Category: Class 8

Students can Download Maths Chapter 3 Geometry Additional Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 3 Geometry Additional Questions

Question 1.
State Pythagoras theorem.
Solution:
In a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.

Question 2.
Write two uses of Pythagoras theorem in our daily life.
Solution:
The Pythagoras theorem is useful in finding the distance and the heights of objects.

Question 3.
Prove Pythagoras theorem.
Solution:
Proof of the Pythagoras theorem using similarity of triangles.

Given: ∠BAC = 90°
To Prove BC² = AB² + AC²
Construction: Draw AD ⊥ BC
Proof:

In ∆ ABC and ∆ DBA
∠B is common and ∠BAC = ∠ADB = 90°
Therefore, ∆ ABC ~ ∆ DBA, (AA similarity)
Hence, the ratio of corresponding sides are equal.
⇒ \(\frac{AB}{DB}\) = \(\frac{BC}{BA}\)
⇒ AB² = BC x DB
Similarity, ∆ ABC ~ ∆ DAC,
⇒ \(\frac{AC}{DC}\) = \(\frac{BC}{AC}\)
⇒ AC² = BC x DB
Adding (1) and (2), we get
AB² + AC² = BC x DB + BC x DC
= BC x (DB + DC) = BC x BC
⇒ AB² + AC² = BC² and hence the proof of the theorem

Question 4.
State the converse of Pythagoras theorem.
Solution:
If in a triangle, the square on the greatest side is equal to the sum of squares on the other two sides, then the triangle is right angled triangle.

Question 5.
Give 2 examples for Pythagorean triplet.
Solution:
(21, 28, 35), (24, 32, 40)

Question 6.
Check whether (2, 3, 5) is Pythagorean triplet.
Solution:

2² + 3² = 4 + 9 = 13
5² = 25
13 ≠ 25
∴ It is not a Pythagorean triplet.

Question 7.
Can the sides that measure 15 cm, 20 cm and 25 cm make a right triangle?
Solution:

20² + 15² = 400 + 225 = 625
25² = 625
20² + 15² = 25²
∴ It is Pythagorean triplet.
∴ The given sides make a right angled triangle.

Question 8.
In the figure, find the value of x.

Solution:
In ABC
12² + 9² = 144 + 81 =225 = 15²
∴ x = 15

Question 9.
In the figure find the value of a and b and verify ∆ ABD is a right angled triangle.
Solution:
Now by altitude – on – hypotenuse theorem,
AB² = BC x BD gives
10² = a x 26
a = \(\frac{100}{26}\) = \(\frac{50}{13}\)

and AD² = CD x BD
24² = b x 26
b = \(\frac{576}{26}\) = \(\frac{288}{13}\)
and in ∆ ABD;
AB² + AD² = BD²
10² + 24² = 576 = 26²
BD = 26
Therefore ∆ ABD is a right angled triangle.

Question 9.
State the altitude-on-Hypotenuse theorem.
Solution:
If an altitude is drawn to the hypotenuse of an right angled triangle then (i) the two triangles are similar to the given triangle and also to each other.

That is ∆ PRQ ~ ∆ PSR ~ ∆ RSQ

1. h² = xy
2. p² = yr and
3. q² = xr, where r = x + y

Students can Download Maths Chapter 3 Geometry Ex 3.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 3 Geometry Ex 3.2

Miscellaneous Practice Problems

Question 1.
The sides of a triangle are 1.2 cm, 3.5 cm and 3.7cm. Is this triangle a right triangle? If so, which side is the hypotenuse?
Solution:
The sides of a triangle are a = 1.2 cm; b = 3.5 cm; c = 3.7 cm
By Pythagoras theorem,
c² = a² + b²
a² + b² = 1.2² + 3
5² – 1.44 +12.25 = 13.69
c² = 3.7² = 13.69
(1) = (2)
Yes, it is a right angled triangle. The hypotenuse c = 3.7 cm.

Question 2.
Rithika buys an LED TV which has a 25 inches screen. If its height is 7 inches, how wide is the screen? Her TV cabinet is 20 inches wide. Will the TV fit into the cabinet? Why?
Solution:

Take the sides of a right angled triangle ∆ ABC as
a = 7 inches
b = 25 inches
c = ?
By Pythagoras theorem,
b² = a² + c²
25 = 7² + c²
c² = 25² – 7² = 625 – 49 = 576
∴ c² = 24²
⇒ c = 24 inches
∴ Width of TV cabinet is 20 inches which is lesser than the width of the screen ie.24 inches. The TV will not fit into the cabinet.

Question 3.
Find the length of the support cable required to support the tower with the floor.

Solution:
From the figure, by Pythagoras theorem,
x² = 20² + 15² = 400 + 225 = 625
x² = 25² ⇒ x = 25ft.
∴ The length of the support cable required to support the tower with the floor is 25 ft.

Question 4.
A ramp is constructed in a hospital as shown. Find the length of the ramp.

Solution:
Take a = 7 ft ; b = 24 ft.
By Pythagoras theorem,
l² = c² + b²
= 7² + 24² = 49 + 576
l = 25ft

Question 5.
In the figure, find MT and AH

Solution:
By Pythagoras theorem,
MT² = MH² + HT²
= 60² + 80² = 3600 + 6400
= 10000 = 100²
∴ MT = 100
By the altitude – on – hypotenuse theorem,
MH² = MA x MT
60² = MA x 100
MA = \(\frac{3600}{100}\) = 36
∴ In ∆ MAH, by Pythagoras theorem,
AH² = MH² – MA²
= 60² – 36² = 3600 – 1296 = 2304 = 48²
∴ AH = 48
∴ Ans MT = 100
AH = 48

Challenging Problems

Question 6.
Mayan travelled 28 km due north and then 21 km due east. What is the least distance that he could have travelled from his starting point?
Solution:
From the figure AC is to be found.

By using Pythagoras theorem,
AC² = AB² + BC²
= 28² + 21² = 784 + 441 = 1225 = 352
∴ AC = 35 km
This is the least distance that he could have travelled from his starting point.

Question 7.
If ∆ APK is an isosceles right angled triangle, right angled at K. Prove that AP² = 2AK².
Solution:
From the figure, ∆ APK is a right triangle.

By using Pythagoras theorem,
AP² = AK² + PK²
= AK² + AK² (since it is an isosceles A)
= 2AK²
Hence it is proved.

Question 8.
The diagonals of the rhombus is 12 cm and 16 cm. Find its perimeter. (Hint: the diagonals of rhombus bisect each other at right angles).
Solution:

Here AO = CO = 8 cm
BO = DO = 6 cm
(∴ the diagonals of rhombus bisect each other at right angles)
∴ In ∆ AOB,
AB² = AO² + OB² = 8² + 6² = 64 + 36
= 100 = 10²
∴ AB = 10
Since it is a rhombus, all the four sides are equal.
AB = BC = CD = DA
∴ Its Perimeter = 10 + 10 + 10 + 10 = 40 cm

Question 9.
In the figure, find AR.

Solution:
∆ AFI, ∆ FRI are right triangles.
By Pythagoras theorem,
AF² = AT² – FT² = 25² – 15²
= 625 – 225 = 400 = 20²
∴ AF = 20 ft ….(1)
FR² = RI² – FI² = 17² – 15² = 289 – 225 = 64 = 8²
FR = 8 ft.
∴ AR = AF + FR = 20 + 8 = 28 ft.

Question 10.
∆ ABC is a right angled triangle in which ∠A = 90° and AM ⊥ BC. Prove that AM = \(\frac{ABxAC}{BC}\). Also if AB = 30 cm and AC = 40 cm, find AM.

Solution:
Given: ∆ ABC is a right angled triangle.
∠A = 90, AM ⊥ BC
To Prove: AM = \(\frac{ABxAC}{BC}\)
Proof:
In the given figure, Area of the triangle A ABC [taking AB as base and AC as height]
= \(\frac{1}{2}\) x AB x AC ….(1)
And also, Area of the triangle ∆ ABC [taking BC as base and AM as height]
= \(\frac{1}{2}\) x BC x AM ……(2)
Since (1) & (2) are the same, we can equate (2) and (1)
i.e. \(\frac{1}{2}\) x BC x AM = \(\frac{1}{2}\) x AB x AC
AM = \(\frac{ABxAC}{BC}\)
Hence proved. If AB = 30 cm, AC = 40 cm
By Pythagoras theorem, BC² = AB² + AC² = 30² + 40² = 900 + 1600 = 2500 = 50²
∴ BC = 50
Using the altitude – on – hypotenuse theorem

Here AC² = MC x BC
40² = MC x 50
∴ BM = BC – MC = 50 – 32 = 18 cm
AM² = BM x MC = 18 x 32 = 576 = 24²
∴ AM = 24 cm

Students can Download Maths Chapter 3 Geometry Ex 3.1 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 3 Geometry Ex 3.1

Question 1.
Fill in the blanks:

Question (i)
If in a ∆ PQR, PR² = PQ² + QR², then the right angle of ∆ PQR is at the vertex ………
Answer:
Q
Hint:

We have also given Lines and Angles concepts and problems for free of cost on our website.

