Category: Class 10

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 11 Trigonometry Ex 11.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 11th Lesson Trigonometry Exercise 11.3

10th Class Maths 11th Lesson Trigonometry Ex 11.3 Textbook Questions and Answers

Question 1.
Evaluate:
i) \(\frac{\tan 36^{\circ}}{\cot 54^{\circ}}\)
Answer:
Given that \(\frac{\tan 36^{\circ}}{\cot 54^{\circ}}\)
= \(\frac{\tan 36^{\circ}}{\cot \left(90^{\circ}-36^{\circ}\right)}\) [∵ cot (90 – θ) = tan θ]
= \(\frac{\tan 36^{\circ}}{\tan 36^{\circ}}\)
= 1

ii) cos 12° – sin 78°
Answer:
Given that cos 12° – sin 78°
= cos 12° – sin(90° – 12°) [∵ sin (90 – θ) = cos θ]
= cos 12° – cos 12° = 0

iii) cosec 31° – sec 59°
Answer:
Given that cosec 31° – sec 59°
= cosec 31° – sec (90° – 31°) [∵ sec (90 – θ) = cosec θ]
= cosec 31° – cosec 31° = 0

iv) sin 15° sec 75°
Answer:
Given that sin 15° sec 75°
= sin 15° . sec (90° – 15°)
= sin 15° . cosec 15° [∵ sec (90 – θ) = cosec θ]
= sin 15° . \(\frac{1}{\sin 15^{\circ}}\) [∵ cosec θ = \(\frac{1}{\sin \theta}\)]
= 1

v) tan 26° tan 64°
Answer:
Given that tan 26° tan 64°
= tan 26° . tan (90° – 26°)
= tan 26° . cot 26° [∵ tan (90 – θ) = cot θ]
= tan 26° . \(\frac{1}{\tan 26^{\circ}}\) [∵ cot θ = \(\frac{1}{\tan \theta}\)]
= 1

Question 2.
Show that
i) tan 48° tan 16° tan 42° tan 74° = 1
Answer:
L.H.S. = tan 48° tan 16° tan 42° tan 74°
= tan 48°. tan 16° . tan(90° – 48°) . tan(90° – 16°)
= tan 48° . tan 16° . cot 48° . cot 16° [∵ tan (90 – θ) = cot θ]
= tan 48° . tan 16° . \(\frac{1}{\tan 48^{\circ}}\) . \(\frac{1}{\tan 16^{\circ}}\) [∵ cot θ = \(\frac{1}{\tan \theta}\)]
= 1 = R.H.S.
∴ L.H.S. = R.H.S.

ii) cos 36° cos 54° – sin 36° sin 54° = 0
Answer:
L.H.S. = cos 36° cos 54° – sin 36° sin54°
= cos (90° – 54°) . cos (90° – 36°) – sin 36° . sin 54° [∵ cos (90 – θ) = sin θ]
= sin 54° . sin 36° – sin 36° . sin 54°
= 0 = R.H.S.
∴ L.H.S. = R.H.S.

Question 3.
If tan 2A = cot (A – 18°), where 2A is an acute angle. Find the value of A.
Answer:
Given that tan 2A = cot (A – 18°)
⇒ cot (90° – 2A) = cot (A – 18°) [∵ tan θ = cot (90 – θ)]
⇒ 90° – 2A = A – 18°
⇒ 108° = 3A
⇒ A = \(\frac{108^{\circ}}{3}\) = 36°
Hence the value of A is 36°.

Question 4.
If tan A = cot B where A and B. are acute angles, prove that A + B = 90°.
Answer:
Given that tan A = cot B
⇒ cot (90° – A) = cot B [∵ tan θ = cot (90 – θ)]
⇒ 90° – A = B
⇒ A + B = 90°

Question 5.
If A, B and C are interior angles of a triangle ABC, then show that \(\tan \left(\frac{\mathbf{A}+\mathbf{B}}{2}\right)=\cot \frac{\mathbf{C}}{2}\)
Answer:
Given A, B and C are interior angles of right angle triangle ABC then A + B + C = 180°.
On dividing the above equation by ‘2’ on both sides, we get 180°

On taking tan ratio on both sides

Hence proved.

Question 6.
Express sin 75° + cos 65° in terms of trigonometric ratios of angles between 0° and 45°.
Answer:
We have sin 75° + cos 65°
= sin (90° – 15°) + cos (90° – 25°)
= cos 15° + sin 25° [∵ sin (90 – θ) = cos θ and cos (90 – θ) = sin θ]

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 6 Progressions Ex 6.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 6th Lesson Progressions Exercise 6.3

10th Class Maths 6th Lesson Progressions Ex 6.3 Textbook Questions and Answers

Question 1.
Find the sum of the following APs:
i) 2, 7, 12,…, to 10 terms.
Answer:
Given A.P: 2, 7, 12, …… to 10 terms
a = 2; d = a₂ – a₁ = 7 – 2 = 5; n = 10
S_n = \(\frac{n}{2}\)[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.
Answer:
Given A.P: -37, -33, -29,…, to 12 terms.
a = -37; d = a₂ – a₁ = (-33) – (-37) = -33 + 37 = 4; 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.
Answer:
Given A.P : 0.6, 1.7, 2.8,…. S₁₀₀
a = 0.6; d = a₂ – a₁ = 1.7 – 0.6 = 1.1; n = 100
S_n = \(\frac{n}{2}\)[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.
Answer:
Given A.P: \(\frac{1}{15}\), \(\frac{1}{12}\), \(\frac{1}{10}\),…, S₁₁
a = \(\frac{1}{15}\); d = a₂ – a₁ = \(\frac{1}{12}\) – \(\frac{1}{15}\) = \(\frac{5-4}{60}\) = \(\frac{1}{60}\); n = 11
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]

2. Find the sums given below =:
i) 7 + 10\(\frac{1}{2}\) + 14 + …. + 84
Answer:
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 = 84
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 = \(\frac{n}{2}\)(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
Answer:
Given A.P: 34 + 32 + 30 + … + 10
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 = \(\frac{-26}{-2}\) = 13
∴ n = 13
Also, S_n = \(\frac{n}{2}\)(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)
Answer:
Given A.P: -5 + (-8) + (-11) + … + (-230)
Here first term, a = -5;
d = a₂ – a₁ = (-8) – (-5) = -8 + 5 = -3 and the last term l = a_n = 10
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
∴ n = 76
Now, S_n = \(\frac{n}{2}\)(a + l)
where a = -5; l = -230
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.
Answer:
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
Now, S_n = \(\frac{n}{2}\)(a + l)
S₁₆ = \(\frac{16}{2}\)(5 + 50)
= 38 × 55
= 440

ii) Given a = 7, a₁₃ = 35, find d and S₁₃.
Answer:
Given: a = 7;
a₁₃ = a + 12d = 35
⇒ 7 + 12d = 35
⇒ 12d = 35 – 7
⇒ n = \(\frac{28}{12}\) = \(\frac{7}{3}\)
Now, S_n = \(\frac{n}{2}\)(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.
Answer:
Given:
a₁₂ = a + 11d = 37
d = 3
So, a₁₂ = a + 11 × 3 = 37
⇒ a + 33 = 37
⇒ a = 37 – 33 = 4
Now, S_n = \(\frac{n}{2}\)(a + l)
S₁₂ = \(\frac{12}{2}\)(4 + 37)
= 6 × 41
= 246

iv) Given a₃ = 15, S₁₀ = 125, find d and a₁₀.
Answer:
Given:
a₃ = a + 2d = 15
⇒ a = 15 – 2d ……… (1)
S₁₀ = 125 but take S₁₀ as 175
i.e., S₁₀ = 175
We know that,

⇒ 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 = 90, find n and a_n.
Answer:
Given a = 2, d = 8, S_n = 90
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]

⇒ 90 = 2n [2n – 1]
⇒ 4n² – 2n = 90
⇒ 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 x 8
= 2 + 32 = 34

vi) Given a_n = 4, d = 2, S_n = -14, find n and a.
Answer:
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
Given a = 2, d = 8, S_n = 90
S_n = \(\frac{n}{2}\)[a + a_n]
-14 = \(\frac{n}{2}\)[(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.
Answer:
Given:
l = a₉ = a + 8d = 28 and S₉ = 144 But,
Now, S_n = \(\frac{n}{2}\)(a + l)
144 = \(\frac{9}{2}\)(a + 28)
⇒ 144 × \(\frac{2}{9}\) = a + 28
⇒ a + 28 = 32
⇒ a = 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?
Answer:
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.
Answer:
Given A.P in which

Substituting d = 4 in equation (1),
we get a + 4 = 14
⇒ a = 14 – 4 = 10

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.
Answer:
Given :
A.P such that S₇ = 49; S₁₇ = 289
We know that,

Substituting d = 2 in equation (1), we get,
a + 3 × 2 = 7
⇒ a = 7 – 6 = 1
∴ a = 1; d = 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 a_n is defined as below:
i) a = 3 + 4n
ii) a_n = 9 – 5n. Also find the sum of the first 15 terms in each case.
Answer:
Given a_n = 3 + 4n
Then a₁ = 3 + 4 × l = 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.

ii) a_n = 9 – 5n
Given: a_n = 9 – 5n.
a₁ = 9 – 5 × l = 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 successive terms is constant (or) starting from the second term, each term is obtained by adding a fixed number ‘-5’ to its preceding term.
Hence {a_n} forms an A.P.

