Class 10

Students can practice TS Class 10 Maths Solutions Chapter 6 Progressions Ex 6.3 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 6 Progressions Exercise 6.3

Question 1.
Find the sum of the following APs :

i) 2, 7, 12, ………., to 10 terms.
Solution:
Given A.P. : 2, 7, 12, ……… to 10 term
a = 2; d = a₂ – a₁ = 7 – 2 = 5; n = 10
s_n = \(\frac{\mathrm{n}}{2}\)[2a + (n – 1)d]
s₁₀ = – \(\frac{10}{2}\)[2 × 2 +(10 – 1)5]
= 5[4 + 9 × 5]
= 5 [4 + 45]
= 5 × 49 = 245

ii) – 37, – 33, – 29, …… to 12 terms.
Solution:
Given A.P :
– 37, – 33, – 29, …., to 12 terms
a = – 37; d = a₂ – a₁ = (- 33) – (- 37)
= – 33 + 37 = 4 and n = 12
S_n = \(\frac{n}{2}\)[2a+ (n – 1)d]
s₁₂ = \(\frac{12}{2}\)[2 × (-37) + (12 – 1)4]
= 6[-74 + 11 × 4]
= 6[-74 + 44] = 6 × (- 30) = -180

iii) 0.6, 1.7, 2.8, …, to 100 terms.
Solution:
Given A.P : 0.6, 1.7, 2.8, …., S₁₀₀
a = 0.6; d = a₂ – a₁ = 1.7 – 0.6 = 1.1 and n = 100
s = \(\frac{\mathrm{n}}{2}\)[2a + (n – 1)d]
s₁₀₀ = \(\frac{100}{2}\)[2 × 0.6 + (100 – 1) × 1.1]
= 50[1.2 + 99 × 1.1]
= 50[1.2 + 108.9]
= 50 × 110.1 = 5505.

iv) \(\frac{1}{15}\), \(\frac{1}{12}\), \(\frac{1}{10}\) ….. to 11 terms.
Solution:

Question 2.
Find the sums given below :

i) 7 + 10\(\frac{1}{2}\) + 14 + ….. + 84
Solution:
Given A.P. : 7 + 10\(\frac{1}{2}\) + 14 + …… + 84
∴ a = 7, d = a₂ – a₁ = 10\(\frac{1}{2}\) – 7 = 3\(\frac{1}{2}\)
and the last term l = a_n
But, a_n = a +(n – 1) d
∴ 84 = 7 + (n – 1) × 3\(\frac{1}{2}\)
⇒ 84 – 7 = (n – 1) × \(\frac{7}{2}\)
⇒ n – 1 = 77 × \(\frac{2}{7}\) = 22
⇒ n = 22 + 1 = 23
Now, S_n = \(\frac{\mathrm{n}}{2}\)(a + l)
Where a = 7; l = 84
S₂₃ = \(\frac{23}{2}\)(7 + 84)
= \(\frac{23}{2}\) × 91 = \(\frac{2093}{2}\) = 1046\(\frac{1}{2}\)

ii) 34 + 32 + 30 + ……. + 10
Solution:
Given A.P : 34 + 32 + 30 + …….. + 10
Here a = 34; d = a₂ – a₁ = 32 – 34 = -2
and the last term ‘l‘ = a_n = 10
But, a_n = a + (n – 1) d
⇒ 10 = 34 + (n – 1)(-2)
⇒ 10 – 34 = -2n + 2
⇒ -2n = -24 – 2
⇒ n = \(\frac{-26}{-2}\) ∴ n = 13
Also, S_n = \(\frac{\mathrm{n}}{2}\)(a + l)
Where a = 34, l = 10
S₁₃ = \(\frac{13}{2}\)(34 + 10)
\(\frac{13}{2}\) × 44 = 13 × 22 = 286

iii) -5 + (-8) + (-11) + …. + (-230).
Solution:
Given AP:
-5 + (-8) + (-11) + …….. + (-230)
Here first term, a = -5;
Last term, l = a_n = -230:
d = a₂ – a₁ = (-8) – (-5)
= -8 + 5 = -3
But, a_n = a + (n – 1) d
∴ (-230) = (-5) + (n – 1) × (-3)
⇒ -230 + 5 = -3n + 3
⇒ -3n + 3 = -225
⇒ -3n = -225 – 3 ⇒ 3n = 228
∴ n = \(\frac{228}{3}\) = 76
Now S_n = \(\frac{\mathrm{n}}{2}\)(a + l)
S₇₆ = \(\frac{76}{2}\)[(-5) + (-230)]
= 38 × (-235) = -8930

Question 3.
In an AP:

i) Given a = 5, d = 3, a_n = 50, find n and S_n.
Solution:
Given:
a = 5; d = 3 ; a_n = a + (n – 1)d = 50
⇒ 50 = 5 + (n – 1)3
⇒ 50 – 5 = 3n – 3
⇒ 3n = 45 + 3
⇒ n = \(\frac{48}{3}\) = 16
S_n = \(\frac{\mathrm{n}}{2}\)(a + l)
S₁₆ = \(\frac{16}{2}\)[5 + 50] = 8 × 55 = 440.

ii) Given a = 7, a₁₃ = 35, find d and S₁₃.
Solution:
Given : a = 7
a₁₃ = a + 12d = 35
⇒ 7 + 12d = 35
⇒ 12d = 35 – 7
⇒ d = \(\frac{28}{12}\) = \(\frac{7}{3}\)
Now, S_n = \(\frac{\mathrm{n}}{2}\)(a + l)
S₁₃ = \(\frac{13}{2}\)[7 + 35]
= \(\frac{13}{2}\) × 42 = 13 × 21 = 273

iii) Given a₁₂ = 37, d = 3, find a and S₁₂.
Solution:
Given : a₁₂ = a + 11d = 37
d = 3
So, a₁₂ = a + 11 × 3 = 37
a + 33 = 37
a = 37 – 33 = 4
Now, S_n = \(\frac{\mathrm{n}}{2}\)(a + l)
S₁₂ = \(\frac{12}{2}\)[4 + 37] = 6 × 41 = 246.

iv) Given a₃ = 15, S₁₀ = 125, find d and a₁₀.
Solution:
Given : a₃ = a + 2d = 15
⇒ a = 15 – 2d —- (1)
S₁₀ = 125 but take S₁₀ as 175
i.e., S₁₀ = 175
We know that,
S_n = \(\frac{\mathrm{n}}{2}\)(a + l)
⇒ S₁₀ = \(\frac{10}{2}\)[a + l]
⇒ 175 = \(\frac{10}{2}\)[2a + 9d]
⇒ 175 = 5[2a + 9d]
⇒ 35 = 2(15 – 2d) + 9d [∵ a = 15 – 2d]
⇒ 35 = 30 – 4d + 9d
⇒ 35 – 30 = 5d ⇒ d = \(\frac{5}{5}\) = 1
Substituting d = 1 in equation (1) we get
a = 15 – 2 × 1 = 15 – 2 = 13
Now, a_n = a + (n – 1) d
a₁₀ = a + 9d = 13 + 9 × 1
= 13 + 9 = 22
∴ a₁₀ = 22 ; d = 1.

v) Given a = 2, d = 8, S_n = 0, find n and a_n.
Solution:
Given : a = 2; d = 8 and S_n = 90
But S_n = \(\frac{\mathrm{n}}{2}\)[2a + (n – 1) d]
⇒ 90 = \(\frac{\mathrm{n}}{2}\)[2 × 2 + (n – 1) × 8]
⇒ 90 = \(\frac{\mathrm{n}}{2}\)[4 + 8n – 8]
⇒ 90 = \(\frac{\mathrm{n}}{2}\)[8n – 4]
⇒ 90 = \(\frac{4 n}{2}\)[2n – 1]
⇒ 90 = 2n[2n – 1]
⇒ 4n² – 2n – 90 = 0
⇒ 2(2n² – n – 45) = 0
⇒ 2n² – n – 45 = 0
⇒ 2n² – 10n + 9n – 45 = 0
⇒ 2n(n – 5) + 9(n – 5) = 0
⇒ (n – 5) (2n + 9) = 0
⇒ n – 5 = 0 (or) 2n + 9 = 0
⇒ n = 5 (or) n = \(\frac{-9}{2}\) (discarded)
∴ n = 5
Now a_n = a₅ = a + 4d = 2 + 4 × 8
= 2 + 32 = 34.

vi) Given a_n = 4, d = 2, S_n = -14, find n and a.
Solution:
Given a_n = a + (n – 1) d = 4 —– (1)
d = 2; S_n = -14
From (1); a + (n – 1) 2 = 4
a = 4 – 2n + 2 ⇒ a = 6 – 2n
S_n = \(\frac{\mathrm{n}}{2}\) [a + a_n]
-14 = \(\frac{\mathrm{n}}{2}\) [(6 – 2n) + 4] [∵ a = 6 – 2n]
-14 × 2 = n(10 – 2n)
⇒ 10n – 2n² = -28
⇒ 2n² – 10n – 28 = 0
⇒ n² – 5n – 14 = 0
⇒ n² – 7n + 2n – 14 = 0
⇒ n(n – 7) + 2(n – 7) = 0
⇒ (n – 7) (n + 2) = 0
⇒ n = 7 (or) n = – 2
∴ n = 7
Now a = 6 – 2n = 6 – 2 × 7
= 6 – 14 = -8
∴ a = – 8; n = 7.

vii) Given l = 28, S = 144 and there are total 9 terms. Find a.
Solution:
Given :
l = a₉ = a + 8d = 28 and S₉ = 144
But, S_n = \(\frac{9}{2}\)(a + l)
144 = \(\frac{9}{2}\)[a + 28]
⇒ 144 × \(\frac{2}{9}\) = a + 28
⇒ a + 28 = 32
⇒ a = 32 – 28 = 4.

Question 4.
The first and the last terms of an A.P are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum ?
Solution:
Given A.P in which a = 17
Last term = l = 350
Common difference, d = 9
We know that, a_n = a + (n – 1) d
350 = 17 + (n – 1)9
⇒ 350 = 17 + 9n – 9
⇒ 9n = 350 – 8
⇒ n = \(\frac{342}{9}\) = 38
Now, S_n = \(\frac{n}{2}\)(a + l)
S₃₈ = \(\frac{38}{2}\)[17 + 350]
= 19 × 367 = 6973
∴ n = 38
S_n = 6973.

Question 5.
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution:
Given A.P in which

Substituting d = 4 in equation (1), we get a + 4 = 14
⇒ a = 14 – 4 = 10
Now, S_n = \(\frac{n}{2}\)[a + l] (or) \(\frac{n}{2}\) [2a (n – 1) d]
∴ S₅₁ = \(\frac{51}{2}\)[2 × 10 + (51 – 1) × 4]
= \(\frac{51}{2}\)[20 + 50 × 4]
= \(\frac{51}{2}\) × (20 + 200)
= \(\frac{51}{2}\) × 220
= 51 × 110 = 5610.

Question 6.
If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms. (T.S. & A.P. Mar. 15, 16)
Solution:
Given :
A.P such that S₇ = 49
S₁₇ = 289
We know that,
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
S₇ = 49 = \(\frac{7}{2}\)[2a + (7 – 1) d]
⇒ \(\frac{49}{7}\) = \(\frac{1}{2}\)[2a + 6d]
⇒ 7 = a + 3d —– (1)
Also, S₁₇ = 289 = \(\frac{17}{2}\) [2a + (17 – 1)d]
\(\frac{289}{17}\) = \(\frac{1}{2}\)(2a + 16d)
⇒ 17 = a + 8d —— (2)

⇒ d = \(\frac{10}{5}\)
Substituting d = 2 in equation (1), we get
a + 3 × 2 = 7
⇒ a = 7 – 6 = 1
∴ a = 1; d = 2
Now, S_n = \(\frac{\mathrm{n}}{2}\)[2a + (n – 1)d]
S_n = \(\frac{\mathrm{n}}{2}\) [2 × 1 + (n -1)2]
= \(\frac{\mathrm{n}}{2}\) [2 + 2n – 2)d] = \(\frac{\mathrm{n} .2 \mathrm{n}}{2}\)
∴ Sum of first n terms S_n = n².
Shortcut : S₇ = 49 = 7²
S₁₇ = 289 = 17²
∴ S_n = n²

Question 7.
Show that a₁, a₂, ……. a_n, …. form an AP where an is defined as below :
i) a_n = 3 + 4n
ii) a_n = 9 – 5n. Also find the sum of the first 15 terms in each case.
Solution:
i) Given a_n = 3 + 4n
Then a₁ = 3 + 4 × 1 = 3 + 4 = 7
a₂ = 3 + 4 × 2 = 3 + 8 = 11
a₃ = 3 + 4 × 3 = 3 + 12 = 15
a₄ = 3 + 4 × 4 = 3 + 16 = 19
Now the pattern is 7, 11, 15, ……..
where a = a₁ = 7; a₂ = 11; a₃ = 15, …. and a₂ – a₁ = 11 – 7 = 4;
a₃ – a₂ = 15 -11 = 4; Here d = 4. Hence a₁, a₂, …., a_n … forms an A.P.
s₁₅ = \(\frac{15}{2}\) [2 × 7 + (15 – 1)4]
= \(\frac{15}{2}\) [14 + 14.4] = \(\frac{15}{2}\) [70]
= 15 × 35 = 525.

ii) a_n = 9 – 5n
Solution:
Given : a_n = 9 – 5n
a₁ = 9 – 5 × 1 = 9 – 5 = 4
a₂ = 9 – 5 × 2 = 9 – 10 = -1
a₃ = 9 – 5 × 3 = 9 – 15 = -6
a₄ = 9 – 5 × 4 = 9 – 20 = -11
………………………………..
Also a₂ – a₁ = -1 – 4 = – 5
a₃ – a₂ = – 6 – (-1) = – 6 + 1 = -5
a₄ – a₃ = – 11 – (- 6)
= -11 + 6
= -5
∴ d = a₂ – a₁ = a₃ – a₂
= a₄ – a₃ = ……. = -5
Thus the difference between any two succes-sive terms is constant.
Hence {a_n} forms A.P.
S₁₅ = \(\frac{15}{2}\)[2 × 4 + (15 – 1) × (-5)]
= \(\frac{15}{2}\) [8 + 14 (-5)] = \(\frac{15}{2}\) × -62
= 15 × – 31 = – 465.

