AP SSC 10th Class Maths Solutions Chapter 6 Progressions Optional Exercise

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

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

10th Class Maths 6th Lesson Progressions Optional Exercise Textbook Questions and Answers

Question 1.
Which term of the AP:
121, 117, 113,…, is the first negative term? [Hint: Find n for an < 0]
Answer:
Given A.P: 121, 117, 113, ……
a = 121 and d = a2 – a1
= 117 – 121 = -4
Let the nth term be the first negative term of the given G.P.
Then, an < 0
⇒ an = a + (n – 1) d < 0
⇒ 121 + (n – 1) (-4) < 0
⇒ 121 – 4n + 4 < 0
⇒ 125 – 4n < 0
⇒ -4n < -125
⇒ 4n >125 125
⇒ n > \(\frac{125}{4}\)
n > 31.25
∴ When n = 32, the term becomes negative, (or)
32 nd term is the first negative term of the given A.P.

AP SSC 10th Class Maths Solutions Chapter 6 Progressions Optional Exercise

Question 2.
The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.
Answer:
Let the 3rd term of AP = a + 2d
and the 7th term of AP = a + 6d
∴ Sum of 3rd and 7th terms = a + 2d + a + 6d = 6
⇒ 2a + 8d = 6
⇒ a + 4d = 3 …… (1)
Now product of above two terms = (a + 2d) (a + 6d) = 8
we can re-write above terms as following
(a + 4d – 2d) (a + 4d + 2d) = 8
⇒ (3 – 2d) (3 + 2d) = 8
⇒ 9 – 4d2 = 8
⇒ 4d2 = 9 – 8 = 1
∴ d2 = \(\frac{1}{4}\)
⇒ d = ± \(\frac{1}{2}\) ……. (2)
Now putting d = \(\frac{1}{2}\) in eq(1) we get
a + 4d = a + 4(\(\frac{1}{2}\)) = 3 ⇒ a = 1
so a = 1, d = \(\frac{1}{2}\)
(or) now putting d = \(\frac{-1}{2}\), we get
a + 4d = a + 4(\(\frac{-1}{2}\)) = 3
⇒ a – 2 = 3
⇒ a = 5
∴ a = 5, d = \(\frac{-1}{2}\)
∴ Sum of sixteen terms =
AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise 10
So S16 = 20 or 76

AP SSC 10th Class Maths Solutions Chapter 6 Progressions Optional Exercise

Question 3.
A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are 2\(\frac{1}{2}\) m apart, what is the length of the wood required for the rungs. [Hint: Number of rungs = \(\frac{250}{25}\) + 1]
AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise 11
Answer:
Given: A ladder with rungs separated by a distance of 25 cm.
Total distance between the rungs = 2\(\frac{1}{2}\) m = 250 m
∴ Number of rungs = \(\frac{250}{25}\) + 1 = 10 + 1 = 11
Length of the bottom and top rungs = 45 cm and 25 cm
where a = 45; last term l = a11 = 25 and d = 2 cm
∴ Sn = \(\frac{n}{2}\)(a + l)
= \(\frac{11}{2}\)(45 + 25)
= \(\frac{11}{2}\) × 70
= 35 × 11
= 385 cm
∴ Length of wood required for total rungs = 385 cm.

Question 4.
The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. And find this value of x. [Hint: Sx-1 = S49 – Sx]
Answer:
Given: Houses with numbers from 1 to 49.
x is a number x such that,
Sx-1 = S49 – Sx
AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise 12
AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise 13
⇒ x(x – 1) + x(x + 1) = 2 × 1225
⇒ x2 – x + x2 + x = 2450
⇒ 2x2 = 2450
⇒ x2 = \(\frac{2450}{2}\) = 1225
⇒ x = √1225 = ±35
∴ x = 35 [∵ x is a counting number]

AP SSC 10th Class Maths Solutions Chapter 6 Progressions Optional Exercise

Question 5.
A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete.
Each step has a rise of \(\frac{1}{4}\) m and a tread of \(\frac{1}{2}\) m. (see Fig.).
Calculate the total volume of concrete required to build the terrace.
[Hint: Volume of concrete required to build the first step = \(\frac{1}{4}\) × \(\frac{1}{2}\) × 50 m ]
AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise 14
Answer:
Length of each step = 50 m = l
Rise/height of each step = \(\frac{1}{4}\) m = h
Tread of each step = \(\frac{1}{2}\) m = b
Given volume of concrete required to build the first step = \(\frac{1}{4}\) × \(\frac{1}{2}\) × 50 m3 = lbh
We can compare the shape of each step with a cuboid.
Volume of the cuboid = l . b . h
Volumes of concrete required to build the 15 steps are
AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise 15
[∵ 1, 2, 3, 15 is in A.P. where a = 1; d = 1, n = 15]
AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise 16
∴ The total volume of concrete required to build the terrace is 750 m3.

AP SSC 10th Class Maths Solutions Chapter 6 Progressions Optional Exercise

Question 6.
150 workers were engaged to finish a piece of work in a certain number of days. Four workers dropped from the work in the second day. Four workers dropped in third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.
[Let the no.of days to finish the work is ‘x’ then 150x = \(\frac{x+8}{2}\)[2 × 150 + (x + 8 – 1) (-4)]
Answer:
Given: Number of workers engaged initially = 150.
4 workers were dropped each day.
Let the total work was to be completed initially was in x days.
∴ Work done by 150 workers in x days = 150.x.
But due to the dropping of 4 workers each day it took 8 more days.
Work done in this case is
150 × 1 + 146 × 1 + 142 × 1 + …. (x + 8) terms,
AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise 17
= (x + 8) (136 – 2x)
= -2x2 + 136x + 1088 – 16x
= -2x2 + 120x + 1088
∴ 150x = -2x2 + 120x + 1088
⇒ 2x2 + 30x – 1088 = 0
⇒ x2 + 15x – 544 = 0
⇒ x2 + 32x – 17x – 544 =0
⇒ x(x + 32) – 17 (x + 32) = 0
⇒ (x + 32) (x – 17) = 0
⇒ x + 32 = 0 (or) x – 17 = 0
⇒ x = – 32 (or) x = 17
x can’t be negative.
∴ x = 17.
i.e., The total work was completed in x + 8 days = 17 + 8 = 25 days.

