Class 10

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 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 6.2 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 6.1 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 7 Coordinate Geometry Ex 7.4 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 7th Lesson Coordinate Geometry Exercise 7.4

10th Class Maths 7th Lesson Coordinate Geometry Ex 7.4 Textbook Questions and Answers

Question 1.
Find the slope of the line joining the two given points.
i) (4,-8) and (5,-2).
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{-2+8}{5-4}\)
= 6

ii) (0, 0) and (√3, 3)
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{3-0}{\sqrt{3}-0}\)
= \(\frac{3}{\sqrt{3}}\)
= \(\frac{\sqrt{3} \times \sqrt{3}}{\sqrt{3}}\)
= √3

iii) (2a, 3b) and (a, -b).
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{-b-3b}{a-2a}\)
= \(\frac{-4b}{-a}\)
= \(\frac{4b}{a}\)

iv) (a, 0) and (0, b).
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{b-0}{0-a}\)
= \(\frac{-b}{a}\)

v) A (-1.4, -3.7), B (-2.4, 1.3).
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{1.3+3.7}{-2.4+1.4}\)
= \(\frac{5.0}{-1}\)
= -5

vi) A (3, -2), B (-6, -2).
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{-2+2}{-6-3}\)
= 0

vii) A (-3\(\frac{1}{2}\), 3), B (-7, 2\(\frac{1}{2}\)).
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{2 \frac{1}{2}-3}{-7+3 \frac{1}{2}}\)
= \(\frac{-\frac{1}{2}}{-3 \frac{1}{2}}\)
= \(\frac{1}{2}\) × \(\frac{2}{7}\)
= \(\frac{1}{7}\)

viii) A(0, 4), B(4, 0)
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{0-4}{4-0}\)
= \(\frac{-4}{4}\)
= -1

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 7 Coordinate Geometry Ex 7.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 7th Lesson Coordinate Geometry Exercise 7.3

10th Class Maths 7th Lesson Coordinate Geometry Ex 7.3 Textbook Questions and Answers

Question 1.
Find the area of the triangle whose vertices are
i) (2, 3), (-1, 0), (2,-4)
Answer:
Given: A (2, 3), B (- 1, 0) and C (2, – 4) are the vertices of a △ABC.
Area of the triangle ABC = \(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\)
= \(\frac{1}{2}|2(0+4)-1(-4-3)+2(3-0)|\)
= \(\frac{1}{2}|8+7+6|\)
= \(\frac{21}{2}\)
= 10\(\frac{1}{2}\) sq.units

ii) (-5, -1), (3, -5), (5, 2)
Answer:
Given: A (- 5, – 1), B (3, – 5) and C (5, 2) are the vertices of △ABC.
Area of the △ABC
= \(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\)
= \(\frac{1}{2}|-5(-5-2)+3(2+1)+5(-1+5)|\)
= \(\frac{1}{2}|35+9+20|\)
= \(\frac{64}{2}\)
= 32 sq.units

iii) (0, 0), (3, 0), (0, 2)
Answer:
Given: O (0, 0), A (3, 0) and B (0, 2) are the vertices of a triangle, △AOB.
Area of the △AOB
= \(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\)
= \(\frac{1}{2}|0(0-2)+3(2-0)+0(0-0)|\)
= \(\frac{1}{2}|6|\)
= 3 sq.units

Or

△AOB = \(\frac{1}{2}\) × OA × OB
= \(\frac{1}{2}\) × 3 × 2
= 3 sq.units

Question 2.
Find the value of ‘K’ for which the points are collinear.
i) (7, -2), (5, 1), (3, K)
Answer:
Given: A (7, – 2), B (5, 1) and C (3, K) are collinear.
∴ Area of △ABC = 0
But area of triangle
\(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\)
⇒ \(\frac{1}{2} \mid 7(1-\mathrm{K})+5(\mathrm{~K}+2)+3(-2-1)\) = 0
⇒ \(|7-7 K+5 K+10-9|\) = 0
⇒ \(|-2 \mathrm{~K}+8|\) = 0
⇒ -2K + 8 = 0
⇒ -2K = -8
⇒ K = \(\frac{8}{2}\)
i.e., K = 4

ii) (8, 1), (K,-4), (2,-5)
Answer:
Given: A (8, 1), B(K, – 4) and C (2, – 5) are collinear.
∴ Area of △ABC = 0
⇒ \(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\) = 0
⇒ \(\frac{1}{2}|8(-4+5)+\mathrm{K}(-5-1)+2(1+4)|\) = 0
⇒ \(|8-6 \mathrm{~K}+10|\) = 0
⇒ \(|18-6 \mathrm{~K}|\) = 0
⇒ 18 – 6K = 0
⇒ 6K = 18
⇒ K = \(\frac{18}{6}\)
i.e., K = 3

iii) (K,K), (2, 3), and (4,-1)
Answer:
A (K, K), B (2, 3) and C (4, – 1) are collinear.
∴ Area of △ABC = 0
⇒ \(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\) = 0
⇒ \(\frac{1}{2}|\mathrm{~K}(3+1)+2(-1-\mathrm{K})+4(\mathrm{~K}-3)|\) = 0
⇒ \(|4K-2-2K+4K-12|\) = 0
⇒ \(|6 \mathrm{~K}-14|\) = 0
⇒ 6K – 14 = 0
⇒ 6K = 14
⇒ K = \(\frac{14}{6}\) = \(\frac{7}{3}\)
∴ K = \(\frac{7}{3}\)

Question 3.
Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.
Answer:

Given: A (0, – 1), B (2, 1) and C (0, 3) are the vertices of △ABC.
Let D, E and F be the midpoints of the sides \(\overline{\mathrm{AB}}\), \(\overline{\mathrm{BC}}\) and \(\overline{\mathrm{AC}}\).