Question:
(ii) If ‘l’ and ‘m’ are the legs and is the hypotenuse of a right angled triangle then, l² = ……….
Answer:
n² – m²
Hint:

Question (iii)
If the sides of a triangle are in the ratio 5:12:13 then, it is ………
Answer:
a right angled triangle.
13² = 169
5² = 25
12² = 144
169 = 25 + 144
132 = 5² + 12²
By Pythagoras theorem, In a right triangle, square of the hypotenuse is equal to the sum of the squares of other two sides.

Question (iv)
If a perpendicular is drawn to the hypotenuse of a right angled triangle, then each of the three pairs of triangles formed are …………
Answer:
Similar.

Question (v)
In the figure BE² = TE x ………

Answer:
EN

Question 2.
Say True or False.

Question (i)
8, 15, 17 is a Pythagorean triplet.
Answer:
True
Hint:
17² = 289
15² = 225
8² = 64
64 + 225 = 289 ⇒ 17² = 15² + 8²

Question (ii)
In a right angled triangle, the hypotenuse is the greatest side.
Answer:
False
Hint:

Question (iii)
One of the legs of a right angled triangle PQR having ∠R = 90° is PQ.
Answer:
False
Hint:

In ∆ PQR, QR and RP are legs and PQ is the hypotenuse

Question (iv)
The hypotenuse of a right angled triangle whose sides are 9 and 40 is 49.
Answer:
False
Hint:

49² = 2401
9² = 81
40² = 1600
40² + 9² = 1600 + 81 + 1681
49² = 2401
2401 ≠ 1681

Question (v)
Pythagoras theorem is true for all types of triangles.
Answer:
False
Hint:
Pyhtagoras theorem is true for only right angled triangles.

Question 3.
Check whether given sides are the sides of right – angled triangles, using Pythagoras theorem,

1. 8, 15, 17
2. 12, 13, 15
3. 30, 40, 50
4. 9, 40, 41
5. 24, 45, 51

Solution:
1. 8, 15, 17
Take a = 8,
b = 15 and
c = 17
Now a² + b² = 8² + 15² = 64 + 225 = 289
17² = 289 = c²
∴ a² + b² = c²
By the converse of Pythagoras theorem, the triangle with given measures is a right angled triangle.
Answer:
yes.

2. 12, 13, 15
Take a = 12
b = 13 and
c = 15
Now a² + b² = 12² + 13² = 144 + 169 = 313
15² = 225 ≠ 313
By the converse of Pythagoras theorem, the triangle with given measures is not a right angled triangle.
Answer:
No.

3. 30, 40, 50
Take a = 30
b = 40 and
c = 50
Now a² + b² = 30² + 40² = 900 + 1600 = 2500
c² = 50² = 2500
∴ a² + b² = c²
By the converse of Pythagoras theorem, the triangle with given measures is a right angled triangle.
Answer:
yes.

4. 9, 40, 41
Take a = 9
b = 40 and
c = 41
Now a² + b² = 9² + 40² = 81 + 1600= 1681
c² = 41² = 1681
∴ a² + b² = c²
By the converse of Pythagoras theorem, the triangle with given measures is a right angled triangle.
Answer:
Yes.

5. 24, 45, 51
Take a = 24
b = 45 and
c = 51 Now
a² + b² = 24² + 45² = 576 + 2025 = 2601
c² = 51² = 2601
a² + b² = c²
By the converse of Pyhtagoreas theorem, the triangle with given measure is a right angled triangle.
Answer:
Yes.

Question 4.
Find the unknown side in the following triangles.

Solution:
(i) From ∆ ABC, by Pythagoras theorem

BC² = AB² + AC²
Take AB² + AC² = 9² + 40² = 81 + 1600 = 1681
BC² = AB² + AC² = 1681 = 41²
BC² = 41² ⇒ BC = 41
∴ x = 41

(ii) From ∆ PQR, by Pythagoras theorem,

PR² = PQ² + QR²
34² = y² + 30²
⇒ y² = 34² – 30²
= 1156 – 900
= 256 = 16²
y² = 16² ⇒ y = 16

(iii) From ∆ XYZ, by Pythagoras theorem,

YZ² = XY² + XZ²
⇒ XY² = YZ² – XZ²
Z² = 39² – 36²
= 1521 – 1296 = 225 = 15²
Z² = 15²
⇒ Z = 15

Question 5.
An isosceles triangle has equal sides each 13 cm and a base 24 cm in length. Find its height.
Solution:
In an isosceles triangle the altitude dives its base into two equal parts. Now in the figure, ∆ ABC is an isosceles triangle with AD as its height.

In the figure, AD is the altitude and Δ ABD is a right triangle.
By Pythagoras theorem,
AB² = AD² + BD²
⇒ AD² = AB² – BD²
= 13² – 12² = 169 – 144 = 25
AD² = 25 = ²
Height: AD = 5 cm

Question 6.
In the figure, find PR and QR.

Solution:
In the figure ∆ PQR is a right triangle.
By Pythagoras theorem,
PR² = PQ² + QR²
(x + 1)² = 7² + x²
x² +2 × x × 1 + 1² = 49+ x²
2x + 1 = 49
2x = 49 – 1 = 48
x = \(\frac{48}{2}\) = 24
∴ PR = x + 1 = 24 + 1 = 25
QR = x = 24

Question 7.
The length and breadth of the screen of an LED – TV are 24 inches and 18 inches. Find the length of its diagonal.

Solution:
The length and breadth of a LED TV form a right angled triangle with its diagonal.
Therefore by Pythagoras theorem,
AC² = AB² + BC²
= 24² + 18² = 576 + 324 = 900 = 30²
∴ AC = 30 ⇒ The length of the diagonal is 30 inches.

Question 8.
Find the distance between the helicopter and the ship.

Solution:
From the figure AS is the distance between the helicopter and the ship.
∆ APS is a right angled triangle, by Pythagoras theorem,
AS² = AP² + PS²
= 80² + 150²
= 6400 + 22500 = 28900 = 170²
∴ The distance between the helicopter and the ship is 170 m

Question 9.
From the figure, 1. If TA = 3 cm and OT = 6 cm, find TG.

Solution:
1. From the figure, if, TA = 3 cm, OT = 6 cm

By Pythagoras theorem,
OA² = OT² – TA² = 6² – 3²
i.e. h² = 36 – 9 = 27 cm.
Now, by altitude – on – hypotenuse theorem
h² = xy
27 = x × 3
x = \(\frac{27}{3}\) = 9 cm
TG = x + 3 = 9 + 3 = 12 cm

Question 10.
If RQ = 15 cm and RP = 20 cm, find PQ, PS and SQ.

