Bhagya

Students can download Maths Chapter 4 Geometry Ex 4.3 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 4 Geometry Ex 4.3

Miscellaneous Practice Problems

Question 1.
What are the angles of an isosceles right-angled triangle?
Solution:
Since it is a right-angled triangle
One of the angles is 90°
The other two angles are equal because it is an isosceles triangle.
The other two angles must be 45° and 45°
Angles are 90°, 45°, 45°.

Question 2.
Which of the following correctly describes the given triangle?

(a) It is a right isosceles triangle
(b) It is an acute isosceles triangle
(c) It is an obtuse isosceles triangle
(d) It is an obtuse scalene triangle
Solution:
(c) It is an obtuse isosceles triangle

Question 3.
Which of the following is not possible?
(a) An obtuse isosceles triangle.
(b) An acute isosceles triangle.
(c) An obtuse equilateral triangle.
(d) An acute equilateral triangle.
Solution:
(c) An obtuse equilateral triangle.

Question 4.
If one angle of an isosceles triangle is 124°, then find the other angles
Solution:
In an isosceles triangle, any two sides are equal. Also, the two angles are equal.
Sum of three angles of a triangle = 180°
Given one angle = 124°
Sum of other two angles = 180° – 124° = 56°
Other angles are = \(\frac{56}{2}\) = 28°
28° and 28°.

Question 5.
The diagram shows a square ABCD. If the line segment joins A and C, then mention the type of triangle so formed.
Solution:

Isosceles right-angled triangles

Question 6.
Draw a line segment AB of length 6 cm. At each end of this line segment AB, draw a line perpendicular to the line segment AB. Are these lines parallel?
Solution:

yes, they are parallel

Challenge Problems

Question 7.
Is a triangle possible with the angles 90°, 90°, and 0°, Why?
Solution:
No, a triangle cannot have more than one right angle.

Question 8.
Which of the following statements is true? Why?
(a) Every equilateral triangle is an isosceles triangle.
(b) Every isosceles triangle is an equilateral triangle
Solution:
“(a)” is true, because an isosceles triangle need not have three equal sides

Question 9.
If one angle of an isosceles triangle is 70°, then find the possibilities for the other two angles.
Solution:
(i) Given one angle = 70°
Also, it is an isosceles triangle.
Another one angle also can be 70°.
Sum of these two angles = 70° + 70° = 140°
We know that the sum of three angles in a triangle = 180°.
Third angle = 180° – 140° = 40°
One possibility is 70°, 70°, and 40°
(ii) Also if one angle is 70°
Sum of other two angles = 180° – 70° = 110°
Both are equal. They are \(\frac{110}{2}\) = 55°.
Another possibility is 70°, 55° and 55°.

Question 10.
Which of the following can be the sides of an isosceles triangle?
(a) 6 cm, 3 cm, 3 cm
(b) 5 cm, 2 cm, 2 cm
(c) 6 cm, 6 cm, 7 cm
(d) 4 cm, 4 cm, 8 cm
Solution:
(c) 6 cm, 6 cm, 7 cm

Question 11.
Study the given figure and identify the following triangles,

(a) equilateral triangle
(b) isosceles triangle
(c) scalene triangles
(d) acute triangle
(e) obtuse triangle
(f) right triangle
Solution:(a) BC = 1 + 1 + 1 + 1 = 4 cm
AB = AC = 4 cm
∆ABC is an equilateral triangle.
(b) ∆ABC and ∆AEF are isosceles triangles.
Since AB = AC = 4 cm Also AE = AF.
(c) In a scalene triangle, no two sides are equal.
∆AEB, ∆AED, ∆ADF, ∆AFC, ∆ABD, ∆ADC, ∆ABF, and ∆AEC are scalene triangles.
(d) In an acute-angled triangle all the three angles are less than 90°.
∆ABC, ∆AEF, ∆ABF, and ∆AEC are acute-angled triangles.
(e) In an obtuse-angled triangle any one of the angles is greater than 90°.
∆AEB and ∆AFC are obtuse angled triangles.
(f) In a right triangle, one of the angles is 90°.
∆ADB, ∆ADC, ∆ADE, and ∆ADF are right-angled triangles.

Question 12.
Two sides of the triangle are given in the table. Find the third side of the triangle.

Solution:
(i) between 3 and 11
(ii) between 0 and 16
(iii) between 4 and 11
(iv) between 4 and 24

Question 13.
Complete the following table:

Solution:
(i) Always acute angles
(ii) Acute angle
(iii) Obtuse angle

Students can download Maths Chapter 4 Geometry Ex 4.2 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 4 Geometry Ex 4.2

Question 1.
Draw a line segment AB = 7 cm and mark a point P on it. Draw a line perpendicular to the given line segment at P.
Solution:

Step 1 : Draw a line AB = 7 cm and take a point P anywhere on the line.
Step 2 : Place the set square on the line in such a way that the vertex which forms right angle coincides with P and one arm of the right angle coincides with the line AB.
Step 3 : Draw a line PQ through P along the other arm of the right angle of the set square.
Step 4 : The line PQ is perpendicular to the line AB at P. That is, PQ ⊥ AB
∠APQ = ∠BPQ = 90°

Question 2.
Draw a line segment LM = 6.5 cm and mark a point X not lying on it. Using a set square construct a line perpendicular to LM through X.
Solution:

Step 1 : Draw a line LM = 6.5 cm and take a point X anywhere above the line LM.
Step 2 : Place one of the arms of the right angle of a set square along the line LM and the other arm of its right angle touches the point X.
Step 3 : Draw a line through the point X meeting LM at Y.
Step 4 : The line XY is perpendicular to the line LM at Y. That is, LM ⊥ XY.

Question 3.
Find the distance between the given lines using a set square at two different points on each of the pairs of lines and check whether they are parallel.

Solution:
They are parallel

Question 4.
Draw a line segment measuring 7.8 cm. Mark a point B above it at a distance of 5 cm. Through B draw a line parallel to the given segment.
Solution:

Step 1 : Draw a line. Mark two points M and N on the line such that MN = 7.8 cm. Mark a point B any where above the line.
Step 2 : Place the set square below B in such a way that one of the edges that form a right angle lies along MN Place the scale along the other edge of the set square.
Step 3 : Holding the scale firmly, Slide the set square along the edge of the scale until the other edge of the set square reaches the point B. Through B draw a line.
Step 4 : The line MN is parallel to AB. That is, MN || AB.

Question 5.
Draw a line and mark a point R above it at a distance of 5.4 cm Through R draw a line parallel to the given line.
Solution:

Step 1 : Using a scale draw a line AB and mark a point Q on the line.
Step 2 : Place the set square in such a way that the vertex of the right angle coincides with Q and one of the edges of right angle lies along AB. Mark the point R such that QR = 5.4 cm
Step 3 : Place the scale and the set square as shown in the figure.
Step 4 : Hold the scale firmly and slide the set square along the edge of the scale until the other edge touches the point R. Draw a line RS through R.
Step 5 : The line RS is parallel to AB. That is, RS || AB.

Students can download Maths Chapter 2 Numbers and Sequences Additional Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Additional Questions

I. Choose the correct answer.

Question 1.
The sum of the exponents of the prime factors in the prime factorisation of 504 is ……….
(1) 3
(2) 2
(3) 1
(4) 6
Answer:
(4) 6

Hint: 504 = 2³ × 3² × 7¹
Sum of the exponents
= 3 + 2 + 1 = 6

Question 2.
If two positive integers a and 6 are expressible in the form a = pq² and b = p³q ; p, q being prime numbers, then L.C.M. of (a, b) is ……………
(i) pq
(2) P² q²
(3) p³ q³
(4) P³ q²
Answer:
(4) P³ q²
Hint: a = p × q² and b = p³ × q
∴ L.C.M. of (a, b) is p³ q²

Question 3.
If n is a natural number then 7³ⁿ – 4³ⁿ is always divisible by …………….
(1) 11
(2) 3
(3) 33
(4) both 11 and 3
Answer:
(4) both 11 and 3
Hint: 7³ⁿ – 4³ⁿ is of the form a²ⁿ – b²ⁿ which is divisible by both a – b and a + b. So 7³ⁿ – 4³ⁿ is divisible by both 7 – 4 = 3 and 7 + 4 = 11.

Question 4.
The value of x when 200 = x (mod 7) is …………………
(1) 3
(2) 4
(3) 54
(4) 12
Answer:
(2) 4
Hint:

200 ≡ x (mod 7)
200 ≡ 4 (mod 7)
The value of x = 4

Question 5.
The common difference of the A.P.
\(\frac { -2 }{ 2b } \) , \(\frac { 1-6b }{ 2b } \), \(\frac { 1-12b }{ 2b } \) is …………….
(1) 2b
(2) -2b
(3) 3
(4) -3
Answer:
(4) -3
Hint:
\(\frac { 1-12b }{ 2b } \) – \(\frac { 1-6b }{ 2b } \) = \(\frac { 1-12b-1+6b }{ 2b } \) = \(\frac { -6b2 }{ 2b } \) = \(\frac { -6b }{ 2b } \) = -3

Question 6.
Which one of the following is not true?
(1) A sequence is a real valued function defined on N.
(2) Every function represents a sequence.
(3) A sequence may have infinitely many terms.
(4) A sequence may have a finite number of terms.
Answer:
(2) Every function represents a sequence.
Hint: A sequence is a function whose domain is the set of natural numbers.