AP SSC 10th Class Maths Solutions Chapter 6 Progressions Optional Exercise

Question 7.
A machine costs Rs. 5,00,000. If the value depreciates 15% in the first year, 13\(\frac{1}{2}\)% in the second year, 12% in the third year and so on. What will be its value at the end of 10 years, when all the percentages will be applied to the original cost?
Answer:
Given: Cost price of a machine = Rs. 5,00,000
Depreciation during the years
AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise 18
Sum of the depreciations =
AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise 19
∴ Cost after 10 years = (100 – 82.5)% of 5,00,000
= 17.5 % of 5,00,000
= \(\frac{17.5 × 500000}{2}\)
= Rs. 87,500
∴ The value at the end of 10 years will be Rs. 87,500.

AP SSC 10th Class Maths Solutions Chapter 11 Trigonometry Optional Exercise

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

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

10th Class Maths 11th Lesson Trigonometry Optional Exercise Textbook Questions and Answers

Question 1.
Prove that
AP SSC 10th Class Maths Solutions Chapter 11 Trigonometry Optional Exercise 2
Answer:
AP SSC 10th Class Maths Solutions Chapter 11 Trigonometry Optional Exercise 1
Hence Proved.

AP SSC 10th Class Maths Solutions Chapter 11 Trigonometry Optional Exercise

Question 2.
Prove that \(\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}\) = \(\frac{1}{\sec \theta-\tan \theta}\) using the identity sec2 θ = 1 + tan2 θ.
Answer:
AP SSC 10th Class Maths Solutions Chapter 11 Trigonometry Optional Exercise 3
AP SSC 10th Class Maths Solutions Chapter 11 Trigonometry Optional Exercise 4
From (1) and (2), we get L.H.S. = R.H.S.
Hence Proved.

AP SSC 10th Class Maths Solutions Chapter 11 Trigonometry Optional Exercise

Question 3.
Prove that (cosec A – sin A) (sec A – cos A) = \(\frac{1}{\tan A+\cot A}\)
Answer:
L.H.S. = (cosec A – sin A) (sec A – cos A)
AP SSC 10th Class Maths Solutions Chapter 11 Trigonometry Optional Exercise 5
Now
AP SSC 10th Class Maths Solutions Chapter 11 Trigonometry Optional Exercise 6
From (1) and (2), we get
L.H.S. = R.H.S.
∴ (cosec A – sin A) (sec A – cos A) = \(\frac{1}{\tan A+\cot A}\)

AP SSC 10th Class Maths Solutions Chapter 11 Trigonometry Optional Exercise

Question 4.
Prove that \(\frac{1+\sec A}{\sec A}\) = \(\frac{\sin ^{2} A}{1-\cos A}\)
Answer:

Question 5.
Prove that \(\left(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\right)\) = \(\left(\frac{1+\tan A}{1+\cot A}\right)^{2}\) = tan2 A
Answer:
AP SSC 10th Class Maths Solutions Chapter 11 Trigonometry Optional Exercise 8
AP SSC 10th Class Maths Solutions Chapter 11 Trigonometry Optional Exercise 9

AP SSC 10th Class Maths Solutions Chapter 11 Trigonometry Optional Exercise

Question 6.
Prove that \(\left(\frac{\sec A-1}{\sec A+1}\right)\) = \(\left(\frac{1-\cos A}{1+\cos A}\right)\)
Answer:
AP SSC 10th Class Maths Solutions Chapter 11 Trigonometry Optional Exercise 10
∴ L.H.S. = R.H.S.

AP SSC 10th Class Maths Solutions Chapter 3 Polynomials InText Questions

AP State Syllabus SSC 10th Class Maths Solutions 3rd Lesson Polynomials InText Questions

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 3 Polynomials InText Questions and Answers.

10th Class Maths 3rd Lesson Polynomials InText Questions and Answers

Do these

Question 1.
State which of the following are polynomials and which are not? Give reasons.   (Page No. 48)
(i) 2x3
(ii) \(\frac{1}{x-1}\)
(iii) 4z2 + \(\frac{1}{7}\)
(iv) m2 – √2 m + 2
(v) p-2 + 1
Answer:
i) 2x3 is a polynomial.
ii) \(\frac{1}{x-1}\) is not a polynomial because its power is negative integer exponent.
iii) 4z2 + \(\frac{1}{7}\) is a polynomial.
iv) m2 – √2 m + 2 is a polynomial.
v) p-2 + 1 is not a polynomial because its power is negative integer exponent.

AP SSC 10th Class Maths Solutions Chapter 3 Polynomials InText Questions

Question 2.
p(x) = x2 – 5x – 6, find the values of p(l), p(2), p(3), p(0), p(-l), p(-2), p(-3).    (Page No. 49)
Answer:
Given polynomial p(x) = x2 – 5x – 6
p(1) = (1)2 – 5(1) – 6 = 1 – 5 – 6 = – 10
p(2) = (2)2 – 5(2) – 6 = 4 – 10 – 6 = -12
p(3) = 32 – 5(3) – 6 = 9 – 15 – 6 = – 12
p(0) = 02 – 5(0) – 6 = – 6
p(-1) = (-1)2 – 5(-1) – 6 = 1 + 5 – 6 = 0
p(-2) = (-2)2 – 5(-2) – 6 = 4 + 10 – 6 = 8
p(-3) = (-3)2 – 5(-3) – 6 = 9 + 15 – 6 = 18

Question 3.
p(m) = m2 – 3m + 1, find the values of p(1)and p(-1).    (Page No. 49)
Answer:
Given polynomial p(m) = m2 – 3m + 1
p(1) = (1)2 – 3(1) + 1 = 1 – 3 + 1 = 2 – 3 = – 1
p(-1) = (-1)2 – 3(-1) + 1 = 1 + 3 + 1 = 5

Question 4.
Let p(x) = x2 – 4x + 3. Find the values of p(0), p(l), p(2), p(3) and obtain zeroes of the polynomial p(x).   (Page No. 50)
Answer:
Given polynomial p(x) = x2 – 4x + 3
p(0) = (0)2 – 4(0) + 3 = 3
p(1) = (1)2 – 4(1) + 3 = 1 – 4 + 3 = 0
p(2) = (2)2 – 4(2) + 3 = 4 – 8 + 3 = – 1
p(3) = (3)2 – 4(3) + 3 = 9 – 12 + 3 = 0
We see that p(1) and p(3) are zeroes of the polynomial p(x).