Solution:

RQ = 15cm
RP = 20 cm and ∆ PQR is a right angled triangle
By Pythagoras theorem, p 20 cm
PQ² = PR² + RQ² = 20² + 15²
= 400 + 225 = 625 = 25²
∴ PQ = 25 cm
Now by altitude – on – hypotenuse theorem,
RQ² = q x r
15² = q x 25
q = \(\frac{225}{25}\) = 9 cm ⇒ SQ = 9 cm
PR² = P x r
20² = P x 25
P = \(\frac{400}{25}\) = 16 cm ⇒ PS = 16 cm
Answer:

1. PQ = 25 cm
2. PS = 16 cm
3. SQ = 9 cm

Objective Type Questions

Question 11.
If ∆ GUT is isosceles and right angled, then ∠TUG is …………

(a) 30°
(b) 40°
(c) 45°
(d) 55°
Answer:
(c) 45°
Hint:
∠U ∠T = 45° (∆ GUT is an isosceles given)
∴ ∠TUG = 45°

Question 12.
The hypotenuse of a right angled triangle of sides 12 cm and 16 cm is ……….
(a) 28 cm
(b)20cm
(c) 24 cm
(d)21cm
Answer:
(b) 20 cm
Hint:
Side take a = 12 cm
b = 16 cm
The hypotenuse c² = a² + b² = 12² + 16^2
2 = 400 ⇒ c = 20 cm

Question 13.
The area of a rectangle of length 21 cm and diagonal 29 cm is ………. cm²
(a) 609
(b) 580
(c) 420
(d) 210
Answer:
(c) 420
Hint:

Length = 21 cm
Diagonal = 29 cm
By the converse of Pythagoras theorem,
AB² + BC² = AC²
21² + x² = 29²
x² = 841 – 441 = 400 = 20²
x = 20 cm
Now area of the rectangle = length x breadth.
i.e. AB x BC = 21 cm x 20 cm = 420 cm²

Question 14.
if the square of the hypotenuse of an isosceles right triangle is 50 cm2, the length of each side is ………..
(c) 10 cm
(a) 25 cm
(b) 5 cm
(c) 10 cm
(d) 20 cm
Answer:
(b) 5 cm
Hint:
By Pythagoras theorem
c² = a² + b²
In an isosceles triangle, a = b
c² = a² + a² = 2a²
⇒ 2a² = 50
a² = 25 ⇒ a = 5cm
∴ The length of each sides a = 5cm, b = 5 cm.

Question 15.
The sides of a right angled triangle are in the ratio 5 : 12 : 13 and its perimeter is 120 units then, the sides are .
(a) 25, 36, 59
(b) 10, 24, 26
(c) 36, 39, 45
(d) 20, 48, 52
Answer:
(d) 20, 48, 52
Hint:
The sides of a right angled triangle are in the ratio 5 : 12 : 13
Take the three sides as 5a, 12a, 13a
Its perimeter is 5a + 12a + 13a = 30a
It is given that 30a = 120 units
a = 4 units
∴ The sides 5a = 5 x 4 = 20 units
12a = 12 x 4 = 48 units
13a = 13 x 4 = 52 units

Students can Download Maths Chapter 2 Algebra Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 2 Algebra Intext Questions

Exercise 2.1
Try These (Text book page no. 32)

Question 1.
Identify which among the following are linear equations.

1. 2 + x = 19 – Linear as degree of the variable x is 1
2. 7x² – 5 = 3 – not linear as highest degree of x is 2
3. 4p³ = 12 – not linear as highest degree ofp is 3
4. 6m + 2 – Linear, but not an equation
5. n = 10 – Linear equation as degree of n is 1
6. 7k – 12= 0 – Linear equation as degree of Hs 1
7. \(\frac{6x}{8}\) + y = 1 – Linear equation as degree ofx & y is 1
8. 5 + y = 3x – Linear equation as degree ofy & x is 1
9. 10p + 2q = 3 – Linear equation a& degree ofp & q is 1
10. x² – 2x-4 – not linear equation as highest degree of x is 2

Think (Text book page no. 32)

Question 1.
Is t(t – 5)= =10 a linear equation? Why?
Solution:
t(t – 5) = 10
= t x t – 5 x t = 10
= t² – 5t = 10
This is not a linear equation as the highest degree of the variable ‘t’ is 2

Question 2.
Is x² = 2x, a linear equation? Why?
Solution:
x² = 2x
= x² – 2x = 0
This is not a linear equations as the highest degree of the variable ‘x’ is 2

Try These (Text book page no. 33)

Convert the following statements into linear equations:

Question 1.
On subtracting 8 from the product of 5 and a number, I get 32.
Solution:
Convert to linear equations:
Given that on subtracting 8 from product of 5 and a, we get 32
5 × x – 8 = 32
5x – 8 = 32

Question 2.
The sum of three consecutive integers is 78.
Solution:
Sum of 3 consecutive integers is 78
Let 1st integer be ‘x’
∴ x + (x + 1) + (x + 2) = 78
∴ x + x + 1 + x + 2 = 78
3x + 3 = 78

Question 3.
Peter had a Two hundred rupee note. After buying 7 copies of a book he was left with ₹ 60.
Solution:
Let cost of one book be ‘x’
∴ Given that 200 – 7 × x = 60
∴ 200 – 7x = 60

Question 4.
The base angles of an isosceles triangle are equal and the vertex angle measures 80°.
Solution:
Let base angles each be equal to x & vertex bottom angle is 80°. Applying triangle
property, sum of all angles is 180°
∴ x + x + 80 = 180°
2x + 80 = 180°

Question 5.
In a triangle ABC, ∠A is 10° more than ∠B. Also ∠C is three times ∠A. Express the equation in terms of angle B
Solution:
Let ∠B = b
Given ∠A = 10° + ∠B = 10 + b
Also given that ∠C = 3 x ∠A = 3 x (10 + 6) = 30 + 3b
Sum of the angles = 180°
∠A + ∠B + ∠C = 180°
10 + b + b + 30 + 3b = 180°
∴ 5b + 40 = 180°

Think (Text book page no. 34)

Question 1.
Can you get more than one solution for a linear equation?
Solution:
Yes, we can get. Consider the below line or equation
x + y = 5
here, when x = 1, y = 4
when x = 2, y = 2
x = 3, y = 2
x = 4, y = 1
Hence, we get multiple solutions for the same linear equation.

Think (Text book page no. 35)

Question 1.
“An equation is multiplied or divided by a non zero number on either side.” Will there be any change in the solution?
Solution:
Not be any change in the solution

Question 2.
“An equation is multiplied or divided by two different numbers on either side”. What will happen to the equation?
Solution:
When an equation is multiplied or divided by 2 different numbers on either side, there will be a change in the equation & accordingly, solution will also change.

Exercise 2.2
Think (Text book page no. 37)

Question 1.
Suppose we take the second piece to be x and the first piece to be (200 – x), how will the steps vary. Will the answer be different?
Solution:
Let 2nd piece be V & 1st piece is 200 – x
Given that 1st piece is 40 cm smaller than hence the other piece
∴ 200 – x = 2 × x – 40

200 + 40 = 2x + x
240 = 3x
∴ x = \(\frac{240}{3}\) = 80
∴ 1st piece = 200 – x = 200 – 80 = 120 cm
2nd piece = x = 80 cm
The answer will not change

Exercise 2.3
Think (Text book page no. 43)

Question 1.
If instead of (4, 3), we write (3, 4) and try to mark it, will it represent ‘M’ again?
Solution:
Let 3, 4 be M, when we mark, we find that it is a different point and not ‘M’

Try These (Text book page no. 45)

Question 1.
Complete the table given below.

Solution:

Question 2.
Write the coordinates of the points marked in the following figure

• A – (-3, 2)
• B – (5, 2)
• C – (5, -3)
• D – (-3, 3)
• E – (-1,4)
• F – (1, 2)
• G – (7, 4)
• H – (0, 2)
• i – (o, 3)
• J – (-3, 0)
• K – (5, 0)
• L – (-1, 0)
• M – (-2, 0)
• N – (-2, -1)
• O – (0, 0)
• P – (-1, -1)
• Q – (1, -1)
• R – (2, -1)
• S – (0, -3)
• T – (7, 0)
• U – (7, -2)

Exercise 2.4
Think (Text book page no. 49)

Question 1.
Which of the points (5, -10) (0, 5) (5, 20) lie on the straight line X = 5?
Solution:
All points on the line X = 5 will have X – coordinate as 5.
Therefore, any point with X – coordinate as 5 will lie on X = 5 line.
Hence the points (5, – 10) & (5,20) will lie on X = 5

Think (Text book page no. 54)

Question 1.
Why is it given that the speed is ‘constant’? If the speed is not constant, will the graph be the same? The graph is named as y = 80 x algebraically. Why?
Solution:

or in other words, the slope of a line in a distance – time graph. We observe that the slope of the graph is constant hence, speed is constant. If the speed is not constant, the graph will be different as slope of the line would change.