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.
Answer:
Given an A.P in which S_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 = l – 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.
Answer:
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 = \(\frac{n}{2}\)(a + l)
S₄₀ = \(\frac{40}{2}\)(6 + 240)
= 20 × 246
= 4920
∴ S₄₀ = 4920

Question 10.
A sum of Rs. 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs. 20 less than its preceding prize, find the value of each of the prizes.
Answer:
Given:
Total/Sum of all cash prizes = Rs. 700
Each prize differs by Rs. 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 = \(\frac{n}{2}\)(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 plant¬ing 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 II 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?
Answer:
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 is an A.P.
Here, a = 3 and l = 36; n = 12
∴ S_n = \(\frac{n}{2}\)(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 semicircles is l₁, l₂, l₃, l₄,….. with centres at A, B, A, B,…, respectively.]
Answer:
Given: l₁, l₂, l₃, l₄,….., l₁₃ are the semicircles 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] [∵ Radii are in A.P. as aj = 0.5 and d = 0.5]
……………………………
r₁₃ = 13 × 0.5 = 6.5
Now, the total length of the spiral = l₁ + l₂ + l₃ + l₄ + ….. + l₁₃ [∵ 13 given]
But circumference of a semi-cirle is πr.
∴ Total length of the spiral = π × 0.5 + π × 1.0 + ………. + π × 6.5
= π × \(\frac{1}{2}\)[l + 2 + 3 + ….. + 13]
[∵ Sum of the first n – natural numbers is \(\frac{n(n+1)}{2}\)
= \(\frac{22}{7} \times \frac{1}{2} \times \frac{13 \times 14}{2}\)
= 11 × 13
= 143 cm.

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?

Answer:
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

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

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)]
Answer:
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 = 8 m (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 = 14 m (5 + 3 × 3)
………………………………
∴ Distance of the tenth ball from the bucket = 5 + 9 × 3 = 5 + 27 = 32 m.
1st ball: Distance covered by the competitor in picking up and dropping it in the bucket = 2 × 5 = 10 m.
2nd ball: Distance covered by the competitor in picking up and dropping it in the bucket = 2 × 8 = 16 m.
3rd ball: Distance covered by the competitor in picking up and dropping it in the bucket = 2 × 11 = 22 m.
………………………………
10th 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 = \(\frac{n}{2}\)[2a + (n – 1)d]
∴ S₁₀ = \(\frac{10}{2}\)[2 × 10 + (10 – 1)6]
= 5 [20 + 54]
= 5 × 74
= 370 m
∴ Total distance = 370 m.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 6 Progressions Ex 6.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 6th Lesson Progressions Exercise 6.1

10th Class Maths 6th Lesson Progressions Ex 6.1 Textbook Questions and Answers

Question 1.
In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
i) The taxi fare after each km when the fare is Rs. 20 for the first km and rises by Rs. 8 for each additional km.
Answer:
Fare for the first km = Rs. 20 = a
Fare for each km after the first = Rs. 8 = d
∴ The fares would be 20, 28, 36, 44, …….
The above list forms an A.P.
Since each term in the list, starting from the second can be obtained by adding ‘8’ to its preceding term.

ii) The amount of air present in a cylinder when a vacuum pump removes \(\frac{1}{4}\)th of the air remaining in the cylinder at a time.
Answer:
Let the amount of air initially present in the cylinder be 1024 lit.
First it removes \(\frac{1}{4}\)th of the volume
i.e., \(\frac{1}{4}\) × 1024 = 256
∴ Remaining air present in the cylinder = 768
At second time it removes \(\frac{1}{4}\)th of 768
i.e., \(\frac{1}{4}\) × 768 = 192
∴ Remaining air in the cylinder = 768 – 192 = 576
Again at third time it removes \(\frac{1}{4}\)th of 576
i.e., \(\frac{1}{4}\) × 576 = 144
Remaining air in the cylinder = 576 – 144 = 432
i.e., the volume of the air present in the cylinder after 1st, 2nd, 3rd,… times is 1024, 768, 576, 432, …..
Here, a₂ – a₁ = 768 – 1024 = – 256
a₃ – a₂ = 576 – 768 = – 192
a₄ – a₃ = 432 – 576 = – 144 .
Thus the difference between any two successive terms is not equal to a fixed number.
∴ The given situation doesn’t show an A.P.

iii) The cost of digging a well, after, every metre of digging, when it costs ? 150 for the first metre and rises by ? 50 for each subsequent metre.
Answer:
Cost for digging the first metre = Rs. 150
Cost for digging subsequent metres = Rs. 50 each.
i.e.,

The list is 150, 200, 250, 300, 350, ……..
Here d = a₂ – a₁ = a₃ – a₂ = a₄ – a₃ = ……. = 50
∴ The given situation represents an A.P.

iv) The amount of money in the account every year, when Rs. 10000 is deposited at compound interest at 8 % per annum.
Answer:
Amount deposited initially = P = Rs. 10,000
Rate of interest = R = 8% p.a [at C.I.]
∴ \(A=P\left(1+\frac{R}{100}\right)^{n}\)

The terms 10800, 11664, 12597.12, ……. a₂ – a₁ = 800
Here, a = 10,000                                     a₃ – a₂ = 864
But, a₂ – a₁ ≠ a₃ – a₂ ≠ a₄ – a₃                a₄ – a₃ = 953.12
∴ The given situation doesn’t represent an A.P.

Question 2.
Write first four terms of the AP, when the first term a and the common difference d are given as follows:
i) a = 10, d = 10
ii) a = -2, d = 0
iii) a = 4, d = – 3
iv) a = – 1, d = 1/2
v) a = – 1.25, d = – 0.25
Answer:

Question 3.
For the following A.Ps, write the first term and the common difference:
i) 3, 1, – 1, – 3,….
ii) – 5, – 1, 3, 7,….
iii) \(\frac{1}{3}\), \(\frac{5}{3}\), \(\frac{9}{3}\), \(\frac{13}{3}\), ……..
iv) 0.6, 1.7, 2.8, 3.9,…
Answer:

Question 4.
Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.
i) 2, 4, 8, 16, …….
ii) 2, \(\frac{5}{2}\), 3, \(\frac{7}{2}\), …….
iii) – 1.2, – 3.2, – 5.2, – 7.2,……
iv) -10,-6, -2, 2, …….
v) 3, 3 + √2, 3 + 2√2, 3 + 3√2, …….
vi) 0.2, 0.22, 0.222, 0.2222, ……
vii) 0, -4, -8, -12, …….
viii) –\(\frac{1}{2}\), –\(\frac{1}{2}\), –\(\frac{1}{2}\), –\(\frac{1}{2}\)
ix) 1, 3, 9, 27,…..
x) a, 2a, 3a, 4a,….
xi) a, a², a³, a⁴, …..
xii) √2, √8, √18, √32, …….
xiii) √3, √6, √9, √12, …….
Answer:

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 6 Progressions Ex 6.5 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 6th Lesson Progressions Exercise 6.5

10th Class Maths 6th Lesson Progressions Ex 6.5 Textbook Questions and Answers

Question 1.
For each geometric progression find the common ratio ‘r’, and then find a_n.
i) 3, \(\frac{3}{2}\), \(\frac{3}{4}\), \(\frac{3}{8}\), …….
Answer:
Given G.P.: 3, \(\frac{3}{2}\), \(\frac{3}{4}\), \(\frac{3}{8}\), …….

ii) 2, -6, 18, -54, …….
Answer:
Given G.P. = 2, -6, 18, -54, …….
a = 2, r = \(\frac{a_{2}}{a_{1}}\) = \(\frac{-6}{2}\) = -3
a_n = a . r^n-1 = 2 × (-3)^n-1
∴ r = -3; a_n = 2(-3)^n-1

iii) -1, -3, -9, -27, ……
Given G.P. = -1, -3, -9, -27, ……
a = -1, r = \(\frac{a_{2}}{a_{1}}\) = \(\frac{-3}{-1}\) = 3
a_n = a . r^n-1 = (-1) × 3^n-1
∴ r = 3; a_n = (-1) × 3^n-1

iv) 5, 2, \(\frac{4}{5}\), \(\frac{8}{25}\), …….
Given G.P. = 5, 2, \(\frac{4}{5}\), \(\frac{8}{25}\), …….
a = 5, r = \(\frac{a_{2}}{a_{1}}\) = \(\frac{2}{5}\)
a_n = a . r^n-1 = 5 × \(\left(\frac{2}{5}\right)^{n-1}\)
∴ r = \(\frac{2}{5}\); a_n = 5\(\left(\frac{2}{5}\right)^{n-1}\)

Question 2.
Find the 10^th and nth term of G.P.: 5, 25, 125,…..
Answer:
Given G.P.: 5, 25, 125,…..
a = 5, r = \(\frac{a_{2}}{a_{1}}\) = \(\frac{25}{5}\) = 5
a_n = a . r^n-1 = 5 × (5)^n-1 = 5^1+n-1 = 5ⁿ
a₁₀ = a . r⁹ = 5 × 5⁹ = 5¹⁰
∴ a₁₀ = 5¹⁰; a_n = 5ⁿ

Question 3.
Find the indicated term of each geometric progression.
i) a₁ = 9; r = \(\frac{1}{3}\); find a₇.
Answer:
a_n = a . r^n-1

ii) a₁ = -12; r = \(\frac{1}{3}\); find a₆.
Answer:
a_n = a . r^n-1

Question 4.
Which term of the G.P.
i) 2, 8, 32,….. is 512?
Answer:
Given G.P.: 2, 8, 32,….. is 512
a = 2, r = \(\frac{a_{2}}{a_{1}}\) = \(\frac{8}{2}\) = 4
Let the n^th term of G.P. be 512

512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 2⁹
∴ 2n – 1 = 9
[∵ bases are equal, exponents are also equal]
∴ 2n = 9 + 1 = 10
n = \(\frac{10}{2}\) = 5
∴ 512 is the 5^th term of the given G.P.

ii) √3, 3, 3√3, …….. is 729?
Answer:
Given G.P.: √3, 3, 3√3, …….. is 729
a = √3, r = \(\frac{a_{2}}{a_{1}}\) = \(\frac{3}{\sqrt{3}}\) = √3
now a_n = a . r^n-1 = 729
⇒ (√3)(√3)^n-1 = 729
⇒ (√3)ⁿ = 3⁶ = (√3)¹²
⇒ n = 12
So 12th term of GP √3, 3, 3√3, …….. is 729.

iii) \(\frac{1}{3}\), \(\frac{1}{9}\), \(\frac{1}{27}\), ……. is \(\frac{1}{2187}\)?
Answer:
Given G.P.: \(\frac{1}{3}\), \(\frac{1}{9}\), \(\frac{1}{27}\), ……. is \(\frac{1}{2187}\)

Let \(\frac{1}{2187}\) be the n^th term of the G.P., then
a_n = a . r^n-1 = \(\frac{1}{2187}\)

[∵ bases are equal, exponents are also equal]
7^th term of G.P is \(\frac{1}{2187}\).