Question 7.
The 8^th term of the sequence 1,1,2,3,5,8, ………. is ……….
(1) 25
(2) 24
(3) 23
(4) 21
Answer:
(4) 21
Hint: In fibonacci sequence
F_n = F_n-1 + F_n-2
F₈ = F₇ + F₆
8^th term = 6^th term + 7^th term
= 8 + (5 + 8) = 21

Question 8.
The next term of \(\frac { 1 }{ 20 } \) in the sequence \(\frac { 1 }{ 2 } \),\(\frac { 1 }{ 6 } \),\(\frac { 1 }{ 12 } \),\(\frac { 1 }{ 20 } \) is ……………
(1) \(\frac { 1 }{ 24 } \)
(2) \(\frac { 1 }{ 22 } \)
(3) \(\frac { 1 }{ 30 } \)
(4) \(\frac { 1 }{ 18 } \)
Answer:
(3) \(\frac { 1 }{ 30 } \)
Hint:
The general term t_n = \(\frac{1}{n(n+1)}\)
⇒ t₅ = \(\frac{1}{5(6)}\) = \(\frac { 1 }{ 30 } \)

Question 9.
If a, 6, c, l, m are in A.P, then the value of a – 46 + 6c – 4l + m is …………
(1) 1
(2) 2
(3) 3
(4) 0
Answer:
(4) 0
Hint: Given, a, b, c, l, m, are in A.P.
a = a; b = a + d; c = a + 2 d; 1 = a + 3d;
m = a + 4d [The general form of A.P.]
a – 4b + 6c – 4l + m
= a – 4(a + d) + 6(a + 2d) – 4 (a + 3d) + a + 4d
= a – 4a – 4d+ 6a + 12 d – 4a – 12d + a + 4d
= a + 6a + a – 4a – 4a
= 8a – 8a
= 0

Question 10.
If a, b, c are in A.P. then \(\frac { a-b }{ b-c } \) is equal to ……………
(1) \(\frac { a }{ b } \)
(2) \(\frac { b }{ c } \)
(3) \(\frac { a }{ c } \)
(4) 1
Answer:
(4) 1
Hint: a, b, c are in A.P.
b – a = c-b [common difference is same t₂ – t₁ = t₃ – t₂]
\(\frac { b-a }{ c-b } \) = 1 ⇒ \(\frac{-(a-b)}{-(b-c)}\) = 1
∴ \(\frac { a-b }{ b-c } \) = 1

Question 11.
If the n^th term of a sequence is 100n + 10, then the sequence is ……
(1) an A.P.
(2) a G..P.
(3) a constant sequence
(4) neither A.P. nor G.P.
Answer:
(1) an A.P.
Hint: t_n = 100/n + 10
t₁ = 100 + 10 = 110
t₂ = 200+ 10 = 210
t₃ = 300+ 10 = 310
∴ The series 110, 210, 310 …………. are in A.P.

Question 12.
If a₁, a₂, a₃, …… are in A.P. such that \(\frac{a_{4}}{a_{7}}=\frac{3}{2}\), then the 13th term of the A.P. is ……………..
(1) \(\frac { 3 }{ 2 } \)
(2) 0
(3) 12a1
(4) 14a1
Answer:
(2) 0
Hint:
\(\frac{a_{4}}{a_{7}}=\frac{3}{2}\)
2a₄ = 3a₇
2(a + 3d) = 3(a + 6d)
2a + 6d = 3a + 18d
0 = a + 12d
[t_n = a + (n – 1)d]
0 = t₁₃

Question 13.
If the sequence a₁, a₂, a₃ ,… is in A.P., then the sequence a₅, a₁₀, a₁₅, …. is …..,
(1) a G.P.
(2) an A.P.
(3) neither A.P nor G.P.
(4) a constant sequence
Answer:
(2) an A.P.
Hint: a₅, a₁₀, a₁₅, ……….. = a + 4d, a + 9d, a + 14d
t₂ – t₁ = a + 9d – (a + 4d) = 5d
t₃ – t₂ = a + 14d – (a + 9d) = 5d
Common difference is 5d
If terms of an A.P are chosen at equal intervals then they form an A.P.

Question 14.
If k + 2,4k – 6, 3k – 2 are the 3 consecutive terms of an A.P, then the value of K is ……………
(1) 2
(2) 3
(3) 4
(4) 5
Answer:
(2) 3
Hint: Here, a = k + 2, b = 4k-6, c = 3k-2
We know that a, b, c are in A.P.
b – a = c – b ⇒ 2b = a + c
2(4k – 6) = k + 2 + 3k – 2
8k – 12 = 4k ⇒ 4k = 12
K = \(\frac { 12 }{ 4 } \) = 3

Question 15.
If a, b, c, l, m, n are in A.P, then 3a + 7, 3b + 7, 3c + 7, 31 + 7, 3m + 7, 3n + 7 form ………..
(1) a G . P.
(2) an A.P.
(3) a constant sequence
(4) neither A.P. nor G.P.
Answer:
(2) an A.P.
Hint: In an A.P, if each term is multiplied by a constant or added by a constant, the resulting sequence is an A.P.

Question 16.
If the third term of a G.P. is 2, then the product of first 5 terms is ……………..
(1) 5²
(2) 2⁵
(3) 10
(4) 15
Answer:
(2) 2⁵
Hint: Consider the 5 terms of the G.P be \(\frac{a}{r^{2}}\),\(\frac { a }{ r } \), a, ar, ar²
Product of 5 terms = \(\frac{a}{r^{2}} \times \frac{a}{r} \times a \times a r \times ar^{2}\), = a⁵, 2⁵ (Given a = 2)

Question 17.
If a, b, c are in G.P., then \(\frac { a-b }{ b-c } \) is equal to …………..
(1) \(\frac { a }{ b } \)
(2) \(\frac { b }{ a } \)
(3) \(\frac { a }{ c } \)
(4) \(\frac { c }{ b } \)
Ans.
(1) \(\frac { a }{ b } \)
Hint: Let the common ratio be “r”

Question 18.
If x, 2x + 2, 3x + 3, are in G.P., then 5x, 10x + 10, 15x + 15, form …………..
(1) anA.P.
(2) a G.P.
(3) a constant sequence
(4) neither A.P. nor a G.P.
Answer:
(2) a G.P.
Hint: If a G.P. is multiplied by a constant then the sequence is also a G.P.
In the given question each term is multiplied by 5

Question 19.
The sequence – 3, – 3, – 3, …… is ……..
(1) an A.P. only
(2) a G.P. only
(3) neither A.P. nor G.P.
(4) both A.P. and G.P.
Answer:
(4) both A.P. and G.P.
Hint: The given sequence is constant.
The sequence is A.P. and G.P.

Question 20.
If the product of the first four consecutive terms of a G.P is 256 and if the common ratio is 4 and the first term is positive, then its 3^rd term is …….
(1) 8
(2) \(\frac { 1 }{ 16 } \)
(3) \(\frac { 1 }{ 32 } \)
(4) 16
Answer:
(1) 8
Hint: The general form of the G.P. is a, ar, ar², ar³, ar⁴, …………..
By data, a (ar) (ar²) (ar³) = 256
a⁴ × r⁶ = 256
[Given r = 4]

Question 21.
In G.P, t₂ = \(\frac { 3 }{ 5 } \) and t₃ = \(\frac { 1 }{ 5 } \) Then the common ratio is ……….
(1) \(\frac { 1 }{ 5 } \)
(2) \(\frac { 1 }{ 3 } \)
(3) 1
(4) 5
Answer:
(2) \(\frac { 1 }{ 3 } \)
Hint: common ratio is
(r) = \(\frac{t_{3}}{t_{2}}=\frac{1}{5} \times \frac{5}{3} \Rightarrow r=\frac{1}{3}\)

Question 22.
If x ≠ 0, then 1 + sec x + sec² x + sec³ x + sec⁴ x + sec⁵ x is equal to ……………
(1) (1 + sec x) (sec² x + sec³ x + sec⁴ x)
(2) (1 + sec x) (1 + sec² x + sec⁴ x)
(3) (1 – sec x) (sec x + sec³ x + sec⁵ x)
(4) (1 + sec x) (1 + sec³ x + sec⁴ x)
Answer:
(2) (1 + sec x) (1 + sec² x + sec⁴ x).
Hint:
1 + sec x + sec² x + sec³ x
+ sec⁴ x + sec⁵ x
= (1 + sex x) + sec² x (1 + sec x) + sec⁴ x (1 + see x)
= (1 + sec x) + sec² x (1 + sec x) + sec⁴ x (1 + secx)
= (1 + sec x) (1 + sec² x + sec⁴ x)

Question 23.
If the n^th term of an A.P. is t_n = 3 – 5n, then the sum of the first n terms is …………….
(1) \(\frac { n }{ 2 } \) [1 – 5n]
(2) n (1 – 5n)
(3) \(\frac { n }{ 2 } \) (1 + 5n)
(4) \(\frac { n }{ 2 } \) (1 + n)
Answer:
(1) \(\frac { n }{ 2 } \) [1 – 5n]
Hint:
t_n =. 3 – 5(n)
t₁ = 3 – 5(1) = -2 ; t₂ = 3 – 10 = -7
a = -2, d = -7 – (-2) = -5
S_n = \(\frac { n }{ 2 } \) [2a + (n – 1)d]
= \(\frac { n }{ 2 } \) [-4 + (n – 1) (-5)]
= \(\frac { n }{ 2 } \) [- 4 -5n + 5] = \(\frac { n }{ 2 } \) [1 – 5n]

Question 24.
The common ratio of the G.P. a^m-n, a^m, a^m+n is …………
(1) a^m
(2) a^-m
(3) aⁿ
(4) a^-n
Answer:
(3) aⁿ

Question 25.
If 1 + 2 + 3 + … + n = k then 1³ + 2³ + ……. + n³ is equal to …………
(1) K²
(2) K³
(3) \(\frac{k(k+1)}{2}\)
(4) (K + 1)³
Answer:
(1) K²

II. Answer the following.

Question 1.
Show that the square of any positive integer of the form 3m or 3m + 1 for some integer m.
Answer:
Let a be any positive integer. Then it is of the form 3q or 3q + 1 or 3q + 2
Case – 1 When a = 3q
a² = (3q)² = 9 q²
= (3q) (3q)
= 3m where m = 3q
Case – 2 When a = 3q + 1
a² = (3q + 1)² = 9q² + 6q + 1
= 3q (3q + 2) + 1
= 3 m + 1
where m = q (3q + 2)
Case – 3 When a = 3q + 2
a² = (3q + 2)² = 9q² + 12q + 4
= 9q² + 12q + 3 + 1
= 3(3q² + 4q + 1) + 1
= 3m + 1
where m = 3q² + 4q + 1
Hence a is of the form 3m or 3m + 1

Question 2.
Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is some integer.
Solution:
Let us start with taking a, where a is a +ve odd integer.
We apply the division algorithm with ‘a’ and ‘b’ = 4.
Since 0 < r < 4, the possible remainders are 0, 1, 2, 3.
That is, a can be 4q, or 4q + 1, or 4q + 2 or 4q + 3, where 1 is the quotient. However, since a is odd, a cannot be 4q or 4q + 2 (since they are both divisible by 2).
Any odd integer is of the form 4q + 1 or 4q + 3.