Students can Download Maths Chapter 2 Algebra Ex 2.4 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 2 Algebra Ex 2.4

Question 1.
Fill in the blanks:

Question (i)
y = px where p ∈ Z always passes through the ………
Answer:
Origin (0, 0)
Hint:
[When we substitute x = 0 in equation, y also becomes zero. (0,0) is a solution]

Question (ii)
The intersecting point of the line x = 4 and y = – 4 is ………
Answer:
4, -4
Hint:
x = 4 is a line parallel to the y – axis and
y = – 4 is a line parallel to the x – axis. The point of intersection is a point that lies on both lines & which should satisfy both the equations. Therefore, that point is (4, -4)

Question (iii)
Scale for the given graph, on the x – axis 1 cm = ……… units y – axis 1 cm = ………. units

Answer:
3 units, 25 units
Hint:
With reference to given graph,
On the x – axis, 1cm = 3 units
y axis, 1cm = 25 units

Question 2.
Say True or False:

Question (i)
The points (1, 1) (2, 2) (3, 3) lie on a straight line.
Answer:
True
Hint:
The points (1, 1), (2, 2), (3, 3) all satisfy the equation y = x
which is straight line. Hence, it is true

Question (ii)
y = – 9x not passes through the origin.
Answer:
False
Hint:
y = – 9x substituting forx as zero, we get y = – 9 x 0 = 0
∴ for x = 0, y = 0. Which means line passes through (0, 0), hence statement is false.

Question 3.
Will a line pass through (2, 2) if it intersects the axes at (2, 0) and (0, 2).
Solution:
Given a line intersects the axis at (2, 0) & (0, 2)
Let line intercept form be expressed as
ax + by = 1 Where a & b are the x & y intercept respectively.
Since the intercept points are (2, 0) & (0, 2)
a = 2, b = 2
∴ 2x + 2y = 1
When the point (2, 2) is considered & substituted in the equation
2x + 2y = 1 we get
2 x 2 + 2 x 2 = 4 ≠ 1
∴ The point (2, 2) does not satisfy the equation. Therefore the line does not pass through (2, 2)
Alternatively graphical method:
as we can see the line doesn’t pass through (2, 2)

Question 4.
A line passing through (4, – 2) and intersects the Y – axis at (0, 2). Find a point on the line in the second quadrant.
Solution:

Line passes through (4, – 2)
y – axis intercept point – (0, 2)
using 2 point formula,
\(\frac { y-y_{ 1 } }{ x-x_{ 1 } } \) = \(\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \)
\(\frac{y-2}{x-0}\) = \(\frac{1-2-2}{4-0}\)
∴ \(\frac{y-2}{x}\) = \(\frac{-4}{4}\) = – 1
y – 2 = – 1 × x

∴ x + y = 2 is the equation of the line.
Any point in II quadrant will have x as negative & y as positive.
So let us take x value as – 2
∴ – 2 + y = 2
y = 2 + 2 = 4
∴ Point in II Quadrant is (-2, 4)

Question 5.
If the points P (5, 3) Q(- 3, 3) R (- 3, – 4) and S form a rectangle then find the coordinate of S.
Solution:
Plotting the points on a graph (approximately)

Steps:

1. Plot P, Q, R approximately on a graph.
2. As it is a rectangle, RS should be parallel to PQ & QR should be parallel to PS
3. S should lie on the straight line from R parallel to x – axis & straight line from P parallel to y – axis
4. Therefore, we get S to be (5, -4)

[Note: We don’t need graph sheet for approximate plotting. This is just for graphical understanding]

Question 6.
A line passes through (6, 0) and (0, 6) and an another line passes through (- 3, 0) and (0, – 3). What are the points to be joined to get a trapezium?
Solution:
In a trapezium, there are 2 opposite sides that are parallel. The other opposite sides are non-parallel.
Now, let us approximately plot the points for our understanding [no need of graph sheet]

1. Plot the points (0, 6), (6, 0), (-3, 0) & (0, – 3)
2. Join (0, 6) & (6, 0)
3. Join (-3, 0) & (0, -3)
4. We find that the lines formed by joining the points are parallel lines.
5. So, for forming a trapezium, we should join (0, 6), (-3, 0) & (0, -3), (6, 0)

Question 7.
Find the point of intersection of the line joining points (- 3, 7) (2, – 4) and (4, 6) (- 5, 7). Also find the point of intersection of these lines and also their intersection with the axis.
Solution:

Equation of line joining 2 points by 2 point formula is given by
\(\frac { y-y_{ 1 } }{ x-x_{ 1 } } \) = \(\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \)
∴ \(\frac{y-7}{x-(-3)}\) = \(\frac{-4-7}{2-(-3)}\)
\(\frac{y-7}{x+3}\) = \(\frac{-11}{2+3}\)
∴ \(\frac{y-7}{x+3}\) = \(\frac{-11}{5}\)
Cross multiplying, we get

Transposing the variables, we get
11x + 5y = 35 – 33 = 2
11x + 5y = 2 – Line 1
Similarly, we should find out equation of second line

∴ 9y – 54 = 13x – 52
9y – 13x = 2 – Line 2
For finding point of intersection, we need to solve the 2 line equation to find a point that will satisfy both the line equations.
∴ Solving for x & y from line 1 & line 2 as below
11x + 5y = 2 ⇒ multiply both sides by 13,
11 x 13x + 5 x 13y = 26
Line 2:
9y – 13x = 2 ⇒ multiply both sides by 11
9 x 11y – 13 x 11x = 22

Substituting this value of y in line 1 we get
11x + 5y = 2
11x + 5 x \(\frac{12}{41}\) = 2
11x = 2 – \(\frac{60}{41}\) = \(\frac{82-60}{41}\) = \(\frac{22}{41}\)
x = \(\frac{2}{41}\)
∴ Point of intersection is (\(\frac{2}{41}\) \(\frac{12}{41}\))
To find point of intersection of the lines with the axis, we should substitute values & check
Line 1:
11x + 5y = 2
Point of intersection of line with x – axis, i.e y coordinate is ‘0’
∴ put y = 0 in above equation
∴ 11x – 5 x 0 = 2
11x + 0 = 2
x = \(\frac{2}{11}\)
∴ Point is (\(\frac{2}{11}\), 0)
Similarly, point of intersection of line with y – axis is when x – coordinate becomes ‘0’
∴ put x = 0 in above equation
∴ 11 x 0 + 5y = 2
∴ 0 + 5y = 2
y = \(\frac{2}{5}\)
∴ Point is (0, \(\frac{2}{5}\))
Similarly for line 2,
9y – 13x = 2
For finding x intercept, i.e point where line meets x axis, we know thaty coordinate becomes ‘0’
∴ Substituting y – 0 in above eqn. we get
9 x 0 – 13x = 2
∴ 0 – 13x = 2
x = \(\frac{-2}{13}\)
∴ Point is (\(\frac{-2}{13}\), 0)
Similarly for y – intercept, x – coordinate becomes ‘0’,
∴ Substituting for x = 0 in above equation, we get
9y – 13 x 0 = 2
9y – 0 = 2
9y = 2
y = \(\frac{2}{9}\)
Point is (0, \(\frac{2}{9}\))

Question 8.
Draw the graph of the following equations:

1. x = – 7
2. y = 6

Solution:

Question 9.
Draw the graph of

1. y = – 3x
2. y = x – 4

Solution:
To draw graph, we need to find out some points.
1. for y = – 3x, let us first substituting values & check
put x = 0
y = – 3 x 0 = 0 ∴ (0,0) is a point
put x = 1
y = – 3 x 1 = – 3 ∴ (1, – 3) is a point
If join these 2 points, we will get the line

2. for y = x – 4
put x = 0
y = 0 – 4 = -4 ∴ (0, – 4) is a point
x = 4
y = 4 – 4 = 0 ∴ (4, 0) is a point
Now let us plot the points & join them on graph

Question 10.
Find the values

Solution:
Let y = x + 3
(i) if x = 0
y = 0 + 3 = 3
∴ y = 3

(ii) y = 0
0 = x + 3
∴ x = – 3

(iii) x = – 2
y = – 2 + 3
∴ y = 1

(iv) y = -3
-3 = x + 3
∴ x = -6

Let 2x + 7 – 6 = 0
(i) x = 0
2 x 0 + y – 6 = 0
∴ 7 = 6

(ii) y = 0
2x + 0 – 6 = 0
2x = 6
x = 3

(iii) x = – 2
2 x (- 2) + y – 6 = 0
y – 10 = 0
y = 10

(iv) y = -3
2x – 3 – 6 = 0
2x = 9
x = \(\frac{9}{2}\)

Question 11.
The following is a table of , values connecting the radii of a few circles and their circumferences (Taking π = \(\frac{22}{7}\)) Illustrate the relation with a graph and find

1. The radius when the circumference is 242 units.
2. The circumference when the radius is 24.5 units.

Solution:

r = 24.5 Circumference?