Question 5.
Find the 12^th term of a G.P. whose 8 term is 192 and the common ratio is 2.
Answer:
Given a G.P. such that a₈ = 192 and r = 2
a_n = a . r^n-1
a₈ = a . (2)^8-1 = 192

= 3 × 2¹⁰ = 3 × 1024 = 3072.

Question 6.
The 4^th term of a geometric progression is \(\frac{2}{3}\) and the seventh term is \(\frac{16}{81}\). Find the geometric series.
Answer:
Given: In a G.P.

Now substituting r = \(\frac{2}{3}\) in equation (1)
we get,

Question 7.
If the geometric progressions 162, 54, 18, ….. and \(\frac{2}{81}\), \(\frac{2}{27}\), \(\frac{2}{9}\),….. have their n^th term equal, find the value of n.
Answer:
Given G.P.: 162, 54, 18, ….. and \(\frac{2}{81}\), \(\frac{2}{27}\), \(\frac{2}{9}\),……

Given that n^th terms are equal
a_n = a . r^n-1

⇒ 3^n-1+n-1 = 81 × 81
⇒ 3^2n-2 = 3⁴ × 3⁴
⇒ 3^2n-2 = 3⁸ [∵ a^m . aⁿ = a^m+n]
⇒ 2n – 2 = 8
[∵ bases are equal, exponents are also equal]
2n = 8 + 2
⇒ n = \(\frac{10}{2}\) = 5
The 5^th terms of the two G.P.s are equal.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 6 Progressions Ex 6.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 6th Lesson Progressions Exercise 6.2

10th Class Maths 6th Lesson Progressions Ex 6.2 Textbook Questions and Answers

Question 1.
Fill in the blanks in the following table, given that ‘a’ is the first term, d the common difference and an the nth term of the A.P:

Answer:

Question 2.
Find the i) 30th term of the A.P.: 10, 7, 4,……
ii) 11th term of the A.P.: -3, –\(\frac{1}{2}\), 2,…
Answer:
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) = – 3
= –\(\frac{1}{2}\) + 3
= \(\frac{-1+6}{2}\)
= \(\frac{5}{2}\)
a_n = a + (n – 1) d
= -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.
Answer:
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.
Answer:
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\(\frac{1}{2}\), find a₂, a₃.
Answer:
Given: a₁ = a = 5 ….. (1)
a₄ = a + 3d = 9\(\frac{1}{2}\) ….. (2)
Solving equations (1) and (2);

⇒ 3d = 4\(\frac{1}{2}\)
⇒ 3d = \(\frac{9}{2}\)
⇒ d = \(\frac{9}{2 \times 3}\) = \(\frac{3}{2}\)
∴ a₂ = a + d = 5 + \(\frac{3}{2}\) = \(\frac{13}{2}\)
a₃ = a + 2d = 5 + 2 × \(\frac{3}{2}\) = 5 + 3 = 8

iv) a₁ = -4; a₆ = 6, find a₂, a₃, a₄, a₅.
Answer:
Given: a₁ = a = -4 ….. (1)
a₆ = a + 5d = 6 ….. (2)
Solving equations (1) and (2);
(-4) + 5d = 6
⇒ 5d = 6 + 4
⇒ 5d = 10
⇒ d = \(\frac{10}{5}\)
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.
Answer:
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?
Answer:
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 = \(\frac{80}{2}\) = 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
Answer:
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
Answer:
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 nth term of the given A.P.
a_n = a + (n – 1) d

⇒ -94 = 36 – 5n + 5
⇒ 5n = 94 + 41
⇒ n = \(\frac{135}{5}\) = 27
∴ 27 terms are there.

Question 6.
Check whether, -150 is a term of the AP: 11, 8, 5, 2…
Answer:
Given: A.P. = 11, 8, 5, 2…
Here a₁ = a = 11;
d = a₂ – a₁ = 8 – 11 = -3
If possible, take – 150 as the n^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 16^th term is 73.
Answer:
Given: An A.P. whose

⇒ -5d = -35
⇒ d = \(\frac{-35}{-5}\) = 7
Substituting d = 7 in the equation (1)
we get,
a + 10 x 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 3rd and the 9^th terms of an A.P are 4 and -8 respectively, which term of this A.P is zero?
Answer:
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.
an = 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 term by 7. Find the common difference.
Answer:
Given an A.P in which a₁₇ = a₁₀ + 7
⇒ a₁₇ – a₁₀ = 7
We know that a_n = a + (n – 1)d

⇒ d = \(\frac{7}{7}\) = 1

Question 10.
Two APs have the same common difference. The difference between their 100^th terms is 100, what is the difference between their 1000^th terms?
Answer:
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^th terms,

∴ The difference between their 1000^th terms is (a – b) = 100.
Note: If the common difference for any two A.Ps are equal then difference between 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?
Answer:
The least three digit number is 100.

∴ The least 3 digit number divisible by 7 is 100 + (7 – 2) = 105
The greatest 3 digit number is 999

∴ The greatest 3 digit number divisible by 7 is 999 – 5 = 994.
∴ 3 digit numbers divisible by 7 are
105, 112, 119,….., 994.
a₁ = a = 105; d = 7; a_n = 994
a_n = a + (n – 1)d
⇒ 994 = 105 + (n – 1)7
⇒ (n – 1)7 = 994 – 105
⇒ (n – 1)7 = 889
⇒ n – 1 = \(\frac{889}{7}\) = 127
∴ n = 127 + 1 = 128
∴ There are 128, 3 digit numbers which are divisible by 7.
(or)
\(\frac{\text { last number – first number }}{7}\)
\(\frac{999-100}{7}\)
≃ 128.4 = 128 numbers divisible by 7.

Question 12.
How many multiples of 4 lie between 10 and 250?
Answer:
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?
Answer:
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 nth 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.
Answer:
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.
Answer:
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 – l)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 AP.
Answer:
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 Rs. 5000 and received an increment of Rs. 200 each year. In which year did his income reach Rs. 7000?
Answer:
Given: Salary of Subba Rao in 1995 = Rs. 5000
Annual increment = Rs. 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 Rs. 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
∴ In 11^th year his salary reached Rs. 7000.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 6 Progressions Ex 6.4 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 6th Lesson Progressions Exercise 6.4

10th Class Maths 6th Lesson Progressions Ex 6.4 Textbook Questions and Answers

Question 1.
In which of the following situations, does the list of numbers involved in the form a G.P.?
i) Salary of Sharmila, when her salary is Rs. 5,00,000 for the first year and expected to receive yearly increase of 10% .
Answer:
Given: Sharmila’s yearly salary = Rs. 5,00,000.
Rate of annual increment = 10 %.

Here, a = a₁ = 5,00,000
a₂ = 5,00,000 × \(\frac{11}{10}\) = 5,50,000
a₃ = 5,00,000 × \(\frac{11}{10}\) × \(\frac{11}{10}\) = 6,05,000
a₄ = 5,00,000 × \(\frac{11}{10}\) × \(\frac{11}{10}\) × \(\frac{11}{10}\) = 6,65,000

Every term starting from the second can be obtained by multiplying its pre¬ceding term by a fixed number \(\frac{11}{10}\).
∴ r = common ratio = \(\frac{11}{10}\)
Hence the situation forms a G.P.

ii) Number of bricks needed to make each step, if the stair case has total 30 steps. Bottom step needs 100 bricks and each successive step needs 2 bricks less than the previous step.
Answer:
Given: Bricks needed for the bottom step = 100.
Each successive step needs 2 bricks less than the previous step.
∴ Second step from the bottom needs = 100 – 2 = 98 bricks.
Third step from the bottom needs = 98 – 2 = 96 bricks.
Fourth step from the bottom needs = 96 – 2 = 94 bricks.
Here the numbers are 100, 98, 96, 94, ….
Clearly this is an A.P. but not G.P.

iii) Perimeter of the each triangle, when the mid-points of sides of an equilateral triangle whose side is 24 cm are joined to form another triangle, whose mid-points in turn are joined to form still another triangle and the process continues indefinitely.