Question 3.
Compute x such that 5⁴ = x (mod 8)
Answer:
5² = 25 = 1 (mod 8)
5⁴ = (5²)² = l² (mod 8)
= 1
5⁴ = 1 (mod 8)

Question 4.
The first term of an A.P. is 6 and the common difference is 5. Find the A.P. and its general term.
Answer:
Given, a = 6, d = 5
General term t_n = a + (n – 1) d
= 6 + (n – 1) 5
= 6 + 5 n – 5
= 5 n + 1
The general form of the A.P. is a, a + d, a + 2d,
The A.P. is 6, 11, 16, 21, ……… 5n + 1.

Question 5.
Which term of the arithmetic sequence 24, 23\(\frac { 1 }{ 4 } \), 22 \(\frac { 1 }{ 2 } \), 21 \(\frac { 3 }{ 4 } \), …….. is 3?
Answer:
Given The A.P is 24, 23 \(\frac { 1 }{ 4 } \), 22 \(\frac { 1 }{ 2 } \), 21 \(\frac { 3 }{ 4 } \), …………..

Question 6.
Determine the AP whose 3rd term is 5 and the 7th term is 9.
Solution:
We have
a₃ = a + (3 – 1)d = a + 2d = 5 ……….. (1)
a₇ = a + (7 – 1)d = a + 6d = 9 ………… (2)
(1) – (2) ⇒ -4d = -4 ⇒ d = 1.
Sub, d = 1 in (1), we get
a + 2(1) = 5
a = 3
Hence the required A.P. is 3, 4, 5, 6, 7.

Question 7.
If a, b, c are in A.P. then prove that (a – c)² = 4 (b² – ac).
Answer:
Given a, b, c are in A.P.
n = 10
∴ b – a = c – b
2 b = a + c
Squaring on both sides,
(a + c)² = (2b)²
a² + c² + 2ac = 4b²
(a – c)² + 2ac + 2ac = 4b²
[a² + c² = (a – c)² + 2ac]
(a – c)² = 4b² – 4ac
(a – c)² = 4(b² – ac)
Hence it is proved.
Aliter: Given a, b, c are in A.P.
b – a = c – b
2b = a + c
To prove, (a – c)² = 4(b² – ac)
L.H.S.= (a – c)²
= a² + c² – 2 ac
= (a + c)² – 2 ac – 2ac
= (a + c)² – 4ac
= (2b)² – 4ac (2b = a + c)
= 4b² – 4ac = 4 (b² – ac) = R.H.S
∴ L.H.S = R.H.S., Hence proved.

Question 8.
If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20th term.
Solution:
Here S₁₄ = 1050
n = 14
a = 10
S_n = \(\frac { n }{ 2 } \) (2a + (n – 1)d)
1050 = \(\frac { 14 }{ 2 } \) (20 + 13d)
= 140 + 91d
910 = 91d
d = 10
a₂₀ = 10 + (20 – 1) × 10
= 20
∴ 20th term = 200.

Question 9.
Find the sum of the first 40 terms of the series 1² – 2² + 3² – 4² + ….
Answer:
The given series is 1² – 2² + 3² – 4² + …. 40 terms
Grouping the terms we get,
(1² – 2²) + (3² – 4²) + (5² – 6²) + ………….. 20 terms
(1 – 4) + (9 – 16) + (25 – 36) + …………… 20 terms
(- 3) + (- 7) + (- 11) + ………………. 20 term
This is an A.P
Here a = – 3, d = – 7 – (-3) = – 7 + 3 = – 4 n = 20
S_n = \(\frac { n }{ 2 } \) [2a + (n – 1) d ]
S₂₀ = \(\frac { 20 }{ 2 } \) [2(-3) + 19(-4)]
= 10 (- 6 – 76) = 10 (- 82) = – 820
∴ Sum of 40 terms of the series is – 820.
Aliter. 1² – 2² + 3² – 4² + …….. + 39² – 40²
= 1² + 3² + 5² + ……. + 39² –
(2² + 4² + 6² + ……….. + 40²)
= 1² + 2² + 3² + 4² + …. + 40²
– (2² + 4² + 6² + …. + 40²)
– (2² + 4² + 6² + …. + 40²)
= 1² + 2² + 3² + …. + 40²
= 2 × 2² (1² + 2² + ….. + 20²)

Question 10.
Find the sum of first 24 terms of the list of numbers whose nth term is given by
a_n = 3 + 2n.
Solution:
a_n = 3 + 2n
a₁ = 3 + 2 = 5
a₂ = 3 + 2 × 2 = 7
a₃ = 3 + 2 × 3 = 9
List of numbers becomes 5, 7, 9. 11, ……….
Here,7 – 5 = 9 – 7 = 11 – 9 = 2 and soon. So, it forms an A.P. with common difference d = 2.
To find S₂₄, we have n = 24, a = 5, d = 2.
S₂₄ = \(\frac { 24 }{ 2 } \) [2 × 5 + (24 – 1) × 2 ]
= 12 [10 + 46] = 672
So, sum of first 24 terms of the list of numbers is 672.

Question 11.
If a clock strikes once at 1 o’clock, twice at 2 o’clock and so on, how many times will it strike in a day?
Answer:
Number of times the clock strikes each hour form an A.P.
Number of times strike in 12 hours is
1 + 2 + 3 + …….. + 12
Here, a = 1, d = 1, n = 12, l = 12
S_n = \(\frac { n }{ 2 } \) (a + l) = \(\frac { 12 }{ 2 } \) (1 + 12)
= 6 × 13 = 78 times
∴ Number of times the clock strike in 24 hours
= 78 × 2 = 156 times.

Question 12.
If the 4^th and 7^th terms of a G.P. are 54 and 1458 respectively, find the G.P.
Answer:
Given, t₄ = 54 and t₇ = 1458

Substituting the value of r = 3 in (1)
a × 27 = 54
a = \(\frac { 54 }{ 27 } \) = 2
∴ The Geometric sequence is 2, 6, 18, 54 ………

Question 13.
Which term of the geometric sequence,
(i) 5, 2, \(\frac { 4 }{ 5 } \), \(\frac { 8 }{ 25 } \), ……… is \(\frac { 128 }{ 15625 } \)?
Answer:
The given G.P. is 5, 2, \(\frac { 4 }{ 5 } \),\(\frac { 8 }{ 25 } \), …….., is \(\frac { 128 }{ 15625 } \)
Here a = 5, r = \(\frac { 2 }{ 5 } \), t_n = \(\frac { 128 }{ 15625 } \)
t_n = a.r^n-1

Question 14.
How many consecutive terms starting from the first term of the series 2 + 6 + 18 + … would sum to 728?
Answer:
The given series is
2 + 6 + 18 + …. + t_n = 728
Here a = 2, r = = 3, S_n = 728

∴ Required number of terms = 6

Question 15.
A geometric series consists of four terms and has a positive common ratio. The sum of the first two terms is 9 and sum of the last two terms is 36. Find the series.
Answer:
Let the four terms of the G.P. be a, ar, ar² and ar³ ……….
Given, a + ar = 9 ……(1)
ar² + ar³ = 36
r² (a + ar) = 36
r² (9) = 36
[from (1)]
r² = =4
r = ± 2
then r = 2
(given common positive ratio)
a + a (2) = 9 from (1)
3a = 9
a = 3
∴ The required series
= 3 + 3(2) + 3 (2²) + 3 (2³) + ……
= 3 + 6 + 12 + 24 + ……..

