# Class 10

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## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.8

Question 1.
Find the square root of the following polynomials by division method
(i) x4 – 12x3 + 42x2 – 36x + 9
(ii) 37x2 – 28x3 + 4x4 + 42x + 9
(iii) 16x4 + 8x2 + 1
(iv) 121x4 – 198x3 – 183x2 + 216x + 144
Solution:
The long division method in finding the square root of a polynomial is useful when the degrees of a polynomial is higher.

Synthetic Division Calculator. The calculator will divide the polynomial by the binomial using synthetic division, with steps shown.

Question 2.
Find the square root of the expression $$\frac{x^{2}}{y^{2}}-10 \frac{x}{y}+27-10 \frac{y}{x}+\frac{y^{2}}{x^{2}}$$
Solution:

Question 3.
Find the values of a and b if the following polynomials are perfect squares
(i) 4x4 – 12x3 + 37x2 + bx + a
(ii) ax4 + bx3 + 361ax2 + 220x + 100
Solution:
(i)

Since it is a perfect square.
Remainder = 0
⇒ b + 42 = 0, a – 49 = 0
b = -42, a = 49

(ii) ax4 + bx3 + 361ax2 + 220x + 100

Since remainder is 0
a = 144
b = 264

Question 4.
Find the values of m and n if the following expressions are perfect squares
(i) $$\frac{1}{x^{4}}-\frac{6}{x^{3}}+\frac{13}{x^{2}}+\frac{m}{x}+n$$
(ii) x4 – 8x3 + mx2 + nx + 16
Solution:
(i)

(ii)

Since remainder is 0,
m = 24, n = -32

## Samacheer Kalvi 10th English Book Answers Solutions Guide

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## Samacheer Kalvi 10th Maths Book Answers Solutions Guide

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## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Ex 6.2

Question 1.
Find the angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of a tower of height 10$$\sqrt{3}$$ m.
Solution:

Question 2.
A road is flanked on either side by continuous rows of houses of height 4$$\sqrt{3}$$ m with no space in between them. A pedestrian is standing on the median of the road facing a row house. The angle of elevation from the pedestrian to the top of the house is 30°. Find the width of the road.
Solution:

Question 3.
To a man standing outside his house, the angles of elevation of the top and bottom of a window are 60° and 45° respectively. If the height of the man is 180 cm and if he is 5 m away from the wall, what is the height of the window? ($$\sqrt{3}$$ = 1.732)
Solution:

Let ‘H’ be the fit of the window. Given that elevation of top of the window is 60°.

Given that elevation of bottom of the window is 45°.

∴ Height of the window = 3.66 m

Question 4.
A statue 1.6 m tall stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 40°. Find the height of the pedestal. (tan 40° = 0.8391, $$\sqrt{3}$$ = 1.732)
Solution:

Let ‘p’ be the fit of the pedestal and d be the distance of statue from point of cabs, on the ground.
Given the elevation of top of the statue from pf on ground is 60°.

Question 5.
A flag pole ‘h’ metres is on the top of the hemispherical dome of radius V metres. A man is standing 7 m away from the dome. Seeing the top of the pole at an angle 45° and moving 5 m away from the dome and seeing the bottom of the pole at an angle 30°. Find
(i) the height of the pole
Solution:

Question 6.
The top of a 15 m high tower makes an angle of elevation of 60° with the bottom of an electronic pole and angle of elevation of 30° with the top of the pole. What is the height of the electric pole?
Solution:

Let BD be tower of height = 15 m
AE be pole of height = ‘p’

Question 7.
A vertical pole fixed to the ground is divided in the ratio 1 : 9 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a place on the ground, 25 m away from the base of the pole, what is the height of the pole?
Solution:

This mile calculator estimates the number of driving miles between two locations in the United States.

Question 8.
A traveler approaches a mountain on highway. He measures the angle of elevation to the peak at each milestone. At two consecutive milestones the angles measured are 4° and 8°. What is the height of the peak if the distance between consecutive milestones is 1 mile, (tan 4° = 0.0699, tan 8° = 0.1405).
Solution:

## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Ex 4.3

Question 1.
A man goes 18 m due east and then 24 m due north. Find the distance of his current position from the starting point?
Solution:
Using Pythagoras theorem

AC2 = AB2 + BC2
= (18)2 + (24)2
= 324 + 576
= 900
AC = $$\sqrt{900}$$ = 30 m
∴ The distance from the starting point is 30 m.

