 10m + 8m
= 28m

Question (iii)

Solution:
Perimeter = PQ + QR + RS + SP
= 15 cm + 12 cm + 15 cm + 12 cm
= 54 cm

Question (iv)

Solution:
Perimeter = MN + NO + OP + PL + LM
= 8 cm + 7 cm + 5 cm + 6 cm + 7 cm
= 33 cm

Question (v)

Solution:
Perimeter = AB + BC + CD + DE + EF + FG + GM + MA
=8m + 2m + 6m + 4m + 6m + 2m + 8m + 8m
= 44 m

Question (vi)

Solution:
Perimeter = LM + MN + NO + OP + PQ + QR + RS + SL
= 3 cm + 3 cm + 3 cm + 3 cm + 3 cm + 3 cm + 3 cm + 3 cm
= 24 cm

2. Find the perimeter of the triangle with sides:

Question (i)
5 cm, 6 cm and 7 cm
Solution:
Perimeter of a triangle
= Sum of lengths of its sides
= 5 cm + 6 cm + 7 cm = 18 cm

Question (ii)
10 m, 12 m, 18 m
Solution:
Sides of triangle
= 10 m, 12 m, 18 m
∴ Perimeter of a triangle
= Sum of lengths of its sides
= 10 m + 12 m + 18 m = 40 m

Question (iii)
4.6 cm, 3.2 cm and 5.8 cm.
Solution:
Sides of triangle
= 4.6 cm, 3.2 cm and 5.8 cm
∴ Perimeter of triangle
= Sum of lengths of its sides
= 4.6 cm + 3.2 cm + 5.8 cm
= 13.6 cm

3. Find the perimeter of an isosceles triangle with 15 cm as length of equal side and 18 cm as base.
Solution:
Sides of isosceles triangle = 15 cm, 15 cm, 18 cm
Area of isosceles triangle = Sum of lengths of its sides
= (15 + 15 + 18) cm = 48 cm

4. Find the perimeter of a square with side:

Question (i)
16 cm
Solution:
Side of square = 16 cm
∴ Perimeter of square = 4 × side
= 4 × 16 cm
= 64 cm

Question (ii)
4.8 mm
Solution:
Side of square = 4.8 mm
∴ Perimeter of square = 4 × side
= 4 × 4.8 mm
= 19.2 mm

Question (iii)
125 cm
Solution:
Side of square = 125 cm
∴ Perimeter of square = 4 × side
= 4 × 125 cm
= 500 cm

Question (iv)
45 m
Solution:
Side of square = 45 m
∴ Perimeter of square = 4 × side
= 4 × 45 m
= 180 m

Question (v)
39 cm.
Solution:
Side of square = 39 cm
∴ Perimeter of square = 4 × side
= 4 × 39 cm
= 156 cm

5. Find the perimeter of a rectangle with:

Question (i)
Length 20 m and breadth 15 m
Solution:
Length of rectangle = 20 m
Breadth of rectangle = 15 m
∴ Perimeter of rectangle = 2 × (length + breadth)
= 2 × (20 + 15) m
= 2 × 35 m
= 70 m

Question (ii)
Length 25 m and breadth 35 m
Solution:
Length of rectangle = 25 m
Breadth of rectangle = 35 m
∴ Perimeter of rectangle = 2 × (length + breadth)
= 2 × (25 + 35) m
= 2 × 60 m
= 120 m

Question (iii)
Length 40 cm and breadth 28 cm
Solution:
Length of rectangle = 40 cm Breadth of rectangle = 28 cm
∴ Perimeter of rectangle = 2 × (length + breadth)
= 2 × (40 + 28) cm
= 2 × 68 cm
= 136 cm

Question (iv)
Length 18.3 cm and breadth 6.8 cm
Solution:
Length of rectangle = 18.3 cm
Breadth of rectangle = 6.8 cm
∴ Perimeter of rectangle
= 2 × (length + breadth)
= 2 × (18.3 + 6.8) cm
= 2 × 25.1 cm = 50.2 cm