Steps :

1. Draw the axis, x – axis = radius & y – axis = circumference
2. Mark the points by choosing scale of 1 cm = 7 units for x axis & 1 cm = 44 units for y – axis
3. Join the points to form a line.
4. Now for r value = 24.5 (mid value between 21 & 28), draw vertical line to touch the main line & from there drop line parallel to x axis and note where it meets y axis.

It meets between 132 & 176. Taking the mid value
i.e \(\frac{132+176}{2}\) = 154, we get circumference of 154 when r = 24.5.
Similarly, for circumference of 242, we get r value to be 38.5

Question 12.
An over-head tank is full with water. Water leaks out from it, at a constant rate of 10 litres per hour. Draw a “time-wastage” graph for this situation and find

1. The water wasted in 150 minutes
2. The time at which 75 litres of water is wasted.

Solution:
From over head tank, water leaks out at 10 l/hr.
∴ Let us see how much leakage happens with time
Time     Leakage
1 hr        10 ltrs.
2 hrs       20 ltrs.
3 hrs       30 ltrs.
4 hrs       40 ltrs.. and so on
150 min = 120 min + 30 min
= 2 hr 30 min = 2 \(\frac{1}{2}\) hrs
Now let us plot a graph between time & water leakage

Water wasted in 150 min (2\(\frac{1}{2}\) hrs) = 25 litres
Time at which 75 litre of water is wasted = 7\(\frac{1}{2}\) hrs = 450 minutes.

Students can Download Maths Chapter 2 Algebra Ex 2.5 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 2 Algebra Ex 2.5

Miscellaneous Practice Problems

Question 1.
The sum of three numbers is 58. The second numberis three times of two-fifth of the first number and the third number is 6 less than the first number. Find the three numbers.
Solution:
Here what we know
a + b + c = 58 (sum of three numbers is 58)
Let the first number be ‘x’
b = a + 3 (the second number is three times of \(\frac{1}{2}\) of the first number)
b = 3 x \(\frac{2}{5}\)x = \(\frac{6}{5}\)x
Third number = x – 6
Sum of the numbers is given as 58.
∴ x + \(\frac{6}{5}\)x + (x – 6) = 58
Multiplying by 5 throughout, we get
5 × x + 6x + 5 × (x – 6) = 58 x 5
5x + 6x + 5x – 30 = 290
∴ 16x = 290 + 30
∴ 16x = 320
∴ x = \(\frac{320}{16}\)x = 20
Answer:
1st number =20
2nd number = 3 x \(\frac{2}{5}\) x 20 = 24
3rd number = 24 – 6 = 14

Question 2.
In triangle ABC, the measure of ∠B is two third the measure of ∠A. The measure of ∠C is 20° more than the measure of ∠A. Find the measures of the three angles.
Solution:
Let angle ∠A be a°
Given that ∠B = \(\frac{2}{3}\) x ∠A = \(\frac{2}{3}\)a
& given ∠C = ∠A + 20 = a + 20
Since A, B & C are angles of a triangle, they add up to 180° (∆ property)
∠A + ∠B + ∠C = 180°
a + \(\frac{2}{3}\) a + a + 20 = 180°
\(\frac{3a+2a+3a}{3}\) + 20 = 180°
\(\frac{8a}{3}\) = 180 – 20 = 160°

∠C = 80°

Question 3.
Two equal sides of an isosceles triangle are 5y – 2 and 4y + 9 units. The third side is 2y + 5 units. Find y and the perimeter of the triangle.
Solution:
Given that 5y – 2 & 4y + 9 are the equal sides of an isosceles triangle.
The 2 sides are equal

∴ 5y – 4y = 9 + 2 (by transposing)
∴ y = 11
∴ 1st side = 5y – 2 = 5 x 11-2 = 55 – 2 = 53
2nd side = 53
3rd side = 2y + 5 = 2 x 11 + 5 = 22 + 5 = 27
Perimeter is the sum of all 3 sides
∴ P = 53 + 53 + 27 = 133 units

Question 4.
In the given figure, angle XOZ and angle ZOY form a linear pair. Find the value of x.

Solution:
Since ∠XOZ & ∠ZOY form a linear pair,
by property, we have their sum to be 180°
∴ ∠XOZ + ∠ZOY = 180°
3x – 2 + 5x + 6 = 180°
8x + 4 = 180 = 8x = 180 – 4
∴ 8x = 176 ⇒ x = \(\frac{176}{8}\) ⇒ x = 22°
XOZ = 3x – 2 = 3 x 22 – 2 = 66 – 2 = 64°
YOZ = 5x + 6 = 5 x 22 + 6
= 110 + 6 = 116

Question 5.
Draw a graph for the following data:

Does the graph represent a linear relation?
Solution:
Graph between side of square & area

When we plot the graph, we observe that it is not a linear relation.

Challenging Problems

Question 6.
Three consecutive integers, when taken in increasing order and multiplied by 2, 3 and 4 respectively, total up to 74. Find the three numbers.
Solution:
Let the 3 consecutive integers be ‘x’, ’x + 1 & ‘x + 2’
Given that when multiplied by 2, 3 & 4 respectively & added up, we get 74

Simplifying the equation, we get
2x + 3x + 3 + 4x + 8 = 74
9x + 11 = 74
9x = 63 ⇒ x = \(\frac{63}{9}\) = 7
First number = 7
Second numbers = x + 1
Third numbers = x + 2
∴ The numbers are 7, 8 & 9

Question 7.
331 students went on a field trip. Six buses were filled to capacity and 7 students had to travel in a van. How many students were there in each bus?
Solution:
Let the number of students in each bus be ‘x’
number of students in 6 buses = 6 × x = 6x
Apart from 6 buses, 7 students went in van
A total number of students is 331
∴ 6x + 7 = 331
∴ 6x = 331 – 7 = 324
∴ x = \(\frac{324}{6}\) = 54
∴ There are 54 students in each bus.

Question 8.
A mobile vendor has 22 items, some which are pencils and others are ball pens. On a particular day, he is able to sell the pencils and ball pens. Pencils are sold for ₹ 15 each and ball pens are sold at ₹ 20 each. If the total sale amount with the vendor is ₹ 380, how many pencils did he sell?
Solution:
Let vendor have ‘p’ number of pencils & ‘b’ number of ball pens
Given that total number of items is 22
∴ p + b = 22
Pencils are sold for ₹ 15 each & ball pens for ₹ 20 each
total sale amount = 15 x p + 20 x b
= 15p + 20b which is given to be 380.
∴ 15p + 20b = 380
Dividing by 5 throughout,
\(\frac{15p}{5}\) + \(\frac{20b}{5}\) = \(\frac{380}{5}\)
⇒ 3p + 4b = 76
Multiplying equation (1) by 3 we get
3 x 7 + 3 x b = 22 x 3
⇒ 3p + 3b = 66
Equation (2) – (3) gives

∴ b = 10
∴ p = 12
He sold 12 pencils

Question 9.
Draw the graph of the lines y = x, y = 2x, y = 3x and y = 5x on the same graph sheet. Is there anything special that you find in these graphs?
Solution:

1. y = x
2. y = 2x
3. y = 3x
4. y = 5x

1. y = x
When x = 1, y = 1
x = 2, y = 2
x = 3, y = 2

2. y = 2x
When x = 1, y = 2
X = 2, y = 4
X = 3, y = 6

3. y = 3x
when x = 1, y = 3
x = 2, y = 6
x = 3, y = 9

4. y = 5x
When x = 1, y = 5
x = 2, y = 10
x = 3, y = 15

When we plot the alijpve points & join the points to form line, we notice that the lines become progressively steeper. In other words, the slope keeps increasing.