AP SSC 10th Class Maths Solutions Chapter 3 Polynomials InText Questions

Question 5.
Check whether -3 and 3 are the zeroes of the polynomial x2 – 9.    (Page No. 50)
Answer:
Given polynomial p(x) = x2 – 9
Zero of the polynomial p(x) = 0
x2 – 9 = 0
⇒ x2 = 9
⇒ x = V9 = ± 3
∴ x = + 3, – 3
∴ Zeroes of the polynomial p(x) are – 3 and 3.

Try these

Question 1.
Write 3 different quadratic, cubic and 2 linear polynomials with different number of terms. (Page No. 48)
Answer:
Quadratic polynomials:
AP SSC 10th Class Maths Chapter 3 Polynomials InText Questions 1
Cubic polynomials :
AP SSC 10th Class Maths Chapter 3 Polynomials InText Questions 2
Linear polynomials :
f(t) = √2 t + 5
g(u) = \(\frac{2}{3}\) u – \(\frac{5}{2}\) and
q(y) = 3y
Yes, we can write polynomials of any degree.

Question 2.
Write a quadratic polynomial and a cubic polynomial in variable x in the general form. (Page No. 49)
Answer:
General form of a quadratic polynomial having variable ‘x’ is
f(x) = ax3 + bx2 + c, a ≠ 0
General form of a cubic polynomial having variable ‘x’ is
f(x) = ax3 + bx2 + cx + d, a ≠ 0

AP SSC 10th Class Maths Solutions Chapter 3 Polynomials InText Questions

Question 3.
Write a general polynomial q(z) of degree n with coefficients that are b0…. bn. What are the conditions on b0…. bn. (Page No. 49)
Answer:
q(z) = b0zn + b1zn-1 + b2zn-2 …….. + bn-1z + bn is a polynomial of n degree where b0, b1, b2,…… bn-1, bn are real coefficients and b0 ≠ 0.

Do this

Question 1.
Draw the graph of i) y = 2x + 5, ii) y = 2x – 5, iii) y = 2x and find the point of intersection on X – axis. Is the x-coordinates of these points also the zero of the polynomial?   (Page No. 52)
Answer:
i) Given that y = 2x + 5
AP SSC 10th Class Maths Chapter 3 Polynomials InText Questions 3
AP SSC 10th Class Maths Chapter 3 Polynomials InText Questions 4
Result: The graph y = 2x + 5 cuts the X – axis at the point (-2.5, 0).
Hence, the zeroes of the polynomial is -2.5.

ii) Given that y = 2x – 5
AP SSC 10th Class Maths Chapter 3 Polynomials InText Questions 5
AP SSC 10th Class Maths Chapter 3 Polynomials InText Questions 6
Result: The graph of y = 2x – 5 cuts the X – axis at the point (2.5, 0).
The zeroes of the polynomial is 2.5 = \(\frac{5}{2}\)

AP SSC 10th Class Maths Solutions Chapter 3 Polynomials InText Questions

iii) Given that y = 2x
AP SSC 10th Class Maths Chapter 3 Polynomials InText Questions 7
AP SSC 10th Class Maths Chapter 3 Polynomials InText Questions 8
Result: The graph passes through the origin.
So, the zeroes of the polynomial y = 2x is zero.

Try these

Question 1.
Draw the graphs of (i) y = x2 – x – 6 (ii) y = 6 – x – x2 and find zeroes in each case. What do you notice? (Page No. 53)
Answer:
i) Given that y = x2 – x – 6
AP SSC 10th Class Maths Chapter 3 Polynomials InText Questions 9
AP SSC 10th Class Maths Chapter 3 Polynomials InText Questions 10
Result: From the graph we observe that 3 and -2 are the intersecting points of X – axis.
So, the zeroes of given quadratic polynomial are 3 and -2.

AP SSC 10th Class Maths Solutions Chapter 3 Polynomials InText Questions

ii) Given that y = 6 – x – x2
AP SSC 10th Class Maths Chapter 3 Polynomials InText Questions 12
AP SSC 10th Class Maths Chapter 3 Polynomials InText Questions 11
Result: From the graph we observe that – 3 and 2 are the intersecting points ol X – axis.
So, the zeroes of given quadratic polynomial are – 3 and 2.

Question 2.
Write three quadratic polynomials that have 2 zeroes each.   (Page No. 55)
Answer:
y = x2 – x – 2 having two zeroes, i.e., (2, 0) and (- 1, 0).
y = 3 – 2x – x2 having two zeroes i.e., (1,0) and (- 3, 0).
y = x2 – 3x – 4 having two zeroes i.e., (-1, 0) and (4, 0)

AP SSC 10th Class Maths Solutions Chapter 3 Polynomials InText Questions

Question 3.
Write one quadratic polynomial that has one zero. (Page No. 55)
Answer:
Quadratic Polynomial y = x2 – 6x + 9 has only one zero i.e., 3.

Question 4.
How will you verify if a quadratic polynomial it has only one zero?  (Page No. 55)
Answer:
If the graph of the given quadratic polynomial touches X – axis at exactly one point, then I can confirm it has only one zero.

Question 5.
Write three quadratic polynomials that have no zeroes for x that are real numbers.  (Page No. 55)
Answer:
The quadratic polynomials y = 2x2 – 4x + 5 and y = – 3x2 + 2x – 1 and y = x2 – 2x + 4 have no zeroes.

Question 6.
Find the zeroes of cubic polynomials
(i) – x3
(ii) x2 – x3
(iii) x3 – 5x2 + 6x
without drawing the graph of the polynomial.  (Page No. 57)
Answer:
i) Given polynomial is y = – x3
f(x) = -x3 ; f(x) = 0
x3 = 0
x = \(\sqrt[3]{0}\) = 0
∴ Zero of the polynomial f(x) is only one i.e., 0.