Question 8.
If the sum of the first n terms of an AP is 4n – n², what is the first term (remember the first term is S₁) ? What is the sum of first two terms ? What is the second term ? Similarly, find the 3^rd, the 10^th and the n^th terms.
Solution:
Given an A.P in which S<sub?n = 4n – n²
Taking n = 1 we get
S₁ = 4 × 1 – 1² = 4 – 1 = 3
n = 2; S₂ = a₁ + a₂ = 4 × 2 – 2²
= 8 – 4 = 4
n = 3; S₃ = a₁ + a₂ + a₃
= 4 × 3 – 3² = 12 – 9 = 3
n = 4; S₄ = a₁ + a₂ + a₃ + a₄
= 4 × 4 – 4² = 16 – 16 = 0
Hence, S₁ = a₁ = 3
∴ a₂ = S₂ – S₁ = 4 – 3 = 1
a₃ = S₃ – S₂ = 3 – 4 = -1
a₄ = S₄ – S₃ = 0 – 3 = -3
So, d = a₂ – a₁ = 1 – 3 = -2
Now, a₁₀ = a + 9d [∴ a_n = a + (n – 1) d]
= 3 + 9 × (- 2)
= 3 – 18 = -15
a_n = 3 + (n – 1) × (-2)
= 3 – 2n + 2 = 5 – 2n.

Question 9.
Find the sum of the first 40 positive integers divisible by 6.
Solution:
The given numbers are the first 40 positive multiples of 6
⇒ 6 × 1, 6 × 2, 6 × 3, ……., 6 × 40
⇒ 6, 12, 18, …… 240
S_n = \(\frac{\mathrm{n}}{2}\) [a + l]
= \(\frac{40}{2}\) [6 + 240]
= 20 × 246 = 4920
∴ S₄₀ = 4920.

Question 10.
A sum of ₹ 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹ 20 less than its preceding prize, find the value of each of the prizes.
Solution:
Given :
Total l Sum of all cash prizes = ₹ 700
Each prize differs by ₹ 20
Let the prizes (in ascending order) be
x, x + 20, x + 40, x + 60, x + 80, x + 100, x + 120
∴ Sum of the prizes = S₇ = \(\frac{\mathrm{n}}{2}\) [a + l]
⇒ 700 = \(\frac{7}{2}\) [x + x + 120]
⇒ 700 × \(\frac{2}{7}\) = 2x + 120
⇒ 100 = x + 60
⇒ x = 100 – 60 = 40
∴ The prizes are 160, 140, 120, 100, 80, 60, 40.

Question 11.
In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g : a section of Class I will plant 1 tree, a section of Class 11 will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students ?
Solution:
Given : Classes : From I to XII
Section : 3 in each class.
∴ Trees planted by each class = 3 × class number

∴ Total trees planted = 3 + 6 + 9 + 12 + ………+ 36; n = 12
∴ S_n = \(\frac{\mathrm{n}}{2}\) [a + l]
S₁₂ = \(\frac{12}{2}\) [3 + 36]
= 6 × 39 = 234
∴ Total plants = 234

Question 12.
A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, … as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles ? (Take π = \(\frac{22}{7}\))

[Hint: Length of successive semi-circles is l₁, l₂, l₃, l₄,……… with centres at A, B, A, B,…., respectively.]
Solution:

Given : l₁; l₂, l₃, l₄, …….l₁₃ are the semi-circles
with centres alternately at A and B; with radii
r₁ = 0.5 cm [1 × 0.5]
r₂ = 1.0 cm [2 × 0.5]
r₃ = 1.5 cm [3 × 0.5]
r₄ = 2.0 cm [4 × 0.5]
………………………..
r₁₃ = 13 × 0.5 = 6.5
[∴ Radii are in A.P. as a₁ = 0.5 and d = 0.5]
Now, the total length of the spiral = l₁ + l₂ +….. + l₁₃ [… 13 given]
But circumference of a semi-circle is πr.
∴ Total length of the spiral = π × 0.5 + π × 1.0 + ……… + π × 6.5
= π × \(\frac{1}{2}\)[1 + 2 + 3 + ……. + 13]
Sum of the first n – natural numbers is \(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\)]

Question 13.
200 logs are stacked in the following manner : 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row ?

Solution:
Given : Total logs = 200
Number of logs stacked in the first row = 20
Number of logs stacked in the second row = 19
Number of logs stacked in the third row = 18
………………………………
The number series is 20, 19, 18, …. is an A.P.
where a = 20 and d = a₂ – a₁ = 19 – 20 = -1
Also, S_n = 200
∴ S_n = \(\frac{n}{2}\) [2a + (n – 1)d]
200 = \(\frac{n}{2}\) [2 × 20 + (n – 1) × (-1)]
200 = \(\frac{n}{2}\) [40 – n + 1]
200 = \(\frac{n}{2}\) [41 – n]
400 = 41n – n²
⇒ n² – 41n + 400 = 0
⇒ n² – 25n – 16n + 400 = 0
⇒ n(n – 25) – 16(n – 25) = 0
⇒ (n – 25) (n – 16) = 0
⇒ n = 25 (or) 16
There can’t be 25 rows as we are starting with 20 logs in the first row.
∴ Number of rows must be 16.
∴ n = 16 and t₁₆ = a + (n – 1) d = 20 + (16 – 1) (-1) = 20 – 15 = 5

Question 14.
In a bucket and ball race, a bucket is placed at the starting point, which is 5 m from the first ball, and the other balls are placed 3 m apart in a straight line. There are ten balls in the line.

A competitor starts from the bucket, picks up the nearest ball, runs back with it, drops it in the bucket, runs back to pick up the next ball, runs to the bucket to drop it in, and she continues in the same way until all the balls are in the bucket. What is the total distance the competitor has to run ?
[Hint : To pick up the first ball and the second ball, the total distance (in metres) run by a competitor is 2 × 5 + 2 × (5 + 3)]
Solution:
Given : Balls are placed at an equal distance of 3 m from one another.
Distance of first ball from the bucket = 5 m
Distance of second ball from the bucket = 5 + 3 = 8m (5 + 1 × 3)
Distance of third ball from the bucket = 8 + 3 = 11 m (5 + 2 × 3)
Distance of fourth ball from the bucket = 11 + 3 = 14m (5 + 3 × 3)
…………………………………………………..
∴ Distance of the tenth ball from the bucket = 5 + 9 × 3 = 5 + 27 = 32m.
1^st ball: Distance covered by the competitor in picking up and dropping it in the bucket = 2 × 5 = 10 m.
2^nd ball : Distance covered by the competitor in picking up and dropping it in the bucket = 2 × 8 = 16 m.
3^rd ball : Distance covered by the competitor in picking up and dropping it in the bucket = 2 × 11 = 22 m.
……………………………………………………
10^th ball: Distance covered by the competitor in picking up and dropping it in the bucket = 2 × 32 = 64 m
Total distance = 10 m + 16 m + 22 m +………….+ 64 m.
Clearly, this is an A.P in which a = 10; d = a₂ – a₁ = 16 – 10 = 6 and n = 10.
S_n = \(\frac{\mathrm{n}}{2}\) [2a + (n – 1)d]
= \(\frac{10}{2}\) [2 × 10 + (10 – 1) × 6]
= 5[20 + 54] = 5 × 74 = 370 m
∴ Total distance = 370 m.

Telangana SCERT TS 10th Class Physical Science Study Material Pdf 2nd Lesson Chemical Equations Textbook Questions and Answers.

TS 10th Class Physical Science 2nd Lesson Questions and Answers Chemical Equations

Improve Your Learning
I. Reflections on concepts

Question 1.
What information do you get from a balanced chemical equation?
Answer:

1. A chemical equation gives information about the reactants and products by means of their symbols and formulae.
2. It gives the ratio of molecules of reactants and products.
3. It gives the relative masses of reactants and products.
4. If the masses are expressed in grams, then the equation also gives the molar ratios of reactants and products.
5. We can calculate the volumes of gases liberated at given condition of temperature and pressure using molar mass and molar volume relationship.
6. Using molar mass and Avagadro’s number we can calculate the number of molecules and atoms of different substances.

Question 2.
Why should we balance a chemical equation?
Answer:
Chemical reactions obeys law of conservation of mass. So, the total number of atoms of each element in the reactants must be equal to the total number of atoms of each element in the products. So we should have to balance chemical equation.

Question 3.
Balance the following chemical equations.
(a) NaOH + H₂SO₄ → Na₂SO₄ + H₂O
Answer:
Step 1: Write unbalanced equation
NaOH + H₂SO₄ → Na₂SO₄ + H₂O

Step 2: Compare number of atoms of each element on both sides. Add the suitable coefficients to balance equation.
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

Step 3 : Make sure the coefficients are reduced to their smallest whole number values.
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

Step 4: Check the answer

(b) KClO₃ → KCl + O₂
Answer:
Step 1: Write unbalanced equation.
KClO₃ → KCl +O₂

Step 2: Add suitable coefficients to balance equation on both sides.
2KClO₃ → 2KCl +3O₂

Step 3: Make sure the coefficients are reduced to their smallest whole number values.
2KClO₃ → 2KCl +3O₂

Step 4: Check the answer.

(c) Hg (NO₃)₂ + Kl → Hg I₂ + KNO₃
Answer:
Step 1: Write unbalanced equation
Hg(NO₃)₂+KI → HgI₂ +KNO₃

Step 2: Add suitable coefficients to balance equation on both sides
Hg(NO₃)₂ +2KI → HgI₂ +2KNO₃

Step 3: Make sure the coefficients are reduced to their smallest whoíe number values.
Hg(NO₃ )₂ +2 Kl → Hg I₂ + 2KNO₃

Step 4: Check the answer

Question 4.
Mention the physical states of the reactants and products of the following chemical reactions and balance the equations.
(a) C₆H₁₂O₆ → C₂H₅OH + CO₂
Answer:
C₆H₁₂O₆(aq) → 2C₂H₅OH(aq)+2CO₂(g)

(b) NH₃ + Cl₂ → N₂ + NH₄Cl^(s)
Answer:
8NH_3(g) +3Cl_2(g) → N_2(g) +6NH₄Cl

(c) Na+H₂O → NaOH+H₂
Answer:
2Na(s)+2H₂O (l) → 2NaOH (aq)+H₂(g)

II. Application Of Concepts

Question 1.
Balance the following chemical equations after writing the symbolic representation.
(a) Calcium hydroxide (s) + Nitric acid (aq) → Water(l) + Calcium nitrate(aq)
(b) Magnesium (s) + Iodine(s) → Magnesium Iodide (s)
Answer:
(a) Calcium hydroxide (s) + Nitric acid (aq) → Water(l) + Calcium nitrate(aq)
Answer:
Ca(OH)_2(sol) + 2HNO_3(so1) → 2H₂O_(l) + Ca(NO₃)_2(aq) Double displacement reaction

(b) Magnesium (s) + Iodine(s) → Magnesium Iodide (s)
Answer:
Mg_(s) + I_2(s) → MgI_2(s)
Chemical combination reaction.

Question 2.
Write the following chemical reactions including the physical states of the substances and balance chemical equations
(a) Sodium hydroxide reacts with hydrochloric acid to form sodium chloride and water.
Answer:
(a) Sodium hydroxide reacts with hydrochloric acid to form sodium chloride and water.
Answer:
NaOH (aq) + HCl (aq) ) → NaCl (aq) +H₂O (l)

(b) Barium chloride reacts with liquid sodium sulphate to leave Barium sulphate as a precipitate and also form liquid sodium chloride.
Answer:
BaCl_2(aq)+ Na₂SO_4(aq) → BaSO₄ ↓ +2 NaCl(aq )

Question 3.
Potassium nitrate and Sodium nitrate reacts separately with copper sulphate solution. Write balanced chemical equations for the above reactions.
Answer:

1. Potassium nitrate reacts with copper sulphate to form potassium sulphate and copper nitrate.
   2KNO₃ + CuSO₄ → K₂SO₄ + Cu(NO₃)₂
   2. Sodium nitrate reacts with copper sulphate to form sodium sulphate and copper nitrate.
   2NaNO₃ + CuSO₄ → Na₂SO₄ + Cu(NO₃)₂

Higher Order Thinking Questions

Question 1.
2 moles of zinc reacts with a cupric chloride solution containing 6.023 × 10²² formula units of CuC12. Calculate the moles of copper obtained.
Zn(s) + CuCl_2(aq) → ZnCl_2(aq) + Cu (s)
Answer:
Given equation i.s
Zn(s) + CuCl_2(aq) → ZnCl_2(aq) + Cu(s)
1 mol + 1 mole → 1 mole + mole
From the above equation it is clear that 1 mole of zinc react with 1 mole of CuCl₂ solution to give 1 mole of copper.
2 moIes of Zn required 2 moles (12.046 × 10²² formula units) of CuCl₂, But only 1 mole (6.023 × 10²² formula units) of CuCl₂ is available.
So, the No. of moles of copper obtained depends on amount of CuCl₂ present.
∴6.023 × 10²² formula units (1 mole) of CuCl₂ products 1 mole of copper.