Question 10.
Consider the number of angles of a convex polygon and the number of sides of that polygon. Tabulate as follows:

Use this draw a graph illustrating the relations hip between the number of angles and the number of sides of a polygon.
Solution:
Angles:

Students can Download Maths Chapter 2 Algebra Additional Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 2 Algebra Additional Questions

Question 1.
A bus runs constantly at a speed of 60 km/hr. Draw a time-distance graph for the situation. Also find the

1. Time taken to cover 360 km.
2. Distance covered in 4\(\frac{1}{2}\) hours.

Solution:
Given the bus runs constantly at a speed of 60 km/hr.
i.e. for 1 hour = 60 km
2 hours = 2 x 60 = 120 km
3 hours = 3 x 60 = 180 km
We can tabulate as below,

Take a suitable scale:

1. Mark the number of hours on the x – axis
2. Mark the distance in km on the y – axis
3. Plot the points (1, 60), (2,120), (3, 180), (4, 240) and (5, 300)
4. Join the points and get a straight line from the graph.
   • Time taken to cover 360 km is 6 hours.
   • Distance covered in 4\(\frac{1}{2}\) hours is

Question 2.
A company dealing with finance gives 15% simple interest on deposits made by senior citizens. Illustrate by a graph the relation between the deposit and the interest gained. Use the graph to complete the following.

1. The annual interest obtainable for investment of ₹ 450
2. The amount a senior citizen has to invest to get an annual simple interest of ₹ 100.

Solution:
Using the formula for calculating the simple interest, the following table of values is prepared.


These are the points to be plotted in the graph sheet, let us take the deposits along x – axis and annual simple interest along y – axis.
We choose the scale as follows.
Then we plot the points and draw the straight line.
From the graph we find.

1. Corresponding to ₹ 450 on the x – axis, we get the interest as ₹ 67.5 on the y – axis.
2. Corresponding to ₹ 100 on the y – axis we get the deposit as ₹ 666 on the x – axis.

Question 3.
The following table gives the quantity of petrol and its cost

Plot the graph.
Solution:

1. Take a suitable scale on both the axes. Here we take on the x – axis
   1 cm = 1 litre, on the y axis 1 cm = 70 rupees.
2. Mark number of litres of petrol along the x – axis
3. Mark the cost of petrol along the y – axis.
4. Plot the points (1, 70), (2, 140), (3, 210), (4, 280), (5, 350), (6, 420) and (7,490)
5. Join the points.

The graph can help us to estimate few more things also.

Students can Download Maths Chapter 2 Algebra Ex 2.3 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 2 Algebra Ex 2.3

Question 1.
Fill in the blanks:

Question (i)
X – axis and Y – axis intersect at ……..
Answer:
Origin (0, 0)

Question (ii)
The coordinates of the point in third quadrant are always ……….
Answer:
negatives

Question (iii)
(0, -5) point lies on ………. axis.
Answer:
y – axis.

Question (iv)
The x coordinate is always ……… on the y – axis.
Answer:
Zero

Question (v)
Coordinates are the same for a line parallel to Y – axis.
Answer:
X.

Question 2.
Say True or False:

Question (i)
(-10, 20) lies in the second quadrant.
Answer:
True
Hint:
(-10, 20)
x = – 10,
y = 20
∴ (- 10, 20) lies in second quadrant – True

Question (ii)
(-9, 0) lies on the x – axis.
Answer:
True
Hint:
(-9, 0) on x – axis, Y – coordinate is always zero.
∴ (-9, 0) lies on x – axis – True

Question (iii)
The coordinates of the origin are (1, 1).
Answer:
False
Hint:
Coordinate of origin is (0, 0), not (1, 1). Hence – False

Question 3.
Find the quadrants without plotting the points on a graph sheet.
(3, -4), (5, 7), (2, 0), (- 3, – 5), (4, – 3), (- 7, 2), (- 8, 0), (0,10), (- 9, 50).
Solution:

• If x & y coordinate are positive – I quad
• If x is positive, y is negative – IV quad
• If x is negative, y is positive – II quad
• If both are negative, then – III quad

Question 4.
Plot the following points in a graph sheet.
A(5, 2), B(- 7, – 3), C(- 2, 4), D(- 1, – 1), E(0, – 5), F(2, 0), G(7, – 4), H(- 4, 0), 1(2,3), J(8, – 4) K (0, 7).
Solution:

Question 5.
Use the grid graph to determine the coordinates where each figure is located.

a) Star ……..
b) Bird ……..
c) Red Circle ……..
d) Diamond ……..
e) Triangle ……..
f) Ant ……..
g) Mango ……..
h) Housefly ……..
i) Medal ……..
j) Spider ……..
Solution:
a) Star (3, 2)
b) bird (-2, 0)
c) Red Circle (-2, -2)
d) Diamond (-2, 2)
e) Triangle (-1, -1)
f) Ant (3, -1)
g) Mango (0, 2)
h) Housefly (2, 0)
i) Medal (-3, 3)
j) Spider (0, -2)

Students can Download Maths Chapter 2 Algebra Ex 2.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 2 Algebra Ex 2.2

Question 1.
Fill in the blanks:

Question (i)
The solution of the equation ax + b = 0 is ………
Answer:
– \(\frac{b}{a}\)
Solution:
ax + b = 0
ax = – b
∴ x = – \(\frac{b}{a}\)

Question (ii)
If a and b are positive integers then the solution of the equation ax = b has to be always ………
Answer:
positive
Hint:
Since a & b are positive integers,
The solution to the equation ax = b is x = – \(\frac{b}{a}\) is also positive.

Question (iii)
One-sixth of a number when subtracted from the number itself gives 25. The number is ……..
Answer:
30
Hint:
Let the number be x.
As per question, when one sixth of number is subtracted from itself it gives 25
x – \(\frac{x}{6}\) = 25
∴ \(\frac{6x-x}{6}\) = 25
∴ \(\frac{5x}{6}\) = 25
∴ x = \(\frac{25×6}{3}\) = 5 x 6 = 30

Question (iv)
If the angles of a triangle are in the ratio 2 : 3 : 4 then the difference between the greatest and the smallest angle is
Answer:
40°
Hint:
Given angles are in the ratio 2 : 3 : 4
Let the angles be 2x, 3x & 4x
Since sum of the angles of a triangle is 180°,
We get
2x + 3x + 4x = 180
∴ 9x = 180
∴ X = \(\frac{180}{9}\) = 20°
∴ The angles are 2x = 2 x 20 = 40°
3x = 3 x 20 = 60°
4x = 4 x 20 = 80°
∴ Difference between greatest & smallest angle is
80° – 40° = 40°

Question (v)
In an equation a + b = 23. The value of a is 14 then the value of b is ……..
Answer:
b = 9
Hint:
Given equation is a + b = 23
a = 14
14 + b = 23
b = 23 – 14 = 9
b = 9

Question 2.
Say True or False

Question (i)
“Sum of a number and two times that number is 48” can be written as y + 2y = 48
Answer:
True
Hint:
Let the number be ‘y’
Sum of number & two times that number is 48
Can be written as y + 2y = 48 – True

Question (ii)
5(3x + 2) = 3(5x – 7) is a linear equation in one variable.
Answer:
True
Hint:
5 (3x + 2) = 3 (5x – 7) is a linear equation in one variable – ‘x’ – True

Question (iii)
x = 25 is the solution of one third of a number is less than 10 the original number.
Answer:
False
Hint:
One third of number is 10 less than original number.
Let number be ‘x’ Therefore let us frame the equation
\(\frac{x}{3}\) = x – 10
∴ x = 3x – 30
3x – x = 30
2x = 30
x = 15 is the solution

Question 3.
One number is seven times another. If their difference is 18, find the numbers.
Solution:
Let the numbers be x & y
Given that one number is 7 times the other & that the difference is 18.
Let x = 7y
also, x – y = 18 (given)
Substituting for x in the above
We get 7y – y = 18
∴ 6y = 18
y = \(\frac{18}{6}\) = 3
x = 7y = 7 x 3 = 21
The number are 3 & 21

Question 4.
The sum of three consecutive odd numbers is 75. Which is the largest among them?
Solution:
Given sum of three consecutive odd numbers is 75
Odd numbers are 1, 3, 5,1,9, 11, 13,……..
∴ The difference between 2 consecutive odd numbers is always 2. or in other words, if one odd number is x, the next odd number would be x + 2 and the next number would be x + 2 + 2 = x + 4
i.e x + 4
Since sum of 3 consecutive odd nos is 75
∴ x + x + 2 + x + 4 = 75
3x + 6 = 75 ⇒ 3x = 75 – 6
3x = 69
x = \(\frac{69}{3}\) = 23
The odd numbers are 23, 23 + 2, 23 + 4
i.e 23, 25, 27
∴ Largest number is 27.