AP SSC 10th Class Maths Solutions Chapter 3 Polynomials InText Questions

ii) Given that y = x2 – x3
f(x) = x2 (1 – x)
f(x) = 0
⇒ x2 (1 – x) = 0
⇒ x2 = 0 and 1 – x = 0
⇒ x = 0 and x = 1
∴ The zeroes of the polynomial f(x) are two i.e., 0 and 1.

iii) Given that x3 – 5x2 + 6x Let f(x) = x3 – 5x2 + 6x
= x(x2 – 5x + 6)
= x(x2 – 2x – 3x + 6)
= x[x(x – 2) – 3(x – 2)]
= x(x – 2) (x – 3)
∴ The zeroes of the polynomial f(x) are x = 0 and x = 2 and x = 3

Do these

Question 1.
Find the zeroes of the quadratic polynomials given below. Find the sum and product of the zeroes and verify relationship to the coefficients of terms in the polynomial. (Page No. 62)
i) p(x) = x2 – x – 6
ii) p(x) = x2 – 4x + 3
iii) p(x) = x2 – 4
iv) p(x) = x2 + 2x + 1
Answer:
i) Given polynomial p(x) = x2 – x – 6
We have x2 – x – 6 = x2 – 3x + 2x – 6
= x(x – 3) + 2(x – 3)
= (x – 3) (x + 2)
So, the value of x2 – x – 6 is zero when x – 3 = 0 or x + 2 = 0
i.e., x = 3 or x = -2
So, the zeroes of x2 – x – 6 are 3 and – 2.
∴ Sum of the zeroes = 3 – 2 = 1
= – \(\frac{\text { Coefficient of } x}{\text { Coefficient of } x^{2}}\) = \(\frac{-(-1)}{1}\) = 1
And product of the zeroes = 3 × (-2) = -6
= \(\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\) = \(\frac{-6}{1}\) = -6

ii) p(x) = x2 – 4x + 3
Answer:
Given polynomial p(x) = x2 – 4x + 3
We have, x2 – 4x + 3 = x2 – 3x – x + 3
= x(x – 3) – 1 (x – 3)
= (x – 3) (x – 1)
So, the value of x2 – 4x + 3 is zero when x – 3 = 0 or x – 1 =0, i.e.,
when x = 3 or x = 1 So, the zeroes of x2 – 4x + 3 are 3 and 1
∴ Sum of the zeroes = 3 + 1 = 4
= – \(\frac{\text { Coefficient of } x}{\text { Coefficient of } x^{2}}\) = \(\frac{-(-4)}{1}\) = 4
And product of the zeroes = 3 × 1 = 3
= \(\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\) = \(\frac{3}{1}\) = 3

AP SSC 10th Class Maths Solutions Chapter 3 Polynomials InText Questions

iii) Given polynomial p(x) = x2 – 4
We have, x2 – 4 = (x – 2) (x + 2)
So, the value of x2 – 4 is zero
when x – 2 = 0 or x + 2 = 0
i.e., x = 2 or x = – 2
So the zeroes of x2 – 4 are 2 and – 2
∴ Sum of the zeroes = 2 + (- 2) = 0
= – \(\frac{\text { Coefficient of } x}{\text { Coefficient of } x^{2}}\) = \(\frac{-0}{1}\) = 0
And product of the zeroes = 2 × (-2) = -4
= \(\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\) = \(\frac{-4}{1}\) = -4

iv) Given polynomial p(x) = x2 + 2x + 1
We have x2 + 2x + 1 = x2 + x + x + 1
= x(x + 1) + l(x + 1)
= (x + 1) (x + 1)
So, the value of x2 + 2x + 1 is zero
when x + 1 = 0 (or) x + 1 = 0, i.e.,
when x = – 1 or – 1
o, the zeroes of x2 + 2x + 1 are – 1 and – 1.
∴ Sum of the zeroes = (-1) + (-1) = -2
= – \(\frac{\text { Coefficient of } x}{\text { Coefficient of } x^{2}}\) = \(\frac{-2}{1}\) = -2
And product of the zeroes = (-1) × (-1) = 1
= \(\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\) = \(\frac{1}{1}\) = 1

Question 2.
If α, β and γ are the zeroes of the given cubic polynomials, find the values as given in the table. (Page No. 66)
AP SSC 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.3 1
Answer:
l) Given polynomial is x3 + 3x2 – x – 2.
Comparing given polynomial with ax3 + bx2 + cx + d,
the values are a = 1, b = 3, c = -l, d = -2
α + β + γ = \(\frac{-b}{a}\) = \(\frac{-3}{1}\) = -3
αβ + βγ + γα = \(\frac{c}{a}\) = \(\frac{-1}{1}\) = -1
αβγ = \(\frac{-d}{a}\) = \(\frac{-(-2)}{1}\) = 2

AP SSC 10th Class Maths Solutions Chapter 3 Polynomials InText Questions

2) Given polynomial is 4x3 + 8x2 – 6x – 2
Compare the polynomial with ax3 + bx2 + cx + d = 0
Then a = 4, b = 8, c = – 6 and d = – 2
α + β + γ = \(\frac{-b}{a}\) = \(\frac{-8}{4}\) = -2
αβ + βγ + γα = \(\frac{c}{a}\) = \(\frac{-6}{4}\) = \(\frac{-3}{2}\)
αβγ = \(\frac{-d}{a}\) = \(\frac{-(-2)}{4}\) = \(\frac{1}{2}\)

3) Given polynomial is x3 + 4x2 – 5x – 2
Compare the polynomial with ax3 + bx2 + cx + d = 0
Then a = 1, b = 4, c = – 5 and d = – 2
α + β + γ = \(\frac{-b}{a}\) = \(\frac{-4}{1}\) = -4
αβ + βγ + γα = \(\frac{c}{a}\) = \(\frac{-5}{1}\) = -5
αβγ = \(\frac{-d}{a}\) = \(\frac{-(-2)}{1}\) = 2

4) Given polynomial is x3 + 5x2 + 4
Compare the polynomial with ax3 + bx2 + cx + d = 0
Then a = 1, b = 5, c = 0 and d = 4
α + β + γ = \(\frac{-b}{a}\) = \(\frac{-5}{1}\) = -5
αβ + βγ + γα = \(\frac{c}{a}\) = \(\frac{0}{1}\) = 0
αβγ = \(\frac{-d}{a}\) = \(\frac{-4}{1}\) = -4

AP SSC 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.3 3

Try this

Question 1.
i) Find a quadratic polynomial with zeroes -2 and \(\frac{1}{3}\). (Page No. 64)
Answer:
Let the quadratic polynomial be ax2 + bx + c, a ≠ 0 and its zeroes be α and β.
Here α = – 2 and β = \(\frac{1}{3}\)
Sum of the zeroes = α + β
= -2 + \(\frac{1}{3}\) = \(\frac{-5}{3}\)
Product of the zeroes = αβ
= \(\frac{1}{3}\) × (-2) = \(\frac{-2}{3}\)
∴ ax2 + bx + c is [x2 – (α + β)x + αβ]
= [x2 – \(\left(\frac{-5}{3}\right)\)x + \(\left(\frac{-2}{3}\right)\)]
the quadratic polynomial will be 3x2 + 5x – 2.