Question 16.
Suppose that five people are ill during the first week of an epidemic and each sick person spreads the contagious disease to four other people by the end of the second week and so on. By the end of 15^th week, how many people will be affected by the epidemic?
Answer:
Number of people affected by the epidemic during each week form a geometric series.
S₁₅ = 5 + (4 × 5) + (4 × 20) + (4 × 80) + …. 15 terms
= 5 + 20 + 80 + 320 + … 15 terms
Here a = 5, r = 4, n = 15

Question 17.
A gardener wanted to reward a boy for his good deeds by giving some mangoes. He gave the boy two choices. He could either have 1000 mangoes at once or he could get 1 mango on the first day, 2 on the second day, 4 on the third day, 8 mangoes on the fourth day and so on for ten days. Which option should the boy choose to get the maximum number of mangoes?
Answer:
I choice
The boy could get 1000 mangoes at once
II choice
The boy receives mangoes daily for ten days
S₁₀ = 1 + 2 + 4 + 8 + ……… 10 terms

Question 18.
Find the value of k if
1³ + 2³ + 3³ + ….. + K³ = 2025
Answer:
Given, 1³ + 2³ + 3³ + … + K³ = 2025

Question 19.
If 1³ + 2³ + 3³ + …… + K³ = 8281, then find 1 + 2 + 3 + … + K.
Answer:
Given, 1³ + 2³ + 3³ + …… + K³ = 8281

1 + 2 + 3 + …… + K = 91

Question 20.
Find the sum of all 11 term of an AP whose middle most term in 30.
Answer:
Let ‘a’ be the first term and ‘d’ be the common difference of the give A.P.
Middle most term = (\(\frac { 11+1 }{ 2 } \))^th = 6^th term
t_n = a + (n – 1)d
t₆ = a + 5d
a + 5d = 30
S_n = \(\frac { n }{ 2 } \) [2a + (n – 1)d]
S₁₁ = \(\frac { 11 }{ 2 } \) [2a + 10d]
= \(\frac { 11 }{ 2 } \) × 2 [a + 5d]
= 11 × 30
= 330
Sum of 11 terms = 330

III. Answer the following.

Question 1.
Use Euclid’s division algorithm to find the HCF of 867 and 255.
Answer:
Here 867 > 255
Applying Euclid’s Lemma to 867 and 255 we get

867 = (255 × 3) + 102
The remainder 102 ≠ 0
Again applying Euclid’s
Lemma to 255 and 102
255 = 102 × 2 + 51
The remainder 51 ≠ 0
Again applying Euclid’s
Lemma to 102 and 51 we get
102 = 51 × 2 + 0
The remainder is 0
∴ HCF of 867 and 255 is 51

Question 2.
Find the number of integer solutions of 5x = 2 (mod 13)
Answer:
5x ≡ 2 (mod 13) can be written as
5x – 2 = 13 k for some integer
5x = 13 k + 2
x = \(\frac{13 k+2}{5}\)
Since 5k is an integer \(\frac{13 k+2}{5}\) cannot be an inter.
There is no integer solution.

Question 3.
Find the 40^th term of A.P. whose 9^th term is 465 and 20^th term is 388.
Answer:
t_n = a + (n – 1) d
t₉ = a + 8d (t₉ = 465)
a + 8d = 465 ……(1)

Substitute the value of d = -7 in (1)
a + 8(-7) = 465
a – 56 = 465
a = 465 + 56 = 521
a = 521, d = -7, n = 40
t₄₀ = 521 + 39(-7)
= 521 – 273
= 248
40^th term of an A.P. is 248

Question 4.
Find the three consecutive terms in an A.P. whose sum is 18 and the sum of their squares is 140.
Answer:
Let the three consecutive terms in an
A.P. be m – d,m,m + d
By the given data,
Sum of threee terms = 18
m – d + m + m + d = 18
3m = 18
m = \(\frac { 18 }{ 3 } \) = 6
Again by the given data,
Sum of their squares = 140
(m – d)² + m² + (m + d)² = 140
m² + d² – 2md + m² + m² + d² + 2md = 140
3m² + 2d² = 140
3(62) + 2d² = 140
3(36) + 2d² = 140
2d² = 140 – 108
2d² = 32
d² = \(\frac { 32 }{ 2 } \) = 16
∴ d = ± 4
when, m = 6, d = + 4
m – d = 6 – 4 = 2
m = 6
m + d = 6 + 4 = 10
when, m = 6, d = -4
m – d = 6-(-4) = 6 + 4= 10
m = 6
m + d = 6 +(-4) = 6 – 4 = 2
∴ The three numbers are 2, 6 and 10 or 10, 6,2.

Question 5.
If m times the m^th term of an A.P. is equal to n times its n^th term, then show that the (m + n)^th term of the A.P. is zero.
Answer:
Given, mt_m = nt_n
m[a + (m – 1) d] = n [a + (n – 1) d]
[we know that t_n = a + (n – 1)d]
m[a + md – d] = n[a + nd – d]
ma + m²d – md = na + n²d – nd
ma – na + m²d – n²d = md – nd
a (m – n) + d (m² – n²) = d(m – n)
a (m – n) + d(m + n)(m – n) = d(m – n) ÷ by (m – n) on both sides,
a + d (m + n) = d
a + d(m + n) – d = 0
a + d(m + n – 1) = 0 ….. (1)
To prove, t_(m+n) = 0
t_(m+n) = a + (m + n – 1)d
t_(m+n) = 0(from(1))
Hence it is proved.

Question 6.
If a, b, c are in A.P. then prove that (a – c)² = 4 (b² – ac).
Answer:
Given a, b, c are in A.P.
∴ b – a = c – b
2 b = a + c
Squaring on both sides,
(a + c)² = (2b)²
a² + c² + 2 ac = 4 b²
(a – c)² + 2 ac + 2 ac = 4 b²
[a² + c² = (a – c)² + 2 ac]
(a – c)² = 4b² – 4ac
(a – c)² = 4 (b² – ac)
Hence it is proved.
Aliter: Given a, b, c are in A.P.
b – a = c – b
2 b = a + c
To prove, (a – c)² = 4(b² – ac)
L.H.S. = (a – c)²
= a² + c² – 2ab
= (a + c)² – 2ac – 2ac
= (a + c)² – 4ac
= (2b)² – 4ac (2b = a + c)
= 4 b² – 4ac = 4 (b² – ac) = R.H.S
∴ L.H.S = R.H.S., Hence proved.

Question 7.
The ratio of the sums of first m and first n terms of an arithmetic series is m² : n² show that the ratio of the m^th and n^th terms is (2m – 1) : (2n -1).
Answer:

n[2a + (m – 1) d] = tn[2a + nd-d]
2 an + mnd- nd = 2 am + mnd— md 2an-2am = nd- md 2 a (n-m) = d(n- m)
÷ by (n – m) on both sides,
2a = d
To prove, t_m : t_n = (2m – 1) : (2n – 1)
L.H.S = t_m : t_n
= a + (m – 1) d : a + (n – 1)d
= a + (m – 1) 2a : a + (n – 1)2a
[Substitute the value of d = 2a]
= a + 2 am – 2 a: a + 2 am – 2a
= 2am – a : 2an – a
= a (2m – 1) : a (2n – 1)
= (2m – 1) : (2n – 1) = R. H. S
= (2m – 1) : (2n – 1)
∴ t_m : t_n
L.H.S = R.H.S
Hence it is proved.

Question 8.
A construction company will be penalised each day for delay in construction of a bridge. The penalty will be ₹4000 for the first day and will increase by ₹1000 for each following day. Based on its budget, the company can afford to pay a maximum of ₹1,65,000 towards penalty. Find the maximum number of days by which the completion of work can be delayed.
Solution:
Penalty for the first day (a) = ₹ 4000
Increased rate for every day (d) = ₹ 1000
Maximum amount of penalty (S_n)
= ₹ 1,65,000
S_n = \(\frac { n }{ 2 } \) [2a + (n-1) 1000]
165000 = \(\frac { n }{ 2 } \) [2(4000) + (n – 1) 1000]
= \(\frac { n }{ 2 } \) [8000 + 1000 n – 1000]
165000 = \(\frac { n }{ 2 } \) (7000 + 1000n)
330000 = 7000 n + 1000 n²
0 = 1000 n² + 7000 n – 330000 ÷ 1000 on both sides,
n² + 7n – 330 = 0
(n + 22) (n – 15) = 0
n = -22 or n = 15
n = -22 (not possible)
∴ Maximum number of days for which the work can be delayed is 15 days.

Question 9.
If the product of three consecutive terms in G.P. is 216 and sum of their products in pairs is 156, find them.
Let the three consecutive terms of G.P. be \(\frac { a }{ r } \), a, ar.
Product of three terms = 216
\(\frac { a }{ r } \) × a × ar = 216
a³ = 216
a³ = 63
a = 6
Sum of their products in pairs = 156

Question 10.
If a, b, c, d are in a geometric sequence, then show that
(a – b + c) (b + c + d) = ab + bc + cd.
Answer:
Given, a, b, c, d are in a geometric sequence.
Let a = a, b = ar, c = ar², d = ar³
To prove, (a – b + c) (b + c + d) = ab + bc + cd
L.H.S. = (a – b + c)(b + c + d)
= (a – ar + ar²) (ar + ar² + ar³ )
= a (1 – r + r²)ar (1 + r + r²)
= a²r (1 – r + r²) (1 + r + r²)
= a²r (1 + r² + r4)
= a²r + a²r³ + a²r⁵
= a (ar) + ar (ar²) + ar² (ar³)
= ab + bc + cd
= R.H.S.
L.H.S. = R.H.S., Hence proved.