Question 2.
There are two paths that one can choose to go from Sarah’s house to James house. One way is to take C street, and the other way requires to take A street and then B street. How much shorter is the direct path along C street? (Using figure).

Solution:
By using Pythagoras theorem
AC2 = AB2 + BC2
= 22 + (1.5)2
= 4 + 2.25
= 6.25
AC = 2.5 miles.
If one chooses C street the distance from James house to Sarah’s house is 2.5 miles
If one chooses A street and B street he has to go 2 + 1.5 = 3.5 miles.
2.5 < 3.5, 3.5 – 2.5 = 1 Through C street is shorter by 1.0 miles.
∴ The direct path along C street is shorter by 1 mile.

To round to the nearest tenth, look at the tenth’s place (right after the decimal).

Question 3.
To get from point A to point B you must avoid walking through a pond. You must walk 34 m south and 41 m east. To the nearest meter, how many meters would be saved if it were possible to make a way through the pond?
Solution:
By using Pythagoras

AC2 = AB2 + BC2
= 342 + 412
= 1156+ 1681
= 2837
AC = 53.26 m
Through B one must walk 34 + 41 = 75 m walking through a pond one must comes only 53.2 m
∴ The difference is (75 – 53.26) m = 21.74 m
∴ To the nearest, one can save 21.74 m.

Question 4.
In the rectangle WXYZ, XY + YZ = 17 cm, and XZ + YW = 26 cm. Calculate the length and breadth of the rectangle?

Solution:
XY + YZ = 17 cm …………. (1)
XZ + YW = 26 cm ………… (2)
(2) ⇒ XZ = 13, YW = 13
(∵ In rectangle diagonals are equal).
(1) ⇒ XY = 5, YZ = 12 XY + YZ = 17
⇒ Using Pythagoras theorem
52 + 122 = 25 + 144 = 169 = 132
∴ In ∆XYZ = 132 = 52 + 122 it is verified
∴ The length is 12 cm and the breadth is 5 cm.

Question 5.
The hypotenuse of a right triangle is 6 m more than twice of the shortest side. If the third side is 2 m less than the hypotenuse, find the sides of the triangle?
Solution:
Let a is the shortest side.
c is the hypotenuse
b is the third side.

∴ The sides of the triangle are 10m, 24m, 26m.
Verification 262 = 102 + 242
676 = 100 + 576 = 676

Question 6.
5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4m high. If the foot of the ladder is moved 1.6 m towards the wall, then find the distance by which the top of the ladder would slide upwards on the wall.
Solution:
Let the distance by which top of the slide moves upwards be assumed as ‘x’.

From the diagram, DB = AB – AD
= 3 – 1.6 ⇒ DB = 1.4 m
also BE = BC + CE
= 4 + x
∴ DBE is a right angled triangle
DB2 + BE2 = DE2 ⇒ (1.4)2 + (4 + x)2= 52
⇒ (4 + x)2 = 25 – 1.96 ⇒ (4 + x)2 = 23.04
⇒ 4 + x = $$\sqrt{23.04}$$ = 4.8
⇒ x = 4.8 – 4 ⇒ x = 0.8 m

Question 7.
The perpendicular PS on the base QR of ∆PQR intersects QR at S, such that QS = 3 SR. Prove that 2PQ2 = 2PR2 + QR2.
Solution:

In ∆PQS,
PQ2 = PS2 + QS2 ………… (1)
In ∆PSR,
PR2 = PS2 + SR2 ……….. (2)
(1) – (2) ⇒ PQ2 – PR2 = QS2 – SR2 …………. (3)

Hence it proved.

Question 8.
In the adjacent figure, ABC is a right-angled triangle with right angle at B and points D, E trisect BC. Prove that 8AE2 = 3AC2 + 5AD2.