Question 2.
1 mole of propane (C₃H₈) on combustion at STP gives A’ kilo Joules of heat energy. Calculate the heat liberated when 2.4 ltrs of propane on combustion at STP.
Answer:
The chemical equation for combination of propane is
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O + A (heat energy)
1 mole of propane gives ‘A kilojoules of heat energy.
(1 mole of any gas occupies 22.4 litres at SW)
i.e., 22.4 ltrs. of propane gives ‘A’ kilojoules of heat energy
⇒ 2.25 ltrs. of propane gives \(\frac{2.24}{22.4} \) x A = \(\frac{1}{10}\) A (0.1 A) heat energy.

Question 3.
Calculate the mass and volume of Oxygen required at STP to convert 2.4 kg of graphite Into carbon dioxide.
Answer:
The chemical equation is
C + O₂ → CO₂ (∴ graphite is also carbon)
12 gm + 32 gm → 44 gm
1 mole + 1 mole → 1 mole
From the above equation
12 gm of Graphite requires 32 gm of oxygen

1 mole of oxygen occupies 22.4 litres
200 moles Is oxygen occupIes 22.4 x 200 4480 litres.

TS 10th Class Physical Science Chemical Equations Intext Questions

Page 20

Question 1.
How do we know a chemical reaction has taken place?
Answer:
a. The original substances lose their characteristic properties. Hence these may be products with different physical states and colours permanently.

b. Chemical changes may be exothermic or endothermic i.e. they may involve liberation of heat energy or absorption of heat energy.

c. They may form an insoluble substance known as precipitate.

d. There may be gas liberation in a chemical change.

Page 21
Activity 1.
Take about 1 g of quick lime (calcium oxide) in a beaker. Add 10 ml of water to this.
Touch the beaker with your finger.

Question 2.
What do you notice?
Answer:
We notice that the beaker is hot when we touch it. The reason is that the calcium oxide (quick lime) reacts with water and in the process heat energy Is released. Calcium oxide dissolves in water producing colourless solution of Ca(OH)₂.

Question 3.
What is the nature of the solution?
Answer:
This solution Is a basic solution because a red litmus paper turns blue when dipped in the above solution

Activity 2
Take about 100 ml of water in a beaker and dissolve a small quantity of sodium sulphate (Na₂SO₄).
Take about 100ml of water In another beaker and dissolve a small quantity of barium chloride (BaCl₂) observe the colours of the solutions obtained.

Question 4.
What are the colours of the above solutions?
Answer:
Colourless solutions

Question 5.
Can you name the solutions obtained?
Answer:
Sodium sulphate solution and barium chloride solution

Question 6.
Add Na₂SO₄ solution to BaCl₂ solution and observe. Do you observe any change on mixing these solutions?
Answer:
A white precipitate of barium sulphate is formed

Activity 3
Take a few zinc granules in a conical flask. Add about 5 ml of dilute hydrochloric acid to the conical flask.

Question 7.
What changes do you notice?
Answer:
Effervescence Is observed

Question 8.
Keep a burning match stick near the mouth of the conical flask. with happens to burning match stick?
Answer:
It is put out. A pop sound is also heard.

Question 9.
Touch the bottom of the conical flask with your fingers. What do you notice?
Answer:
It Is hot

Question 10.
Is there any change in temperature?
Answer:
Temperature increases.

Page 22

Question 11.
Can you write a chemical reaction in any other shorter way other than the way we disused above?
Answer:
The reaction of calcium oxide with water can be written using symbols of elements.
CaO + H₂O → Ca(OH)₂

Page 23
Question 12.
Is the number of atoms of each element are equal on both sides?
Answer:
The number of atoms of each element on both sides is equal.

Question 13.
Sodium sulphate reacts with barium chloride to give white precipitate, barium sulphate.
Na₂SO₄ + BaCl₂ → BaSO₄ ↓ + NaCl

Question 14.
Do the atoms of each element on left side equal to the atoms of the elements on the right side of the equation?
Answer:
No, the sodium and chlorine atoms are not balanced.

Page 25

Question 15.
Is it a balanced equation as per rules?
Answer:
Yes. But the coefficients are not the smallest numbers.

Question 16.
How do you say?
Answer:
Though the equation is balanced, the coefficients are not the smallest whole numbers. It would be necessary to divide all coefficients of equation by 2 to reach the final equation.
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Think And Discuss
You have brushed the wall with an aqueous suspension of Ca(OH)₂. After two days the wall turned to white colour.

Question 1.
What are the steps involved in whitewashing of walls? (P.No. 23)
Answer:
A solution of slaked lime [Ca(OH)₂] is prepared by adding water to quick lime [CaoO]. When Ca(OH)₂ is applied to the wall ¡t reacts with carbon dioxide in air to form a thin layer of calcium carbonate giving a shiny finish to the walls.

Question 2.
Write the balanced chemical reactions using the appropriate symbols. (P.No. 23)
Answer:
CaO_(s) + H₂O_(l) → Ca(OH)_2(Aq) + Q_{(heat energy)}

TS 10th Class Physical Science Chemical Equations Activities

Activity 1

Question 1.
How can you demonstrate action of quick lime with water? What is the nature of the product? (2 Marks)
Answer:

1. Take about 1 g of quick lime (Calcium oxide) in a beaker. Add 10 ml of water to this. Touch the beaker with your fingers
2. We will notice that the beaker is hot when we touch it. Hence that’s an exothermic reaction.
3. The reason is that the Calcium oxide reacts with water and in that process heat energy is released.
4. Calcium oxide dissolves In water producing colourless solution, This solution turns red litmus to blue. Hence the product is a base.
   CaO + H₂O → Ca(OH)₂ + Q

Activity 2

Question 2.
Explain the reaction between Sodium sulphate and Barium chloride.
Answer:

1. luke about 100 ml of water in a beaker and dissolve a small quantity of sodium sulphate (Na₂SO₄)
2. Take about 100 ml of water in another beaker and dissolve a small quantity of Barium chloride (BaCl₂).
3. These two solutions are colourless.
4. Add Na₂SO₄ solution to BaCl₂ solution and observe.
5. Sodium sulphate solution on mixing with Barium chloride solution alarms a precipitate of Barium sulphate and also soluble Sodium chloride.
   Na₂SO_4(aq) + BaCl_(aq) → BaSO_4(s)↓ + 2NaCl_(aq)
6. This is a double displacement reaction.

Question 3.
Explain the reaction of Zinc with HCl and write a balanced equation.
(OR)
Write the required material and experimental procedure for the experiment, “Hydrochloric acid reacts with ‘Zn’pieces and liberates H₂”.
Answer:
Material required :
(i) Conical flask
(ii) Zinc granules
(iii) HCl
(iv) Matchbox.

1. Take a few zinc granules in a conical flask.
2. Add about 5 ml of dilute Hydrochloric acid to the conical flask.
3. We observe a gas evolving out.
4. Now keep a burning match stick near the mouth of the conical flask.
5. The match stick is put out with a ‘pop’ sound.
6. I Touch the bottom of the conical flask. We feel hot. Hence it is an exothermic reaction.

Zn_(s) + 2HCl(l) → ZnCl_2(s) + H_2(g) ↑

Students can practice TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise

Question 1.
Some points are plotted on a plane. Each point is joined with all remaining points by line segments. Find the number of points if the number of line segments are 10.
Solution:
Number of distinct line segments that can be formed out of n-points = \(\frac{\mathrm{n}(\mathrm{n}-1)}{2}\)
Given : No. of line segments \(\frac{\mathrm{n}(\mathrm{n}-1)}{2}\) = 10
⇒ n² – n = 20
⇒ n² – n – 20 = 0
⇒ n² – 5n + 4n – 20 = 0
⇒ n(n- 5) + 4(n – 5) = 0
⇒ (n – 5) (n + 4) = 0
⇒ n – 5 = 0 (or) n + 4 = 0
⇒ n = 5 (or) – 4
∴ n = 5 [n – can’t be negative]

Question 2.
A two digit number is such that the product of the digits is 8. When 18 is added to the number, they interchange their places. Determine the number.
Solution:
Let the digit in the units place = x
Let the digit in the tens place = y
∴ The number = 10y + x
By interchanging the digits the number becomes 10x + y
By problem (10x + y) – (10y + x) = 18
⇒ 9x + 9y = 18
⇒ 9(x – y) = 18
⇒ x – y = \(\frac{18}{9}\) = 2
⇒ y = x – 2
(i.e.) digit in the tens place = x – 2
digit in the units place = x
Product of the digits = (x – 2) x
By problem x² – 2x = 8
x² – 2x – 8 = 0
⇒ x² – 4x + 2x – 8 = 0
⇒ x(x – 4) + 2(x – 4) = 0
⇒ (x – 4) (x + 2) = 0
⇒ x – 4 = 0 (or) x + 2 = 0
⇒ x = 4 (or) x = -2
x = 4 [ ∵ x can’t be negative]
∴ The number is 24.

Question 3.
A piece of wire 8m in length, cut into two pieces and each piece is bent into a square. Where should the cut in the wire be made if the sum of the areas of these squares is to be 2 m² ?
[Hint: x + y = 8, (\(\frac{x}{4}\))² + (\(\frac{y}{4}\))² = 2
⇒ (\(\frac{x}{4}\))² + \(\left(\frac{8-x}{4}\right)\) = 2
Solution:
Let the length of the first piece = x m
Then length of the second piece = (8 – x) m
∴ Side of the 1^st square = \(\frac{x}{4}\) m and
Side of the second square = \(\frac{8-x}{4}\) m
Sum of the areas = 2 m²
\(\left(\frac{x}{4}\right)^2\) + \(\left(\frac{8-x}{4}\right)^2\) = 2
⇒ x² + 64 + x² – 16x = 16 × 2 = 32
⇒ 2x² – 16x + 64 = 32
⇒ 2x² – 16x + 32 = 0
⇒ 2(x² – 8x + 16) = 0
⇒ x² – 4x – 4x + 16 = 0
⇒ x(x – 4) – 4(x – 4) = 0
⇒ (x – 4) (x – 4) = 0
∴ x = 4
The cut should be made at the centre making two equal pieces of length 4m, 4m.

Question 4.
Vinay and Praveen working together can paint the exterior of a house in 6 days. Vinay by himself can complete the job in 5 days less than Praveen. How long will it take Vinay to complete the job by himself ?
Solution:
Let the time taken by Vinay to complete the job = x days
Then the time taken by Praveen to complete the job = x + 5 days
Both worked for 6 days to complete a job.
∴ Total work done by them is
\(\frac{6}{x}\) + \(\frac{6}{x+5}\) = 1
⇒ 6(2x + 5) = x² + 5x
⇒ x² – 7x – 30 = 0
⇒ x² – 10x + 3x – 30 = 0
⇒ x(x – 10) + 3(x – 10) = 0
⇒ x – 10 = 0 (or) x + 3 = 0
⇒ x = 10(or) x = -3
x = 10 (∵ x can’t be negative)
∴ Time taken by Vinay = x = 10 days
Time taken by Praveen = x + 5 = 15 days.