Question (v)
Length 0.125 m and breadth 15 cm.
Solution:
Length of rectangle
= 0.125 m = 12.5 cm
Breadth of rectangle = 15 cm
∴ Perimeter of rectangle = 2 × (length + breadth)
= 2 × (12.5 + 15) cm
= 2 × 27.5 cm
= 55 cm

6. Find the perimeter of a regular hexagon with side:

Question (i)
5 cm
Solution:
Side of a regular hexagon = 5 cm
Perimeter of a regular hexagon = 6 × side
= 6 × 5 cm
= 30 cm

Question (ii)
12 cm
Solution:
Side of a regular hexagon = 12 cm
Perimeter of a regular hexagon = 6 × side
= 6 × 12 cm
= 72 cm

Question (iii)
7.2 cm.
Solution:
Side of a regular hexagon = 7.2 cm
Perimeter of a regular hexagon = 6 × side
= 6 × 7.2 cm
= 43.2 cm

7. Find the perimeter of an equilateral triangle with side:

Question (i)
10 cm
Solution:
Side of an equilateral triangle
= 10 cm
∴ Perimeter of an equilateral triangle = 3 × side
= 3 × 10 cm
= 30 cm

Question (ii)
8 m
Solution:
Side of an equilateral triangle = 8 m
∴ Perimeter of an equilateral triangle = 3 × side
= 3 × 8 m
= 24 m

Question (iii)
24 m
Solution:
Side of an equilateral triangle = 24 m
∴ Perimeter of an equilateral triangle = 3 × side
= 3 × 24 m
= 72 m

Question (iv)
5.6 m
Solution:
Side of an equilateral triangle = 5.6 m
∴ Perimeter of an equilateral triangle = 3 × side
= 3 × 5.6 m
= 16.8 m

Question (v)
12.1 cm.
Solution:
Side of an equilateral triangle = 12.1 cm
∴ Perimeter of an equilateral triangle = 3 × side
= 3 × 12.1 cm
= 36.3 cm

8. If the perimeter of a triangle is 48 cm and two sides are 12 cm and 17 cm. Find the third side.
Solution:
Perimeter of a triangle = 48 cm Sum of length of two sides = (12 + 17) cm = 29 cm
∴ Third side = 48 cm – 29 cm
= 19 cm

9. Find the side of an equilateral triangle, if the perimeter is:

Question (i)
45 cm
Solution:
Given perimeter of an equilateral triangle = 45 cm
Perimeter of an equilateral triangle = 3 × (side of the triangle)
⇒ 45 cm = 3 × side
⇒ Side of the triangle
= 15cm

Question (ii)
69 mm
Solution:
Given perimeter of an equilateral triangle = 69 mm
Perimeter of an equilateral triangle = 3 × (side of the triangle)
⇒ 69 = 3 × (side of the triangle)
⇒ Side of the triangle = \(\frac{69 \mathrm{~mm}}{3}\)
= 23 mm

Question (iii)
117 cm.
Solution:
Given perimeter of an equilateral triangle = 117 cm
Perimeter of an equilateral triangle = 3 × (side of the triangle)
⇒ 117 = 3 × (side of the triangle)
⇒ Side of the triangle = \(\frac{117 \mathrm{~cm}}{3}\)
= 39 cm

10. Find the side of a square if the perimeter is:

Question (i)
52 cm
Solution:
Given Perimeter of a square = 52 cm
Perimeter of a square = 4 × (side of square)
⇒ Side of square
= \(\frac{\text { perimeter of a square }}{\text { 4 }}\)
= \(\frac{52 \mathrm{~cm}}{4}\)
= 13 cm

Question (ii)
60 cm
Solution:
Given perimeter of a square = 60 cm
Side of a square
= \(\frac{\text { perimeter of a square }}{\text { 4 }}\)
= \(\frac{60 \mathrm{~cm}}{4}\)
= 15 cm

Question (iii)
112 cm.
Solution:
Given perimeter of a square = 112 cm
Side of a square
= \(\frac{\text { perimeter of a square }}{\text { 4 }}\)
= \(\frac{112 \mathrm{~cm}}{4}\)
= 28 cm

11.