AP SSC 10th Class Maths Solutions Chapter 3 Polynomials InText Questions

ii) What is the quadratic polynomial whose sum of zeroes is \(\frac{-3}{2}\) and the product of zeroes is -1.
Answer:
Let the quadratic polynomial be ax2 + bx + c and its zeroes be α and β.
Here α + β = \(\frac{-3}{2}\) and αβ = -1
Thus, the polynomial formed = x2 – (α + β)x + αβ
= x2 – \(\left(\frac{-3}{2}\right)\)x + (-1)
= x2 + \(\frac{3x}{2}\) – 1
The other polynomials are (x2 + \(\frac{3x}{2}\) – 1)
then the polynomial is 2x2 + 3x – 2.

AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise

AP State Syllabus SSC 10th Class Maths Solutions 5th Lesson Quadratic Equations Optional Exercise

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 5 Quadratic Equations Optional Exercise Textbook Questions and Answers.

10th Class Maths 5th Lesson Quadratic Equations Optional Exercise Textbook Questions and Answers

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

AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise

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

AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise

Question 3.
A piece of wire 8m in length is cut into twp pieces and each piece is bent into a square. Where should the cut in the wire be made if the sum of the areas of these squares is to be 2 m2?
AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise 1
Answer:
Let the length of the first peice = x m
Then length of the second piece = 8 – x m
∴ Side of the 1st square = \(\frac{x}{4}\) m and
Side of the second square = \(\frac{8-x}{4}\) m
sum of the areas = 2 m2
AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise 2
⇒ x2 + 64 + x2 – 16x = 16 × 2 = 32
⇒ 2x2 – 16x + 64 = 32
⇒ 2x2 – 16x + 32 = 0
⇒ 2(x2 – 8x + 16)= 0
⇒ x2 – 8x + 16 = 0
⇒ x2 – 4x – 4x + 16 = 0
⇒ x(x – 4) – 4(x – 4) = 0
⇒ (x – 4) (x – 4) = 0
∴ x = 4
∴ The cut should be made at the centre making two equal pieces of length 4 m, 4 m.

Question 4.
Vinay and Praveen working together can paint the exterior of a house in 6 days. Vinay by himself can complete the job in 5 days less than Praveen. How long will it take Vinay to complete the job by himself?
Answer:
Let the time taken by Vinay to complete the job = x days
Then the time taken by Praveen to complete the job = x + 5 days
Both worked for 6 days to complete a job.
∴ Total Work done by them is
AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise 3
⇒ 6(2x + 5) = x2 + 5x
⇒ x2 – 7x – 30 = 0
⇒ x2 – 10x + 3x – 30 = 0
⇒ x(x – 10) + 3(x – 10) = 0
⇒ (x – 10) (x + 3) = 0
⇒ x – 10 = 0 (or) x + 3 = 0
⇒ x = 10 (or) x = -3
∴ x = 10 (∵ x can’t be negative)
∴ Time taken by Vinay = x = 10 days
Time taken by Praveen = x + 5 = 15 days.

AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise

Question 5.
Show that the sum of the roots of a quadratic equation ax2 + bx + c = 0 is \(\frac{-b}{a}\).
Answer:
Let the Q.E. = ax2 + bx + c = 0 (a ≠ 0)
⇒ ax2 + bx = -c
AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise 4
AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise 5
∴ Sum of roots of a Q.E. is \(\frac{-b}{a}\)

Question 6.
Show that the product of the roots of a quadratic equation ax2 + bx + c = 0 is \(\frac{c}{a}\).
Answer:
Let the Q.E. = ax2 + bx + c = 0 (a ≠ 0)
⇒ ax2 + bx = -c
AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise 6
AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise 7

AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise

Question 7.
The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is 2\(\frac{16}{21}\) find the fraction.
Answer:
Let the numerator = x
then denominator = 2x + 1
Then the fraction = \(\frac{x}{2x+1}\)
Its reciprocal = \(\frac{2x+1}{x}\)
AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise 8
105x2 + 84x + 21 = 116x2 + 58x
11x2 – 26x – 21 = 0
11x2 – 33x + 7x – 21 = 0
11x (x – 3) + 7 (x – 3) = 0
(x – 3) (11x + 7) = 0
⇒ x – 3 = 0 (or) 11x + 7 = 0
⇒ x = 3 (or) \(\frac{-7}{11}\)
∴ x = 3
Numerator = 3;
Denominator = 2 × 3 + 1 = 7
Fraction = \(\frac{3}{7}\).

Question 8.
A ball is thrown vertically upwards from the top of a building of height 29.4m and with an initial velocity 24.5m/sec. If the height H of the ball from the ground level is given by H = 29.4 + 24.5t – 4.9t2, then find the time taken by the ball to reach the ground.
Answer:
AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise 9
Initial velocity ‘U’ = 24.5
height of the ball from the ground can be expressed as
H = 29.4 + 24.5 t – 4.9 t2
The ball has to reach the ground in ‘t’ seconds, which means Height from ground H = 0
So 29.4 + 24.5t – 4.9t2 = 0 = H
⇒ 4.9 t2 – 24.5t – 29.4 = 0
⇒ 4.9 [t2 – 5t – 6] = 0
∴ t2 – 5t – 6 = 0
⇒ t2 – 6t + t – 6 = 0
⇒ t(t – 6) + 1 (t – 6) = 0
(t – 6) (t + 1) = 0
⇒ t – 6 = 0
∴ t = 6 or t + 1 = 0
⇒ t = -1 but ‘t’ cannot be negative
So t = 6
it means in 6 seconds of time the ball reaches ground.