Question 11.
Find the sum of the first n terms of the series 0.4 + 0.94 + 0.994 + ………..
Answer:
Given series is 0.4 + 0.94 + 0.994 + ……………. + n terms.
S_n = 0.4 + 0.94 + 0.994 + ……….. + n terms

Question 12.
Find the total area of 12 squares whose sides are 12 cm, 13 cm,… 23 cm respectively.
Answer:
Given, the sides of 12 squares are 12 cm, 13 cm, 14 cm,… 23 cm
Total area of 12 squares
= 12² + 13² + 14² + … + 23²
= (1² + 2² + 3² … + 23²) – (1² + 2² + … + 11²)

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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.7

Question 1.
Find the quotient and remainder of the following.
(i) 4x³ + 6x² – 23x + 18) ÷ (x + 3)
Solution:

∴ The quotient = 4x² – 6x – 5
The remainder = 33

(ii) (8y³ – 16y² + 16y – 15) ÷ (2y – 1)
Solution:

∴ The quotient = 4y² – 6y + 5
The remainder = -10

(iii) (8x³ – 1) ÷ (2x – 1)
Solution:

∴ The quotient = 4x² + 2x + 1
The remainder = 0

(iv) (-18z + 14z² + 24z³ + 18) ÷ (3z + 4)
Solution:

∴The quotient = 8z² – 6z + 2
The remainder = 10

Question 2.
The area of a rectangle is x² + 7x + 12. If its breadth is (x + 3) then find its length.
Solution:
Let the length of the rectangle be “l”
The breadth of the rectangle = x + 3
Area of the rectangle = length × breadth
x² + 7x + 12 = l(x + 3)

Length of the rectangle = x + 4

Question 3.
The base of a parallelogram is (5x + 4). Find its height if the area is 25x² – 16.
Solution:
Let the height of the parallelogram be “h”.
Base of the parallelogram = 5x + 4
Area of a parallelogram = 25x² – 16
∴ Base x Height = 25x² – 16
(5x + 4) × h = 25x²– 16

Height of the parallelogram = 5x – 4

Question 4.
The sum of (x + 5) observations is (x³ + 125). Find the mean of the observations.
Solution:
Sum of the observation = x³ + 125
Number of observation = x + 5

Mean = x² – 5x + 25

Question 5.
Find the quotient and remainder for the following using synthetic division:
(i) (x³ + x² – 7x – 3) ÷ (x – 3)
Solution:
p(x) = x³ + x² – 7x – 3
d(x) = x – 3 [p(x) = d(x) × q(x) + r]

x – 3 = 0
x = 3
Hence the quotient = x² + 4x + 5
Remainder = 12

(ii) (x³ + 2x² – x – 4) ÷ (x + 2)
Solution:
p(x) = x³ + 2x² -x – 4
d(x) = x + 2

x + 2 = 0
x = -2
The quotient = x² – 1
Remainder = -2

(iii) (3x³ – 2x² + 7x – 5) ÷ (x + 3)
Solution:
p(x) = 3x³ – 2x² + 7x – 5
d(x) = x + 3

x + 3 = 0
x = -3
The quotient = 3x² – 11x + 40
Remainder = -125

(iv) (8x⁴ – 2x² + 6x + 5) ÷ (4x + 1)
Solution:
p(x) = 8x⁴ – 2x² + 6x + 5
d(x) = 4x + 1

Question 6.
If the quotient obtained on dividing (8x⁴ – 2x² + 6x – 7) by (2x + 1) is (4x³ + px² -qx + 3), then find p, q and also the remainder.
Solution:
p(x) = 8x⁴ – 2x² + 6x – 7
d(x) = 2x + 1

2x + 1 = 0
2x = -1
x = –\(\frac{1}{2}\)
The quotient = \(\frac{1}{2}\) [8x³ – 4x² + 6]
= 4x³ – 2x² + 3
= 4x³ – 2x² + 0x + 3
The given quotient is = 4x³ + px² – qx + 3
(compared with the given quotient)
The value of p = -2 and q = 0
Remainder = -10

Question 7.
If the quotient obtained on dividing 3x³ + 11x² + 34x + 106 by x – 3 is 3x² + ax + b, then find a, b and also the remainder.
Solution:
p(x) = 3x³ + 11x² + 34x + 106
d(x) = x – 3

x – 3 = 0
x = 3
The quotient is = 3x² + 20x + 94
The given quotient is = 3x² + ax + b
Compared with the given quotient
The value of a = 20 and b = 94
The remainder = 388

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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.16

1. In the

write (i) The number of elements
(ii) The order of the matrix
(iii) Write the elements a₂₂, a₂₃, a₂₄, a₃₄, a₄₃, a₄₄.
Answer:
(i) The number of elements is 16
(ii) The order of the matrix is 4 × 4
(iii) Elements corresponds to

Question 2.
If a matrix has 18 elements, what are the possible orders it can have? What if it has 6 elements?
Answer:
The possible orders of the matrix having 18 elements are

The possible orders of the matrix having 6 elements are

Question 3.
Construct a 3 × 3 matrix whose elements are given by
(i) a_ij = |i – 2j|
Answer:
a_ij = |i – 2j|
The general 3 × 3 matrices is

a₁₁ = |1 – 2(1)| = |1 – 2| = | – 1| = 1
a₁₂ = |1 – 2(2)| = |1 – 4| = | – 3| = 3
a₁₃ = |1 – 2(3)| = |1 – 6| = | – 5| = 5
a₂₁ = |2 – 2(1)| = |2 – 2| = 0 = 0
a₂₂ = |2 – 2(2)| = |2 – 4| = | – 2| = 2
a₂₃ = |2 – 2(3)| = |2 – 6| = | – 4| = 4
a₃₁ = |3 – 2(1)| = |3 – 2| = | 1 | = 1
a₃₂ = |3 – 2(2)| = |3 – 4| = | – 1 | = 1
a₃₃ = |3 – 2(3)| = |3 – 6| = | – 3 | = 3
The required matrix

(ii) a_ij = \(\frac{(i+j)^{3}}{3}\)
Answer:
a₁₁ = \(\frac{(1+1)^{3}}{3}\) = \(\frac{2^{3}}{3}\) = \(\frac { 8 }{ 3 } \)
a₁₂ = \(\frac{(1+2)^{3}}{3}\) = \(\frac { 27 }{ 3 } \) = 9
a₁₃ = \(\frac{(1+3)^{3}}{3}\) = \(\frac { 64 }{ 3 } \) = \(\frac { 64 }{ 3 } \)
a₂₁ = \(\frac{(2+1)^{3}}{3}\) = \(\frac { 27 }{ 3 } \) = 9
a₂₂ = \(\frac{(2+2)^{3}}{3}\) = \(\frac { 64 }{ 3 } \) = \(\frac { 64 }{ 3 } \)
a₂₃ = \(\frac{(2+3)^{3}}{3}\) = \(\frac { 125 }{ 3 } \) = \(\frac { 125 }{ 3 } \)
a₃₁ = \(\frac{(3+1)^{3}}{3}\) = \(\frac { 64 }{ 3 } \) = \(\frac { 64 }{ 3 } \)
a₃₂ = \(\frac{(3+2)^{3}}{3}\) = \(\frac { 125 }{ 3 } \) = \(\frac { 125 }{ 3 } \)
a₃₃ = \(\frac{(3+3)^{3}}{3}\) = \(\frac { 216 }{ 3 } \) = 72
The required matrix

Question 4.
If  then find the tranpose of A.
Answer:

transpose of A = (A^T)

Question 5.
If then find the tranpose of – A
Answer:

Transpose of – A = (-A^T) =

Question 6.
If A = then verify (AT)T = A
Answer:

Hence it is verified

Question 7.
Find the values of x, y and z from the following equations
(i)

Answer:
Since the given matrices are equal then all the corresponding elements are equal.
y = 12, z = 3, x = 3
The value of x = 3, y = 12 and z = 3

(ii)

Answer:
x + y = 6 ……(1)
5 + z = 5
z = 5 – 5 = 0
xy = 8
y = \(\frac { 8 }{ x } \)
Substitute the value of y = \(\frac { 8 }{ x } \) in (1)
x + \(\frac { 8 }{ x } \) = 6
x² + 8 = 6x
x² – 6x + 8 = 0
(x – 4) (x – 2) = 0
∴ x – 4 = 0 or x – 2 = 0
x = 4 or x = 2
y = \(\frac { 8 }{ 4 } \) = 2 or y = \(\frac { 8 }{ 2 } \) = 4

∴ The value of x, y and z are 4, 2, 0 (or) 2, 4, 0

(iii)

Solution:
x + y + z = 9 ……….(1)
x + z = 5 ……….(2)
y + z = 7 ……….(3)

Substitute the value of y = 4 in (3)
y + z = 7
4 + z = 7
z = 7 – 4
= 3
Substitute the value of z = 3 in (2)
x + 3 = 5
x = 5 – 3
= 2
∴ The value of x = 2 , y = 4 and z = 3

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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Ex 4.5

Question 1.
If in triangles ABC and EDF, \(\frac { AB }{ DE } \) = \(\frac { BC }{ FD } \) then they will be similar, when ……….
(1) ∠B = ∠E
(2) ∠A = ∠D
(3) ∠B = ∠D
(4) ∠A = ∠F
Answer:
(3) ∠B = ∠D
Hint:

Question 2.
In ∆LMN, ∠L = 60°, ∠M = 50°. If ∆LMN ~ ∆PQR then the value of ∠R is ……………
(1) 40°
(2) 70°
(3) 30°
(4) 110°
Answer:
(2) 70°
Hint:
Since ∆LMN ~ ∆PQR
∠N = ∠R
∠N = 180 – (60 + 50)
= 180 – 110°
∠N = 70° ∴ ∠R = 70°

Question 3.
If ∆ABC is an isosceles triangle with ∠C = 90° and AC = 5 cm, then AB is ………
(1) 2.5 cm
(2) 5 cm
(3) 10 cm
(4) 5 \(\sqrt { 2 }\) cm
Answer:
(4) 5 \(\sqrt { 2 }\) cm
Hint:

AB² = AC² + BC²
AB² = 5² + 5²
(It is an isosceles triangle)
AB = \(\sqrt { 50 }\) = \(\sqrt{25 \times 2}\)
AB = 5 \(\sqrt { 2 }\)

Question 4.
In a given figure ST || QR, PS = 2 cm and SQ = 3 cm. Then the ratio of the area of ∆PQR to the area of ∆PST is …………….