Answer:
Given: An equilateral triangle whose perimeter = 24 cm.
Side of the equilateral triangle = \(\frac{24}{3}\) = 8 cm.
[∵ All sides of equilateral are equal] ……. (1)
Now each side of the triangle formed by joining the mid-points of the above triangle in step (1) = \(\frac{8}{2}\) = 4 cm
[∵ A line joining the mid-points of any two sides of a triangle is equal to half the third side.]
Similarly, the side of third triangle = \(\frac{4}{2}\) = 2 cm
∴ The sides of the triangles so formed are 8 cm, 4 cm, 2 cm,
a = 8

Thus each term starting from the second; can be obtained by multiplying its preceding term by a fixed number \(\frac{1}{2}\).
∴ The situation forms a G.P.

Question 2.
Write three terms of the G.P. when the first term ‘a’ and the common ratio ‘r’ are given.
i) a = 4 ; r = 3.
Answer:
The terms are a, ar, ar², ar³, ……..
∴ 4, 4 × 3, 4 × 3² , 4 × 3² , ……
⇒ 4, 12, 36, 108, ……

ii) a = √5 ; r = \(\frac{1}{5}\)
Answer:
The terms are a, ar, ar², ar³, ……..

iii) a = 81 ; r = –\(\frac{1}{3}\)
Answer:
The terms of a G.P are:
a, ar, ar², ar³, ……..

⇒ 81, -27, 9,

iv) a = \(\frac{1}{64}\); r = 2.
Answer:
Given: a = \(\frac{1}{64}\); r = 2.

∴ The G.P is \(\frac{1}{64}\), \(\frac{1}{32}\), \(\frac{1}{16}\), …….

Question 3.
Which of the following are G.P. ? If they are G.P, write three more terms,
i) 4, 8, 16, ……
Answer:
Given: 4, 8, 16, ……
where, a₁ = 4; a₂ = 8; a₃ = 16, ……
\(\frac{a_{2}}{a_{1}}=\frac{8}{4}=2\)
\(\frac{a_{3}}{a_{2}}=\frac{16}{8}=2\)
∴ r = \(\frac{a_{2}}{a_{1}}=\frac{a_{3}}{a_{2}}\) = 2
Hence 4, 8, 16, … is a G.P.
where a = 4 and r = 2
a₄ = a . r³ = 4 × 2³ = 4 × 8 = 32
a₅ = a . r⁴ = 4 × 2⁴ = 4 × 16 = 64
a₆ = a . r⁵ = 4 × 2⁵ = 4 × 32 = 128

ii) \(\frac{1}{3}\), –\(\frac{1}{6}\), \(\frac{1}{12}\), …….
Answer:
Given: t₁ = \(\frac{1}{3}\), t₂ = –\(\frac{1}{6}\), t₃ = \(\frac{1}{12}\), ….
\(\frac{1}{3}\), –\(\frac{1}{6}\), \(\frac{1}{12}\), …….

Hence the ratio is common between any two successive terms.
∴ \(\frac{1}{3}\), –\(\frac{1}{6}\), \(\frac{1}{12}\), ……. is G.P.
where a = \(\frac{1}{3}\) and r = –\(\frac{1}{2}\)

iii) 5, 55, 555, ……..
Answer:
Given: t₁ = 5, t₂ = 55, t₃ = 555, ….

∴ 5, 55, 555, …….. is not a G.P.

iv) -2, -6, -18, ……
Given: t₁ = -2, t₂ = -6, t₃ = -18

∴ -2, -6, -18, is a G.P.
where a = -2 and r = 3
a_n = a . r^n-1 =
a₄ = a . r³ = (-2) × 3³ = -2 × 27 = -54
a₅ = a . r⁴ = (-2) × 3⁴ = -2 × 81 = -162
a₆ = a . r⁵ = (-2) × 3⁵ = -2 × 243 = -486

v) \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{6}\), …….
Answer:

i.e., \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{6}\), ….. is not a G.P.

vi) 3, -3², 3³, ……
Given: t₁ = 3, t₂ = -3², t₃ = 3³, ……


i.e., every term is obtained by multiplying its preceding term by a fixed number -3.
3, -3², 3³, …… forms a G.P,
where a = 3; r = -3
a_n = a . r^n-1
a₄ = a . r³ = 3 × (-3)³ = 3 × (-27) = -81
a₅ = a . r⁴ = 3 × (-3)⁴ = 3 × 81 = 243
a₆ = a . r⁵ = 3 × (-3)⁵ = 3 × (-243) = -729

vii) x, 1, \(\frac{1}{x}\), …….
Answer:
Given: t₁ = x, t₂ = 1, t₃ = \(\frac{1}{x}\), ……

Hence x, 1, \(\frac{1}{x}\), …. forms a G.P.
where a = x; r = \(\frac{1}{x}\)

viii) \(\frac{1}{\sqrt{2}}\), -2, \(\frac{8}{\sqrt{2}}\), …….
Answer:
Given: t₁ = \(\frac{1}{\sqrt{2}}\), t₂ = -2, t₃ = \(\frac{8}{\sqrt{2}}\), ……

ix) 0.4, 0.04, 0.004, ……..
Answer:
Given: t₁ = 0.4, t₂ = 0.04, t₃ = 0.004, ……

∴ 0.4, 0.04, 0.004, …….. forms a G.P.
where a = 0.4; r = \(\frac{1}{10}\)

Question 4.
Find x so that x, x + 2, x + 6 are consecutive terms of a geometric progression.
Answer:
Given x, x + 2 and x + 6 are in G.P. but read it as x, x + 2 and x + 6.
∴ r = \(\frac{\mathrm{t}_{2}}{\mathrm{t}_{1}}\) = \(\frac{\mathrm{t}_{3}}{\mathrm{t}_{2}}\)
⇒ \(\frac{x+2}{x}\) = \(\frac{x+6}{x+2}\)
⇒(x + 2)² = x(x + 6)
⇒ x² + 4x + 4 = x² + 6x
⇒ 4x – 6x = – 4 = -2x = -4
∴ x = 2

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 7 Coordinate Geometry Ex 7.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 7th Lesson Coordinate Geometry Exercise 7.1

10th Class Maths 7th Lesson Coordinate Geometry Ex 7.1 Textbook Questions and Answers

Question 1.
Find the distance between the following pair of points,
(i) (2, 3) and (4, 1)
Answer:
Distance = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
= \(\sqrt{(4-2)^{2}+(1-3)^{2}}\)
= \(\sqrt{4+4}\)
= √8 = 2√2 units

ii) (- 5, 7) and (-1, 3)
Answer:
Distance = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
= \(\sqrt{(-1+5)^{2}+(3-7)^{2}}\)
= \(\sqrt{4^{2}+(-4)^{2}}\)
= \(\sqrt{16+16}\)
= √32 = 4√2 units

iii) (- 2, -3) and (3, 2)
Answer:

iv) (a, b) and (- a, – b)
Answer:

Question 2.
Find the distance between the points (0, 0) and (36, 15).
Answer:
Given: Origin O (0, 0) and a point P (36, 15).
Distance between any point and origin = \(\sqrt{x^{2}+y^{2}}\)
∴ Distance = \(\sqrt{36^{2}+15^{2}}\)
= \(\sqrt{1296+225}\)
= \(\sqrt{1521}\)
= 39 units

∴ 1521 = 3² × 13²
\(\sqrt{1521}\) = 3 × 13 = 39

Question 3.
Verify that the points (1, 5), (2, 3) and (-2, -1) are collinear or not.
Answer:
Given: A (1, 5), B (2, 3) and C (- 2, – 1)

Here the sum of no two segments is equal to third segment.
Hence the points are not collinear.
!! Slope of AB, m₁ = \(\frac{3-5}{2-1}\) = -2
Slope of BC, m₂ = \(\frac{-1-3}{-2-2}\) = 1
m₁ ≠ m₂
Hence A, B, C are not collinear.

Question 4.
Check whether (5, -2), (6, 4) and (7,-2) are the vertices of an isosceles triangle.
Answer:
Let A = (5, – 2); B = (6, 4) and C = (7, – 2).

Now we have, AB = BC.
∴ △ABC is an isosceles triangle,
i.e., given points are the vertices of an isosceles triangle.

Question 5.
In a classroom, 4 friends are seated at the points A, B, C and D as shown in figure. Jarina and Phani walk into the class and after observing for a few minutes Jarina asks Phani “Don’t you think ABCD is a square?” Phani disagrees. Using distance formula, find which of them is correct. Why?

Answer:
Given: Four friends are seated at A, B, C and D where A (3, 4), B (6, 7), C (9, 4) and D (6, 1).
Now distance

BD = \(\sqrt{(6-6)^{2}+(1-7)^{2}}\) = √36 = 6
Hence in □ ABCD four sides are equal
i.e., AB = BC = CD = DA
= 3√2 units
and two diagonals are equal.
i.e., AC = BD = 6 units.
∴ □ ABCD forms a square.
i.e., Jarina is correct.

Question 6.
Show that the following points form an equilateral triangle A(a, 0), B(- a, 0), C(0, a√3).
Answer:
Given: A (a, 0), B (- a, 0), C (0, a√3).

Now, AB = BC = CA.
∴ △ABC is an equilateral triangle.

Question 7.
Prove that the points (-7, -3), (5, 10), (15, 8) and (3, -5) taken in order are the corners of a parallelogram.
Answer:
To show that the given points form a parallelogram.
We have to show that the mid points of each diagonal are same. Since diagonals of a parallelogram bisect each other.
Now let A(-7, -3), B(5, 10), C(15, 8) and D(3, -5)
Then midpoint of diagonal

∴ (1) = (2)
Hence the given are vertices of a parallelogram.