Question 5.
Show that the sum of roots of a quadratic equation is \(\frac{-\mathbf{b}}{\mathbf{a}}\).
Solution:
Let the Q.E. = ax² + bx + c = 0 (a ≠ 0)
⇒ ax² + bx = -c
⇒ x² + \(\frac{\mathrm{b}}{\mathrm{a}}\)x = \(\frac{-\mathrm{c}}{\mathrm{a}}\)
⇒ x² + 2.\(x \cdot \frac{b}{2 a}\) = \(-\frac{c}{a}\)

sum of the roots

∴ Sum of roots of a Q.E. is \(\frac{-b}{a}\)

Question 6.
Show that the product of the roots of a Q.E is \(\frac{\mathbf{c}}{\mathrm{a}}\).
Solution:
Let the Q.E. = ax² + bx + c = 0(a ≠ 0)
⇒ ax² + bx = -c


Product of the roots

Question 7.
The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is 2\(\frac{16}{21}\), find the fraction.
Solution:
Let the numerator = x
then denominator = 2x + 1
Then the fraction = \(\frac{x}{2 x+1}\)
Its reciprocal = \(\frac{2 x+1}{x}\)

⇒ 105x² + 84x + 21 = 116x² + 58x
⇒ 11x² – 26x – 21 = 0
⇒ 11x² – 33x + 7x – 21 = 0
⇒ 11x (x – 3) + 7(x – 3) = 0
⇒ (x – 3)(11x + 7) = 0
⇒ x – 3 = 0 (or) 11x + 3 = 0
⇒ x = 3 (or) \(\frac{-3}{11}\)
∴x = 3
Numerator = 3;
Denominator = 2 × 3 + 1 = 7
Fraction = \(\frac{3}{7}\)

Students can practice 10th Class Maths Solutions Telangana Chapter 5 Quadratic Equations InText Questions to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 5 Quadratic Equations InText Questions

Try This

Question 1.
Check whether the following equations are quadratic or not.
i) x² – 6x – 4 = 0
ii) x³ – 6x² + 2x – 1 = 0
iii) 7x = 2x²
iv) x² + \(\frac{1}{\mathrm{x}^2}\) = 2
v) (2x + 1) (3x + 1) = 6(x – 1)(x – 2)
vi) 3y² = 192 (Page No. 102)
Solution:
i) x² – 6x – 4 = 0
Yes, It’s a quadratic equation.

ii) x³ – 6x² + 2x – 1 = 0
No, It is not a quadratic equation.
[∵ degree is 3]

iii) 7x = 2x²
Yes, It’s a quadratic equation.

iv) x² + \(\frac{1}{x^2}\) = 2
No, It is not a quadratic equation.
[∵ degree is 4]

v) (2x + 1) (3x + 1) = 6(x – 1) (x – 2)
No, It is not a quadratic equation.
[coefficient of x² on both sides is same i.e., 6]

vi) 3y² = 192
Yes, It’s a quadratic equation.

Try This

Question 1.
Verify whether 1 and \(\frac{3}{2}\) are the roots of the equation 2x² – 5x + 3 = 0. (Page No. 107)
Solution:
Let the given
Q.E. be p(x) = 2x² – 5x + 3
Now p(1) = 2(1)² – 5(1) + 3
= 2 – 5 + 3 = 0
∴ 1 is a root of 2x² – 5x + 3 = 0
also p(\(\frac{3}{2}\)) = 2(\(\frac{3}{2}\))² – 5(\(\frac{3}{2}\)) + 3
= 2 × \(\frac{9}{4}\) – \(\frac{15}{2}\) + 3
= \(\frac{9}{2}\) + 3 – \(\frac{15}{2}\) = \(\frac{9+6-15}{2}\) = 0
∴ \(\frac{3}{2}\) is also a root of 2x² – 5x + 3 = 0

Do This

Question 1.
Solve the equations by completing the square. (Page No. 113)

i) x² – 10x + 9 = 0
Solution:
Given x² – 10x + 9 = 0
⇒ x² – 10x = – 9
⇒ x² – 2. x. 5 = – 9
⇒ x² – 2.x. 5 + 5² = -9 + 5²
⇒ (x – 5)² = 16
∴ x – 5 = ± 4
∴ x – 5 = 4 (or) x – 5 = – 4
⇒ x = 9 (or) x = 1
⇒ x = 9 (or) 1

ii) x² – 5x + 5 = 0
Solution:
Given : x² – 5x + 5 = 0
⇒ x² – 5x = -5

iii) x² + 7x – 6 = 0
Solution:
x² + 7x – 6 = 0
x² + 7x = 6
x² + 2.x.\(\frac{7}{2}\) = 6
x² + 2.x.\(\frac{7}{2}\) + (\(\frac{7}{2}\))² = 6 + (\(\frac{7}{2}\))²

Think – Discuss

Question 1.
We have three methods to solve a quadratic equation. Among these three, which method would you like to use ? Why ? (Page No. 115)
Solution:
If the Q.E. has district and real roots, we use factorisation. If Q.E. has no real roots, we use quadratic formula.

Try This

Question 1.
Explain the benefits of evaluating the discriminate of a quadratic equation before attempting to solve it. What does its value signifies ? (Page No. 122)
Solution:
The discriminant b² – 4ac of a Q.E. ax² + bx + c = 0 give the clear idea about the nature of Q the roots of the Q.E. If the discriminant D = b² – 4ac > 0, the Q.E. has distinct and real roots.

If b² – 4ac = 0, the Q.E. had equal roots. If b² – 4ac < 0, the Q.E. has no real roots. By the 3 value of the discriminant, we can state the nature of the roots of a Q.E. without actually finding them.

Question 2.
Write three quadratic equations one having two distinct real solutions, one having no real solution and one having exactly one real solution. (Page No. 122)
Solution:

Students can practice 10th Class Maths Study Material Telangana Chapter 9 Tangents and Secants to a Circle InText Questions to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle InText Questions

Do This

Question 1.
Draw a circle with any radius. Draw four tangents at different points. How many tangents can you draw to this circle ? (AS₃, AS₅) (Page No. 226)
Solution:

Let ‘O’ be the centre of the circle with radius ‘OA’. l, m, n, p and q be the tangents to the circle at A, B, C, D and E. We can draw a tangent at each point on the circle, i.e., infintely many tangents can be drawn to a circle.

Question 2.
How many tangents you can draw to circle from a point away from it ? (AS₃) (Page No. 226)
Solution:

We can draw only two tangents from an exterior point.

Question 3.
In the below figure which are tangents to the circles ? (AS₃) (Page No. 226)
Solution:

P and M are the tangents to the circles.

Question 4.
Draw a circle and a secant PQ of the circle on a paper as shown below. Draw various lines parallel to the secant on both sides of it. What happens to the length of chord coming closer and closer to the center of the circle ?
What is the longest chord ? How many tangents can you draw to a circle, which are parallel to each other. (Page No. 227)
Solution:

The length of the chord increases as it comes closer to the centre of the circle.

i) What is the longest chord ?  (Page No. 227)
Solution:
Diameter is the longest of all chords.

ii) How many tangents can you draw to a circle which are parallel to each other ? (Page No. 227)
Solution:
Only one tangent can be drawn parallel to a given tangent. To a circle, we can draw infinetely pairs of parallel tangents.

Try This

Question 1.
How can you prove the converse of the above theorem. “If a line in the plane of a circle is perpendicular to the radius at its end point on the circle, then the line is tangent to the circle”. (Page No. 228)
Solution:
Given : Circle with centre ‘O’, a point ‘A’ on the circle and the line AT’ perpendicular to OA’.
RTP : ‘AT’ is a tangent to the circle at A.
Construction :

Suppose ‘AT’ is not a tangent then ‘AT’ produced either way if necessary, will meet the circle again. Let it do so at R join OR

Proof : Since OA = OP (radii)
∴ ∠OAP = ∠OPA
But ∠OPA = 90°
∴ Two angles of a triangle are right angles which is impossible.
∴ Our supposition is false, hence AT is a tangent.

We can find some more results using the above theorem.

1. Since there can be only one perpendicular OP at the point R it follows that one and only are tangent can be drawn to a circle at a given point on the circumference.
2. Since there can be only one perpendicular to XY at the point R it follows that the perpendicular to a tangent at its point of contact passes through the centre.

Question 2.
How can you draw the tangent to a circle at a given point when the centre of the circle is not known ? (Page No. 229) Hint : Draw equal angles ∠QPX and ∠PRQ. Explain the construction.
Solution:

Steps of construction :

1. Take a point ‘P’ and draw a chord PR through P
2. Construct ∠PRQ and measure it.
3. Construct ∠QPX at P equal to ∠PRQ.
4. Extend PX on otherside. XY is the required tangent at R

Note : Angle between a tangent and chord is equal to angle in the alternate segment.

Try This

Question 1.
Use pythagorus theorem and write proof for the statement “the lengths of tangents drawn from an external point to a circle are equal”. (Page No. 231)
Solution:
Given : Two tangents PA and PB to a circle with centre ‘O’, from an exterior point P.
R.T.P : PA = PB

Proof : In ∆OAP; ∠OAP = 90°
∴ AP² = OP² – OA² [∵ square of the hypotenuse is equal to the sum of squares on the other 2 sides → Pythagoras theorem]
⇒ (AP)² = (OP)² – (OB)²
[∵ OA = OB, radii of the same circle]
⇒ AP² = BP²
[∴ In a ∆OBP; OB² + BP² = OP²
⇒ BP² = OP² – OB²]
⇒ AP² = BP²
∴ PA = PB (CPCT)
Hence proved.

Question 2.
Draw a pair of radii OA and OB such that ∠BOA = 120°. Draw the bisector of ∠BOA and draw lines perpendicular to OA and OB at A and B. These lines meet on the bisector of ∠BOA at a point which is the external point and the perpendicular lines are the required tangents. Construct and justify. (Page No. 235)
Solution:
Justification :
OA ⊥ PA
OB ⊥ PB
Also in ∆OAP, ∆OBP
OA = OB
∠OAP = ∠OBP
OP = OP
∴ ∆OAP ≅ ∆OBP
∴ PA = PB [Q.E.D]

Do This

Question 1.
Shankar made the following pictures also

What shapes can they be broken into that we can find area easily ? (Page No. 237)
Solution:

Into two rectangles.

A rectangle and a circle

A cone and a secant

A rectangle and 2 segments

Make some more pictures and think of the shapes they can be divided into different parts.
Solution:

A cone and segment

A rectangle and a segment

A square and four segments

A cone and segment
A rectangle and a segment A square and four segments.

Do This

Question 1.
Find the area of sector, whose radius is 7 cm with the given angle : (Page No. 239)
(i) 60°
(ii) 30°
(iii) 72°
(iv) 90°
(v) 120°
Solution:

Question 2.
The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 10 minutes. (Page No. 239)
Solution:
Angle made by minute hand in 360°
1 min = \(\frac{360^{\circ}}{60^{\circ}}\) = 6°
Angle made by minute hand in
10 min = 10 × 6° = 60°
The area swept by minute hand is in the shape of a sector with radius.
r = 14 cm and angle x° = 60°

Area (A) = \(\frac{x^{\circ}}{360^{\circ}}\) × πr²
= \(\frac{60^{\circ}}{360^{\circ}}\) × \(\frac{22}{7}\) × 14 × 14
= \(\frac{616}{6}\) ⇒ 102.66 cm²
∴ Area swept by the minute hand in 10 minutes = 102.66 cm²

Try This

Question 1.
How can you find the area of major seg¬ment using are a of minor segment ?
Solution:

Area of the major segment = (Area of the circle) – (Area of the minor segment)

Students can practice 10th Class Maths Study Material Telangana Chapter 9 Tangents and Secants to a Circle Optional Exercise to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Optional Exercise

Question 1.
Prove that the angle between the two tan-gents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.
Solution:
PA and PB are tangents to the circle with centre ‘0’ from and external point P.
Join OA, OB and OR The angle between the tangents is ∠APB.
We have to prove that ∠AOB + ∠APB = 180°
PA is a tangent to the circle with centre ‘O’.

A is the point of contact. AO is the radius drawn through the point of contact i.e., A
∴ ∠PAO = 90°
PB is a tangent to the circle with centre ‘O’.
B is the point of contact. BO is the radius drawn through the point of contact i.e., B
∴ ∠PBO = 90°
OAPB is a quadrilateral.
The sum of the angles of a quadrilateral is 360°
∠APB + ∠PBO + ∠AOB + ∠PAO = 360°
∠APB + ∠AOB + ∠PAO + ∠PBO = 360°
∠APB + ∠AOB + 90° + 90° = 360°
∴ ∠APB + ∠AOB = 360° – 90° – 90°
= 360° – 180° = 180°
∠APB and ∠AOB are supplementary.

Question 2.
PQ is a chord of length 8cm of a circle of radius 5cm. The tangents at P and Q intersect at a point T (See figure). Find the length of TP.
Solution:
PQ is a chord of length 8cm of a circle with centre ‘O’.
Radius of the circle = 5 cm.
⇒ OP = OQ = 5cm
The tangents at P and Q meet at T. Join OT.
OT is the perpendicular bisector of PQ.

In triangle OPT, ∠OPT = 90°
∴ PT² = OT² – OP² …………….. (1)
In ∆ PRO, ∠R = 90°
∴ OP² = OR² + PR²
⇒ OR² = OP² – PR²
= 5² – 4²
= 25 – 16 = 9
∴ OR = \(\sqrt{9}\) = 3 cm
In the right triangle OPT,
PR is perpendicular to the hypotenuse OT.
∴ PR² = TR. RO
4² = TR × 3 ⇒ TR = \(\frac{4^2}{3}\) = \(\frac{16}{3}\)
OT = OR + TR = \(\frac{3}{1}\) + \(\frac{16}{3}\) = \(\frac{25}{3}\)
From (1), PT² = \(\left(\frac{25}{3}\right)^2\) – (5)²
= \(\frac{625}{9}\) – \(\frac{25}{1}\)
= \(\frac{625-225}{9}\) = \(\frac{400}{9}\)
∴ PT = \(\sqrt{\frac{400}{9}}\) = \(\frac{20}{3}\) cm

Question 3.
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Solution:
ABCD is a quadrilateral, circumscribing a circle whose centre is O.
The opposite sides AB and CD are subtend angles ∠AOB and ∠COD at the centre.