Area of a triangle ABC =
\(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\)
= \(\frac{1}{2}|0(1-3)+2(3+1)+0(-1-1)|\)
= \(\frac{1}{2}|8|\)
= 4 sq.units
Area of △DEF = \(\frac{1}{2}|1(2-1)+1(1-0)+0(0-2)|\)
= \(\frac{1}{2}|1+1|\)
= \(\frac{2}{2}\)
= 1 sq.units
Ratio of areas = △ABC : △DEF = 4 : 1.
△ADF ≅ △BED ≅ △DEF ≅ △CEF
∴ △ABC : △DEF = 4 : 1

Question 4.
Find the area of the quadrilateral whose vertices taken inorder are (-4, -2), (-3, -5),(3, -2) and (2, 3).
Answer:

Given: A (- 4, – 2), B (- 3, – 5), C (3, – 2) and D (2, 3) are the vertices of the quadrilateral ▱ ABCD.
Area of ▱ ABCD = △ABC + △ACD.
Area of a triangle =
\(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\)

Question 5.
Find the area of the triangle formed by the points by using Heron’s formula.
i) (1, 1), (1, 4) and (5, 1)
ii) (2, 3), (-1,3) and (2, -1)
Answer:
i) (1, 1) (1, 4) and (5, 1)
let A (1, 1) B(l, 4) and C(5, 1) are the vertices then length of sides can be calculated using the formula
\(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
now

now formula for area of triangle using Heron’s formula = △ = \(\sqrt{s(s-a)(s-b)(s-c)}\)
where s = \(\frac{a+b+c}{2}\)
∴ s = \(\frac{3+4+5}{2}\) = \(\frac{12}{2}\) = 6
∴ △ = \(\sqrt{6(6-5)(6-4)(6-3)}\)
= \(\sqrt{6 \times 1 \times 2 \times 3}\)
= \(\sqrt{6 \times 6}\)
= 6 sq. units
∴ area of given triangle = 6 sq units

ii) let the vertices of given triangle A (2, 3), B (-l, 3) and C (2, -1)

∴ a = 5, b = 4, c = 3 units
now from using Heron’s formula area of triangle
= △ = \(\sqrt{s(s-a)(s-b)(s-c)}\)
where s = \(\frac{a+b+c}{2}\)
= \(\frac{5+4+3}{2}\)
= \(\frac{12}{2}\) = 6
∴ △ = \(\sqrt{6(6-5)(6-4)(6-3)}\)
= \(\sqrt{6 \times 1 \times 2 \times 3}\)
= \(\sqrt{6 \times 6}\)
= 6 sq. units
∴ area of given triangle = 6 sq units

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 7 Coordinate Geometry Ex 7.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 7th Lesson Coordinate Geometry Exercise 7.2

10th Class Maths 7th Lesson Coordinate Geometry Ex 7.2 Textbook Questions and Answers

Question 1.
Find the coordinates of the point which divides the line segment joining the points (-1, 7) and (4, -3) in the ratio 2 :3.
Answer:
Given points P (-1, 7) and Q (4, – 3). Let ‘R’ be the required point which divides \(\overline{\mathrm{PQ}}\) in the ratio 2:3. Then

Question 2.
Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).
Answer:
Given points A (4, – 1) and B (- 2, – 3) Let P and Q be the points of trisection
of \(\overline{\mathrm{AB}}\), then AP = PQ = QB.

∴ P divides \(\overline{\mathrm{AB}}\) internally in the ratio 1 : 2.

Also, Q divides \(\overline{\mathrm{AB}}\) in the ratio 2 : 1 internally.

Question 3.
Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
Answer:
Let the point (-1, 6) divides the line segment joining the points (-3, 10) and (6, -8) in a ratio of m : n

⇒ 6m – 3n = -(m + n) = -m – n
⇒ 6m + m = – n + 3n
⇒ 7m = 2n
⇒ \(\frac{m}{n}=\frac{2}{7}\)
⇒ m : n = 2 : 7
∴ The point (-1, 6) divides the given line segment in a ratio of 2 : 7.

Question 4.
If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
Answer:

Given: ▱ ABCD is a parallelogram where A (1, 2), B (4, y), C (x, 6) and D (3, 5).
In a parallelogram, diagonals bisect each other.
i.e., the midpoints of the diagonals coincide with each other.
i.e.,midpoint of AC = midpoint of BD

⇒ 1 + x = 7 and 8 = y + 5
⇒ x = 7 – 1 and y = 8 – 5
∴ x = 6 and y = 3.

Question 5.
Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, -3) and B is (1, 4).
Answer:

Given:
A circle with centre ‘C’ (2, -3). \(\overline{\mathrm{AB}}\) is a diameter where
B = (1, 4); A = (x, y).
C is the midpoint of AB.
[∵ Centre of a circle is the midpoint of the diameter]

4 = x + 1 and – 6 = y + 4
⇒ x = 4 – 1 = 3 and y = -6 – 4 = -10
A (x, y) = (3, -10)

Question 6.
If A and B are (-2, -2) and (2, -4) respectively. Find the coordinates of P such that AP = \(\frac{3}{7}\) AB and P lies on the segment AB.
Answer:
Given: A (- 2, – 2) and B (2, – 4)
P lies on AB such that AP = latex]\frac{3}{7}[/latex] AB

i.e., P divides \(\overline{\mathrm{AB}}\) in the ratio 3 : 4 By section formula,

Question 7.
Find the coordinates of points which divide the line segment joining A (-4, 0) and B (0, 6) into four equal parts.
Answer:
Given, A (- 4, 0) and B (0, 6).
Let P, Q and R be the points which divide AB into four equal parts.