Question 5.
The length of a rectangle is \(\frac{1}{3}\)rd of its breadth. If its perimeter is 64 m, then find the length and breadth of the rectangle.
Solution:

Let length & breadth of rectangle be l and ‘6’ respectively
Given that length is \(\frac{1}{3}\) of breadth,
∴ l = \(\frac{1}{3}\) x b ⇒ l = \(\frac{b}{3}\) ⇒ b = 3l …..(1)
Also given that perimeter is 64 m
Perimeter = 2 x (l + b)
2 x l + 2 x b = 64
Substituting for vahie of b from (1), we get
2l + 2(3l) = 64
∴ 2l + 6l = 64
8l = 64
∴ l = \(\frac{64}{8}\) = 8 m
b = 3l = 3 x 8 = 24 m
length l = 8 m & breadth h = 24 m

Question 6.
A total of 90 currency notes, consisting only of X5 and ?10 denominations, amount to ₹ 500. Find the number of notes in each denomination.
Solution:
Let the number of ₹ 5 notes be ‘x’
And number of ₹ 10 notes be ly ’
Total numbers of notes is x + y = 90 (given)
The total value of the notes is 500 rupees.
Value of one ₹ 5 rupee note is 5
Value of x ₹ 5 rupee notes is 5 × x = 5x
Value ofy ₹ 10 rupee notes is 10 × y = 10y
∴ The total value is 5x + 10y which is 500
we have 2 equations:
x + y = 90
5x + 1oy = 500
Multiplying both sides of (1) by 5, we get
5 × x + 5 x y = 90 x 5
5x + 5y = 450
Subtracting (3) from (2), we get

∴ y = \(\frac{50}{5}\) = 10
Substitute y = 10 in equation (1)
x + y = 90 ⇒ x + 10 = 90
⇒ x = 90 – 10 ⇒ x = 80
There are ₹ 5 denominations are 80 numbers and ₹ 10 denominations are 10 numbers

Question 7.
At present, Thenmozhi’s age is 5 years more than that of Murali’s age. Five years ago, the ratio of Thenmozhi’s age to Murali’s age was 3:2. Find their present ages.
Solution:
Let present ages ofThenmozhi & Murali be l’ & ‘m’
Given that at present
Thenmozhi’s age is 5 years more than Murali
∴ t = m + 5
5 years ago, Thenmozhi’s age would be t – 5
& Murali’s age would be m – 5
Ratio of their ages is given as 3 : 2
∴ \(\frac{t-5}{m-5}\) = \(\frac{3}{2}\) [∴ By cross multiplication]

2(t – 5) = 3(m – 5)
2 x t – 2 x 5 = 3 x m – 3 x 5 ⇒ 2t – 10 = 3m – 15
Substituting for t from (1)
2(m + 5) – 10 = 3m – 15
2m + 10 – 10 = 3m – 15
2m = 3m – 15
3m – 2m = 15
m = 15
t = m + 5 = 15 + 5 = 20
∴ Present ages of Thenmozhi & Murali are 20 & 15

Question 8.
A number consists of two digits whose sum is 9. If 27 is subtracted from the original number, its digits are interchanged. Find the original number.
Solution:
Let the units/digit of a number be ‘u’ & tens digit of the number be ‘t’
Given that sum of it’s digits is 9
∴ t + u = 9 ….(1)
If 27 is subtracted from original number, the digits are interchanged
The number is written as 10t + u
[Understand: Suppose a 2 digit number is 21,
it can be written as 2 x 10 + 1
∴ 32 = 3 x 10 + 2
45 = 4 x 10 + 5
tu = t x 10 + u = 10t + u]
Given that when 27 is subtracted, digits interchange
10t + u – 27 = 10u + t (number with interchanged digits)
∴ By transposition & bringing like variables together
10t – t + 10u =27
∴ 9t – 9u = 27
Dividing by ‘9’ throughout, we get
\(\frac{9t}{9}\) – \(\frac{9u}{9}\) = \(\frac{27}{9}\) ⇒ t – u = 3 ….(2)
Solving (1) & (2)

t = \(\frac{12}{2}\) = 6
∴ u = 3
t = 6 substitute in (1)
t + u = 9
⇒ 6 + u = 9
⇒ u = 9 – 6 = 3
Hence the number is 63.

Question 9.
The denominator of a fraction exceeds its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, we get \(\frac{3}{2}\). Find the original fraction.
Solution:
Let the numerator & denominator be ‘n’ & ‘d’
Given that denominator exceeds numerator by 8
∴ d = n + 8
If numerator increased by 17 & denominator decreased by 1,
it becomes (n + 17) & (d – 1), fraction is \(\frac{3}{2}\).
i.e = \(\frac{n+17}{d-1}\) = \(\frac{3}{2}\) by cross multiplying, we get
\(\frac{n+17}{d-1}\) = \(\frac{3}{2}\)
2(n + 17) = 3(d – 1)
2n + 2 x 17 = 3d – 3

∴ 34 + 3 = 3d – 2n
∴ 3d – 2n = 37
Substituting eqn. (1) in (2), we get,
3 x (n + 8) – 2n = 37
3n + 3 x 8 – 2n = 37

∴ n = 37 – 24 = 13
d = n + 8 = 13 + 8 = 21
The fraction is \(\frac{n}{d}\) = \(\frac{13}{21}\)

Question 10.
If a train runs at 60 km/hr it reaches its destination late by 15 minutes. But, if it runs at 85 kmph it is late by only 4 minutes. Find the distance to be covered by the train.
Solution:
Let the distance to be covered by train be ‘d’
Using the formula, time take (t) = \(\frac{Distance}{Speed}\)
Case 1:
If speed = 60 km/h
The time taken is 15 minutes more than usual (t + \(\frac{15}{60}\))
Let usual time taken be ‘t’ hrs.
Caution:
Since speed is given in km/hr, we should take care to maintain all units such as time should be in hour and distance should be in km.
Given that in case 1, it takes 15 min. more
15 m = \(\frac{15}{60}\) hr = \(\frac{1}{4}\) hr.
∴ Substituting in formula
\(\frac{Distance}{Speed}\) = time
∴ \(\frac{d}{60}\) = t + \(\frac{1}{4}\)
Since usually it takes ‘t’ hr, but when running at 60 km/h, it takes 15 min (\(\frac{1}{4}\)) extra.
Multiplying by 60 on both sides
d = 60 x t + 60 x \(\frac{1}{4}\) = 60t + 15 …..(1)
Case 2:
Speed is given as 85 km/h
Time taken is only 4 min (\(\frac{4}{60}\) hr) more than usual time
time taken = (t + \(\frac{1}{15}\)) hr. (\(\frac{4}{60}\) = \(\frac{1}{15}\))
Using the formula,
\(\frac{Distance}{Speed}\) = time
∴ \(\frac{d}{85}\) = t + \(\frac{1}{15}\)
Multiplying by 85 on both sides
\(\frac{d}{85}\) x 85 = 85 x t + 85 x \(\frac{1}{15}\)
∴ d = 85 t + \(\frac{17}{3}\)
From (1) & (2), we will solve for ‘t’
Equating & eliminating ‘d we get

By transposing, we get
15 – \(\frac{17}{3}\) = 85t – 60t
\(\frac{45-17}{3}\) = 25t
∴ 25t = \(\frac{28}{3}\)
∴ t = \(\frac{28}{3×25}\) = \(\frac{28}{75}\) hr (\(\frac{28}{75}\) x 60 = 22.4 min)
Substituting this value of ‘t’ in eqn. (1), we get
d = 60t+ 15 = 60 x \(\frac{28}{75}\) + 15 = \(\frac{1680}{75}\) + 15 = 22.4 + 15
= 37.4 km

Objective Type Questions

Question 11.
Sum of a number and its half is 30 then the number is ……..
(a) 15
(b) 20
(c) 25
(d) 40
Answer:
(b) 20
Hint:
Let number be V
half of number is \(\frac{x}{2}\)
Sum of number and it’s half is given by
x + \(\frac{x}{2}\) = 30 [Multiplying by 2 on both sides]
2x + x = 30 x 2
3x = 60
x = \(\frac{60}{3}\) = 20

Question 12.
The exterior angle of a triangle is 120° and one of its interior opposite angle 58°, then the other opposite interior angle is ………
(a) 62°
(b) 72°
(c) 78°
(d) 68°
Answer:
(a) 62°

Hint:
As per property of ∆, exterior angle is equals to sum of interior opposite angles Let the other interior angle to be found be ‘x’
∴ x + 58 = 120°
∴ x = 120 – 58 = 62°