Similarly, the other pair of opposite sides AD and BC subtend angles ∠AOD and ∠BOC at the centre.
We have to prove that
∠AOB + ∠COD = 180°
and ∠AOD + ∠BOC = 180°
Join OR OQ, OR and OS.
We know that the tangents drawn from an ex¬ternal point to a circle subtend equal angles at the centre.
AP and AS are the tangents to the circle from A.
∴ ∠AOP = ∠AOS ……………. (1)
BP and BQ are the tangents to the circle from B.
∴ ∠BOP = ∠BOQ …………….. (2)
CQ and CR are the tangents to the circle from C.
∴ ∠COQ = ∠COR ………………. (3)
DR and DS are the tangents to the circle from D.
∴ ∠DOR = ∠DOS …………………. (4)
Adding (1), (2), (3) and (4), we get
∠AOP + ∠BOP + ∠COQ + ∠DOR + ∠AOS + ∠BOQ + ∠COR + ∠DOS = 360°
(∵ Sum of the angles formed at’O’is equal to 360°)
(∠AOP + ∠BOP) + (∠COR + ∠DOR) + (∠BOQ + ∠COQ) + (∠AOS + ∠DOS) = 360° 2∠AOP + 2∠COR + 2 ∠BOQ + 2 ∠DOS = 360°
⇒ ∠AOP + ∠COR + ∠BOQ + ∠DOS = 180°
⇒∠AOP + ∠BOQ + ∠COR + ∠DOS = 180°
⇒ ∠AOP + ∠BOP + ∠COR + ∠DOR = 180°
(∵ ∠BOQ = ∠BOP and ∠DOS = ∠DOR
⇒ ∠AOB + ∠COD = 180°
In ∆^les BOP & BOQ)
Similarly, we can prove that
∠BOC + ∠AOD = 180°

Question 4.
Draw a line segment AB of length 8cm. Taking A as centre, draw a circle of radius 4cm and taking B as centre, draw another circle of radius 3cm. Construct tangents to each circle from the centre of the other circle.
Solution:

1. Draw a line segment AB = 8 cm.
2. Take A as centre and draw a circle (C₁) of radius 4cm.
3. Take B as centre and draw a circle (C₂) of radius 3 cm.
4. Draw the perpendicular bisector PQ of AB.
5. Draw a circle with AB as diameter (C₃). This circle intersects the circle with A as centre in T₁ and T₂. Join BT₁ and BT₂. These are the tangents to the circle C₁.
6. The circle with AB as diameter (C₃) intersects the circle with B as centre (C₂) in R₁ and R₂. Join AR₁ and AR₂. These are the tangents to the circle C₂.

Question 5.
Let ABC be a right angled triangle in which AB = 6cm, BC = 8cm and ∠B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
Solution:
Construction :

1. Draw a line segment AB = 6cm.
   Make ∠ABX = 90°.
   Mark a point C on \(\overrightarrow{\mathrm{BX}}\) such that BC = 8cm. Join AC.
2. Draw BD ⊥ AC intersecting AC in D.
3. Draw the perpendicular bisectors of BC and CD. Let them intersect at ‘O’.
4. Take ‘O’ as centre and OB = OD = OC as radius, draw a circle passing through B, D and C
5. Join AO. Find M the mid point of AO.
6. Take M as centre and radius as (AM = MO) draw a circle passing through A and C, intersecting the first circle at T₁ and T₂.
7. Join AT₁ and AT₂.
8. AT₁ and AT₂ are the required tangents.

Question 6.
Find the area of the shaded region in the figure, given in which two circles with centres A and B touch each other at the point C. If AC = 8 cm and AB = 3 cm.

Solution:
AC denotes the radius of the bigger circle.
AC = 8cm. (Given)
AB denotes the distance between the centres of two circles.
AB = 3cm (Given)
∴ BC denotes the radius of the smaller circle
BC = AC – AB = 8 – 3 = 5 cm
The Area of the shaded region = Area of the bigger circle – Area of the smaller circle =
= π(8)² – π(5)²
= π[8² – 5²]
= π[64 – 25]
= \(\frac{22}{7}\) × 39
= \(\frac{858}{7}\)
= 122.57 cm²

Question 7.
ABCD is a rectangle with AB = 14 cm and BC = 7cm. Taking DC, BC and AD as diameters, three semi-circles are drawn as shown in the figure. Find the area of the shaded region.

Solution:
ABCD is a rectangle.
Length of the rectangle AB = CD = 14 cm
Breadth of the rectangle AD = BC = 7cm
Area of the rectangle ABCD
= length × breadth
= 14 × 7
= 98 cm²
Two semi-circles are drawn on AD and BC clearly, They are of the same diameters.
Hence, Area of two semi-circles becomes area of one complete circle with diameter 7 cm.
So, radius of the circle (r) = \(\frac{7}{2}\) cm
Area of the circle = πr²
= \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\)
= 38.5 cm² ……………….. (1)
Area a of the semi-circle whose diameter is 14 cm = \(\frac{1}{2}\)πr² ; Here, r = 7cm
= \(\frac{1}{2}\) × \(\frac{22}{7}\) × 7 × 7
= 77 cm²
Area of the shaded region in the rectangle
= Area of the rectangle – Area of the semicircle
= 98 – 77
= 21 cm² ……………… (2)
Area of the total shaded region
= Area of the circle (shaded) + Area of the shaded region in the rectangle.
= (38.5 + 21) cm²
= 59.5 cm²

Students can practice 10th Class Maths Study Material Telangana Chapter 9 Tangents and Secants to a Circle Ex 9.3 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Exercise 9.3

Question 1.
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : (use π = 3.14)
(i) Minor segment
(ii) Major segment. (A.P. Mar. ’16, June ’15)
Solution:

Given :
Angle subtended by the chord = 90°
Radius of the circle = 10 cm
Area of the minor segment = Area of the sector POQ – Area of ∆POQ
Area of the sector = \(\frac{\mathrm{x}^{\circ}}{360^{\circ}}\) × πr²
= \(\frac{90}{360}\) × 3.14 × 10 × 10 = 78.5
Area of the triangle POQ
= \(\frac{1}{2}\) × Base × Height
= \(\frac{1}{2}\) × 10 × 10 = 50
∴ Area of the minor segement
⇒ 78.5 – 50 = 28.5 cm²
Area of the major segment = Area of the circle – Area of the minor segment
⇒ 3.14 × 10 × 10 – 28.5
⇒ 314 – 28.5 cm²
⇒ 285.5 cm²

Question 2.
A chord of a circle of radius 12 cm. subtends an angle of 120° at the centre. Find the area of the corresponding minor segment of the circle, (use π = 3.14 and \(\sqrt{3}\) = 1.732).
Solution:
Radius of the circle r = 12 cm
Area of the sector = \(\frac{\mathrm{x}^{\circ}}{360^{\circ}}\) πr²
Here x = 120°

= \(\frac{120^{\circ}}{360^{\circ}}\) × 3.14 × 12 × 12
= 150.72
Drop a perpendicular from ‘O’ to the chord ‘PQ’
∆OPM = ∆OQM [∵ OP = OQ, ∠P = ∠Q; angles opposite to equal sides OP & OQ, ∠OMP = ∠OMQ by A.A.S]
∴ ∆OPQ = ∆OPM + ∆OQM
= 2(∆OPM)
Area of ∆OPM = \(\frac{1}{2}\) × PM × OM
But cos 30° = \(\frac{\mathrm{PM}}{\mathrm{OP}}\)
[∴ In ∆OPQ ∠POQ = 120° ∠OPQ = ∠OQP = \(\frac{180-120^{\circ}}{2}\) = 120°]
∴ PM = \(\frac{12 \times \sqrt{3}}{2}\) = 6\(\sqrt{3}\)
Also sin 30° = \(\frac{\mathrm{OM}}{\mathrm{OP}}\)
⇒ \(\frac{1}{2}\) = \(\frac{\mathrm{OM}}{12}\) ⇒ OM = \(\frac{12}{2}\) = 6
∴ ∆OPM = \(\frac{1}{2}\) × 6\(\sqrt{3}\) × 6
= 18 × 1.732 = 31.176 cm²
∴ ∆OPQ = 2 × 31.176 = 62.352 cm²
Area of the minor segment PQ = (Area of the sector) – (Area of the ∆OPQ)
= 150.72 – 62.352 = 88.368 cm²

Question 3.
A car has two wipers which do not overlap. Each wiper has a blade of length 25cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades (use π = \(\frac{22}{7}\))
Solution:

Angle made by the each blade = 115°
It is evident that each wiper sweeps a sector of a circle of radius 25cm and sector angle 115°.
Area of a sector = \(\frac{\theta^{\circ}}{360^{\circ}}\) × πr²
Total area cleaned at each sweep of the blades
= 2 × \(\frac{\theta^{\circ}}{360^{\circ}}\) × πr²
= 2 × \(\frac{115^{\circ}}{360^{\circ}}\) × \(\frac{22}{7}\) × 25 × 25
= 2 × \(\frac{23}{72}\) × \(\frac{22}{7}\) × 25 × 25
= \(\frac{23}{36}\) × \(\frac{22}{7}\) × 25 × 25
= 1254.96 cm²

Question 4.
Find the area of the shaded region in the figure, where ABCD is a square of side 10cm and semicircles are drawn with each side of the square as diameter. (use π = 3.14).

Solution:
Let us mark the four unshaded regions as I, II, III and IV.

Area of I + Area of III = Area of ABCD – Areas of 2 semicircles with radius 5 cm.
= 10 × 10 – 2 × \(\frac{1}{2}\) × 7π × 5²
= 100 – 3.14 × 25
⇒ 100 – 78.5 ⇒ 21.5 cm²
Similarly, Area of II + Area of IV = 21.5 cm²
So, area of the shaded region
= are a of ABCD – Area of unshaded region
⇒ 100 – 2 × 21.5
= 100 – 43 = 57 cm²

Question 5.
Find the area of the shaded region in the figure, if ABCD is a square of side 7cm and APD and BPC are semi-circles. (use π = \(\frac{22}{7}\))

Solution:
Given,
ABCD is a square of side 7 cm
Area of the shaded region
= Area of ABCD – Area of 2 semi-circles with 7 radius (\(\frac{7}{2}\) = 3.5 cm)
APD and BPC are semi-circles 1 22
= 7 × 7 – 2 × \(\frac{1}{2}\) × \(\frac{22}{7}\) × 3.5 × 3.5
= 49 – 38.5 = 10.5 cm²
∴ Area of shaded region = 10.5 cm²

Question 6.
In figure, OACB is a quadrant of a circle with centre ‘O’ and radius 3.5 cm. If OD = 2 cm, find the area of the shaded region, (use π = \(\frac{22}{7}\)).

Solution:
Given OACB is a quadrant of a circle radius 3.5 cm OD = 2 cm
Area of the shaded region = Area of the sector – Area of ABOD
= \(\frac{\mathrm{x}^{\circ}}{360^{\circ}}\) × πr² – \(\frac{1}{2}\) OB . OD
= \(\frac{99^{\circ}}{360^{\circ}}\) × \(\frac{22}{7}\) × 3.5 × 3.5 – \(\frac{1}{2}\) × 3.5 × 2
= \(\frac{1}{4}\) × \(\frac{22}{7}\) × 3.5 × 3.5 – 3.5
= \(\frac{1}{4}\) × \(\frac{22}{7}\) × 12.25 – 3.5
= 9.625 – 3.5 = 6.125 cm²
Area of shaded region = 6.125 cm²

Question 7.
AB and CD are respectively arcs of two concentric circles of radii 21cm and 7cm with centre ‘O’ (see figure) If ∠AOB = 30°, find the area of the shaded region. (Use π = \(\frac{22}{7}\))

Solution:
Area of the shaded region
= Area of sector OAB – Area of the sector COD
= \(\frac{30^{\circ}}{360^{\circ}} \) × \(\frac{22}{7}\) × 21 × 21 – \(\frac{30^{\circ}}{360^{\circ}} \) × \(\frac{22}{7}\) × 7 × 7
= \(\frac{30^{\circ}}{360^{\circ}} \) × \(\frac{22}{7}\) (21 × 21 – 7 × 7)
= \(\frac{1}{12}\) × \(\frac{22}{7}\) (441 – 49)
= \(\frac{1}{6}\) × \(\frac{11}{7}\) × 392
= \(\frac{1}{6}\) × 11 × 56
= \(\frac{11 \times 28}{3}\)
= 102.67 cm²

Question 8.
Calculate the area of the designed region in figure, common between the two quadrants of the circles of radius 10 cm. each, (use π = 3.14)

Solution:
Radius of the circle (r) = 10 cm
Area of the designed region
= 2 [Area of quadrant ABYD – Area of ∆ABD]
= 2 [\(\frac{1}{4}\) × πr² – \(\frac{1}{2}\) × Base × Height]
= 2 [(\(\frac{1}{4}\) × 3.14 × 10 × 10) – (\(\frac{1}{2}\) × 10 × 10)]
= 2 [78.5 – 50]
⇒ 2 × 28.5
= 57 cm²

Students can practice 10th Class Maths Study Material Telangana Chapter 10 Mensuration Optional Exercise to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 10 Mensuration Optional Exercise

Question 1.
A golf ball has diameter equal to 4.1cm. Its surface has 150 dimples each of radius 2mm. Calculate total surface area which is exposed to the surroundings (Assume that the dimples are all hemispherical) [π = \(\frac{22}{7}\)]
Solution:
Area exposed = surface area of the ball – total area of 150 dimples with radius 2 mm
= 4πr² – 150 × πr²
= 4 × \(\frac{22}{7}\) × \(\frac{4.1}{2}\) × \(\frac{4.1}{2}\) – 150 × \(\frac{22}{7}\) × \(\frac{2}{10}\) × \(\frac{2}{10}\)
[∵ 2 mm = \(\frac{2}{10}\) cm]
= 52.831 – 18.85 = 33.972 cm²

Question 2.
A cylinder of radius 12 cm. contains water to a depth of 20 cm. when a spherical iron ball is dropped in to the cylinder and thus the level of water is raised by 6.75 cm. Find the radius of the ball. [π = \(\frac{22}{7}\)]
Solution:
Rise in the water level is seen as a cylinder of radius ‘r’ = r₁ = 12 cm
Height, h = 6.75 cm
Volume of the rise = Volume of the spherical iron ball dropped
πr₁²h = \(\frac{4}{3}\) πr₂³
r₁²h = \(\frac{4}{3}\) π₂³
12 × 12 × 6.75 cm³ = \(\frac{4}{3}\) × r₂³ cm³
r₂³ = \(\frac{3}{4}\) × 12 × 12 × 6.75
= 9 × 12 × 6.75
= 108 × 5.75
r₂³ = 729
r₂³ = 9 × 9 × 9
∴ 729 = (3 × 3) × (3 × 3) × (3 × 3)
∴ Radius of the ball r = r₂ = 9 cm.