P divides \(\overline{\mathrm{AB}}\) in the ratio 1 : 3, Q → 1 : 1 and R → 3 : 1 Use section formula to find P, Q and R.
Then, Q is the midpoint of \(\overline{\mathrm{AB}}\)
P is the midpoint of \(\overline{\mathrm{AQ}}\)
R is the midpoint of \(\overline{\mathrm{QB}}\)

Question 8.
Find the coordinates of the points which divides the line segment joining A(-2, 2) and B(2, 8) into four equal parts.
Answer:
Given, A (- 2, 2) and B (2, 8).
Let P, Q and R be the points which divide AB into four equal parts.

Then, Q is the midpoint of \(\overline{\mathrm{AB}}\)
P is the midpoint of \(\overline{\mathrm{AQ}}\)
R is the midpoint of \(\overline{\mathrm{QB}}\)

Question 9.
Find the coordinates of the point which divide the line segment joining the points (a + b, a-b) and (a-b, a + b) in the ratio 3 : 2 internally.
Answer:
Given : A (a + b, a – b) and B (a – b, a + b).
Let P (x, y) divides \(\overline{\mathrm{AB}}\) in the ratio 3 : 2 internally.
Section formula

Question 10.
Find the coordinates of centroid of the triangle with following vertices:
i) (-1, 3), (6, -3) and (-3, 6)
Answer:
Given: △ABC in which- A (- 1, 3), B (6, -3) and C (-3, 6)

ii) (6, 2), (0, 0) and (4, -7)
Answer:
Given: The three vertices of a triangle are A (6, 2), B (0, 0) and C (4, – 7).
Centroid (x, y)

iii) (1,-1), (0, 6) and (-3, 0)
Answer:
Given: (1, -1), (0, 6) and (-3, 0) are the vertices of a triangle.
Centroid (x, y)

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 8 Similar Triangles Ex 8.4 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 8th Lesson Similar Triangles Exercise 8.4

10th Class Maths 8th Lesson Similar Triangles Ex 8.4 Textbook Questions and Answers

Question 1.
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
Answer:

Given : □ ABCD is a rhombus.
Let its diagonals AC and BD bisect each other at ‘O’.
We know that “the diagonals in a rhombus are perpendicular to each other”.
In △AOD; AD² = OA² + OD² ………. (1)
[Pythagoras theorem]
In △COD; CD² = OC² + OD² ………. (2)
[Pythagoras theorem]
In △AOB; AB² = OA² + OB² ………. (3)
[Pythagoras theorem]
In △BOC; BC² = OB² + OC² ………. (4)
[Pythagoras theorem]
Adding the above equations we get AD² + CD² + AB² + BC² = 2 (OA² + OB2 + OC² + OD²)

Question 2.
ABC is a right triangle right angled at B. Let D and E be any points on AB and BC respectively. Prove that AE² + CD² = AC² + DE².

Answer:
Given: In △ABC; ∠B = 90°
D and E are points on AB and BC.
R.T.P.: AE² + CD² = AC² + DE²
Proof: In △BCD, △BCD is a right triangle right angled at B.
∴ BD² + BC² = CD² ……… (1)
[∵ Pythagoras theorem states that hypotenuse² = side² + side²]
In △ABE; ∠B = 90°
Adding (1) and (2), we get
BD² + BC² + AB² + BE² – CD² + AE²
(BD² + BE²) + (AB² + BC²) = CD² + AE²
DE² + AC² – CD² + AE² [Q.E.D.]
[∵ (i) In △DBE, ∠B = 90° and DE² = BD² + BE²
(ii) In △ABC, ∠B = 90° and AB² + BC²]

Question 3.
Prove that three times the square of any side of an equilateral triangle is equal to four times the square of the altitude.
Answer:
Given: △ABC, an equilateral triangle;
AD – altitude and the side is a units, altitude h units.
R.T.P: 3a² = 4h²

Proof: In △ABD, △ACD
∠B = ∠C [∵ 60°]
∠ADB = ∠ADC [∵ 90°]
∴ ∠BAD = ∠DAC [∵ Angle sum property]
Also, BA = CA
∴ △ABD s △ACD (by SAS congruence condition)
Hence, BD = CD = \(\frac{1}{2}\)BC = \(\frac{a}{2}\) [∵ c.p.c.t]
Now in △ABD, AB² = AD² + BD²
[∵ Pythagoras theorem]
a² = h² + \(\left(\frac{a}{2}\right)^{2}\)
a² = h² + \(\frac{a^{2}}{4}\)
h² = \(\frac{4 a^{2}-a^{2}}{4}\)
∴ h² = \(\frac{3 a^{2}}{4}\)
⇒ 4h² = 3a² (Q.E.D)

Question 4.
PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM² = QM . MR.
Answer:

Given: In △PQR, ∠P = 90° and PM ⊥ QR.
R.T.P : PM² = QM . MR
Proof: In △PQR; △MPR
∠P = ∠M [each 90°]
∠R = ∠R [common]
∴ △PQR ~ △MPR ……… (1)
[A.A. similarity]
In △PQR and △MQP,
∠P = ∠M (each 90°)
∠Q = ∠Q (common)
∴ △PQR ~ △MQP ……… (2)
[A.A. similarity]
From (1) and (2),
△PQR ~ △MPR ~ △MQP [transitive property]
∴ △MPR ~ △MQP
\(\frac{MP}{MQ}\) = \(\frac{PR}{QP}\) = \(\frac{MR}{MP}\)
[Ratio of corresponding sides of similar triangles are equal]
\(\frac{PM}{QM}\) = \(\frac{MR}{PM}\)
PM . PM = MR . QM
PM² = QM . MR [Q.E.D]

Question 5.
ABD is a triangle right angled at A and AC ⊥ BD.
Show that (i) AB² = BC BD
(ii) AD² = BD CD
(iii) AC² = BC DC.