Question 8.
Show that the points (-4, -7), (-1, 2), (8, 5) and (5, -4) taken in order are the vertices of a rhombus. And find its area.
(Hint: Area of rhombus = \(\frac{1}{2}\) × product of its diagonals)
Answer:
Given in ▱ ABCD , A(-4, – 7), B (- 1, 2), C (8, 5) and D (5,-4)


∴  In ▱ ABCD, AB = BC = CD = AD [from sides are equal]
Hence ▱ ABCD is a rhombus.
Area of a rhombus = \(\frac{1}{2}\) d₁d₂
= \(\frac{1}{2}\) × 12√2 × 6√2
= 72 sq. units.

Question 9.
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer.
i) (-1,-2), (1,0), (-1,2), (-3,0)
Answer:
Let A (- 1, -2), B (1, 0), C (- 1, 2), D (- 3, 0) be the given points. Distance formula

In ▱ ABCD, AB = BC = CD = AD – four sides are equal.
AC = BD – diagonals are equal.
Hence, the given points form a square,

ii) (-3, 5), (1, 10), (3, 1), (-1,-4).
Answer:
Let A(-3, 5), B(l,10), C(3, 1), D(-l, -4) then

In ▱ ABCD, \(\overline{\mathrm{AB}}\) = \(\overline{\mathrm{CD}}\) and \(\overline{\mathrm{BC}}\) = \(\overline{\mathrm{AD}}\) (i.e., both pairs of opposite sides are equal) and \(\overline{\mathrm{AC}}\) ≠ \(\overline{\mathrm{BD}}\).
Hence ▱ ABCD is a parallelogram,
i.e., The given points form a parallelogram.

iii) (4, 5), (7, 6), (4, 3), (1, 2).
Answer:
Let A (4, 5), B (7, 6), C (4, 3) and D (1, 2) be the given points.
Distance formula

In ▱ ABCD, AB = CD and BC = AD (i.e., both pairs of opposite sides are equal) and AC ≠ BD.
Hence ▱ ABCD is a parallelogram, i.e., The given points form a parallelogram.

Question 10.
Find the point on the X-axis which is equidistant from (2, -5) and (-2,9).
Answer:
Given points, A (2, – 5), B (- 2, 9).
Let P (x, 0) be the point on X – axis which is equidistant from A and B. i.e., PA = PB.
Distance formula = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)

But PA = PB.
⇒ \(\sqrt{x^{2}-4 x+29}=\sqrt{x^{2}+4 x+85}\)
Squaring on both sides, we get
x² – 4x + 29 = x² + 4x + 85
⇒ – 4x – 4x = 85 – 29
⇒ – 8x = 56
⇒ x = \(\frac{56}{-8}\) = -7
∴ (x, 0) = (- 7, 0) is the point which is equidistant from the given points.

Question 11.
If the distance between two points (x, 7) and (1, 15) is 10, find the value of x.
Answer:
Formula for distance between two points = \(\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}\)
Now distance between (x, 7) and (1,15) is 10.
∴ \(\sqrt{(x-1)^{2}+(7-15)^{2}}\) = 10
∴ (x – l)² + (-8)² = 10²
⇒ (x – l)² = 100 – 64 = 36
∴ x – 1 = √36 = ± 6
∴ x – 1 = 6 or x – 1 = -6
⇒ x = 6 + 1 = 7 or x = -6 + 1 = -5
∴ x = 7 or x = – 5

Question 12.
Find the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.
Answer:
Given: P (2, – 3), Q (10, y) and
\(\overline{\mathrm{PQ}}\) = 10.
Distance formula = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)

⇒ y² + 6y + 73 = 100
⇒ y² + 6y – 27 = 0
⇒ y² + 9y – 3y – 27 = 0
⇒ y (y + 9) – 3 (y + 9) = 0
⇒ (y + 9) (y – 3) = 0
⇒ y + 9 = 0 or y – 3 = 0
⇒ y = -9 or y = 3
⇒ y = – 9 or 3.

Question 13.
Find the radius of the circle whose centre is (3, 2) and passes through (-5,6).
Answer:

Given: A circle with centre A (3, 2) passing through B (- 5, 6).
Radius = AB
[∵ Distance of a point from the centre of the circle]
Distance formula = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)

Question 14.
Can you draw a triangle with vertices (1, 5), (5, 8) and (13, 14)? Give reason.
Answer:
Let A (1, 5), B (5, 8) and C (13, 14) be the given points.
Distance formula

Here, AC = AB + BC.
∴ △ABC can’t be formed with the given vertices.
[∵ Sum of the any two sides of a triangle must be greater than the third side].

Question 15.
Find a relation between x and y such that the point (x, y) is equidistant from the points (-2, 8) and (-3, -5).
Answer:
Let A (- 2, 8), B (- 3, – 5) and P (x, y). If P is equidistant from A, B, then PA = PB.
Distance formula =


Squaring on both sides we get, x² + y² + 4x – 16y + 68
= x² + y² + 6x +10y + 34
⇒ 4x – 16y – 6x – 10y = 34-68
⇒ – 2x – 26y = -34
⇒ x + 13y = 17 is the required condition.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 10 Mensuration Ex 10.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 10th Lesson Mensuration Exercise 10.3

10th Class Maths 10th Lesson Mensuration Ex 10.3 Textbook Questions and Answers

Question 1.
An iron pillar consists of a Cylindrical portion of 2.8 m. height and 20 cm. in diameter and a cone of 42 cm. height surmounting it. Find the weight of the pillar if 1 cm³ of iron weighs 7.5 g.
Answer:
Volume of the iron pillar = Volume of the cylinder + Volume of the cone
Cylinder:

Radius = \(\frac{d}{2}\) = \(\frac{20}{2}\) = 10 cm
Height = 2.8 m = 280 cm
Volume = πr²h
= \(\frac{22}{7}\) × 10 × 10 × 280
= 88000 cm³
Cone:
Radius ‘r’ = \(\frac{d}{2}\) = \(\frac{20}{2}\) = 10 cm
height ‘h’ = 42 cm
Volume = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 10 × 10 × 42
= 4400 cm³
∴ Total volume = 88000 + 4400 = 92400 cm³
∴ Total weight of the pillar at a weight of 7.5 g per 1 cm³ = 92400 × 7.5
= 693000 gms
= \(\frac{693000}{1000}\) kg
= 693 kg.

Question 2.
A toy is made in the form of hemisphere surmounted by a right cone whose circular base is joined with the plane surface of the hemisphere. The radius of the base of the cone is 7 cm. and its volume is 3/2 of the hemisphere. Calculate the height of the cone and the surface area of the toy correct to 2 places of decimal.
(Take π = \(3 \frac{1}{7}\))
Answer:

Given r = 7 cm and
Volume of the cone = \(\frac{3}{2}\) volume of the hemisphere
\(\frac{1}{3}\)πr²h = \(\frac{3}{2}\) × \(\frac{2}{3}\) × πr³
∴ h = 3r
= 3 × 7 = 21 cm
Surface area of the toy = C.S.A. of the cone + C.S.A. of hemisphere
Cone:
Radius (r) = 7 cm
Height (h) = 21 cm
Slant height l = \(\sqrt{r^{2}+h^{2}}\)
= \(\sqrt{7^{2}+21^{2}}\)
= \(\sqrt{49+441}\)
= √490
= 22.135 cm.
∴ C.S.A. = πrl
= \(\frac{22}{7}\) × 7 × 22.135 = 486.990 cm²
Hemisphere:
Radius (r) = 7 cm
C.S.A. = 2πr²
= 2 × \(\frac{22}{7}\) × 7 × 7
= 308 cm²
C.S.A. of the toy = 486.990 + 308 = 794.990 cm²

Question 3.
Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 7 cm.
Answer:
Radius of the cone with the largest volume that can be cut out from a cube of edge 7 cm = \(\frac{7}{2}\) cm

Height of the cone = edge of the cube = 7 cm
∴ Volume of the cone V = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\) × 7
= 89.83 cm³.

Question 4.
A cylindrical tub of radius 5 cm and length 9.8 cm is full of water. A solid in the form of right circular cone mounted on a hemisphere is immersed into the tub. The radius of the hemi¬sphere is 3.5 cm and height of cone outside the hemisphere is 5 cm. Find the volume of water left in the tub. (Take π = \(\frac{22}{7}\))
Answer:

The tub is in the shape of a cylinder, thus
Radius of the cylinder (r) = 5 cm
Length of the cylinder (h) = 9.8 cm
Volume of the cylinder (V) = πr²h
= \(\frac{22}{7}\) × 5 × 5 × 9.8
Volume of the tub = 770 cm³.
Radius of the hemisphere (r) = 3.5 cm
Volume of the hemisphere = \(\frac{2}{3}\)πr³
= \(\frac{2}{3}\) × \(\frac{22}{7}\) × 3.5 × 3.5 × 3.5
= \(\frac{22 \times 12.25}{3}\)
= \(\frac{269.5}{3}\)
Radius of the cone (r) = 3.5 cm
Height of the Cone (h) = 5 cm
Volume of the cone V = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 3.5 × 3.5 × 5
= \(\frac{192.5}{3}\)
Volume of the solid = Volume of the hemisphere + Volume of the cone
= \(\frac{269.5}{3}\) + \(\frac{192.5}{3}\) = \(\frac{462}{3}\) = 154 cm³
Now, when the solid is immersed in the tub, it replaces the water whose volume is equal to volume of the solid itself.
Thus the volume of the water replaced = 154 cm³.
∴ Volume of the water left in the tub = Volume of the tub – Volume of the solid = 770 – 154 = 616 cm³.