Question (i)
The perimeter of rectangular field is 260 m. If its length is 80 m then find its breadth.
Solution:
Given perimeter of rectangular field = 260 m
and Length of the rectangular field = 80 m
∴ Perimeter of rectangular field = 2 × (length + breadth)
⇒ 260 = 2 × (80 + breadth)
⇒ \(\frac {260}{2}\) = 80 + breadth
⇒ 80 + breadth = 130
⇒ breadth = 130 – 80 = 50 m
Hence breadth of rectangular field = 50 m

Question (ii)
The perimeter of a rectangular garden is 140 m. If its breadth is 45 m then find its length.
Solution:
Given perimeter of rectangular garden = 140 m
and breadth of rectangular garden = 45 m
∴ Perimeter of rectangular garden = 2 × (length + breadth)
⇒ 140 = 2 × (length + 45)
⇒ \(\frac {260}{2}\) = length + 45
⇒ length = 70 – 45 = 25 m
Hence length of rectangular garden = 25 m

Question (iii)
The perimeter of a rectangle is 114 cm. If its length is 32 cm then find its breadth.
Solution:
Given perimeter of rectangle = 114 cm
and length of rectangle = 32 cm
∴ Perimeter of rectangle
= 2 × (length + breadth)
⇒ 114 = 2 × (32 + breadth)
⇒ \(\frac {114}{2}\) = 32 + breadth
⇒ breadth = 57 – 32 = 25 cm
Hence breadth of rectangle = 25 cm

12. The side of a triangular field are 15 m, 20 m and 18 m. Find the total distance travelled by a boy in taking 2 complete rounds of this field.
Solution:
Sides of a triangular fields = 15 m, 20 m and 18 m
Distance covered in one round of a triangular field = Perimeter of rectangular field = Sum of the length of the sides of a rectangular field
= 15 m + 20 m + 18 m
= 53 m
∴ Distance covered in taking 2 complete rounds of this field
= 2 × 53 m
= 106 m

13. Find the cost of fencing a square field of side 26 m at the rate of ₹ 3 per metre.
Solution:
Given side of the square field = 26 m
∴ Perimeter of the square field = 4 × side
= 4 × 26 m = 104 m
Perimeter of fencing = 104 m
Cost of 1 m of fencing = ₹ 3
Cost of 104 m of fencing
= 104 × ₹ 3
= ₹ 312

14. Mani runs around a square park of side 75 m. Kush runs around a rectangular park of length 60 m and breadth 45 m. Who covers less distance?
Solution:
Side of a square park = 75 m
Perimeter of square park = 4 × (side)
= 4 × (75 m) = 300 m
∴ Distance covered by Mani = 300 m
Length of rectangular park = 60 m
Breadth of rectangular park = 45 m
Perimeter of rectangular park = 2 (length + breadth)
= 2 × (60 + 45) m
= 2 × (105) m
= 210 m
∴ Distance covered by Kush = 210 m
Kush covers less distance.

15. Find the cost of framing a rectangular whiteboard with length 240 cm and breadth 150 cm at the rate of ₹ 6 per cm.
Solution:
Length of rectangular white board = 240 cm
Breadth of rectangular white board = 150 cm
Perimeter of rectangular white board = 2 × (length + breadth)
= 2 × (240 + 150) cm
= 2 × (390) cm
= 780 cm
Cost of fencing 1 cm = ₹ 6 × 780
= ₹ 4680

16. If length of a rectangle is ‘a’ units and breadth is 5 units. Find the perimeter of the rectangle.
Solution:
Given length of rectangle = ‘a’ units,
and Breadth of rectangle = 5 units
Perimeter of rectangle = 2 × (length + breadth)
= 2 × (a + 5) units
= 2 (a + 5) units

17. Fill in the blanks:

Question (i)
The sum of lengths of all sides of a polygon is called ……………. .
Solution:
perimeter

Question (ii)
Perimeter of Square = ……………. × side.
Solution:
4

Question (iii)
Perimeter of Rectangle = 2 × (………. +………) .
Solution:
length, breadth

Question (iv)
Side of a square = (……………) ÷ 4.
Solution:
perimeter

Question (v)
Perimeter of an equilateral triangle = …………….. × side.
Solution:
3.

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.