Solution:
Since D and E are the points of trisection of BC,
therefore BD = DE = CE
Let BD = DE = CE = x
Then BE = 2x and BC = 3x
In right triangles ABD, ABE and ABC, (using Pythagoras theorem)
We have AD2 = AB2 + BD2
⇒ AD2 = AB2 + x2 ……………. (1)
AE2 = AB2 + BE2
⇒ AB2 + (2x)2
⇒ AE2 = AB2 + 4x2 ………… (2)
and AC2 = AB2 + BC2 = AB2 + (3x)2
AC2 = AB2 + 9x2
Now 8 AE2 – 3 AC2 – 5 AD2 = 8 (AB2 + 4x2) – 3 (AB2 + 9x2) – 5 (AB2 + x2)
= 8AB2 + 32x2 – 3AB2 – 27x2 – 5AB2 – 5x2
= 0
∴ 8 AE2 – 3 AC2 – 5 AD2 = 0
8 AE2 = 3 AC2 + 5 AD2.
Hence it is proved.

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

Question 1.
Use Euclid’s algorithm to find the HCF of 4052 and 12756.
Solution:
Since 12576 > 4052 we apply the division lemma to 12576 and 4052, to get HCF
12576 = 4052 × 3 + 420.
Since the remainder 420 ≠ 0, we apply the division lemma to 4052
4052 = 420 × 9 + 272.
We consider the new divisor 420 and the new remainder 272 and apply the division lemma to get
420 = 272 × 1 + 148, 148 ≠ 0.
∴ Again by division lemma
272 = 148 × 1 + 124, here 124 ≠ 0.
∴ Again by division lemma
148 = 124 × 1 + 24, Here 24 ≠ 0.
∴ Again by division lemma
124 = 24 × 5 + 4, Here 4 ≠ 0.
∴ Again by division lemma
24 = 4 × 6 + 0.
The remainder has now become zero. So our procedure stops. Since the divisor at this stage is 4.
∴ The HCF of 12576 and 4052 is 4.

Question 2.
If the HCF of 65 and 117 is in the form (65m – 117) then find the value of m.
By Euclid’s algorithm 117 > 65
117 = 65 × 1 + 52
52 = 13 × 4 × 0
65 = 52 × 1 + 13
H.C.F. of 65 and 117 is 13
65m – 117 = 13
65 m = 130
m = $$\frac { 130 }{ 65 }$$ = 2
The value of m = 2

The smallest number to appear on both lists is 60, so 60 is the least common of 15 and 20.

Question 3.
Find the LCM and HCF of 6 and 20 by the prime factorisation method.
Solution:
We have 6 = 21 × 31 and
20 = 2 × 2 × 5 = 22 × 51
You can find HCF (6, 20) = 2 and LCM (6, 20) = 2 × 2 × 3 × 5 = 60. As done in your earlier classes. Note that HCF (6, 20) = 21 = product of the smallest power of each common prime factor in the numbers.
LCM (6, 20) = 22 × 31 × 51 = 60.
= Product of the greatest power of each prime factor, involved in the numbers.

Common Multiples of 16: 16, 32, 48, 64, 80,… Hence the Least common multiple of 12 and 16 is 48. The LCM of 12 and 16 is 48.

Question 4.
Prove that $$\sqrt { 3 }$$ is irrational.
Let us assume the opposite, (1) $$\sqrt { 3 }$$ is irrational.
Hence $$\sqrt { 3 }$$ = $$\frac { p }{ q }$$
Where p and q(q ≠ 0) are co-prime (no common factor other than 1)

Hence, 3 divides p2
So 3 divides p also …………….. (1)
Hence we can say
$$\frac { p }{ 3 }$$ = c where c is some integer
p = 3c
Now we know that
3q2 = p2
Putting = 3c
3q2 = (3c)2
3q2 = 9c2
q2 = $$\frac { 1 }{ 3 }$$ × 9c2
q2 = 3c2
$$\frac{q^{2}}{3}$$ = C2
Hence 3 divides q2
So, 3 divides q also ……………. (2)
By (1) and (2) 3 divides both p and q
By contradiction $$\sqrt { 3 }$$ is irrational.

Question 5.
Which of the following list of numbers form an AP? If they form an AP, write the next two terms:
(i) 4, 10, 16, 22, …
(ii) 1, -1,-3, -5,…
(iii) -2, 2, -2, 2, -2, …
(iv) 1, 1, 1, 2, 2, 2, 3, 3, 3,…
Solution:
(i) 4, 10, 16, 22, …….
We have a2 – a1 = 10 – 4 = 6
a3 – a2 = 16 – 10 = 6
a4 – a3 = 22 – 16 = 6
∴ It is an A.P. with common difference 6.
∴ The next two terms are, 28, 34

(ii) 1, -1, -3, -5
t2 – t1 = -1 – 1 = -2
t3 – t2 = -3 – (-1) = -2
t4 – t3 = -5 – (-3) = -2
The given list of numbers form an A.P with the common difference -2.
The next two terms are (-5 + (-2)) = -7, -7 + (-2) = -9.