Question 5.
In the adjacent figure, the height of a solid cylinder is 10 cm and diameter 7 cm. Two equal conical holes of radius 3 cm and height 4 cm are cut off as shown in the figure. Find the volume of the remaining solid.

Answer:
Volume of the remaining solid = Volume of the given solid – Total volume of the two conical holes
Radius of the given cylinder (r) = \(\frac{d}{2}\) = \(\frac{7}{2}\) = 3.5 cm
Height of the cylinder (h) = 10 cm
Volume of the cylinder (V) = πr²h
= \(\frac{22}{7}\) × 3.5 × 3.5 × 10
= \(\frac{2695}{7}\)
= 385 cm³.
Radius of each conical hole, ‘r’ = 3 cm
Height of the conical hole, h = 4 cm
Volume of each conical hole,
V = \(\frac{1}{3}\)πr²h = \(\frac{1}{3}\) × \(\frac{22}{7}\) × 3 × 3 × 4
= \(\frac{792}{21}\)
= \(\frac{264}{7}\)
Total volume of two conical holes = 2 × \(\frac{264}{7}\) = \(\frac{528}{7}\) cm³
Hence, the remaining volume of the solid

Question 6.
Spherical marbles of diameter 1.4 cm. are dropped into a cylindrical beaker of diameter 7 cm., which contains some water. Find the number of marbles that should be dropped into the beaker, so that water level rises by 5.6 cm.
Answer:
Rise in the water level is seen in cylindrical shape of Radius = Beaker radius
= \(\frac{d}{2}\) = \(\frac{7}{2}\) = 3.5 cm

Height ‘h’ of the rise = 5.6 cm.
∴ Volume of the ‘water rise’ = πr²h
= \(\frac{22}{7}\) × 3.5 × 3.5 × 5.6
= \(\frac{22 \times 12.25 \times 5.6}{7}\)
= 215.6
Volume of each marble dropped = \(\frac{4}{3}\)πr³
Where radius r = \(\frac{d}{2}\) = \(\frac{1.4}{2}\) = 0.7 cm
∴ V = \(\frac{4}{3}\) × \(\frac{22}{7}\) × 0.7 × 0.7 × 0.7
= 1.4373 cm³
∴ Volume of the ‘rise’ = Total volume of the marbles.
Let the number of marbles be ‘n’ then n × volume of each marble = volume of the rise.
n × 1.4373 = 215.6
= \(\frac{215.6}{1.4373}\)
∴ Number of marbles = 150.

Question 7.
A pen stand is made of wood in the shape of cuboid with three conical depressions to hold the pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depression is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand.

Answer:
Volume of the wood in the pen stand = Volume of cuboid – Total volume of three depressions.
Length of the cuboid (l) = 15 cm
Breadth of the cuboid (b) = 10 cm
Height of the cuboid (h) = 3.5 cm
Volume of the cuboid (V) = lbh = 15 × 10 × 3.5 = 525 cm³.
Radius of each depression (r) = 0.5 cm
Height / depth (h) = 1.4 cm
Volume of each depressions V = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 0.5 × 0.5 × 1.4
= \(\frac{7.7}{3 \times 7}\) = \(\frac{1.1}{3}\) cm³
Total volume of the three depressions = 3 × \(\frac{1.1}{3}\)
= 1.1 cm³
∴ Volume of the wood = 525 – 1.1 = 523.9 cm³

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 8 Similar Triangles Ex 8.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 8th Lesson Similar Triangles Exercise 8.3

10th Class Maths 8th Lesson Similar Triangles Ex 8.3 Textbook Questions and Answers

Question 1.
Equilateral triangles are drawn on the three sides of a right angled triangle. Show that the area of the triangle on the hypotenuse is equal to the sum of the areas of triangles on the other two sides.
Answer:

Let △PQR is a right angled triangle, ∠Q = 90°
Let PQ = a, QR b and
PR = hypotenuse = c
Then from Pythagoras theorem we can
say a² + b² = c² ……… (1)
△PSR is an equilateral triangle drawn on hypotenuse
∴ PR = PS = RS = c,
Then area of triangle on hypotenuse
= \(\frac{\sqrt{3}}{4}\)c² ……… (2)
△QRU is an equilateral triangle drawn on the side ‘QR’ = b
∴ QR = RU = QU = b
Then area of equilateral triangle drawn on the side = \(\frac{\sqrt{3}}{4}\)b² …….. (3)
△PQT is an equilateral triangle drawn on another side ‘PQ’ = a
∴ PQ = PT = QT = a
Area of an equilateral triangle drawn an another side ‘PQ’ = \(\frac{\sqrt{3}}{4}\)a² …….. (4)
Now sum of areas of equilateral triangles on the other two sides

= Area of equilateral triangle on the hypotenuse.
Hence Proved.

Question 2.
Prove that the area of the equilateral triangle described on the side of a square is half the area of the equilateral triangles described on its diagonal.
Answer:

Let PQRS is square whose side is ‘a’ units then PQ = QR = RS = SP = ‘a’ units.
Then the diagonal
\(\overline{\mathrm{PR}}\) = \(\sqrt{a^{2}+a^{2}}\) = a√2 units.
Let △PRT is an equilateral triangle, then PR = RT = PT = a√2
∴ Area of equilateral triangle constructed on diagonal

Let △QRZ is another equilateral triangle whose sides are
\(\overline{\mathrm{QR}}\) = \(\overline{\mathrm{RZ}}\) = \(\overline{\mathrm{QZ}}\) = ‘a’ units
Then the area of equilateral triangle constructed on one side of square = \(\frac{\sqrt{3}}{4}\)a² ……. (2)
∴ \(\frac{1}{2}\) of area of equilateral triangle on diagonal = \(\frac{1}{2}\left(\frac{\sqrt{3}}{2} a^{2}\right)\) = \(\frac{\sqrt{3}}{4}\)a² = area of equilateral triangle on the side of square.
Hence Proved.

Question 3.
D, E, F are midpoints of sides BC, CA, AB of △ABC. Find the ratio of areas of △DEF and △ABC.
Answer:

Given in △ABC D, E, F are the midpoints of the sides BC, CA and AB.
In △ABC, EF is the line join of mid-points of two sides AB and AC of △ABC.
Thus FE || BC [∵ \(\frac{AF}{FB}\) = \(\frac{AE}{EC}\) Converse of B.P.T.]
Similarly DE divides AC and BC in the same ratio, i.e., DE || AB.
Now in □ BDEF, both pairs of opposite sides (BD || EF and DE || BF) are parallel.
Hence □ BDEF is a parallelogram where DF is a diagonal.
∴ △BDF ≃ △DEF ……… (1)
Similarly we can prove that
△DEF ≃ △CDE ……… (2)
[∵ CDFE is a parallelogram]
Also, △DEF ≃ △AEF …….. (3)
[∵ □ AEDF is a parallelogram]
From (1), (2) and (3)
△AEF ≃ △DEF ≃ BDF ≃ △CDE
Also, △ABC = △AEF + △DEF + △BDF + △CDE = 4 . △DEF
Hence, △ABC : △DEF = 4 : 1.

Question 4.
In △ABC, XY || AC and XY divides the triangle into two parts of equal area. Find the ratio of \(\frac{AX}{XB}\).
Answer:

Given: In △ABC, XY || AC.
XY divides △ABC into two points of equal area.
In △ABC, △XBY
∠B = ∠B
∠A = ∠X
∠C = ∠Y
[∵ XY || AC; (∠A, ∠X) and ∠C, ∠Y are the pairs of corresponding angles]
Thus △ABC ~ △XBY by A.A.A similarity condition.
Hence \(\frac{\Delta \mathrm{ABC}}{\Delta \mathrm{XBY}}=\frac{\mathrm{AB}^{2}}{\mathrm{XB}^{2}}\)
[∵ The ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides]

⇒ \(\frac{AX}{XB}\) + 1 = √2
⇒ \(\frac{AX}{XB}\) = √2 – 1
Hence the ratio \(\frac{AX}{XB}\) = \(\frac{√2 – 1}{1}\).

Question 5.
Prove that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
Answer:

Given: △ABC ~ △XYZ
R.T.P: \(\frac{\Delta \mathrm{ABC}}{\Delta \mathrm{XYZ}}=\frac{\mathrm{AD}^{2}}{\mathrm{XW}^{2}}\)
Proof : We know that the ratio of areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

Hence the ratio of areas of two similar triangles is equal to the squares of ratio of their corresponding medians.