Question 3.
A solid toy is in the form of a right circular cylinder with a hemispherical shape at one end and a cone at the other end. Their common diameter is 4.2 cm. and height of the cylindrical and conical portion are 12 cm. and 7 cm. respectively. Find the volume of the solid toy. [π = \(\frac{22}{7}\)]
Solution:

Volume of the toy
= volume of the hemisphere + volume of the cylinder + volume of the cone.
= \(\frac{2}{3}\) πr³ + πr²h₁ + \(\frac{1}{3}\) πr²h₂
= πr²(\(\frac{2}{3}\)r + h₁ + \(\frac{\mathrm{h}_2}{3}\))
= \(\frac{22}{7}\) × \(\frac{4.2}{2}\) × \(\frac{4.2}{2}\)[\(\frac{2}{3}\) × \(\frac{4.2}{2}\) + 12 + \(\frac{7}{3}\)]
= 11 × 0.6 × 2.1 [1.4 + 12 + \(\frac{7}{3}\)]
= 13.86[13.4 + \(\frac{7}{3}\)]
= 13.86 × \(\left[\frac{40.2+7}{3}\right]\)
= \(\frac{13.86 \times 47.2}{3}\) = 218.064 cm³

(or)

Hemisphere :
Radius = \(\frac{\text { diameter }}{2}\) = \(\frac{4.2}{2}\) = 2.1 cm
V = \(\frac{2}{3}\) πr³ = \(\frac{2}{3}\) × \(\frac{22}{7}\) × 2.1 × 2.1 × 2.1
= 19.404 cm³

Cylinder :
Radius, r = \(\frac{\mathrm{d}}{2}\) = \(\frac{4.2}{2}\) = 2.1 cm
height, h = 12 cm
V = πr²h = \(\frac{22}{7}\) × 2.1 × 2.1 × 12
= 166.32 cm³

Cone :
Radius, r = \(\frac{\mathrm{d}}{2}\) = \(\frac{4.2}{2}\) = 2.1cm
Height, h = 7 cm
V = \(\frac{1}{3}\) πr²h = \(\frac{1}{3}\) × \(\frac{22}{7}\) × 2.1 × 2.1 × 7
= 32.34 cm³
∴ Total volume = 19.404 + 166.32 + 32.34
= 218.064 cm³.

Question 4.
Three metal cubes with edges 15 cm., 12 cm. and 9 cm. respectively are melted together and formed into a single cube. Find the diagonal of this cube.
Solution:
Edges l₁ = 15 cm, l₂ = 12 cm, l₁₃ = 9 cm.

Volume of the resulting cube = Sum of the volumes of the three given cubes
L³ = l³₁h
L³ = l₁³ + l₂³ + l₃³
L³ = 15³ + 12³ + 9³
L³ = 3375 + 1728 + 729
L³ = 5832 = 18 × 18 × 18
∴ Edge of the new cube l = 18 cm
Diagonal = \(\sqrt{3 l^2}\) = \(\sqrt{3 \times 18^2}\)
= \(\sqrt{3 \times 324}\) = \(\sqrt{972}\) = 31.176 cm.

Question 5.
A hemi-spherical bowl of internal diameter 36 cm. contains a liquid. This liquid is to be filled in cylindrical bottles of radius 3 cm. and height 6 cm. How many bottles are required to empty the bowl ?
Solution:
Let the number of bottles required = n.
Then total valume of a ’n’ bottles = volume of the hemispherical bowl.
n. πr₁²h = \(\frac{2}{3}\) πr₂²h
Bottle :
Radius, r₁ = 3 cm
Height, h = 6 cm
Volume, V = πr₁²h
= \(\frac{22}{7}\) × 3 × 3 × 6 = \(\frac{1188}{7}\)
∴ Total volume of n bottles = n × \(\frac{1188}{7}\) cm³.
Bowl :

∴ 72 bottles are required to empty the bowl.

Students can practice TS Class 10 Maths Solutions Chapter 6 Progressions Ex 6.2 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 6 Progressions Exercise 6.2

Question 1.
Fill in the blanks in the following table, given that a is the first term, d the common difference and a_n the n^th term of the A.P :

Solution:

Question 2.
Find the
i) 30^th term of the A.P. : 10, 7, 4, …… (A.P. June 15)
ii) 11^th term of the A.P. : -3, \(\frac{-1}{2}\) ,2, …….
Solution:
i) Given A.P. = 10, 7, 4, ….
a₁ = 10; d = a₂ – a₁ = 7 – 10 = – 3
a_n = a + (n – 1) d
a₃₀ = 10 + (30 – 1)(-3)
= 10 + 29 × (- 3)
= 10 – 87
= -77

ii) Given A.P. = -3, \(\frac{-1}{2}\), 2,……..
a₁ = -3; d = a₂ – a₁ = \(\frac{-1}{2}\) – (-3)
= \(\frac{-1}{2}\) + 3 = = \(\frac{-1+6}{2}\) = \(\frac{5}{2}\)
a_n = a + (n – 1)d
a₁₁ = -3 + (11 – 1) × (\(\frac{5}{2}\))
= -3 + 10 × \(\frac{5}{2}\)
= -3 + 5 × 5
= -3 + 25 = 22

Question 3.
Find the respective terms for the following APs.

i) a₁ = 2; a₃ = 26, find a₂.
Solution:
Given : a₁ = a = 2 —— (1)
a₃ = a + 2d = 26 —— (2)
Equation (2) – equation (1)
⇒ (a + 2d) – a = 26 – 2
⇒ 2d = 24
d = \(\frac{24}{2}\) = 12
Now a₂ = a + d = 2 + 12 = 14

ii) a₂ = 13; a₄ = 3, find a₁, a₃.
Solution:
Given : a₂ = a + d = 13 —— (1)
a₄ = a + 3d = 3 —- (2)
Solving equations (1) and (2);

Substituting, d = – 5 in equation (1) we get
a + (-5) = 13
∴ a = 13 + 5 = 18
i.e., a₁ = 18
a₃ = a + 2d = 18 + 2(-5)
= 18 – 10 = 8

iii) a₁ = 5; a₄ = 9\(\frac{1}{2}\), a₂, a₃.
Solution:
Given : a₁ = a = 5 —– (1)
a₄ = a + 3d = 9\(\frac{1}{2}\) —– (2)
Solving equations (1) and (2);

iv) a₁ = – 4; a₆ = 6, find a₂, a₃, a₄, a₅.
Solution:
Given : a₁ = a = – 4 —– (1)
a₆ = a + 5d = 6 —– (2)
Solving equations (1) and (2);
(-4) + 5d = 6 ⇒ 5d = 6 + 4
⇒ 5d = 10 ⇒ d = \(\frac{10}{5}\) = 2
Now, a₂ = a + d = -4 + 2 = -2;
a₃ = a + 2d = -4 + 2 × 2
= -4 + 4 = 0;
a₄ = a + 3d = (-4) + 3 × 2
= -4 + 6 = 2;
a₅ = a + 4d = -4 + 4 × 2
= -4 + 8 = 4

v) a₂ = 38; a₆ = -22, find a₁, a₃, a₄, a₅.
Solution:
Given : a₂ = a + d = 38 —– (1)
a₆ = a + 5d = -22 —– (2)
Subtracting (2) from (1) we get

Now substituting, d = -15 in equation (1), we get
a + (-15) = 38 ⇒ a = 38 + 15 = 53 Thus,
a₁ = a = 53;
a₃ = a + 2d = 53 + 2 × (-15) = 53 – 30 = 23;
a₄ = a + 3d = 53 + 3 × (-15) = 53 – 45 = 8;
a₅ = a + 4d = 53 + 4 × (-15) = 53 – 60 = -7

Question 4.
Which term of the AP :
3, 8, 13, 18, ……. is 78 ?
Solution:
Given : 3, 8, 13, 18,
Here a = 3; d = a₂ – a₁ = 8 – 3 = 5
Let ’78’ be the nth term of the given A.P.
∴ a_n = a + (n – 1) d
78 = 3 + (n – 1)5
78 = 3 + 5n – 5
⇒ 5n = 78 + 2 ⇒ n = \(\frac{80}{5}\) = 16
∴ 78 is the 16th term of the given A.P.

Question 5.
Find the number of terms in each of the following APs :
i) 7, 13, 19, ……., 205 (A.P. Mar. ’16)
Solution:
Given : A.P : 7, 13, 19, ………
Here a₁ = a = 7; d = a₂ – a₁ = 13 – 7 = 6
Let 205 be the n^th term of the given A.P.
Then, a_n= a + (n – 1) d
205 = 7 + (n – 1) 6
⇒ 205 = 7 + 6n – 6
⇒ 205 = 6n + 1
⇒ 6n = 205 – 1 = 204
∴ n = \(\frac{204}{6}\) = 34
∴ 34 terms are there.

ii) 18, 15\(\frac{1}{2}\), 13, ………, -47.
Solution:
Given A.P = 18, 15\(\frac{1}{2}\), 13, ….
Here a₁ = a = 18
d = a₂ – a₁ = 15\(\frac{1}{2}\) – 18 = -2\(\frac{1}{2}\) = –\(\frac{5}{2}\)
Let ‘-47’ be the n^th term of the given A.P.
a_n = a + (n – 1) × d
⇒ -47 = \(\frac{18 \times 2+(n-1)(-5)}{2}\)
⇒ – 94 = 36 – 5n + 5
⇒ 5n = 94 + 41 ⇒ n = \(\frac{135}{5}\) = 27
i.e., 27 terms are there.

Question 6.
Check whether, -150 is a term of the AP : 11, 8, 5, 2, ….
Solution:
Given A.P. = 11, 8, 5, 2,
Here a = a₂ – a₁ = 8 – 11 = -3
If possible, take -150 as the n,sup>th term of the given A.P.
a_n = a + (n – 1) d
-150 = 11 + (n – 1) × (- 3)
⇒ -150 = 11 – 3n + 3
⇒ 14 – 3n = -150
⇒ 3n = 14 + 150 = 164
∴ n = \(\frac{164}{3}\) = 54\(\frac{2}{3}\)
Here n is not an integer.
∴ -150 is not a term of the given A.P.

Question 7.
Find the 31^st term of an A.P. whose 11^th term is 38 and the 16th term is 73.
Solution:
Given : An A.P. whose

Substituting d = 7 in the equation (1) we get,
a + 10 × 7 = 38
⇒ a + 70 = 38
⇒ a = 38 – 70 = – 32
Now, the 31^st term = a + 30d
= (- 32) + 30 × 7
= – 32 + 210 = 178

Question 8.
If the 3^rd and the 9^th terms of an A.P are 4 and – 8 respectively, which term of this A.P is zero ?
Solution:
Given : An A.P whose

Substituting d = – 2 in equation (1) we get a + 2 × (- 2) = 4
⇒ a – 4 = 4
⇒ a = 4 + 4 = 8
Let n^th term of the given A.P be equal to zero.
a_n = a + (n – 1) d
⇒ 0 = 8 + (n – 1) × (-2)
⇒ 0 = 8 – 2n + 2
⇒ 10 – 2n = 0
⇒ 2n = 10 and n = \(\frac{10}{2}\) = 5
∴ The 5^th term of the given A.P is zero.

Question 9.
The 17^th term of an A.P exceeds its 10^th term by 7. Find the common difference.
Solution:
Given an A.P. in which a₁₇ = a₁₀ + 7
⇒ a₁₇ – a₁₀ = 7
We know that a_n = a + (n – 1) d

Question 10.
Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms ?
Solution:
Let the first A.P be : a, a + d, a + 2d,
Second A.P be : b, b + d, b + 2d, b + 3d,….
Also, general term, a_n = a + (n – 1) d
Given that, a₁₀₀ – b₁₀₀ = 100
⇒ a + 99d – (b + 99d) = 100
⇒ a – b = 100
Now the difference between their 1000 terms,

∴ The difference between their 1000th terms is (a – b) = 100.