Answer:
Given: In △ABD; ∠A = 90° AC ⊥ BD
R.T.P.:
i) AB² = BC . BD
Proof: In △ABD and △CAB,
∠BAD = ∠ACB [each 90°]
∠B = ∠B [common]
∴ △ABD ~ △CBA
[by A.A. similarity condition]
Hence, \(\frac{AB}{BC}\) = \(\frac{BD}{AB}\) = \(\frac{AD}{AC}\)
[∵ Ratios of corresponding sides of similar triangles are equal]
\(\frac{AB}{BD}\) = \(\frac{BC}{AB}\)
⇒ AB . AB = BC . BD
∴ AB² = BC . BD

ii) AD² = BD . CD
Proof: In △ABD and △CAD
∠BAD = ∠ACD [each 90°]
∠D = ∠D (common)
∴ △ABD ~ △CAD [A.A similarity]
Hence, \(\frac{AB}{AC}\) = \(\frac{BD}{AD}\) = \(\frac{AD}{CD}\)
⇒ \(\frac{BD}{AD}\) = \(\frac{AD}{CD}\)
AD . AD = BD . CD
AD² = BD . CD [Q.E.D]

iii) AC² = BC . DC
Proof: From (i) and (ii)
△ACB ~ △DCA
[∵ △BAD ~ △BCA ~ △ACD
Hence, \(\frac{AC}{DC}\) = \(\frac{BC}{AC}\) = \(\frac{AB}{AD}\)
\(\frac{AC}{DC}\) = \(\frac{BC}{AC}\)
AC . AC = BC . DC
AC² = BC . DC [Q.E.D]

Question 6.
ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC².
Answer:

Given: In △ABC; ∠C = 90°; AC = BC.
R.T.P.: AB² = 2AC²
Proof: In △ACB; ∠C = 90°
Hence, AC² + BC² = AB²
[Square of the hypotenuse is equal to sum of the squares of the other two sides – Pythagoras theorem]
⇒ AC² + AC² = AB² [∵ AC = BC given]
⇒ AB² = 2AC² [Q.E.D.]

Question 7.
‘O’ is any point in the interior of a triangle ABC.
OD ⊥ BC, OE ⊥ AC and OF ⊥ AB, show that
i) OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE²
ii) AF² + BD² + CE² = AE² + CD² + BF².

Answer:
Given: △ABC; O’ is an interior point of △ABC.
OD ⊥ BC, OE ⊥ AC, OF ⊥ AB.
R.T.P.:
i) OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE²
Proof: In OAF, OA² = AF² + OF² [Pythagoras theorem]
⇒ OA² – OF² = AF² …….. (1)
In △OBD,
OB² = BD² + OD²
⇒ OB² – OD² = BD² …….. (2)
In △OCE, OC² = CE² + OE²
OC² – OE² = CE² ……… (3)
Adding (1), (2) and (3) we get,
OA² – OF² + OB² – OD² + OC² – OE² = AF² + BD² + CE²
OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE² ……… (4)

ii) AF² + BD² + CE² = AE² + CD² + BF²
In △OAE,
OA² = AE² + OF² ……… (1)
⇒ OA² – OE² = AE²
In △OBF, OB² = BF² + OF²
OB² – OF² = BF² ……… (2)
In △OCD, OC² = OD² + CD²
OC² – OD² = CD² ……… (3)
Adding (1), (2) and (3) we get
OA² – OE² + OB² – OF² + OC² – OD² = AE² + BF² + CD²
⇒ OA² + OB² + OC² – OD² – OE² – OF² = AE² + CD² + BF²
⇒ AF² + BD² + CE² = AE² + CD² + BF² [From problem (i)]

Question 8.
A wire attached to vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
Answer:

Height of the pole AB = 18 m.
Length of the wire AC = 24 m.
Distance beween the pole and the stake be ‘d’ meters.
By Pythagoras theorem,
Hypotenuse² = side² + side²
24² = 18² + d²
d² = 24² – 18² = 576 – 324 = 252
= \(\sqrt{36 \times 7}\)
∴ d = 6√7 m.

Question 9.
Two poles of heights 6 m and 11m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.
Answer:

Let the height of the first pole AB = 6 m.
Let the height of the second pole CD = 11 m.
Distance between the poles AC = 12 m.
From the figure □ ACEB is a rectangle.
∴ AB = CE = 6 m
ED = CD – CE = 11 – 6 = 5 m
Now in △BED; ∠E = 90°; DE = 5 m; BE = 12 m
BD² = BE² + DE²
[hypotenuse² = side² + side² – Pythagoras theorem]
= 12² + 5²
= 144 + 25
BD² = 169
BD = √l69 = 13m
∴ Distance between the tops of the poles = 13 m.

Question 10.
In an equilateral triangle ABC, D is a point on side BC such that BD = \(\frac{1}{3}\) BC. Prove that 9AD² = 7AB².
Answer:

In △ABE, ∠E = 90°
⇒ \(\overline{\mathrm{AB}}\) is hypotenuse.
∴ AB² = AE² + BE²
AE² = AB² – BE²
⇒ AE² = AB² – \(\left(\frac{BC}{2}\right)^{2}\)
= AE² = AB² – \(\left(\frac{AB}{2}\right)^{2}\) (∵ AB = BC)
⇒ AE² = \(\frac{3}{4}\)AB² ……… (1)
In △ADE, ∠E = 90°
⇒ \(\overline{\mathrm{AD}}\) is hypotenuse.
⇒ AD² = AE² + DE²
⇒ AE² = AD² + DE²

⇒ 28 AB² = 36 AD²
⇒ 7 AB² = 9 AD²
Hence proved.