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

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

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

10th Class Maths 9th Lesson Tangents and Secants to a Circle Optional Exercise Textbook Questions and Answers

Question 1.
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line – segment joining the points of contact at the centre.
Answer:
AP SSC 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Optional Exercise 1
Given: A circle with centre ‘O’.
Two tangents \(\overleftrightarrow{\mathrm{PQ}}\) and \(\overleftrightarrow{\mathrm{PT}}\) from an external point P. Let Q, T be the points of contact.
R.T.P: ∠P and ∠QOT are supplementary.
Proof: OQ ⊥ PQ
[∵ radius is perpendicular to the tangent at the point of contact] also OT ⊥ PT
∴ ∠OQP + ∠OTP = 90° + 90° = 180° Nowin oPQOT,
∠OTP + ∠TPQ + ∠PQO + ∠QOT
= 360° (angle sum property)
180° + ∠P + ∠QOT = 360°
∠P + ∠QOT = 360°- 180° = 180° Hence proved. (Q.E.D.)

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

Question 2.
PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T (See figure). Find the length of TP.
AP SSC 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Optional Exercise 3
Answer:
AP SSC 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Optional Exercise 4
Given: PQ = 8
⇒ PR = 4
⇒ PO2 = PR2 + OR2
⇒ 25 = 16 + OR2
⇒ OR = 3
Now let RT = x and PT in △OPT, ∠P = 90°
∴ OT is hypotenuse.
∴ OT2 = OP2 + PT2
(Pythagoras theorem)
(3 + x)2 = 52 + y2 …….. (1)
and in △PRT, ∠R = 90°
∴ \(\overline{\mathrm{PT}}\) is hypotenuse.
∴ PT2 = PR2 + RT2
y2 = 42 + x2 …….. (2)
Now putting the value of y2 = 42 + x2 in equation (1) we got
(3 + x)2 = 52 + x2 + 42
9 + x2 + 6x = 25 + 16 + x2
6x = 25 + 16 – 9 = 25 + 7 = 32
⇒ x = \(\frac{32}{6}\) = \(\frac{16}{3}\)
Now from equation (2), we get
AP SSC 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Optional Exercise 5

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

Question 3.
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Answer:
AP SSC 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Optional Exercise 6
Given: Let a circle with centre ‘O’ touches the sides AB, BC, CD and DA of a quadrilateral ABCD at the points P, Q, R and S respectively.
R.T.P: ∠AOB + ∠COD = 180°
∠AOD + ∠BOC = 180°
Construction: Join OP, OQ, OR and OS.
Proof: Since the two tangents drawn from an external point of a circle subtend equal angles.
At the centre,
∴ ∠1 = ∠2
∠3 = ∠4 (from figure)
∠5 = ∠6
∠7 = ∠8
Now, ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°
[∵ Sum of all the angles around a point is 360°]
So, 2 (∠2 + ∠3 + ∠6 + ∠7) = 360°
and 2 (∠1 + ∠8 + ∠4 + ∠5) = 360°
(∠2 + ∠3) + (∠6 + ∠7) = \(\frac{360}{2}\) = 180°
Also, (∠1 + ∠8) + (∠4 + ∠5) = \(\frac{360}{2}\) = 180°
So, ∠AOB + ∠COD = 180°
[∵ ∠2 + ∠3 = ∠AOB;
∠6 + ∠7 = ∠COD
∠1 + ∠8 = ∠AOD
and ∠4 + ∠5 = ∠BOC [from fig.]]
and ∠AOD + ∠BOC = 180°

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

Question 4.
Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.
Answer:
AP SSC 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Optional Exercise 2
Steps of construction:

  1. Draw a line segment AB of length 8 cm.
  2. With A and B as centres and 4 cm, 3 cm as radius draw two circles.
  3. Draw the perpendicular bisectors \(\stackrel{\leftrightarrow}{\mathrm{XY}}\) of AB. Let \(\stackrel{\leftrightarrow}{\mathrm{XY}}\) and AB meet at M.
  4. Taking M as centre and MA or MB as radius draw a circle which cuts the circle with centre A at P and Q and circle with centre B at R, S.
  5. Join BP, BQ and AR, AS.

Question 5.
Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
Answer:
AP SSC 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Optional Exercise 7
Steps of construction:

  1. Draw AABC such that AB = 6 cm; ∠B = 90° and BC – 8 cm.
  2. Drop a perpendicular BD from B on AC.
  3. Draw the circumcircle to ABCD. Let ‘E’ be its centre.
  4. Join AE and draw its perpendicular bisector \(\stackrel{\leftrightarrow}{\mathrm{XY}}\). Let it meet AE at M.
  5. Taking M as centre and MA or ME as radius draw a circle, which’ cuts the circumcircle of △BCD at P and B.
  6. Join AP and extend AB, which are the required tangents.

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

Question 6.
Find the area of the shaded region in the figure, given in which two circles with centres A and B touch each other at the point C. If AC = 8 cm. and AB = 3 cm.
AP SSC 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Optional Exercise 8
Answer:
Given: Two circles with centres A and B, whose radii are 8 cm and 5 cm.
[∵ AC = 8 cm, AB = 3 cm ⇒ BC = 8 – 3 = 5 cm]
Area of the shaded region = (Area of the larger circle) – (Area of the smaller circle)
AP SSC 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Optional Exercise 9

Question 7.
ABCD is a rectangle with AB = 14 cm. and BC = 7 cm. Taking DC, BC and AD as diameters, three semicircles are drawn as shown in the figure. Find the area of the shaded region.
AP SSC 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Optional Exercise 10
Answer:
Given AB = 14 cm, AD = BC = 7 cm Area of the shaded and unshaded region
= (2 × Area of the semi-circles with AD as diameter) + Area of the rectangle
AP SSC 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Optional Exercise 11

AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 4th Lesson Pair of Linear Equations in Two Variables Optional Exercise

10th Class Maths 4th Lesson Pair of Linear Equations in Two Variables Optional Exercise Textbook Questions and Answers