Question 6.
△ABC ~ △DEF. BC = 3 cm, EF = 4 cm and area of △ABC = 54 cm². Determine the area of △DEF.
Answer:

Given: △ABC ~ △DEF
BC = 3 cm
EF = 4 cm
△ABC = 54 cm²
∴ △ABC ~ △DEF, we have
\(\frac{\Delta \mathrm{ABC}}{\Delta \mathrm{DEF}}=\frac{\mathrm{BC}^{2}}{\mathrm{EF}^{2}}\)
[∵ The ratio of two similar triangles is equal to the ratio of the squares of the corresponding sides].
\(\frac{54}{\Delta \mathrm{DEF}}\) = \(\frac{3^{2}}{4^{2}}\)
∴ △DEF = \(\frac{54 \times 16}{9}\) = 96 cm²

Question 7.
ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1 cm and BP = 3 cm, AQ =1.5 cm, CQ = 4.5 cm. Prove that area of △APQ = \(\frac{1}{16}\) (area of △ABC).
Answer:

Given: △ABC and \(\overline{\mathrm{PQ}}\) – a line segment meeting AB in P and AC in Q.
AP = 1 cm; AQ =1.5 cm;
BP = 3 cm; CQ = 4.5 cm.
\(\frac{AP}{PQ}\) = \(\frac{1}{3}\) ……… (1)
\(\frac{AQ}{QC}\) = \(\frac{1.5}{4.5}\) = \(\frac{1}{3}\) ……..(2)
From (1) and (2),
\(\frac{AP}{BP}\) = \(\frac{AQ}{CQ}\)
[i.e., PQ divides AB and AC in the same ratio – By converse of Basic pro-portionality theorem]
Hence, PQ || BC.
Now in △APQ and △ABC
∠A = ∠A (Common)
∠P = ∠B [∵ Corresponding angles for the parallel lines PQ and BC]
∠Q = ∠C
∴ △APQ ~ △ABC [∵ A.A.A similarity condition]
Now, \(\frac{\Delta \mathrm{APQ}}{\Delta \mathrm{ABC}}=\frac{\mathrm{AP}^{2}}{\mathrm{AB}^{2}}\)
[∵ Ratio of two similar triangles is equal to the ratio of the squares of their corresponding sides].
= \(\frac{1^{2}}{(3+1)^{2}}\) = \(\frac{1}{16}\) [∵ AB = AP + BP = 1 + 3 = 4 cm]
∴ △APQ = \(\frac{1}{16}\) (area of △ABC) [Q.E.D]

Question 8.
The areas of two similar triangles are 81 cm² and 49 cm² respectively. If the altitude of the bigger triangle is 4.5 cm. Find the corresponding altitude of the smaller triangle.

Answer:
Given: △ABC ~ △DEF
△ABC = 81 cm²
△DEF = 49 cm²
AX = 4.5 cm
To find: DY
We know that,
\(\frac{\Delta \mathrm{ABC}}{\Delta \mathrm{DEF}}=\frac{\mathrm{AX}^{2}}{\mathrm{DY}^{2}}\)
[∵ Ratio of areas of two similar triangles is equal to ratio of the squares of their corresponding altitudes]

∴ DY = 3.5 cm.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 10 Mensuration Ex 10.4 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 10th Lesson Mensuration Exercise 10.4

10th Class Maths 10th Lesson Mensuration Ex 10.4 Textbook Questions and Answers

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 the cylinder.
Answer:
Given, sphere converted into cylinder.
Hence volume of the sphere = volume of the cylinder.
Sphere:
Radius, r = 4.2 cm
Volume V = \(\frac{4}{3}\)πr³
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × 4.2 × 4.2 × 4.2
= 4 × 22 × 0.2 × 4.2 × 4.2
= 4 x 22 x 0.2 x 4.2 x 4.2
= 310.464
Cylinder:
Radius, r = 6 cm
Height h = h say
Volume = πr²h
= \(\frac{22}{7}\) × 6 × 6 × h
= \(\frac{22 \times 36}{7} h\)
= \(\frac{792}{7} h\)
Hence, \(\frac{792}{7} h\) = 310.464
h = \(\frac{310.464 \times 7}{792}\) = 2.744cm
!! π can be cancelled on both sides i.e., sphere = cylinder

Question 2.
Three metallic spheres of radii 6 cm., 8 cm. and 10 cm. respectively are melted together to form a single solid sphere. Find the radius of the resulting sphere.
Answer:
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 ‘r’.
Then volume of the resultant sphere = sum of the volumes of the three small spheres.


∴ 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 20 m deep well with diameter 7 m. is dug and the earth got by digging is evenly spread out to form a rectangu¬lar platform of base 22 m. × 14 m. Find the height of the platform.
Answer:
Volume of earth taken out = πr²h
= \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\) × 20
= 770 m
Let height of plot form = H m.
∴ 22 × 14 × H = \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\) × 20
H = \(\frac{35}{14}\) = \(\frac{5}{2}\) = \(2 \frac{1}{2} \mathrm{~m}\)
∴ The height of the plat form is \(2 \frac{1}{2} \mathrm{~m}\)

Question 4.
A well of diameter 14 m. is dug 15 m. deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 7 m to form an embankment. Find the height of the embankment
Answer:
Volume of the well = Volume of the embank
Volume of the cylinder = Volume of the embank
Cylinder :
Radius r = \(\frac{d}{2}\) = \(\frac{14}{2}\) = 7 cm
Height/depth, h = 15 m
Volume V = πr²h
= \(\frac{22}{7}\) × 7 × 7 × 15
= 22 × 7 × 15
= 2310 m³
Embank:

Let the height of the embank = h m
Inner radius ‘r’ = Radius of well = 7 m
Outer radius, R = inner radius + width
= 7m + 7m = 14 m
Area of the base of the embank = (Area of outer circle) – (Area of inner circle)
= πR² – πr²
= π(R² – r²)
= \(\frac{22}{7}\)\(\left(14^{2}-7^{2}\right)\)
= \(\frac{22}{7}\) × (14+7) × (14-7)
= \(\frac{22}{7}\) × 21 × 7
= 462 m²
∴ Volume of the embank = Base area × height
= 462 × h = 462 h m³
∴ 462 h m³ = 2310 m³
h = \(\frac{2310}{462}\) = 5 m.

Question 5.
A container shaped like a right circular cylinder having diameter 12 cm. and height 15 cm. 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 the number of such cones which can be filled with ice-cream.
Answer:
Let the number of cones that can be filled with the ice-cream be ‘n’.
Then total volume of all the cones with a hemi spherical top = Volume of the ice-cream

Ice-cream cone = Cone + Hemisphere = πr²h

Cone:
Radius = \(\frac{d}{2}\) = \(\frac{6}{2}\) = 3 cm
Height, h = 12 cm
Volume V = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 3 × 3 × 12
= \(\frac{22}{7}\) × 36
= \(\frac{792}{7}\)
Hemisphere:
Radius = \(\frac{d}{2}\) = \(\frac{6}{2}\) = 3 cm
Volume V = \(\frac{2}{3}\)πr³
= \(\frac{2}{3}\) × \(\frac{22}{7}\) × 3 × 3 × 3
= \(\frac{44 \times 9}{7}\)
= \(\frac{396}{7}\)
∴ Volume of each cone with ice-cream = \(\frac{792}{7}\) + \(\frac{396}{7}\) = \(\frac{1188}{7}\) cm³
Cylinder:
Radius = \(\frac{d}{2}\) = \(\frac{12}{2}\) = 6 cm
Height, h = 15 cm
Volume V = πr²h
= \(\frac{22}{7}\) × 6 × 6 × 15
= \(\frac{22 \times 36 \times 15}{7}\)
= \(\frac{11880}{7}\)
∴ \(\frac{11880}{7}\) = n × \(\frac{11880}{7}\)
⇒ n = \(\frac{11880}{7}\) × \(\frac{7}{1188}\) = 10
∴ n = 10.

Question 6.
How many silver coins, 1.75 cm in diameter and thickness 2 mm., need to be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?
Answer:
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 coin is a cylinder and V = πr²h]

∴ 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, 1/4 of the water flows out. Find the number of lead shots dropped into the vessel.
Answer:

Let the number of lead shots dropped = n
Then total volume of n lead shots = \(\frac{1}{4}\) volume of the conical vessel.
Lead shots:
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
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 200

∴ 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 \(\frac{d}{2}\) cm and height 3 cm. Find the number of cones so formed.
Answer:
Let the no. of small cones = n Then,
total volume of n cones = Volume of sphere Diameter = 28 cm.
Cones:
Radius r = \(\frac{d}{2}\)

Height, h = 3 cm

Total volume of n-cones = n . \(\frac{154}{9}\) cm³
Sphere:
Radius = \(\frac{d}{2}\) = \(\frac{28}{2}\) = 14 cm

No. of cones formed = 672.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 9 Tangents and Secants to a Circle Ex 9.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 9th Lesson Tangents and Secants to a Circle Exercise 9.3

10th Class Maths 9th Lesson Tangents and Secants to a Circle Ex 9.3 Textbook Questions and Answers

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
Answer:

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 = \(\frac{x}{360}\) × πr²
\(\frac{90}{360}\) × 3.14 × 10 × 10 = 78.5
Area of the triangle = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × 10 × 10 = 50
∴ Area of the minor segment = 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 √3 = 1.732)
Answer:
Radius of the circle r = 12 cm.
Area of the sector = \(\frac{x}{360}\) × πr²
Here, x = 120°

\(\frac{120}{360}\) × 3.14 × 12 × 12 = 150.72
Drop a perpendicular from ‘O’ to the chord PQ.
△OPM = △OQM [∵ OP = OQ ∠P = ∠Q; angles opp. to equal sides OP & OQ; ∠OMP = ∠OMQ by A.A.S]
∴ △OPQ = △OPM + △OQM = 2 . △OPM
Area of △OPM = \(\frac{1}{2}\) × PM × OM

= 18 × 1.732 = 31.176 cm
∴ △OPQ = 2 × 31.176 = 62.352 cm²
∴ Area of the minor segment
= (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 25 cm. sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades. (use π = \(\frac{22}{7}\))
Answer:
Angle made by the each blade = 115°
Total area swept by two blades
= Area of the sector with radius 25 cm and angle 115°+ 115° = 230°
= Area of the sector = \(\frac{x}{360}\) × πr²
= \(\frac{230}{360}\) × \(\frac{22}{7}\) × 25 × 25
= 1254.96
≃  1255 cm²

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

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

Area of I + Area of II
= Area of ABCD – Areas of two semicircles with radius 5 cm
= 10 × 10 – 2 × \(\frac{1}{2}\) × π × 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 = Area of ABCD – Area of unshaded region
= 100 – 2 × 21.5 = 100 – 43 = 57 cm²

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

Answer:
Given,
ABCD is a square of side 7 cm.
Area of the shaded region = Area of ABCD – Area of two semicircles with radius \(\frac{7}{2}\) = 3.5 cm
APD and BPC are semicircles.
= 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 cm

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 = \(\frac{22}{7}\)).
Answer:
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 △BOD

= 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 21 cm. and 7 cm. with centre O (See figure). If ∠AOB = 30°, find the area of the shaded region. (use π = \(\frac{22}{7}\)).