Question 11.
In the given figure, ABC is a triangle right angled at B. D and E are points on BC trisect it. Prove that 8 AE² = 3 AC² + 5 AD².

Answer:
In △ABC, ∠B=90°
⇒ \(\overline{\mathrm{AD}}\) is hypotenuse.
AC2 = AB2 + BC2
3AC² = 3AB² + 3BC² …….. (1)
In △ABD, ∠B = 90°
⇒ AD is hypotenuse.
∴ AD² = AB² + BD² = AB² + \(\left(\frac{BC}{3}\right)^{2}\)
⇒ AD² = AB² + \(\frac{\mathrm{BC}^{2}}{9}\)
⇒ 5 AD² = 5 AB² + \(\frac{5 \mathrm{BC}^{2}}{9}\) …….. (2)
(1) + (2)
3 AC² + 5 AD² = 3 AB² + 3 BC² + 5 AB² + \(\frac{5}{9} \mathrm{BC}^{2}\)
= 8AB² + \(\frac{32}{9} \mathrm{BC}^{2}\) ……… (3)
Now in △ABE, ∠B = 90°
⇒ \(\overline{\mathrm{AE}}\) is hypotenuse.
⇒ AE² = AB² + BE² = AB² + \(\left(\frac{2}{3} BC\right)^{2}\)
= AB² + \(\frac{4}{9} \mathrm{BC}^{2}\)
⇒ AE² = 8AB² + \(\frac{32}{9} \mathrm{BC}^{2}\) ……… (4)
∴ RHS of (3) and (4) are equal.
∴ LHS of (3) and (4) are equal.
∴ 8 AE² = 3 AC² + 5 AD².
Hence proved.

Question 12.
ABC is an isosceles triangle right angled at B. Equilateral triangles ACD and ABE are constructed on sides AC and AB. Find the ratio between the areas of △ABE and △ACD.
Answer:

Given: △ABC, AB = BC and ∠B = 90°
△ABE on AB; △ACD on AC are equiangular triangles.
Let equal sides of the isosceles right triangle, AB = BC = a (say)
Then, in △ABC, ∠B = 90°
AC² – AB² + BC²
[hypotenuse² = side² + side² – Pythagoras theorem] = a² + a² = 2a²
Since, △ABE ~ △ACD
\(\frac{\Delta \mathrm{ABE}}{\Delta \mathrm{ACD}}\) = \(\frac{\mathrm{AB}^{2}}{\mathrm{AC}^{2}}\)
[∵ Ratio of areas of two similar tri-angles is equal to the ratio of squares of their corresponding sides]
= \(\frac{a^{2}}{2 a^{2}}\) = \(\frac{1}{2}\)
△ABE : △ACD = 1 : 2.

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 8 Similar Triangles Ex 8.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 8th Lesson Similar Triangles Exercise 8.2

10th Class Maths 8th Lesson Similar Triangles Ex 8.2 Textbook Questions and Answers

Question 1.
In the given figure, ∠ADE = ∠B
i) Show that △ABC ~ △ADE
ii) If AD = 3.8 cm, AE = 3.6 cm, BE = 2.1 cm, BC = 4.2 cm, find DE.

Answer:
i) Given: △ABC and ∠ADE = ∠B
R.T.P: △ABC ~ △ADE.
Proof: In △ABC and △ADE
∠A = ∠A [∵ Common]
∠B = ∠ADE [∵ Given]
∴ ∠C = ∠AED [∵ By Angle Sum property of triangles] △ABC ~ △ADE by AAA similarity condition.]

ii) AD = 3.8 cm, AE = 3.6 cm, BE = 2.1 cm, BC = 4.2 cm, find DE.
To find DE; △ABC ~ △ADE
Hence,
\(\frac{AB}{AD}\) = \(\frac{BC}{DE}\) = \(\frac{AC}{AE}\)
[∵ Ratios of corresponding sides are equal]

Question 2.
The perimeters of two similar triangles are 30 cm and 20 cm respectively. If one side of the first triangle is 12 cm, determine the corresponding side of the second triangle.
Answer:

Given: △ABC ~ △PQR
Perimeter of △ABC = 30 cm.
Perimeter of △PQR = 20 cm.
AB = 12 cm.
To find: \(\overline{\mathrm{PQ}}\)
Ratio of perimeters = 30 : 20 = 3 : 2
Let the length of the side corresponding to the side with length 12 cm be x.
Then 30 : 20 : : 12 : x
30x = 20 x 12
\(x = \frac{20 \times 12}{30}\) = 8 cm

Question 3.
A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/sec. If the lamp-post is 3.6 m above the ground, find the length of her shadow after 4 seconds.
Answer:

Given:
A lamp-post \(\overline{\mathrm{AB}}\) of height = 3.6 m
= 360 cm.
Speed of the girl = 1.2 m/sec.
Distance travelled in 4 sec = Speed x Time = 1.2 × 4 = 4.8 m = 480 cm.
\(\overline{\mathrm{CD}}\), height of the girl = 90 cm.
Let the length of the shadow at a distance of 4.8 m from the lamp post = x cm.
From the figure,
△ABE ~ △DCE
[∵ ∠B = ∠C = 90°
∠E = ∠C common
(A.A. similarity)]
Hence,
\(\frac{AB}{DC}\) = \(\frac{BE}{CE}\) = \(\frac{AE}{DE}\)
∴ \(\frac{360}{90}\) = \(\frac{480+x}{x}\)
⇒ 4 = \(\frac{480+x}{x}\)
⇒ 4x = 480 + x
⇒ 4x – x = 480
⇒ 3x = 480
⇒ x = 160 cm = 1.6 m
∴ Length of the shadow = 1.6 m