Note : If the common difference for any two A.Ps are equal then difference n^th terms of two A.Ps is same for all natural values of n.

Question 11.
How many three-digit numbers are divisible by 7 ?
Solution:
The list of three digit numbers divisible by 7 , is,
105, 112, 119, ………, 994
Here a = 105; d = 7; a_n = 994
a_n = a + (n – 1) d
994 = 105 + (n – 1) 7
(n – 1)7 = 889
n -1 = \(\frac{889}{7}\)
n – 1 = 127
∴ n = 127 + 1 = 128
There are 128 three digit numbers which are divisible by 7.

Question 12.
How many multiples of 4 lie between 10 and 250 ?
Solution:
Given numbers : 10 to 250

∴ Multiples of 4 between 10 and 250 are
First term : 10 + (4 – 2) = 12
Last term : 250 – 2 = 248
∴ 12, 16, 20, 24, ………, 248
a = a₁ = 12; d = 4: a_n = 248
a_n = a + (n – 1) d
248 = 12 + (n – 1) × 4
⇒ (n – 1) 4 = 248 – 12
⇒ n – 1 = \(\frac{236}{4}\) = 59
∴ n = 59 + 1 = 60
There are 60 numbers between 10 and 250 which are divisible by 4.

Question 13.
For what value of n, are the n^th terms of two APs : 63, 65, 67, … and 3, 10, 17, …. equal ?
Solution:
Given : The first A.P is 63, 65, 67, ….
where a = 63, d = a₂ – a₁
⇒ d = 65 – 63 = 2
and the second A.P is 3, 10, 17, ….
where a = 3, d = a₂ – a₁ = 10 – 3 = 7
Suppose the n^th terms of the two A.Ps are equal, where a_n = a + (n – 1) d
⇒ 63 + (n -1) 2 = 3 + (n – 1) 7
⇒ 63 + 2n – 2 = 3 + 7n – 7
⇒ 61 + 2n = 7n – 4
⇒ 7n – 2n = 61 +4
⇒ 5n = 65
⇒ n = \(\frac{65}{5}\) = 13
∴ 13^th terms of the two A.Ps are equal.

Question 14.
Determine the AP whose third term is 16 and the 7^th term exceeds the 5^th term by 12.
Solution:
Given : An A.P in which
a₃ = a + 2d = 16 —– (1)
and a₇ = a₅ + 12
i.e., a + 6d = a + 4d + 12
⇒ 6d – 4d = 12
⇒ 2d = 12
⇒ d = \(\frac{12}{2}\) = 6
Substituting d = 6 in equation (1) we get
a + 2 × 6 = 16
⇒ a = 16 – 12 = 4
∴ The series/A. P is a, a + d, a + 2d, a + 3d, …
⇒ 4, 4 + 6, 4 + 12, 4 + 18, …
⇒ A.P: 4, 10, 16, 22,….

Question 15.
Find the 20^th term from the end of the AP : 3, 8, 13,…….., 253.
Solution:
Given : An A.P : 3, 8, 13, …., 253
Here a = a₁ = 3
d = a₂ – a₁ = 8 – 3 = 5
a_n = 253, where 253 is the last term
a_n = a + (n – 1) d
∴ 253 = 3 + (n – 1) 5
⇒ 253 = 3 + 5n – 5
⇒ 5n = 253 + 2
⇒ n = \(\frac{255}{5}\) = 51
∴ The 20^th term from the other end would be
1 + (51 – 20) = 31 + 1 = 32
∴ a₃₂ = 3 + (32 – 1) × 5
= 3 + 31 × 5
= 3 + 155
= 158

Question 16.
The sum of the 4^th and 8^th terms of an AP is 24 and the sum of the 6^th and 10^th terms is 44. Find the first three terms of the (A.P. Mar. ’15)
Solution:
Given an A.P in which
a₄ + a₈ = 24 ⇒ a + 3d + a + 7d = 24
⇒ 2a + 10d = 24
⇒ a + 5d = 12 —– (1)
and a₆ + a₁₀ = 44
⇒ a + 5d + a + 9d = 44
⇒ 2a + 14d = 44
⇒ a + 7d = 22 —– (2)

Also a + 5d = 12
⇒ a + 5(5) = 12
⇒ a + 25 = 12
⇒ a = 12 – 25 = -13
∴ The A.P is a , a + d, a + 2d, ……..
i.e., – 13, (- 13 + 5), (- 13 + 2 × 5)
⇒ -13, -8, -3, ………

Question 17.
Subba Rao started work in 1995 at an annual salary of ₹ 5,000 and received an increment of ₹ 200 each year. In which year did his income reach Rs. 7000?
Solution:
Given : Salary of Subba Rao in 1995 = ₹ 5000
Annual increment = ₹ 200
i.e., His salary increases by Rs. 200 every year.

Clearly 5000, 5200, 5400, ……… forms an A.P in which a = 5000 and d = 200.
Now suppose that his salary reached ₹ 7000
after x – years,
i.e., a_n = 7000
But, a_n = a + (n – 1) d
7000 = 5000 + (n – 1) 200
⇒ 7000 – 5000 = (n – 1) 200
⇒ n – 1 = \(\frac{2000}{200}\)
= 10
⇒ n = 10 + 1 = 11
∴ In 11^th year his salary reached ₹ 7000.

Students can practice 10th Class Maths Study Material Telangana Chapter 10 Mensuration Ex 10.4 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 10 Mensuration Exercise 10.4

Question 1.
A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of cylinder. (AS4)
Solution:
Radius of the sphere(r) = 4.2 cm
Volume of the sphere = \(\frac{4}{3}\) πr³
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × 4.2 × 4.2 × 4.2
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × \(\frac{42}{10}\) × \(\frac{42}{10}\) × \(\frac{42}{10}\) cm³
Radius of the cylinder (r) = 6 cm
Height of the cylinder (h) = ?
Volume of the cylinder = πr²h
= \(\frac{22}{7}\) × 6 × 6 × h cm³
By problem, volume of the metallic sphere = volume of the cylinder
∴ \(\frac{22}{7}\) × 6 × 6 × h = \(\frac{4}{3}\) × \(\frac{22}{7}\) × \(\frac{42}{10}\) × \(\frac{42}{10}\) × \(\frac{42}{10}\)
∴ h = \(\frac{4}{3}\) × \(\frac{22}{7}\) × \(\frac{42}{10}\) × \(\frac{42}{10}\) × \(\frac{42}{10}\) × \(\frac{7}{22}\) × \(\frac{1}{6 \times 6}\) = 2.74 cm
Height of the cylinder = 2.74 cm.

Question 2.
Three metallic spheres of radii 6 cm; 8 cm. and 10 cm respectively are melted to from a single solid sphere. Find the radius of resulting sphere. (AS₁)
Solution:
Given : Radii of the three spheres
r₁ = 6 cm
r₂ = 8 cm
r₃ = 10 cm
These three are melted to form a single sphere. Let the radius of the resulting sphere be V Then volume of the resultant sphere = sum of the the volumes of the three small spheres.
⇒ \(\frac{4}{3}\)πr³ = \(\frac{4}{3}\)πr₁³ + \(\frac{4}{3}\)πr₂³ + \(\frac{4}{3}\)πr₃³
⇒ \(\frac{4}{3}\)πr³ = \(\frac{4}{3}\)π(r₁³ + r₂³ + r₃³)
r³ = r₁³ + r₂³ + r₃³
= 6³ + 8³ + 10³
= 216 + 512 + 1000 = 1728.
∴ 1728 = (2 × 2 × 3) × (2 × 2 × 3) × (2 × 2 × 3)
r³ = 12 × 12 × 12
r³ = 12³
∴ r = 12
Thus the radius of the resultant sphere = 12 cm.

Question 3.
A 20m deep well of diameter 7m is dug and the earth got by digging is evenly spread out to form a platform 22 m of base 14 m. Find the height of the platform. (AS₄)
Solution:
Diameter of the well (d) = 7 m
Radius the well (r) = \(\frac{7}{2}\) m
Depth of the well (h) = 20 m
Quantity of the earth dig out = Volume of the cylindrical well
= πr²h = \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\) × 20
= 770 cm³
Area of the platform on which the earth is spread out = 22 × 14 = 308 cm²
Let the height of the platform be = x cm
∴ 308 × x = 770
⇒ x = \(\frac{770}{308}\) = 2.5 m
Hence, the height of the platform = 2.5m

Question 4.
A well of diameter 14 m. is dug 15m. deep. The earth taken out of it has been spread evenly to form circular embankment of width 7m. Find the height of the embankment. (AS₄)
Solution:
Diameter of the well (d) = 14m
Radius of the well = \(\frac{\mathrm{d}}{2}\) = \(\frac{14}{2}\) = 7m
Depth of the well (h) = 15m

Quantity of the earth dug out = πr²h
= \(\frac{22}{7}\) × 7 × 7 × 15
Area of circular embankment = π(R + r) (R – r)
= 7(14 + 7)(14 – 7)
= \(\frac{22}{7}\) × 21 × 7 = 462 m²
Let the height of the embankment be ‘x’ m.
462 × x = \(\frac{22}{7}\) × 7 × 7 × 15
∴ x = \(\frac{22}{7}\) × 7 × 7 × 15 × \(\frac{1}{462}\)
Hence, the height of the embankment = 5m.

Question 5.
A Container shaped like a right circular cylinder having diameter 12 cm. and height 15cm. is full of ice cream. The ice cream is to be filled into cones of height 12 cm. and diameter 6 cm. having a hemispherical shape on the top. Find die number of such cones which can be filled with the ice cream. (AS₄)
Solution:
Diameter of the container (d) = 12 km
Radius of the cylindrical container (r) = 6 cm
Height of the cylindrical container (h) = 15 cm

Volume of the cylinder = πr²h
= \(\frac{22}{7}\) × 6 × 6 × 15 cm³ = \(\frac{11880}{7}\) cm³
Diameter of the base of the cone(d) = 6 cm
Radius of the base of the cone (r) = \(\frac{\mathrm{d}}{2}\) = \(\frac{6}{2}\) = 3 cm
Volume of the conical part = \(\frac{1}{3}\) × πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 3 × 3 × 12
= \(\frac{792}{7}\) cm³
Radius of the hemisphere(r) = 3 cm

Radius of the hemisphere (r) = 3 cm
Volume of the hemi-sphere = \(\frac{2}{3}\) πr³
= \(\frac{2}{3}\) × \(\frac{22}{7}\) × 3 × 3 × 3 = \(\frac{396}{7}\) cm³
Total volume of the conical part and hemispherical part
= \(\frac{792}{7}\) + \(\frac{396}{7}\) = \(\frac{1188}{7}\) cm³
Number of cones which can be filled with ice cream = \(\frac{11880}{7}\) + \(\frac{1188}{7}\)
(Q Number of icecream cones = \(\frac{\text { Volume of the cylinder }}{\text { Total volume of ice – cream cone }} \)
= \(\frac{11880}{7}\) × \(\frac{7}{1188}\) = 10

Question 6.
How many silver coins, 1.75 cm in diameter and thickness 2mm, need to be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm ? (AS₄)
Solution:
Let the number of silver coins needed to melt = n
then total volume of n coins = volume of the cuboid.
n × πr²h = lbh
[∵ The shape of the coins is a cylinder then v = πr²h]
n × \(\frac{22}{7}\) × \(\left[\frac{1.75}{2}\right]^2\) × \(\frac{2}{10}\) = 5.5 × 10 × 3.5
[∵ 2mm = \(\frac{2}{10}\) cm, r = \(\frac{\mathrm{d}}{2}\)]
n × \(\frac{22}{7}\) × \(\frac{1.75}{2}\) × \(\frac{1.75}{2}\) × \(\frac{2}{10}\) = 55 × 3.5
n = 55 × 3.5 × \(\frac{7 \times 2 \times 2 \times 10}{22 \times 1.75 \times 1.75 \times 2}\)
= \(\frac{55 \times 35 \times 7 \times 4}{22 \times 2 \times 1.75 \times 1.75}\) = \(\frac{5 \times 35 \times 7}{1.75 \times 1.75}\)
= \(\frac{175 \times 7}{1.75 \times 1.75}\) = \(\frac{100 \times 1}{0.25}\) = 400
∴ 400 Silver coins are needed.