(iii) -2, 2,-2, 2,-2
t2 – t1 = 2-(-2) = 4
t3 – t2 = -2 -2 = -4
t4 – t3 = 2 – (-2) = 4
It is not an A.P.

(iv) 1, 1, 1, 2, 2, 2, 3, 3, 3
t2 – t1 = 1 – 1 = 0
t3 – t2 = 1 – 1 = 0
t4 – t3 = 2 – 1 = 1
Here t2 – t1 ≠ t3 – t2
∴ It is not an A.P.

Question 6.
Find n so that the nth terms of the following two A.P.’s are the same.
1, 7,13,19,… and 100, 95,90,…
The given A.P. is 1, 7, 13, 19,….
a = 1, d = 7 – 1 = 6
tn1 = a + (n – 1)d
tn1 = 1 + (n – 1) 6
= 1 + 6n – 6 = 6n – 5 … (1)
The given A.P. is 100, 95, 90,….
a = 100, d = 95 – 100 = – 5
tn2 = 100 + (n – 1) (-5)
= 100 – 5n + 5
= 105 – 5n …..(2)
Given that, tn1 = tn2
6n – 5 = 105 – 5n
6n + 5n = 105 + 5
11 n = 110
n = 10
∴ 10th term are same for both the A.P’s.

Question 7.
In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 is the third, and so on. There are 5 rose plants in the last row. How many rows are there in the flower bed?
The number of rose plants in the 1st, 2nd, 3rd,… rows are
23, 21, 19,………….. 5
It forms an A.P.
Let the number of rows in the flower bed be n.
Then a = 23, d = 21 – 23 = -2, l = 5.
As, an = a + (n – 1)d i.e. tn = a + (n – 1)d
We have 5 = 23 + (n – 1)(-2)
i.e. -18 = (n – 1)(-2)
n = 10
∴ There are 10 rows in the flower bed.

Question 8.
Find the sum of the first 30 terms of an A.P. whose nth term is 3 + 2n.
Given,
tn = 3 + 2n
t1 = 3 + 2 (1) = 3 + 2 = 5
t2 = 3 + 2 (2) = 3 + 4 = 7
t3 = 3 + 2 (3) = 3 + 6 = 9
Here a = 5,d = 7 – 5 = 2, n = 30
Sn = $$\frac { n }{ 2 }$$ [2a + (n – 1)d]
S30 = $$\frac { 30 }{ 2 }$$ [10 + 29(2)]
= 15 [10 + 58] = 15 × 68 = 1020
∴ Sum of first 30 terms = 1020

Question 9.
How many terms of the AP: 24, 21, 18, . must be taken so that their sum is 78?
Solution:
Here a = 24, d = 21 – 24 = -3, Sn = 78. We need to find n.
We know that,
Sn = $$\frac { n }{ 2 }$$ (2a + (n – 1)d)
78 = $$\frac { n}{ 2 }$$ (48 + 13(-3))
78 = $$\frac { n}{ 2 }$$ (51 – 3n)
or 3n2 – 51n + 156 = 0
n2 – 17n + 52 = 0
(n – 4) (n – 13) = 0
n = 4 or 13
The number of terms are 4 or 13.

Question 10.
The sum of first n terms of a certain series is given as 3n2 – 2n. Show that the series is an arithmetic series.
Solution:
Given, Sn = 3n2 – 2n
S1 = 3 (1)2 – 2(1)
= 3 – 2 = 1
ie; t1 = 1 (∴ S1 = t1)
S2 = 3(2)2 – 2(2) = 12 – 4 = 8
ie; t1 + t2 = 8 (∴ S2 = t1 + t2)
∴ t2 = 8 – 1 = 7
S3 = 3(3)2 – 2(3) = 27 – 6 = 21
t1 + t2 + t3 = 21 (∴ S3 = t1 + t2 + t3)
8 + t3 = 21 (Substitute t1 + t2 = 8)
t3 = 21 – 8 ⇒ t3 = 13
∴ The series is 1,7,13, …………. and this series is an A.P. with common difference 6.