(1) 25 : 4
(2) 25 : 7
(3) 25 : 11
(4) 25 : 13
Answer:
(1) 25 : 4
Hint. Area of ∆PQR : Area of ∆PST
\(\frac{\text { Area of } \Delta \mathrm{PQR}}{\text { Area of } \Delta \mathrm{PST}}=\frac{\mathrm{PQ}^{2}}{\mathrm{PS}^{2}}=\frac{5^{2}}{2^{2}}=\frac{25}{4}\)
Area of ∆PQR : Area of ∆PST = 25 : 4

Question 5.
The perimeters of two similar triangles ∆ABC and ∆PQR are 36 cm and 24 cm respectively. If PQ =10 cm, then the length of AB is ………….
(1) 6 \(\frac { 2 }{ 3 } \) cm
(2) \(\frac{10 \sqrt{6}}{3}\)
(3) 66 \(\frac { 2 }{ 3 } \) cm
(4) 15 cm
Answer:
(4) 15 cm
Hint:

Question 6.
If in ∆ABC, DE || BC. AB = 3.6 cm, AC = 2.4 cm and AD = 2.1 cm then the length of AE is ………..
(1) 1.4 cm
(2) 1.8 cm
(3) 1.2 cm
(4) 1.05 cm
Answer:
(1) 1.4 cm
Hint:

Question 7.
In a ∆ABC, AD is the bisector of ∠BAC. If AB = 8 cm, BD = 6 cm and DC = 3 cm.
The length of the side AC is ………….
(1) 6 cm
(2) 4 cm
(3) 3 cm
(4) 8 cm
Answer:
(2) 4 cm
Hint:

Question 8.
In the adjacent figure ∠BAC = 90° and AD ⊥ BC then ………..

(1) BD.CD = BC²
(2) AB.AC = BC²
(3) BD.CD = AD²
(4) AB.AC = AD²
Answer:
(3) BD CD = AD²
Hint:

Question 9.
Two poles of heights 6 m and 11m stand vertically on a plane ground. If the distance between their feet is 12 m, what is the distance between their tops?
(1) 13 m
(2) 14 m
(3) 15 m
(4) 12.8 m
Answer:
(1) 13 m
Hint:

AC² (Distance between the two tops)
= AE² + EC²
= 5² + 122
= 25 + 144= 169
AC = \(\sqrt { 169 }\) = 13 cm

Question 10.
In the given figure, PR = 26 cm, QR = 24 cm, ∠PAQ = 90°, PA = 6 cm and QA = 8 cm. Find ∠PQR.

(1) 80°
(2) 85°
(3) 75°
(4) 90°
Answer:
(4) 90°
Hint.
PR = 26 cm, QR = 24 cm, ∠PAQ = 90°
In the ∆PQR,

In the right ∆ APQ
PQ² = PA² + AQ²
= 6² + 8²
= 36 + 64 = 100
PQ = \(\sqrt { 100 }\) = 10
∆ PQR is a right angled triangle at Q. Since
PR² = PQ² + QR²
∠PQR = 90°

Question 11.
A tangent is perpendicular to the radius at the
(1) centre
(2) point of contact
(3) infinity
(4) chord
Solution:
(2) point of contact

Question 12.
How many tangents can be drawn to the circle from an exterior point?
(1) one
(2) two
(3) infinite
(4) zero
Answer:
(2) two

Question 13.
The two tangents from an external points P to a circle with centre at O are PA and PB.
If ∠APB = 70° then the value of ∠AOB is ……….
(1) 100°
(2) 110°
(3) 120°
(4) 130°
Answer:
(2) 110°
Hint.

∠OAP = 90°
∠APO = 35°
∠AOP = 180 – (90 + 35)
= 180 – 125 = 55
∠AOB = 2 × 55 = 110°

Question 14.
In figure CP and CQ are tangents to a circle with centre at 0. ARB is another tangent touching the circle at R. If CP = 11 cm and BC = 7 cm, then the length of BR is …….

(1) 6 cm
(2) 5 cm
(3) 8 cm
(4) 4 cm
Answer:
(4) 4 cm
Hint.
BQ = BR = 4 cm (tangent of the circle)
PC = QC = 11 cm (tangent of the circle)
QC = 11 cm
QB + BC = 11
QB + 7 = 11
QB = 11 – 7 = 4 cm
BR = BQ = 4 cm

Question 15.
In figure if PR is tangent to the circle at P and O is the centre of the circle, then ∠POQ is ………….

(1) 120°
(2) 100°
(3) 110°
(4) 90°
Answer:
(1) 120°
Hint.
Since PR is tangent of the circle.
∠QPR = 90°
∠OPQ = 90° – 60° = 30°
∠OQB = 30°
In ∆OPQ
∠P + ∠Q + ∠O = 180°
30 + 30° + ∠O = 180°
(OP and OQ are equal radius)
∠O = 180° – 60° = 120°

Students can download Maths Chapter 5 Coordinate Geometry Ex 5.3 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.3

Question 1.
Find the mid-points of the line segment joining the points.
(i) (-2, 3) and (-6, -5)
(ii) (8, -2) and (-8, 0)
(iii) (a, b) and (a + 2b, 2a – b)
(iv) (\(\frac{1}{2},\frac{-3}{7}\)) and (\(\frac{3}{2},\frac{-11}{7}\))
Solution:

Question 2.
The centre of a circle is (-4, 2). If one end of the diameter of the circle is (-3, 7) then find the other end.
Solution:
Let the other end of the diameter B be (a, b)

∴ \(\frac{-3+a}{2}\) = -4
-3 + a = -8
a = -8 + 3
a = -5
\(\frac{7+b}{2}\)
7 + b = 4
b = 4 – 7 ⇒ b = -3
The other end of the diameter is (-5, -3).

Question 3.
If the mid-point (x, y) of the line joining (3, 4) and (p, 7) lies on 2x + 2y + 1 = 0, then what will be the value of p?
Solution:

3 + p + 11 + 1 = 0 ⇒ p + 15 = 0 ⇒ p = -15
The value of p is -15.

Question 4.
The mid-point of the sides of a triangle are (2, 4), (-2, 3) and (5, 2). Find the coordinates of the vertices of the triangle.
Solution:
Let the vertices of the ΔABC be A (x₁ y₁), B (x₂, y₂) and C (x₃, y₃)

\(\frac{x_{2}+x_{3}}{2}\) = -2 ⇒ x₂ + x₃ = -4 → (3)
\(\frac{y_{2}+y_{3}}{2}\) = 3 ⇒ y₂ + y₃ = 6 → (4)

\(\frac{x_{1}+x_{3}}{2}\) = 5 ⇒ x₁ + x₃ = 10 → (5)
\(\frac{y_{1}+y_{3}}{2}\) = 2 ⇒ y₁ + y₃ = 4 → (6)
By adding (1) + (3) + (5) we get
2x₁ + 2x₂ + 2x₃ = 4 – 4 + 10
2(x₁ + x₂ + x₃) = 10 ⇒ x₁ + x₂ + x₃ = 5
From (1) x₁ + x₂ = 4 ⇒ 4 + x₃ = 5
x₃ = 5 – 4 = 1
∴ The vertices of the ΔABC are
A (9, 3)
B (-5, 5), C (1, 1)
From (3) x₂ + x₃ = -4 ⇒ x₁ + (-4) = 5
x₁ = 5 + 4 = 9
From (5) ⇒ x₁ + x₃ = 8
x₂ + 10 = 5
x₂ = 5 – 10 = -5
∴ x₁ = 9, x₂ = -5, x₃ = 1
By adding (2) + (4) + (6) we get
2y₃ + 2y₂ + 2y₃ = 8 + 6 + 4
2(y₁ +y₂ + y₃) = 18 ⇒ y₁ + y₂ + y₃ = 9
From (2) ⇒ y₁ + y₂ = 8
8 + y₃ = 9 ⇒ y₃ = 9 – 8 = 1
From (4)
y₂ + y₃ = 6 ⇒ y₁ + 6 = 9
y₁ = 9 – 6 = 3
From (6)
y₁ + y₃ = 4 ⇒ y₂ + 4 = 9
y₂ = 9 – 4 = 5
∴ y₁ = 3, y₂ = 5, y₃ = 1

Question 5.
O (0, 0) is the centre of a circle whose one chord is AB, where the points A and B (8, 6) and (10, 0) respectively. OD is the perpendicular from the centre to the chord AB. Find the coordinates of the mid-point of OD.
Solution:
Note: Since OD is perpendicular to AB, OD bisect the chord
D is the mid-point of AB

Question 6.
The points A(-5, 4) , B(-1, -2) and C(5, 2) are the vertices of an isosceles right-angled triangle where the right angle is at B. Find the coordinates of D so that ABCD is a square.
Solution:
Since ABCD is a square
Mid-point of AC = mid-point of BD
Let the point D be (a, b)

But mid-point of BD = Mid-point of AC
(\(\frac{-1+a}{2}, \frac{-2+b}{2}\)) = (0, 3)
\(\frac{-1+a}{2}\) = 0
-1 + a = 0
a = 1
(\(\frac{-2+b}{2}\))
-2 + b = 6
b = 6 + 2 = 8
∴ The vertices D is (1, 8).