Question 1.
i) \(\frac{2x}{a}\) + \(\frac{y}{b}\) = 2
\(\frac{x}{a}\) – \(\frac{y}{b}\) = 4
Answer:
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise 1
Substituting x = 2a in the equation (1) we get
\(\frac{2}{a}\)(2a) + \(\frac{y}{b}\) = 2
⇒ 4 + \(\frac{y}{b}\) = 2
⇒ \(\frac{y}{b}\) = -2
⇒ y = -2b
∴ The solution (x, y) = (2a, -2b)

AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise

ii) \(\frac{x+1}{2}\) + \(\frac{y-1}{3}\) = 8
\(\frac{x-1}{3}\) + \(\frac{y+1}{2}\) = 9
Answer:
Given: \(\frac{x+1}{2}\) + \(\frac{y-1}{3}\) = 8
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise 2
⇒ 3x + 2y + 1 = 48
⇒ 3x + 2y = 47 …… (1)
and \(\frac{x-1}{3}\) + \(\frac{y+1}{2}\) = 9
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise 3
⇒ 2x + 3y + 1 = 54
⇒ 2x + 3y = 53 …… (2)
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise 4
⇒ y = \(\frac{-65}{-5}\) = 13
Substituting y = 13 in equation (1) we get
3x + 2(13) = 47
⇒ 3x = 47 – 26
⇒ 3x = 21
⇒ x = \(\frac{21}{3}\) = 7
∴ The solution (x, y) = (7, 13)

iii) \(\frac{x}{7}\) + \(\frac{y}{3}\) = 5
\(\frac{x}{2}\) – \(\frac{y}{9}\) = 6
Answer:
Given: \(\frac{x}{7}\) + \(\frac{y}{3}\) = 5 and \(\frac{x}{2}\) – \(\frac{y}{9}\) = 6
⇒ \(\frac{3x+7y}{21}\) = 5 and \(\frac{9x-2y}{18}\) = 6
⇒ 3x + 7y = 105 …….. (1) and
9x – 2y = 108 …….. (2)
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise 5
⇒ y = \(\frac{207}{23}\) = 9
Substituting y = 9 in equation (1) we get
3x + 7(9) = 105
⇒ 3x = 105 – 63
⇒ 3x = 42
⇒ x = \(\frac{42}{3}\) = 14
∴ The solution (x, y) = (14, 9)

AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise

iv) √3x + √2y = √3
√5x + √3y= √3
Answer:
Given that √3x + √2y = √3 …… (1)
√5x + √3y = √3 …… (2)
By following elimination method
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise 6
Now again following elimination method
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise 7
∴ The solution x = \(\frac{3-\sqrt{6}}{3-\sqrt{10}}\) and y = \(\frac{3-\sqrt{15}}{3-\sqrt{10}}\)

v) \(\frac{ax}{b}\) + \(\frac{by}{a}\) = a + b
ax – by = 2ab
Answer:
Given: \(\frac{ax}{b}\) + \(\frac{by}{a}\) = a + b ……. (1)
ax – by = 2ab …….. (2)
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise 8
Substituting y = – a in equation (2)
we get ax – b(-a) = 2ab
ax + ab = 2ab
ax = 2ab – ab = ab
⇒ x = \(\frac{ab}{a}\) = b
∴ (x, y) = (b, -a)

AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise

vi) 2x + 3y = 17
2x+2 – 3y+1 = 5
Answer:
Given: 2x + 3y = 17 and
2x+2 – 3y+1 = 5
Take 2x = a and 3y = b then the given equations reduce to
2x + 3y = 17 ⇒ a + b = 17 …… (1)
2x . 22 – 3y . 3 = 5 ⇒ 4a – 3b = 5 …… (2)
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise 9
Substituting b = 9 in equation (1) we get
a + 9 = 17 ⇒ a = 17 – 9 = 8
But a = 2x – 8 and b = 3y = 9
⇒ 2x = 23 and 3y = 32
⇒ x = 3 and y = 2
∴ The solution (x, y) is (3, 2)

AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise

Question 2.
Animals in an experiment are to be kept on a strict diet. Each animal is to receive among other things 20g of protein and 6g of fat. The laboratory technicians purchased two food mixes, A and B. Mix A has 10% protein and 6% fat. Mix B has 20% protein and 2% fat. How many grams of each mix should be used?
Answer:
Let x gms of mix A and y gms of mix B are to be mixed, then
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise 10
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise 11
Substituting y = 60 in equation (1)
we get x + 2 × 60 = 200
⇒ x + 120 = 200
⇒ x = 200 – 120 = 80 gm
∴ Quantity of mix. A = 80 gms.
Quantity of mix. B = 60 gms.

AP SSC 10th Class Maths Solutions Chapter 13 Probability Optional Exercise

AP State Syllabus SSC 10th Class Maths Solutions 13th Lesson Probability Optional Exercise

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 13 Probability Optional Exercise Textbook Questions and Answers.

10th Class Maths 13th Lesson Probability Optional Exercise Textbook Questions and Answers

Question 1.
Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on
(i) the same day?
(ii) consecutive days?
(iii) different days?
Answer:
Shyam and Ekta can visit the shop in the following combination:
AP SSC 10th Class Maths Solutions Chapter 13 Probability Optional Exercise 1
(T Tu) (W, Tu) (Th, Tu) (F, Tu) (S, Tu) (Tu, W) (W, W) (Th, W) (F, W) (S, W) (Tu, Th) (W, Th) (Th, F) (F, Th) (S, Th) (Tu, F) (W, F) (Th, S) (F, F) (S, F) (Tu, S) (W, S) (Th, Th) (F, S) (S, S)
∴ Number of total outcomes = 5 × 5 = 52 = 25 [also from the above table]
i) Number of favourable outcomes to that of visiting on the same day
(Tu, Tu), (W, W), (Th, Th), (F, F), (S, S) = 5
∴ Probability of visiting the shop on the same day = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\) = \(\frac{5}{25}\) = \(\frac{1}{5}\)
ii) Number of outcomes favourable to consecutive days
(Tu, W), (W, Th), (Th, F), (F, S), (W, Tu), (Th, W), (F, Th), (S, F) = 8
∴ Probability of visiting the shop on consecutive days = \(\frac{8}{25}\)
iii) If P(E) is the probability of visiting the shop on the same day,
then P(\(\overline{\mathrm{E}}\)) is the probability of visiting the shop not on the same day.
i.e., P(\(\overline{\mathrm{E}}\)) is the probability of visiting the shop on different days.
Such that P(E) + P(\(\overline{\mathrm{E}}\)) = 1
P(\(\overline{\mathrm{E}}\)) = 1 – P(E) = 1 – \(\frac{1}{5}\) = \(\frac{4}{5}\)