Question 4.
CM and RN are respectively the medians of similar triangles △ABC and △PQR. Prove that
i) △AMC ~ △PNR
ii) \(\frac{CM}{RN}\) = \(\frac{AB}{PQ}\)
iii) △CMB ~ △RNQ

Answer:
Given : △ABC ~ △PQR
CM is a median through C of △ABC.
RN is a median through R of △PQR.
R.T.P:
i) △AMC ~ △PNR.
Proof: In △AMC and △PNR,
\(\frac{AC}{PR}\) = \(\frac{AM}{PN}\) and ∠A = ∠P [∵ In △ABC, △PQR and M, N are the mid-points of AB and PQ]
∴ △AMC ~ △PNR
[∵ SAS similarity condition]

ii) \(\frac{CM}{RN}\) = \(\frac{AB}{PQ}\)
Proof: From (i) we have
△AMC ~ △PNR
Hence \(\frac{AC}{PQ}\) = \(\frac{AM}{PN}\) = \(\frac{CM}{RN}\)
[∵ Ratio of corresponding sides of two similar triangles are equal]
Thus, \(\frac{CM}{RN}\) = \(\frac{AM×2}{PN×2}\)
[Multiplying both numerator and the denominator by 2]
\(\frac{CM}{RN}\) = \(\frac{AB}{PQ}\) [2AM = AB; 2PN = PQ]

iii) △CMB ~ △RNQ
Proof: In △CMB and △RNQ
∠B = ∠Q [Corresponding angles of △ABC and △PQR]
Also, \(\frac{BC}{RQ}\) = \(\frac{BM}{QN}\)

Thus, △CMB ~ △RNQ by S.A.S similarity condition.

Question 5.
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point ‘O’. Using the criterion of similarity for two triangles, show that \(\frac{OA}{OC}\) = \(\frac{OB}{OD}\).
Answer:

Given : □ ABCD, AB || DC
The diagonals AC and BD intersect at ‘O’.
R.T.P: \(\frac{OA}{OC}\) = \(\frac{OB}{OD}\)
Construction: Draw EF || AB, passing through ‘O’.
Proof: In △ACD, OE || CD [∵ Construction]
Hence \(\frac{OA}{OC}\) = \(\frac{EA}{ED}\) …….. (1)
(∵ Line drawn parallel to one side of a triangle divides other two sides in the same ratio – Basic proportionality theorem)
Also in △ABD, EO || AB [Construction] Hence,
\(\frac{EA}{ED}\) = \(\frac{OB}{OD}\) ……… (2)
(∵ Basic proportionality theorem) From (1) and (2), we have
\(\frac{OA}{OC}\) = \(\frac{OB}{OD}\)
∴ Hence proved.

Question 6.
AB, CD, PQ are perpendicular to BD. AB = x, CD = y and PQ = z, prove that \(\frac{1}{x}+\frac{1}{y}=\frac{1}{z}\).

Answer:
Given ∠B = ∠Q = ∠D = 90°
Thus, AB || PQ || CD.
Now in △BQP, △BDC
∠B = ∠B (Common)
∠Q = ∠D (90°)
∠P = ∠C [∵ Angle Sum property of triangles]
∴ △BQP ~ △BDC
(by A.A.A similarity condition)
Hence \(\frac{BQ}{BD}\) = \(\frac{PQ}{CD}\)
[∵ Ratio of corresponding sides is equal] Also in △DQP and △DBA
∠D = ∠D (Common)
∠Q = ∠B (90°)
∴ △DQP ~ △DBA (by A.A. similarity condition)
\(\frac{QD}{BD}\) = \(\frac{PQ}{AB}\)
[ Ratio of corresponding sides is equal]
Adding (1) and (2), we get

Question 7.
A flag pole 4 m tall casts a 6 m., shadow. At the same time, a nearby building casts a shadow of 24 m. How tall is the building?
Answer:
Given: 4 m length flag pole casts a shadow 6 m.
Let x m length/tall building casts a shadow 24 m.

Let AB be the length of flag pole = 4 m.
Shadow of AB = BC = 6 m.
PQ be the building = x m (say)
QR, the shadow of the building = 24 m
From the figure,
∠A = ∠P
∠B = ∠Q
∴ △ABC ~ △PQR by A.A. similarity condition
Hence \(\frac{AB}{PQ}\) = \(\frac{BC}{QR}\)
[∵ Ratio of corresponding angles is equal]
\(\frac{4}{6}\) = \(\frac{x}{24}\)
x = \(\frac{24 \times 4}{6}\) = 16 m
∴ Height of the building = 16 m.

Question 8.
CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of △ABC and △FEG respectively. If △ABC ~ △FEG then show that
i) \(\frac{CD}{GH}\) = \(\frac{AC}{FG}\)
ii) △DCB ~ △HGE
iii) △DCA ~ △HGF
Answer:

Given: △ABC ~ △FEG.
CD is the bisector of ∠C and GH is the bisector of ∠G.
R.T.P.:
i) \(\frac{CD}{GH}\) = \(\frac{AC}{FG}\)
In △ACD and △FGH
∠A = ∠F
[∵ Corresponding angles of △ABC and △FEG]
∠ACD = ∠FGH
[∵ ∠C = ∠G ⇒ \(\frac{1}{2}\)∠C = \(\frac{1}{2}\)∠G ⇒ ∠ACD = ∠FGH]
∴ By A.A. similarity condition, △ACD ~ △FGH
\(\frac{AC}{FG}\) = \(\frac{CD}{GH}\) = \(\frac{AD}{FH}\)
[∵ Ratio of the Corresponding angles is equal]
⇒ \(\frac{AC}{FG}\) = \(\frac{CD}{GH}\) [Q.E.D]

ii) △DCB ~ △HGE
In △DCB and △HGE,
∠B = ∠E
[∵ Corresponding angles of △ABC and △FEG]
∠DCB = ∠HGE
[∵ ∠C = ∠G ⇒ \(\frac{1}{2}\)∠C = \(\frac{1}{2}\)∠G ⇒ ∠DCB = ∠HGE]
∴ △DCB ~ △HGE . (by A.A. similarity condition)

iii) △DCA ~ △HGF
In △DCA and △HGF
∠A = ∠F
\(\frac{1}{2}\)∠C = \(\frac{1}{2}\)∠G ⇒ ∠DCA = ∠HGF
[∵ Corresponding angles of the similar triangles]
∴ △DCA ~ △HGF
[ A.A. similarity condition]

Question 9.
AX and DY are altitudes of two similar triangles △ABC and △DEF. Prove that AX : DY = AB : DE.
Answer:

Given: △ABC ~ △DEF.
AX ⊥ BC and DY ⊥ EF.
R.T.P.: AX : DY = AB : DE.
Proof: In △ABX and △DEY ∠B = ∠E [∵ Corresponding angles of △ABC and △DEF]
∠AXB = ∠DYE [given]
∴ △ABX ~ △DEY
(by A.A. similarity condition)
Hence \(\frac{AB}{DE}\) = \(\frac{BX}{EY}\) = \(\frac{AX}{DY}\)
[∵ Ratios of corresponding sides of similar triangles are equal]
⇒ AX : DY = AB : DE [Q.E.D.]

Question 10.
Construct a triangle shadow similar to the given △ABC, with its sides equal to \(\frac{5}{3}\) of the corresponding sides of the triangle ABC.
Answer:

Steps of construction :

1. Draw a △ABC with certain measures.
2. Draw a ray \(\overrightarrow{\mathrm{BX}}\) making an acute angle with BC on the side opposite to vertex A.
3. Locate 8 points (B₁, B₂, …., B₈) on \(\overrightarrow{\mathrm{BX}}\) such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅ = B₅B₆ = B₆B₇ = B₇B₈.
4. Join B₅, C.
5. Draw a line parallel to B₅C through which it intersects BC extended at C’.
6. Draw a line parallel to AC through ‘C’ which meets \(\overrightarrow{\mathrm{BA}}\) produced at A’.
7. △A’BC’ is the required triangle.

Question 11.
Construct a triangle of sides 4 cm, 5 cm and 6 cm. Then, construct a triangle similar to it,whose sides are 2/3 of the corresponding sides of the first triangle.
Answer:

Steps of construction:

1. Draw △ABC with AB = 4 cm, BC = 5 cm and CA = 6 cm.
2. Draw a ray BX making an acute angle with BC on the side opposite to vertex A.
3. Mark three points B₁, B₂ and B₃ on \(\overrightarrow{\mathrm{BX}}\) such that BB₁ = B₁B₂ = B₂B₃.
4. Join B₃, C.
5. Draw a line parallel to B₃C through B₂ meeting BC at C’.
6. Draw a line parallel to BA through C’ meeting BA at A’.
7. △A’BC’ is the required triangle.

Question 12.
Construct an isosceles triangle whose base is 8 cm and altitude is 4 cm. Then, draw another triangle whose sides are 1\(\frac{1}{2}\) times the corresponding sides of the isosceles triangle.
Answer:

Steps of construction:

1. Draw AABC in which BC = 8 cm and altitude AD = 4 cm.
2. Draw a ray BX making an acute angle with BC on the side opposite to vertex A.
3. Mark three points B₁, B₂ and B₃ such that BB₁ = B₁B₂ = B₂B₃.
4. Join B₂C.
5. Draw a line parallel to B₂C through B₃ meeting BC produced C’.
6. Draw a line paral1e1 to AC through C’ meeting BA produced at A’.
7. △A’BC’ is the required triangle.

 

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 8 Similar Triangles Ex 8.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 8th Lesson Similar Triangles Exercise 8.1

10th Class Maths 8th Lesson Similar Triangles Ex 8.1 Textbook Questions and Answers

Question 1.
In △PQR, ST is a line such that \(\frac{PS}{SQ}\) = \(\frac{PT}{TR}\) and also ∠PST = ∠PRQ.
Prove that △PQR is an isosceles triangle.
Answer:
Given : In △PQR,
\(\frac{PS}{SQ}\) = \(\frac{PT}{TR}\) and ∠PST= ∠PRQ.

R.T.P: △PQR is isosceles.
Proof: \(\frac{PS}{SQ}\) = \(\frac{PT}{TR}\)
Hence, ST || QR (Converse of Basic proportionality theorem)
∠PST = ∠PQR …….. (1)
(Corresponding angles for the lines ST || QR)
Also, ∠PST = ∠PRQ ……… (2) given
From (1) and (2),
∠PQR = ∠PRQ
i.e., PR = PQ
[∵ In a triangle sides opposite to equal angles are equal]
Hence, APQR is an isosceles triangle.

Question 2.
In the given figure, LM || CM and LN || CD. Prove that \(\frac{AM}{AB}\) = \(\frac{AN}{AD}\).

Answer:
Given : LM || CB and LN || CD In △ABC, LM || BC (given) Hence,

Adding ‘1’ on both sides.

From (1) and (2)
∴ \(\frac{AM}{AB}\) = \(\frac{AN}{AD}\).

Question 3.
In the given figure, DE || AC and DF || AE. Prove that \(\frac{BF}{FE}\) = \(\frac{BE}{AC}\).