Question 7.
A vessel is in the form of an inverted cone. Its height is 8 cm. and the radius of its top is 5 cm. It is filled with water up to the rim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, \(\frac{1}{4}\) of the water flows out. Find the number of lead shots dropped into the vessel. (AS₄)
Solution:
Let the number of lead shots dropped = n
then the total volume of n – lead shots
= \(\frac{1}{4}\) volume of the conical vessel

Leadshots : Radius (r) = 0.5 cm
Volume (V) = \(\frac{4}{3}\) πr³
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × 0.5 × 0.5 × 0.5
Total volume of n-shots
= n × \(\frac{4}{3}\) × \(\frac{22}{7}\) × 0.125
Cone : Radius (r) = 5 cm
height (h) = 8 cm
Volume (V) = \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 5 × 5 × 8
= \(\frac{1}{3}\) × \(\frac{1}{3}\) × 200
\(\frac{1}{4}\)th volume = \(\frac{1}{4}\) × \(\frac{1}{3}\) × \(\frac{22}{7}\) × 200
∴ n × \(\frac{4}{3}\) × \(\frac{22}{7}\) × 0.125 = \(\frac{1}{4}\) × \(\frac{1}{3}\) × \(\frac{22}{7}\) × 200
n = \(\frac{1}{4}\) × \(\frac{1}{3}\) × \(\frac{22}{7}\) × 200 × \(\frac{3}{4}\) × \(\frac{7}{22}\) × \(\frac{1}{0.125}\)
= \(\frac{200 \times 1}{4 \times 4 \times 0.125}\) = \(\frac{200}{2}\) = 100
∴ Number of lead shots = 100

Question 8.
A solid metallic sphere of diameter 28 cm is melted and recast into a number of smaller cones, each of diameter 4\(\frac{2}{3}\) cm and height 3 cm. Find the number of cones so formed. (AS₄)
Solution:
Diameter of the solid metallic sphere (d) = 28 cm
∴ Radius of the sphere (r) = \(\frac{\mathrm{d}}{2}\) = \(\frac{2 r}{2}\) = 14 cm
Volume of the sphere = \(\frac{4}{3}\) pr³
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × 14 × 14 × 14
= \(\frac{176 \times 196}{3}\) cm³
Cone : Diameter of the cone = 4\(\frac{2}{3}\) cm
= \(\frac{14}{3}\) cm
∴ Radius of the cone (r) = \(\frac{14}{3}\) × \(\frac{1}{2}\) = \(\frac{7}{3}\) cm
Height of the cone (h) = 3 cm
Volume of the cone = \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × \(\frac{7}{3}\) × \(\frac{7}{3}\) × 3
= \(\frac{154}{9}\) cm³
Metallic sphere is recast into smaller cones.
Therefore, the number of cones formed
= \(\frac{176 \times 196}{3}\) ÷ \(\frac{154}{9}\)
= \(\frac{176 \times 196}{3}\) × \(\frac{9}{154}\) = 672
Hence, 672 smaller cones are formed.

Students can practice 10th Class Maths Study Material Telangana Chapter 9 Tangents and Secants to a Circle Ex 9.2 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Exercise 9.2

Question 1.
Choose the correct answer and give justification for each.
(i) The angle between a tangent to a circle and the radius drawn at the point of contact is ______
(a) 60°
(b) 30°
(c) 45°
(d) 90°
Solution:
[ d ]
According to theorem (The tangent at any point of a circle is perpendicular to the radius through the point of contact) tangent is perpendicular to the radius at the point of contact.
Therefore, the correct answer is option
(d) i.e – 90°

(ii) From a point Q, the length of the tangent to a circle is 24cm, and the distance of Q from the centre is 25cm. The radius of the circle is
(a) 7 cm
(b) 12 cm
(c) 15 cm
(d) 24.5 cm
Solution:
[ a ]

In the figure, PQ is the tangent to the circle with centre ‘O’, from an external point Q. P is the point of contact. PO is the radius drawn throught the point of contact P.
∴ ∠OPQ = 90°
In the right triangle OPQ,
OQ² = OP² + PQ²(By pythagoras theorem)
Here, OQ = 25 cm; PQ = 24 cm
⇒ OP² = OQ² – PQ²
= 25² – 24²
= 625 – 576 = 49
∴ OP = \(\sqrt{49}\) = 7 cm.
The radius of the circle is 7cm
The correct answer is option (a), i.e 7 cm.

(iii) If AP and AQ are the two tangents of a circle with centre ‘O’ so that ∠POQ = 110°, then ∠PAQ is equal to
(a) 60°
(b) 70°
(c) 80°
(d) 90°
Solution:
[ b ]
AP is a tangent to the circle with centre ‘O’. P is the point of contact. PO is the radius drawn through R
∴ ∠OPA = 90°
Similarly, AQ is a tangent. Q is the point of contact. QO is the radius drawn through Q.
∴ ∠OQA = 90°

In the quadrilateral OQAR the sum of the interior angles is equal to 360°.
⇒ ∠OPA + ∠PAQ + ∠OQA + ∠POQ = 360°
⇒ 90° + ∠PAQ + 90° + 110° = 360°
⇒ 290° + ∠PAQ = 360°
⇒ ∠PAQ = 360° – 290° = 70°
The correct answer is option (b) i.e. 70°

(iv) If tangents PA and PB from a point P to a circle with centre ‘O’ are inclined to each other at an angle of 80°, then ZPOA is equal to
a) 50°
b) 60°
c) 70°
d) 80°
Solution:
[ None ]
PA is a tangent to the circle with centre ‘O’ from an external point R A is the point of contact.
AO is the radius drawn through A.
∴ ∠OAP = 90°
The centre of the circle ‘O’ lies on the bisector of the angle between the two tangents.

∴ ∠OPA = \(\frac{1}{2}\) ∠APB = \(\frac{1}{2}\) × 80° = 40°
Now, in ∆ OAP, ∠OAP = 90°, ∠OPA = 40°
The sum of the angles in a triangle = 180°
∴ ∠OAP + ∠OPA + ∠POA = 180°
90° + 40° + ∠POA = 180°
130° + ∠POA = 180°
∴ ∠POA = 180° – 130° = 50°.
The correct answer is option (a), i.e 50°

(v) In the figure XY and X1Y1 are two parallel tangents to a circle with centre ‘O’ and another tangent AB with point of contact C intersecting XY at ‘A’ and X¹Y¹ at B then ∠AOB =
a) 80°
b) 100°
c) 90°
d) 60°
Solution:
[ C ]

Question 2.
Two concentric circles of radii 5 cm and 3 cm are drawn. Find the length of the chord of the larger circle which touches the smaller circle. (A.P. Mar. ’15)
Solution:
Given : Two circles of radii 3 cm and 5 cm with common centre.

Let AB be a tangent to the inner / small circle and chord to the larger circle.
Let ‘P’ be the point of contact.
Construction : Join OP and OB.
In ∆OPB, ∠OPB = 90°
(radius is perpendicular to the tangent)
OP = 3 cm
OB = 5 cm
Now, (OB)² = (OP)² + (PB)²
[(Hypotenuse)² = (side)² + (side)², pythagorus theorem]
5² = 3² + (PB)²
(PB)² = 25 – 9 = 16
PB = \(\sqrt{16}\) = 4 cm
Now, AB = 2 × PB [∵ The perpendicular drawn from the cnetre of the circle to a chord, bisect it]
AB = 2 × 4 = 8 cm
∴ The length of the chord of the large circle which touches the smaller circle is 8 cm.

Question 3.
Prove that the parallelogram circumscribing a circle is a rhombus. (A.P. Mar. ’16)
Solution:

Given : A circle with centre O’. A parallelogram ABCD, circumscribing the given circle.
Let P Q, R, S be the points of contact.
Required to prove : ABCD is a rhombus
Proof : AP = AS …………… (1)
[∵ tangents drawn from an external point to a circle are equal]
BP = BQ ……………… (2)
CR = CQ ……………… (3)
DR = DS ……………… (4)
Adding (1), (2), (3) and (4) we get
AP + BP + CR + DR = AS + BQ + CQ + DS
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
AB + DC = AD + BC
AB + AB = AD + AD
[∴ Opposite sides of a parallelogram all equal]
2AB = 2AD
AB = AD
Hence, AB = CD and AD = BC
[∴ Opposite sides of a parallelogram]
∴ AB = BC = CD = AD
Thus ◊ ABCD is a rhombus

Question 4.
A triangle ABC is drawn to circumscribe a circle of radius 3 cm. such that the segments BD and DC into which BC is divided by the point of contact D are of length 9 cm. and 3 cm. respectively. Find the sides AB and AC.

Solution:
The given figure can also be drawn as

Given : Let ∆ABC be the given triangle circumscribing the given circle with centre ‘O’ and radius 3 cm
i.e., the circle touches the sides BC, CA and AB at D, E, F respectively.
It is given that BD = 9 cm
CD = 3 cm
∴ Length of 2 tangents drawn from an external point to circle are equal.
∴ BF = BD = 9 cm; AF = AE = x cm (say)
The sides of the triangle are 12 cm, (9 + x) cm, (3 + x) cm
Perimeter = 2S = 12 + 9 + x + 3 + x
⇒ 2S = 24 + 2x
S = 12 + x
⇒ S – a = 12 + x – 12 = x
⇒ S – b = 12 + x – 3 – x = 9
⇒ S – c = 12 + x – 9 – x = 3
∴ Area of the triangle ∆ABC
= \(\sqrt{S(S-a)(S-b)(S-c)}\)
= \(\sqrt{(12+x)(x)(9)(3)}\)
= \(\sqrt{27\left(x^2+12 x\right)}\) …………….. (1)
But ∆ABC = ∆OBC + ∆OCA + ∆OAB
= \(\frac{1}{2}\) × BC × OD + \(\frac{1}{2}\) × CA × OE + \(\frac{1}{2}\) AB × OF
= \(\frac{1}{2}\) (12 × 3) + \(\frac{1}{2}\) (3 + x) × 3 + \(\frac{1}{2}\) (9 + x) × 3
= \(\frac{1}{2}\) [36 + 9 + 3x + 27 + 3x]
= \(\frac{1}{2}\) [72 + 6x] ⇒ 36 + 3x ……………. (2)
From (1) and (2),
\(\sqrt{27\left(x^2+12 x\right)}\) = 36 + 3x
Squaring on both sides we get,
27(x² + 12x) = (36 + 3x)²
27x² + 324x = 1296 + 9x² + 216x
⇒ 18x² + 108x – 1296 = 0
⇒ x² + 6x – 72 = 0
⇒ x² + 12x – 6x – 72 = 0
x(x + 12) – 6(x + 12) = 0
(x – 6) (x + 12) = 0
x = 6 or -12
But ‘x’ can’t be negative hence, x = 6
AB = 9 + 6 = 15 cm
AC = 3 + 6 = 9 cm

Question 5.
Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths. Verify by using phythagorus theorem. (A.P. June ’15)
Solution:
Steps of construction :

1. Draw a circle with centre ‘O’ and radius = 6 cm
2. Take a point P outside the circle such that OP = 10 cm, join OP.
3. Draw the perpendicular bisector to OP which bisects it at M.
4. Taking M as centre and PM or MO as ra-dius draw a circle. Let the circle intersects the given circle at A and B.
5. Join P to A and B.
6. PA and PB are the required tangents of lengths 8 cm each.

Proof : In ∆OAP
(OA)² + (AP)² = 6² + 8²
⇒ 36 + 64 = 100
(OP)² = 10² = 100
∴ (OA)² + (AP)² = (OP)²
Hence AP is a tangent.
Similarly, BP is a tangent.

Question 6.
Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.
Solution:

Construction :

1. Mark a point ‘O’ on the plane of the paper and draw a circle with ‘O’ as centre and 4 cm as radius
2. Taking ‘O’ as centre, draw another circle of radius 6 cm.
3. Mark a point B on the bigger circle. Join OB. Draw the perpendicular bisector of OB to intersect OB at C.
4. Now, Take ‘C’ as centre and CO = CB as radius, draw a circle which intersects the smaller circle at P and Q.
5. Join BP and BQ.
6. BP and BQ are the required tangents
7. By actual measurement, we find BP = BQ = 4.5 cm

Verification :
In ∆ BPO, ∠P = 90°
OP = 4cm, OB = 6cm By Pythagoras Theorem,
OB² = PB² + OP²
⇒ PB² = OB² – OP²
= 6² – 4² = 36 – 16 = 20
∴ PB = \(\sqrt{20}\) = 4.47 cm = 4.5 cm (approximately)

Question 7.
Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle and measure them. Write conclusion.
Solution:

Construction :

1. Draw a circle with the help of a bangle.
2. Take two non-parallel chords AB and CD of this circle.
3. Draw the perpendicular bisectors of AB and CD. Let these intersect at O. Then, O is the centre of the circle drawn.
4. Take a point P outside the circle. Join OR Draw the perpendicular bisector of OR Let M be its midpoint.
5. Draw a circle with M as centre and MO = MP as radius.
6. Join PQ and PR.
7. So, PQ and PR are the required tangents.

Conclusion : Tangents drawn from an external point to a circle are equal.

Question 8.
In a right triangle ABC, a circle with a side AB as diameter is drawn to intersect the hypotenuse AC in P. Prove that the tangent to the circle at P bisects the side BC.

Solution:
Let ABC be a right triangle, right angled at P.
Consider a circle with diameter AB.
From the figure, the tangent to the circle at B meets BC in Q.
Now QB and QP are 2 tangents to the circle from the same point P.
∴ QB = QP ……………. (1)
Also, ∠QPC = ∠QCP
∴ PQ = QC
From (1) and (2);
QB = QC.
Hence proved.

Question 9.
Draw a tangent to a given circle with center O from a point ‘R’ outside the circle. How many tangents can be drawn to the circle from that point ?
[Hint: The distance of two points to the point of contact is the same].
Solution:

Only 2 tangents can be drawn from a given point outside the circle.