Answer:
Given, AB and CD are the arcs of two concentric circles.
Radii of circles = 21 cm and 7 cm and ∠AOB = 30°
We know that,
Area of the sector = \(\frac{x}{360}\) × πr²
Area of the shaded region = Area of the OAB – Area of OCD

∴ Area of shaded region = 102.66 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)
Answer:
Mark two points P, Q on the either arcs.
Let BD be a diagonal of ABCD
Now the area of the segment

= 28.5 + 28.5 = 57 cm²

Side of the square = 10 cm
Area of the square = side × side
= 10 × 10 = 100 cm²
Area of two sectors with centres A and C and radius 10 cm.
= 2 × \(\frac{\pi r^{2}}{360}\) × x = 2 × \(\frac{x}{360}\) × \(\frac{22}{7}\) × 10 × 10
= \(\frac{1100}{7}\)
= 157.14 cm²
∴ Designed area is common to both the sectors,
∴ Area of design = Area of both sectors – Area of square
= 157 – 100 = 57 cm²
(or)
\(\frac{1100}{7}\) – 100 = \(\frac{1100-700}{7}\)
= \(\frac{400}{7}\)
= 57.1 cm²

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 10 Mensuration Ex 10.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 10th Lesson Mensuration Exercise 10.2

10th Class Maths 10th Lesson Mensuration Ex 10.2 Textbook Questions and Answers

Question 1.
A toy is in the form of a cone mounted on a hemisphere. The diameter of the base and the height of the cone are 6 cm and 4 cm respectively. Determine the surface area of the toy. (Use π = 3.14)
Answer:

Diameter of the base of the cone d = 6 cm.
∴ Radius of the base of the cone
r = \(\frac{d}{2}\) = \(\frac{6}{2}\) = 3 cm
Height of the cone = h = 4 cm
Slant height of the cone l = \(\sqrt{r^{2}+h^{2}}\)
= \(\sqrt{3^{2}+4^{2}}\)
= \(\sqrt{9+16}\)
= √25
= 5 cm
∴ C.S.A of the cone = πrl
= \(\frac{22}{7}\) × 3 × 5
= \(\frac{330}{7}\) cm²
Radius of the hemisphere = \(\frac{d}{2}\) = \(\frac{6}{2}\) = 3 cm
C.S.A. of the hemisphere = 2πr²
= 2 × \(\frac{22}{7}\) × 3 × 3
= \(\frac{396}{7}\)
Hence the surface area of the toy = C.S.A. of cone + C.S.A. of hemisphere
= \(\frac{330}{7}\) + \(\frac{396}{7}\)
= \(\frac{726}{7}\) ≃ 103.71 cm².

Question 2.
A solid is in the form of a right circular cylinder with a hemisphere at one end and a cone at the other end. The radius of the common base is 8 cm and the heights of the cylindrical and conical portions are 10 cm and 6 cm respectively. Find the total surface area of the solid. [Use π = 3.14]
Answer:
Total surface area = C.S.A. of the cone + C.S.A. of cylinder + C.S.A of the hemisphere.

Cone:
Radius (r) = 8 cm
Height (h) = 6 cm
Slant height l = \(\sqrt{r^{2}+h^{2}}\)
= \(\sqrt{8^{2}+6^{2}}\)
= \(\sqrt{64+36}\)
= √100
= 10 cm
C.S.A. = πrl
= \(\frac{22}{7}\) × 8 × 10
= \(\frac{1760}{7}\) cm²
Cylinder:
Radius (r) = 8 cm;
Height (h) = 10 cm
C.S.A. = 2πrh
= 2 × \(\frac{22}{7}\) × 8 × 10
= \(\frac{3520}{7}\) cm²
Hemisphere:
Radius (r) = 8 cm
C.S.A. = 2πr²
= 2 × \(\frac{22}{7}\) × 8 × 8
= \(\frac{2816}{7}\) cm²
∴ Total surface area of the given solid
= \(\frac{1760}{7}\) + \(\frac{3520}{7}\) + \(\frac{2816}{7}\)
T.S.A. = \(\frac{8096}{7}\) = 1156.57 cm².

Question 3.
A medicine capsule is ih the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the capsule is 14 mm. and the width is 5 mm. Find its surface area.
Answer:

Surface area of the capsule = C.S.A. of 2 hemispheres + C.S.A. of the cylinder
i) Now for Hemisphere:
Radius (r) = \(\frac{d}{2}\) = \(\frac{5}{2}\) = 2.5 mm
C.S.A of each hemisphere = 2πr²
C.S.A of two hemispheres
= 2 × 2πr² = 4πr²
= 2 × \(\frac{22}{7}\) × \(\frac{5}{2}\) × \(\frac{5}{2}\)
= \(\frac{550}{7}\)
= 78.57 mm².

ii) Now for Cylinder:
Length of capsule = AB =14 mm
Then height (length) cylinder part = 14 – 2(2.5)
h = 14 – 5 = 9 mm
Radius of cylinder part (r) = \(\frac{5}{2}\)
Now C.S.A of cylinder part = 2πrh
= 2 × \(\frac{22}{7}\) × \(\frac{5}{2}\) × 9
= \(\frac{900}{7}\)
= 141.428 mm²
Now total surface area of capsule
= 78.57 + 141.43 = 220 mm²

Question 4.
Two cubes each of volume 64 cm³ are joined end to end together. Find the surface area of the resulting cuboid.
Answer:
Given, volume of the cube.
V = a³ = 64 cm³
∴ a³ = 4 × 4 × 4 = 4³ , Hence a = 4 cm
When two cubes are added, the length of cuboid = 2a = 2 × 4 = 8 cm,
breadth = a = 4 cm.
height = a = 4 cm is formed.

∴ T.S.A. of the cuboid
= 2 (lb + bh + lh)
= 2(8 × 4 + 4 × 4 + 8 × 4)
= 2(32 + 16 + 32)
= 2 × 80
= 160 cm²
∴ The surface area of resulting cuboid is 160 cm².

Question 5.
A storage tank consists of a circular cylinder with a hemisphere stuck on either end. If the external diameter of the cylinder be 1.4 m. and its length be 8 m. Find the cost of painting it on the outside at rate of Rs. 20 per m².
Answer:
Total surface area of the tank = 2 × C.S.A. of hemisphere + C.S.A. of cylinder.

Hemisphere:
Radius (r) = \(\frac{d}{2}\) = \(\frac{1.4}{2}\) = 0.7 m
C.S.A. of hemisphere = 2πr²
= 2 × \(\frac{22}{7}\) × 0.7 × 0.7
= 3.08 m².
2 × C.S.A. = 2 × 3.08 m² = 6.16 m²
Cylinder:
Radius (r) = \(\frac{d}{2}\) = \(\frac{1.4}{2}\) = 0.7 m
Height (h) = 8 m
C.S.A. of the cylinder = 2πrh
= 2 × \(\frac{22}{7}\) × 0.7 × 8
= 35.2 m²
∴ Total surface area of the storage tank = 35.2 + 6.16 = 41.36 m²
Cost of painting its surface area @ Rs. 20 per sq.m, is
= 41.36 × 20 = Rs. 827.2.

Question 6.
A hemisphere is cut out from one face of a cubical wooden block such that the diameter of the hemisphere is equal to the length of the cube. Determine the surface area of the remaining solid.
Answer:
Let the length of the edge of the cube = a units

T.S.A. of the given solid = 5 × Area of each surface + Area of hemisphere
Square surface:
Side = a units
Area = a² sq. units
5 × square surface = 5a² sq. units
Hemisphere:
Diameter = a units;
Radius = \(\frac{a}{2}\)
C.S.A. = 2πr²
= 2π\(\left(\frac{a}{2}\right)^{2}\)
= 2π\(\frac{a^{2}}{4}\) = \(\frac{\pi \mathrm{a}^{2}}{2}\) sq. units
Total surface area = 5a² + \(\frac{\pi \mathrm{a}^{2}}{2}\) = a²\(\left(5+\frac{\pi}{2}\right)\) sq. units.

Question 7.
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure. If the height of the cylinder is 10 cm and its base radius is of 3.5 cm, find the total surface area of the article.

Answer:
Surface area of the given solid = C.S.A. of the cylinder + 2 × C.S.A. of hemisphere.
If we take base = radius
Cylinder:
Radius (r) = 3.5 cm
Height (h) = 10 cm
C.S.A. = 2πrh
= 2 × \(\frac{22}{7}\) × 3.5 × 10
= 220 cm²
Hemisphere:
Radius (r) = 3.5 cm
C.S.A. = 2πr²
= 2 × \(\frac{22}{7}\) × 3.5 × 3.5
= 77 cm²
2 × C.S.A. = 2 × 77 = 154 cm²
∴ T.S.A. = 220 + 154 = 374 cm².