Answer:
In △ABC, DE || AC
Hence \(\frac{BE}{EC}\) = \(\frac{BD}{DA}\) ………. (1)
[∵ A line drawn parallel to one side of a triangle divides the other two sides in the same ratio – Basic proportionality theorem]
Also in △ABE, DF || AE
Hence \(\frac{BF}{FE}\) = \(\frac{BD}{DA}\) ………. (2)
From (1) and (2), \(\frac{BF}{FE}\) = \(\frac{BE}{AC}\) Hence proved.

Question 4.
Prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side (Using Basic proportionality theorem).
Answer:

Given: In △ABC; D is the mid-point of AB.
A line ‘l’ through D, parallel to BC, meeting AC at E.
R.T.P: E is the midpoint of AC.
Proof:
DE || BC (Given)
then
\(\frac{AD}{DB}\) = \(\frac{AE}{EC}\)(From Basic Proportional theorem)
Also given ‘D’ is mid point of AB.
Then AD = DB.
⇒ \(\frac{AD}{DB}\) = \(\frac{DB}{DB}\) = \(\frac{AE}{EC}\) = 1
⇒ AE = EC
∴ ‘E’ is mid point of AC
∴ The line bisects the third side \(\overline{\mathrm{AC}}\).
Hence proved.

Question 5.
Prove that a line joining the mid points of any two sides of a triangle is parallel to the third side. (Using converse of Basic proportionality theorem)
Answer:
Given: △ABC, D is the midpoint of AB and E is the midpoint of AC.
R.T.P : DE || BC.
Proof:

Since D is the midpoint of AB, we have AD = DB ⇒ \(\frac{AD}{DB}\) = 1 ……. (1)
also ‘E’ is the midpoint of AC, we have AE = EC ⇒ \(\frac{AE}{EC}\) = 1 ……. (2)
From (1) and (2)
\(\frac{AD}{DB}\) = \(\frac{AE}{EC}\)
If a line divides any two sides of a triangle in the same ratio then it is parallel to the third side.
∴ DE || BC by Basic proportionality theorem.
Hence proved.

Question 6.
In the given figure, DE || OQ and DF || OR. Show that EF || QR.

Answer:
Given: △PQR, DE || OQ; DF || OR
R.T.P: EF || QR
Proof:
In △POQ;
\(\frac{PE}{EQ}\) = \(\frac{PD}{DO}\) ……. (1)
[∵ ED || QO, Basic proportionality theorem]
In △POR; \(\frac{PF}{FR}\) = \(\frac{PD}{DO}\) ……. (2) [∵ DF || OR, Basic Proportionality Theorem]
From (1) and (2),
\(\frac{PE}{EQ}\) = \(\frac{PF}{FR}\)
Thus the line \(\overline{\mathrm{EF}}\) divides the two sides PQ and PR of △PQR in the same ratio.
Hence, EF || QR. [∵ Converse of Basic proportionality theorem]

Question 7.
In the given figure A, B and C are points on OP, OQ and OR respec¬tively such that AB || PQ and AC || PR. Show that BC || QR.

Answer:
Given:
In △PQR, AB || PQ; AC || PR
R.T.P : BC || QR
Proof: In △POQ; AB || PQ
\(\frac{OA}{AP}\) = \(\frac{OB}{BQ}\) ……… (1)
(∵ Basic Proportional theorem)
and in △OPR, Proof: In △POQ; AB || PQ
\(\frac{OA}{AP}\) = \(\frac{OC}{CR}\) ……… (2)
From (1) and (2), we can write
\(\frac{OB}{BQ}\) = \(\frac{OC}{CR}\)
Then consider above condition in △OQR then from (3) it is clear.
∴ BC || QR [∵ from converse of Basic Proportionality Theorem]
Hence proved.

Question 8.
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at point ‘O’. Show that\(\frac{AO}{BO}\) = \(\frac{CO}{DO}\).
Answer:
Given: In trapezium □ ABCD, AB || CD. Diagonals AC, BD intersect at O.
R.T.P: \(\frac{AO}{BO}\) = \(\frac{CO}{DO}\)

Construction:
Draw a line EF passing through the point ‘O’ and parallel to CD and AB.
Proof: In △ACD, EO || CD
∴ \(\frac{AO}{CO}\) = \(\frac{AE}{DE}\) …….. (1)
[∵ line drawn parallel to one side of a triangle divides other two sides in the same ratio by Basic proportionality theorem]
In △ABD, EO || AB
Hence, \(\frac{DE}{AE}\) = \(\frac{DO}{BO}\)
[∵ Basic proportionality theorem]
\(\frac{BO}{DO}\) = \(\frac{AE}{ED}\) …….. (2) [∵ Invertendo]
From (1) and (2),
\(\frac{AO}{CO}\) = \(\frac{BO}{DO}\)
⇒ \(\frac{AO}{BO}\) = \(\frac{CO}{DO}\) [∵ Alternendo]

Question 9.
Draw a line segment of length 7.2 cm and divide it in the ratio 5 : 3. Measure the two parts.
Answer:

Steps of construction:

1. Draw a line segment \(\overline{\mathrm{AB}}\) of length 7.2 cm.
2. Construct an acute angle ∠BAX at A.
3. Mark off 5 + 3 = 8 equal parts (A₁, A₂, …., A₈) on \(\stackrel{\leftrightarrow}{\mathrm{AX}}\) with same radius.
4. Join A₈ and B.
5. Draw a line parallel to \(\stackrel{\leftrightarrow}{\mathrm{A}_{8} \mathrm{~B}}\) at A₅ meeting AB at C.
6. Now the point C divides AB in the ratio 5:3.
7. Measure AC and BC. AC = 4.5 cm and BC = 2.7 cm.