Question 13.
What sum of money will earn ₹ 500 as simple interest in 1 year at 5% per annum?
(a) 50000
(b) 30000
(c) 10000
(d) 5000
Answer:
(c) 10000
Hint:
Let sum of money be ‘P’
Time period (n) is given as 1 yr.
Rate of simple interest (r) is given as 5% p.a
∴ As per formula for simple interest.
S.I = \(\frac{Pxrxn}{100}\) = \(\frac{px5x1}{100}\)
P x 5 x 1 = 500 x 100
∴ P = \(\frac{500×100}{5}\) = 100 x 100 = 10,000

Question 14.
The product of LCM and HCF of two numbers is 24. If one of the number is 6, then the other number is ………
(a) 6
(b) 2
(c) 4
(d) 8
Answer:
(c) 4
Hint:
Product of LCM & HCF of 2 numbers is always product of the numbers. [this is property]
Product of LCM & HCF is given as 24.
∴ Product of the 2 nos. is 24
Given one number is 6.
Let other number be ‘x’
∴ 6 × x = 24
∴ x = \(\frac{24}{6}\) = 4

Students can Download Maths Chapter 2 Algebra Ex 2.1 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 2 Algebra Ex 2.1

Question 1.
Fill in the blanks

Question (i)
The value of x in the equation x +5 12 ¡s ……….
Answer:
7
Hint:
Given,
x + 5 = 12
x = 12 – 5 = 7 (by transposition method)
Value of x is 7

Question (ii)
The value ofy in the equation y – 9 = (-5) + 7 is ……….
Answer:
11
Hint:
Given,
y – 9 = (- 5) + 7
y – 9 = 7 – 5 (re-arranging)
y – 9 = 2
∴ y = 2 + 9 = 11 (by transposition method)

Question (iii)
The value of m in the equation 8m = 56 is ………
Answer:
7
Hint:
Given,
8m = 56
Divided by 8 on both sides
\(\frac{8xm}{8}\) = \(\frac{56}{8}\)
∴ m = 7

Question (iv)
The value ofp in the equation \(\frac{2p}{3}\) = 10 is ……….
Answer:
1
Hint:
Given,
\(\frac{2p}{3}\) = 10
Multiplying by 3 on both sides,

∴ p = 15

Question (v)
The linear equation in one variable has ……… Solution.
Answer:
one.

Question 2.
Say True or False.

Question (i)
The shifting of a number from one side of an equation to other is called transposition.
Answer:
True

Question (ii)
Linear equation in one variable has only one variable with power 2.
Answer:
False
[Linear equation in one variable has only one variable with power one – correct statement]

Question 3.
Match the following :

(A) (i), (ii), (iv), (iii), (v)
(B) (iii), (iv), (i), (ii), (v)
(C) (iii), (i), (iv), (v), (ii)
(D) (iii), (i), (v), (iv), (ii)
Answer:
(C) (iii),(i), (iv), (v), (ii)
Hint:
a. \(\frac{x}{2}\) = 10,
multiplying by 2 on both sides, we get
\(\frac{x}{2}\) x 2 = 10 x 2 ⇒ x = 20

b. 20 = 6x – 4
by transposition ⇒ 20 + 4 = 6x
6x = 24
dividing by 6 on both sides,
\(\frac{6x}{6}\) = \(\frac{24}{6}\) ⇒ x = 4

c. 2x – 5 = 3 – x
By transposing the variable ‘x’, we get
2x – 5 + x = 3
by transposing – 5 to other side,
2x + x = 3 + 5
∴ 3x = 8

∴ x = \(\frac{8}{3}\)

d. 7x – 4 – 8x = 20
by transposing – 4 to other side,
7x – 8x = 20 + 4
– x = 24
∴ x = – 24
\(\frac{4}{11}\) – x = \(\frac{-7}{11}\)
Transposing \(\frac{4}{11}\) to other side,
– x = \(\frac{-7}{11}\)\(\frac{-4}{11}\) = \(\frac{-7-4}{11}\) = \(\frac{-11}{11}\) = – 1
∴ – x = – 1 ⇒ x = 1

Question 4.
Find x:

Question (i)
\(\frac{2x}{3}\) – 4 = \(\frac{10}{3}\)
Solution:
Transposing -4 to other side, it becomes +4
∴ \(\frac{2x}{3}\) = \(\frac{10}{3}\) + 4
Taking LCM & adding,
\(\frac{2x}{3}\) = \(\frac{10}{3}\) + \(\frac{4}{1}\) = \(\frac{10+12}{3}\) = \(\frac{22}{3}\)
\(\frac{2x}{3}\) = \(\frac{22}{3}\)
Multiplying by 3 on both sides

⇒ 2x = 22
dividing by 2 on both sides,
We get \(\frac{2x}{2}\) = \(\frac{22}{2}\)
∴ x = 11

Question (ii)
y + \(\frac{1}{6}\) – 3y = \(\frac{2}{3}\)
Solution:
Transposing \(\frac{1}{6}\) to the other side,
y – 3y = \(\frac{2}{3}\) – \(\frac{1}{6}\)
Taking LCM,
– 2y = \(\frac{2}{3}\) – \(\frac{1}{6}\) = \(\frac{2×2-1}{6}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
∴ – 2y = \(\frac{1}{2}\) ⇒ 2y = – \(\frac{1}{2}\)
dividing by 2 or both sides.

Question (iii)
\(\frac{1}{3}\) – \(\frac{x}{3}\) = \(\frac{7x}{12}\) + \(\frac{5}{4}\)
Transposing \(\frac{-x}{3}\) to the other side, it becomes + \(\frac{x}{3}\)
∴ \(\frac{1}{3}\) = \(\frac{7x}{12}\) + \(\frac{5}{4}\) + \(\frac{x}{3}\)
Transposing \(\frac{5}{4}\) to the other side, it becomes \(\frac{-5}{4}\)
\(\frac{1}{3}\) + \(\frac{5}{4}\) = \(\frac{7x}{12}\) + \(\frac{x}{3}\)
Multiply by 12 throughout
[we look at the denominators 3,4, 12, 3 and take the LCM, which is 12]

4 – 15 = 7x + x × 4
-11 = 7x + 4x
11x = – 11
x = -1

Question 5.
Find x:

Question (i)
-3(4x + 9) = 21
Solution:

Expanding the bracket,
-3 × 4x + (-3) × 9 = 21
-12x + (-27) = 21
-12x – 27 = 21
Transposing – 27 to other side, it becomes +27
-12x = 21 + 27 = 48
12x = 48 ⇒ 12x = -48
Dividing by 12 on both sides

⇒ x = – 4

Question (ii)
20 – 2 ( 5 – p) = 8
Solution:

Expanding the bracket,
20 – 2 x 5 – 2 x (-p) = 8
20 – 10 + 2 + p = 8 (-2 x -P = 2p)
10 + 2p = 8 transporting 10 to other side
2P = 8 – 10 = -2
∴ 2p = -2
∴ p = -1

Question (iii)
(7x – 5) – 4(2 + 5x) = 10(2 – x)
Solution:

Expanding the brackets,
7x – 5 – 4 × 2 – 4 × 5x = 10 × 2 + 10 × (-x)
7x – 5 – 8 – 20x = 20 – 10x
7x – 13 – 20x = 20 – 10x
Transposing 10x & -13, we get
7x – 13 – 20x + 10x = 20
7x – 20x + 10x = 20 + 13,
Simplifying,
-3x = 33
∴ 3x = -33
x = \(\frac{-33}{3}\) = -11
x = -11

Question 6.
Find x and m:

Question (i)
\(\frac{3x-2}{4}\) – \(\frac{(x-3)}{5}\) = -1
Solution:
\(\frac{3x-2}{4}\) – \(\frac{(x-3)}{5}\)
Taking LCM on LHS, [LCM of 4 & 5 is 20]

∴ 11x + 2 = -20
∴ 11x = – 20 – 2 = – 22
x = \(\frac{-22}{11}\) = -2
x = -2

Question (ii)
\(\frac{m+9}{3m+15}\) = \(\frac{5}{3}\)
Solution:
\(\frac{m+9}{3m+15}\) = \(\frac{5}{3}\)
Cross multiplying, we get

∴ (m + 9) x 3 = 5 x (3m + 15)
m x 3 + 9 x 3 = 5 x 3m + 5 x 15

Transporting 3m & 75, we get
27 – 75 = 15m – 3m
-48 = 12m

⇒ m = -4