Question 7.
The points A (-3, 6), B (0, 7) and C (1, 9) are the mid points of the sides DE, EF and FD of a triangle DEF. Show that the quadrilateral ABCD is a parallellogram.
Solution:
Let D be (x₁ y₁), E (x₂, y₂) and F (x₃, y₃)


x₁ + x₂ = -6 → (1)
y₁ + y₂ = 12 → (2)

x₁ + x₃ = 2 → (5)
y₁ + y₃ = 18 → (6)
Add (1) + (3) + (5)
2x₁ + 2x₂ + 2x₃ = -6 + 0 + 2
2 (x₁ + x₂ + x₃) = -4
x₁ + x₂ + x₃ = -2
From (1) x₁ + x₂ = -6
x3 = -2 + 6 = 4
From (3) x₂ + x₃ = 0
x₁ = -2 + 0
x₁ = -2
From (5) x₁ + x₃ = 2
x₂ + 2 = -2
x₂ = -2 -2 = -4
∴ x₁ = -2, x₂ = -4, x₃= 4
Add (2) + (4) + 6
2y₁ + 2y₂ + 2y₃ = 12 + 14 + 18
2(y₁ + y₂ + y₃) = 44
y₁ + y₂+ y₃ = 44/2 = 22
From (2) y₁ + y₂ = 12
12 + y₃ = 22 ⇒ y₃ = 22 – 12 = 11
From (4) y₂ + y₃ = 14
y₁ + 14 = 22 ⇒ y₁ = 22 – 14
y₁ = 8
From (6)
y₁ +y₃ = 18
y2 + 18 = 22 ⇒ = 22 – 18
= 4
∴ y₁ = 8, y₂ = 4, y₃ = 11
The vertices D is (-2, 8)

Mid-point of diagonal AC = Mid-point of diagonal BD
The quadrilateral ABCD is a parallelogram

Question 8.
A(-3, 2) , B(3, 2) and C(-3, -2) are the vertices of the right triangle, right angled at A. Show that the mid point of the hypotenuse is equidistant from the vertices.
Solution:

AD = CD = BD = \(\sqrt{13}\)
Mid-point of BC is equidistance from the centre.

Students can download Maths Chapter 5 Coordinate Geometry Ex 5.5 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.5

Question 1.
Find the centroid of the triangle whose vertices are
(i) (2, -4), (-3, -7) and (7, 2)
Solution:
Let the vertices of a triangle be A (2, -4), B (-3, -7) and C (7, 2) Centroid

Centroid is (2, -3)

(ii) (-5, -5), (1, -4) and (-4, -2)
Solution:
Let the vertices of a triangle be A (-5, -5), B (1, -4) and C (-4, -2)

Question 2.
If the centroid of a triangle is at (4, -2) and two of its vertices are (3, -2) and (5, 2) then find the third vertex of the triangle.
Solution:
Let the vertices of a triangle be A (3, -2), B (5, 2) and C (x₃, y₃)
Centroid of a triangle is (4, -2)

∴ \(\frac{8+x_{3}}{3}\) = 4
8 + x₃ = 12
x₃ = 12 – 8
= 4
and
\(\frac{y_{3}}{3}\) = -2
y₃ = -6
∴ The third vertex is (4, -6)

Question 3.
Find the length of median through A of a triangle whose vertices are A(-1, 3), B (1, -1) and C (5, 1).
Solution:
AD is the median of the ΔABC
D is the mid-point of BC

Length of the median AD is 5 units.

Question 4.
The vertices of a triangle are (1, 2), (h, -3) and (-4, k). If the centroid of the triangle is at the point (5, -1) then find the value of \(\sqrt{(h+k)^{2}+(h+3k)^{2}}\)
Solution:
Let the vertices A (1, 2), B (h, -3) and C (-4, k)


\(\frac{-3+h}{3}\) = 5
-3 + h = 15
h = 15 + 3 = 18
and
\(\frac{-1+k}{3}\) = -1
-1 + k = -3
k = -3 + 1
k = -2

Question 5.
Orthocentre and centroid of a triangle are A(-3, 5) and B(3, 3) respectively. If C is the circumcentre and AC is the diameter of this circle, then find the radius of the circle.
Solution:
Let PQR be any triangle orthocentre, centroid and circumcentre.

A orthocentre is (-3, 5)
B centroid is (3, 3)
C orthocentre is (a, 6)
Also \(\frac{AB}{BC}\) = \(\frac{2}{1}\)
B divides AC in the ratio 2 : 1
A line divides internally in the ratio point P is (\(\frac{mx_{2}+nx_{1}}{m+n}\), \(\frac{my_{2}+ny_{1}}{m+n}\))
m = 2
x₁ = 3
y₁ = 5
amd
n = 1
x₂ = a
y₂ = b
∴ The point B (\(\frac{2a-3}{2+1}\), \(\frac{2b+5}{2+1}\))
(3, 3) = (\(\frac{2a-3}{3}\), \(\frac{2b+5}{3}\))
\(\frac{2a-3}{3}\) = 3
2a – 3 = 9
2a = 9 + 3
2a = 12
a = \(\frac{12}{2}\) = 6
and
\(\frac{2b+5}{3}\)
2b + 5 = 9
2b = 9 – 5
2b = 4
b = \(\frac{4}{2}\) = 2
∴ Orthocentre C is (6, 2)

Question 6.
ABC is a triangle whose vertices are A (3, 4), B (-2, -1) and C (5, 3). If G is the centroid and BDCG is a parallelogram then find the coordinates of the vertex D.
Solution:
The vertices of a triangle are A (3, 4), B (-2, -1) and C (5, 3)

The point G is (2, 2)
Let the vertices D be (a, b)
Since BDCG is a parallelogram
Mid-point of BC = Mid-point of DG

\(\frac{2+a}{2}\) = \(\frac{3}{2}\)
4 + 2a = 6
2a = 6 – 4
2a = 2
a = 1
and
\(\frac{2+b}{2}\) = 1
2 + b = 2
b = 2 – 2 = 0
The vertices D is (1, 0).

Question 7.

Solution:
In ΔABC, Let A (x₁, y₁), B (x₂, y₁) and C (x₃, y₃)

x₁ + x₂ = 3 → (1)
y₁ + y₂ = 10 → (2)

x₂ + x₃ = 14 → (3)
y₂ + y₃ = 10 → (4)

x₁ + x₃ = 3 → (5)
y₁ + y₃ = 10 → (6)
Add (1) + (3) + (5) We get
2x₁ + 2x₂ + 2x₃ = 30
2(x₁ + x₂ + x₃) = 30
x₁ + x₂ + x₃ = 15
From (1), x₁ + x₂ = 3
∴ x₃ = 12
From (3), x₂ + x₃ = 14
∴ x₁ = 1
From (5), x₁ + x₃ = 13
∴ x₂ = 2
Add (2), (4) and (6) we get
2y₁ + 2y₂ + 2y₃ = -12
2(y₁ + y₂ + y₃) = – 12
∴ y₁ + y₂ + y₃ = -6
From (2), y₁ + y₂ = 10
∴ y₃ = -16
From (4) y₂ + y₃ = -9
∴ y₁ = 3
From (6) y₁ + y₃ = -13
∴ y₂ = 7
The vertices of the A are A (1, 3), B (2, 7) and C (12, -16)

The Centroid of Δ is (5, -2)

Students can download Maths Chapter 3 Algebra Ex 3.18 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.18

Question 1.
Find the order of the product matrix AB if

Answer:
Given A = [a_ij]_p×q and B = [a_ij]_q×r
Order of product of AB = p × r
Order of product of BA is not defined. Number columns in r is not equal to the number of rows in P.
∴ Product BA is not defined.

Question 2.
A has ‘a’ rows and ‘a + 3 ’ columns. B has ‘6’ rows and ‘17 – b’ columns, and if both products AB and BA exist, find a, b?
Solution:
A has a rows, a + 3 columns.
B has b rows, 17 – b columns
If AB exists a × a + 3
b × 17 – b
a + 3 = 6 ⇒ a – 6 = -3 ………… (1)
If BA exists 6 × 17-6
a × a + 3
17 – 6 = a ⇒ a + 6 = 17 …………. (2)
(1) + (2) ⇒ 2a = 14 ⇒ a = 7
Substitute a = 7 in (1) ⇒ 7 – b = -3 ⇒ b = 10
a = 7, b = 10

Question 3.
A has ‘a’ rows and ‘a + 3 ’ columns. B has rows and ‘b’ columns, and if both products AB and BA exist, find a,b?
Answer:

1. Order of matrix AB = 3 × 3
2. Order of matrix AB = 4 × 2
3. Order of matrix AB = 4 × 2
4. Order of matrix AB = 4 × 1
5. Order of matrix AB = 1 × 3

Question 4.

find AB, BA and check if AB = BA?
Answer:

Question 5.
Given that


verify that A(B + C) = AB + AC
Answer:

From (1) and (2) we get
A (B + C) = AB + AC

Question 6.
Show that the matrices

satisfy commutative property AB = BA
Answer:

From (1) and (2) we get
AB = BA. It satisfy the commutative property.

Question 7.

Show that (i) A(BC) = (AB)C
(ii) (A-B)C = AC – BC
(iii) (A-B)^T = A^T – B^T
Answer:



From (1) and (2) we get
A(BC) = (AB)C

From (1) and (2) we get
(A – B) C = AC – BC


From (1) and (2) we get
(A-B)^T = A^T – B^T

Question 8.

then snow that A² + B² = I.
Answer:

Question 9.

prove that AA^T = I.
Answer:

AA^T = I
∴ L.H.S. = R.H.S.

Question 10.
Verify that A² = I when

Answer:

∴ L.H.S. = R.H.S.

Question 11.

show that A² – (a + d)A = (bc – ad)I₂.
Answer:

L.H.S. = R.H.S.
A² – (a + d) A = (bc – ad)I₂

Question 12.

verify that (AB)^T = B^T A^T
Answer:


From (1) and (2) we get, (AB)^T = B^T A^T

Question 13.

show that A² – 5A + 7I₂ = 0
Answer:


L.H.S. = R.H.S.
∴ A² – 5A + 7I₂ = 0

Students can download Maths Chapter 3 Algebra Ex 3.15 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.15

Question 1.
Graph the following quadratic equations and state their nature of solutions.
(i) x² – 9x + 20 = 0
(ii) x² – 4x + 4 = 0
(iii) x² + x + 7 = 0
(iv) x² – 9 = 0
(v) x² – 6x + 9 = 0
(vi) (2x – 3) (x + 2) = 0

(i) x² – 9x + 20 = 0
Answer:
Let y = x² – 9x + 20
(i) Prepare the table of values for y = x² – 9x + 20

(ii) Plot the points (-1, 30) (0,20) (1, 12) (2, 6) (3,2), (4, 0), (5, 0), (6,2) (omit the high value)
(iii) Join the points by a free hand smooth curve.
(iv) The roots of the equation are the X-coordinates of the intersecting points of the curve with X-axis (4, 0) and (5, 0)
There are two points of intersection with the X-axis at 4 and 5. The solution set is 4 and 5. The quadratic equation has real and unequal roots.
(v) Since there is two point of intersection with X-axis (different solution)
∴ The equation x² – 9x + 20 = 0 has real and unequal roots.

(ii) x² – 4x + 4 = 0
Answer:
Let y = x² – 4x + 4
(i) Prepare the table of values for y = x² – 4x + 4

(ii) Plot the points (-3,25) (-2,16) (-1, 9) (0,4) (1,-1) (2, 0), (3,1) and (4, 4)
(iii) Join the points by a free hand smooth curve.
(iv) The roots of the equation are the X-coordinates of the intersecting points of the curve with X-axis (2, 0) which is 2.
(v) Since there is only one point of intersection with X-axis (2, 0).
∴ The solution set is 2.
The Quadratic equation x² – Ax + 4 = 0 has real and equal roots.

(iii) x² + x + 7 = 0
Answer:
Let y = x² + x + 7
(i) Prepare the table of values for y = x² + x + 7

(ii) Plot the points (-4,19) (-3,13) (-2, 9) (-1, 7) (0, 7) (1, 9), (2,13) (3,19) and (4,27)
(iii) Join the points by a free hand smooth curve.
(iv) The solution of the given quadratic equation are the X-coordinates of the intersecting points of the parabola with the X-axis.
(v) The curve does not intersecting the X-axis. There is no solution set.
The equation x² + x + 7 = 0 has no real roots.

(iv) x² – 9 = 0
Answer:
Let y = x² – 9
(i) Prepare the table of values for y = x² – 9

(ii) Plot the points (-4, 7) (-3, 0) (-2, -5) (-1, -8) (0, -9) (1, -8), (2, -5) (3, 0) (4, 7)
(iii) Join the points by a free hand smooth curve.
(iv) The curve intersect the X-axis at -3 and 3.
The solution is (-3, 3).
(v) Since there are two points of intersection -3 and 3 with the X-axis the quadratic equation has real and unequal roots.

(v) x² – 6x + 9 = 0
Answer:
Let y = x² – 6x + 9
(i) Prepare a table of values for y = x² – 6x + 9

(ii) Plot the points (-2,25) (-1,16) (0,9) (1,4) (2,1) (3,0), (4,1) and (5,4) on the graph using suitable scale (omit the points (-4, 49) and (-3, 36)
(iii) Join the points by a free hand smooth curve.
(iv) The X – coordinates of the point of intersection of the curve with X-axis are the roots of the , given equation, provided they intersect.
The solution is 3.
(v) Since there is only one point of intersection with X-axis the quadratic equation x² – 6x + 9 = 0 has real and equal roots.

(vi) (2x – 3) (x + 2) = 0
Answer:
y = (2x – 3) (x + 2)
= 2x² + 4x – 3x – 6
= 2x² + x – 6

(i) Prepare a table of values for y from x – 4 to 4

(ii) Plot the points (-4, 22) (-3, 9) (-2, 0) (-1, -5) (0, -6) (1, -3), (2, 4), (3, 15) and (4, 30).
(iii) Join the points by a free hand smooth curve.
(iv) The curve intersect the X – axis at (-2, 0) and (1\(\frac { 1 }{ 2 } \), 0)
∴ The solution set is (-2,1\(\frac { 1 }{ 2 } \))
(v) Since there are two points of intersection with X – axis, the quadratic equation has real and un – equal roots.

Question 2.
Draw the graph of y = x² – 4 and hence solve x² – x – 12 = 0
Answer:
(i) Draw the graph of y = x² – 4 by preparing the table of values as below.

(ii) Plot the points for the ordered pairs (-4, 12) (-3, 5) (-2, 0) (-1, -3) (0, -4) (1, -3), (2, 0), (3, 5) and (4, 12). Draw the curve with the suitable scale.
(iii) To solve x² – x – 12 = 0 subtract x² – x – 12 from y = x² – 4

The equation y = x + 8 represents a straight line. Prepare a table for y = x + 8

(iv) Mark the point of intersection of the curve and the straight line is (-3, 5) and 4, 12)
∴ The solution set is (-3, 4) for x² – x – 12 = 0.

Question 3.
Draw the graph of y = x² + x and hence solve x² + 1 = 0
Answer:
Let y = x² + x
(i) Draw the graph of y = x² + x by preparing the table.

(ii) Plot the points (-4, 12), (-3, 6), (-2, 2), (-1, 0), (0, 0), (1, 2), (2, 6), (3, 12) and (4, 20).
(iii) Join the points by a free hand to get smooth curve.
(iv) To solve x² + 1 = 0, subtract x² + 1 = 0 from x² + x we get.

The equation represent a straight line. Draw a line y = x – 1

Observe the graph of y = x² + 1 does not interset the parabola y = x² + x.
This x² + 1 has no real roots.

Question 4.
Draw the graph of y = x² + 3x + 2 and use it to solve x² + 2x + 1 = 0.
(i) Draw the graph of y = x² + 3x + 2 preparing the table of values as below.

(ii) Plot the points (-4, 6), (-3, 2), (-2, 0), (-1, 0), (0, 2), (1, 6), (2, 12), (3, 20) (4, 30).
(iii) To solve x² + 2x + 1 = 0 subtract x² + 2x + 1 = 0 from y = x² + 3x + 2

(iv) Draw the graph of y = x + 1 from the table

The equation y = x + 1 represent a straight line.
This line intersect the curve at only one point (-1, 0). The solution set is (-1).

Question 5.
Draw the graph of y = x² + 3x – 4 and hence use it to solve x² + 3x – 4 = 0
Answer:
Let y = x² + 3x – 4
(i) Draw the graph of y = x² + 3x – 4

(ii) Plot the points (-5, 6), (-4, 0), (-3, -4), (-2, -6), (-1, -6), (0, -4), (1, 0), (2, 6), (3, 14) on the graph using suitable scale.
(iii) Join the points by a free hand smooth curve.
The smooth curve is the graph of y = x² – 4x + 4
(iv) To solve x + 3x – 4 = 0, subtract x² + 3x – 4 = 0 from y = x² + 3x – 4.
y = 0
∴ The point of intersection with the x – axis is the solution set.
The solution set is -4 and 1.

Question 6.
Draw the graph of y = x² – 5x – 6 and hence solve x², – 5x – 14 = 0
Answer:
Let y = x² – 5x – 6
(i) Draw the graph of y = x² – 5x – 6 by preparing the table of values as below.

(ii) Plot the points (-3, 18), (-2, 8), (-1, 0), (0, -6), (1, -10), (2, -12), (3, -12), (4, -10), (5,-6), (6, 0) and (7, 8).
(iii) Join the points by a free hand to get smooth curve.
(iv) To solve x² – 5x – 14 = 0, subtract x² – 5x – 14 = 0 from y = x² – 5x – 6.

The equation y = 8 represent a straight line draw a straight line through y = 8 intersect the curve at two places. From the two points draw perpendicular line to the X – axis it will intersect at -2 and 7.
The solution is -2 and 7

Question 7.
Draw the graph of y = 2x² – 3x – 5 and hence solve 2x² – 4x – 6 = 0
Answer:
(i) Draw the graph of y = 2x² – 3x – 5 by preparing the table of values given below.

(ii) Plot the points (-3, 22), (-2, 9), (-1, 0), (0, -5), (1,-6), (2, -3), (3, 4), (4, 15) on the graph sheet using suitable scale.
(iii) To solve 2x² – 4x – 6 = 0 subtract 2x² – 4x – 6 = 0 from y = 2x² – 3x – 5

(iv) y = x + 1 represent a straight line.

The straight line intersect the curve at (-1, 0) and (3, 4). From the two point draw perpendicular lines to the X – axis it will intersect at -1 and 3.
The solution set is (-1, 3)

Question 8.
Draw the graph of y = (x – 1) (x + 3) and hence solve x² – x – 6 = 0
Answer:
y = (x – 1) (x + 3)
y = x² + 2x – 3
(i) Draw the graph of y = x² + 2x – 3 by preparing the table of values given below

(ii) Plot the points (-4, 5), (-3, 0), (-2, -3), (-1, -4), (0, -3), (1, 0), (2, 5), (3, 12) and (4, 21) on the graph sheet using suitable scale.
(iii) To solve x² – x – 6 = 0 subtract x² – x – 6 = 0 from y = x2 + 2x – 3
(iv) Draw the graph of y = 3x + 3 by preparing the table.

(v) The straight line cuts the curve at (-2, -3) and (3, 12). Draw perpendicular lines from the point to X – axis.
The line cut the X – axis at -2 and 3.
The solution set is (-2, 3)