AP SSC 10th Class Maths Solutions Chapter 13 Probability Optional Exercise

Question 2.
A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.
Answer:
Number of red balls in the bag = 5 As the probability of blue balls is double the probability of red balls, we have that number of blue balls is double the number of red balls.
∴ Blue balls = 5 × 2 = 10.
[!! Let the number of blue balls = x
Number of red balls = 5
Total no. of balls = x + 5
Total outcomes in drawing a ball at random = x + 5
Number of outcomes favourable to red ball = 5
∴ P(R) = \(\frac{5}{x+5}\)
from the problem,
P(B) = 2 × \(\frac{5}{x+5}\) = \(\frac{10}{x+5}\)
Also \(\frac{5}{x+5}\) + \(\frac{10}{x+5}\) = 1
[∵ P(E) + P(\(\overline{\mathrm{E}}\)) = 1]
⇒ \(\frac{5+10}{x+5}\) = 1
⇒ \(\frac{15}{x+5}\) = 1
⇒ x + 5 = 15]

Question 3.
A box contains 12 balls out of which x are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball? If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x.
Answer:
Number of black balls = x
Total number of balls in the box = 12
Probability of drawing a black ball = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\) = \(\frac{x}{12}\) …….. (1)
When 6 more black balls are placed in the box, number of favourable outcomes to black ball becomes = x + 6.
Total number of balls in the box becomes = 12 + 6 = 18.
Now the probability of drawing a black ball become = \(\frac{x+6}{18}\) …….. (1)
By Problem,
\(\frac{x+6}{18}\) = 2. \(\frac{x}{12}\)
⇒ \(\frac{x+6}{18}\) = \(\frac{x}{6}\)
⇒ 6(x + 6) = 18(x)
⇒ 6x + 36 = 18x
⇒ 18x – 6x = 36
⇒ 12x = 36
⇒ x = \(\frac{36}{12}\) = 3
Check:
AP SSC 10th Class Maths Solutions Chapter 13 Probability Optional Exercise 2
and hence proved.

AP SSC 10th Class Maths Solutions Chapter 13 Probability Optional Exercise

Question 4.
A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is 2/3. Find the number of blue marbles in the jar.
Answer:
Total number of marbles in the jar = 24.
Let the number of green marbles = x.
Then number of blue marbles = 24 – x.
Probability of drawing a green marbles = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\) = \(\frac{x}{24}\)
By problem,
\(\frac{x}{24}\) = \(\frac{2}{3}\)
⇒ 3x = 24 × 2
x = \(\frac{24 \times 2}{3}\) = 16
∴ Number of green marbles = 16
Number of blue marbles = 24 – 1 = 8
∴ Probability of picking blue marble = \(\frac{8}{24}\) = \(\frac{1}{3}\)
(OR)
P(B) = P(E) – P(G) = 1 – \(\frac{2}{3}\) = \(\frac{1}{3}\)
[!! P(G) = \(\frac{2}{3}\)
P(G) + P(B) = 1
∴ P(B) = 1 – P(G) = 1 – \(\frac{2}{3}\) = \(\frac{1}{3}\)
Number of blue marbles in the jar = \(\frac{1}{3}\) × 24 = 8]

AP Board 6th Class Social Studies Solutions Chapter 7 Emergence of Kingdoms and Republics

SCERT AP Board 6th Class Social Solutions 7th Lesson Emergence of Kingdoms and Republics Textbook Questions and Answers.

AP State Syllabus 6th Class Social Studies Solutions 7th Lesson Emergence of Kingdoms and Republics

6th Class Social Studies 7th Lesson Emergence of Kingdoms and Republics Textbook Questions and Answers

Improve Your Learning

Question 1.
What do you mean by Gana? How were they different from the kingdoms ruled by the kings?
Answer:
The term ‘gana’ means people of equal status. Sangha means ‘assembly’. Gana Sangha means an assembly of equal-status people. They cover a small area that was ruled by a superior group among them. These gana sanghas practiced “all are equal” traditions.
A kingdom means a territory and was ruled by a king or queen. In a kingdom, a family which rules for a long, period becomes a dynasty.

Question 2.
Why did the Rajas of Mahajanapadas build forts?
Answer:

  1. The rajas of Mahajanapadas built forts to protect their capital city.
  2. Huge walls of wood, brick or stone were built around the cities.
  3. Forts were probably built because people were afraid of attacks from other kings and needed protection.
  4. Some rulers wanted to show their wealth and strength by building large, tall, and impressive walls around their cities.
  5. The land and the people living inside the forts could be controlled more easily by the king.

AP Board 6th Class Social Studies Solutions Chapter 7 Emergence of Kingdoms and Republics

Question 3.
Can you point out the difference between the way villages are managed today and in the time of Mahajanapadas?
Answer:
Nowadays, regular elections are taking place in villages which promotes democracy in our country. But in earlier times there was a ruler who controlled the village which did not help the democracy, but it helped in the monarch system which went for a long time in our country.

Question 4.
Find out how the craftspersons are; taxed by the government today? Was it the same in the times of Mahajanapadas?
Answer:
Craft persons have to pay taxes at the time of Mahajanapadas. Sometimes they have to work free of charge for the king for one day of every month.
But today if a craft person earns money more than the specified amount by the government they have to pay the tax for the excess amount they earned. There is no chance to work instead of paying tax.

Question 5.
Through what sources do you know about Mahajanapadas?
Answer:
We can know much about those villages and towns from two kinds of sources from archaeological excavations in different places and from the books composed during that period:

Question 6.
Write the agricultural practices that led to an increase in agricultural production in the time of Mahajanapadas.
Answer:
Two major changes were practiced during the Mahajanapadas period.

  1. Iron ploughshares were used. Heavy, clayey soil could be turned over better than with a wooden ploughshare, so that more grain could be produced.
  2. People of Mahajanapadas began transplanting paddy. Instead of scattering seeds on the ground, from which plants would sprout, saplings were grown and then planted in the fields. Production developed due to this since many plants survived.

AP Board 6th Class Social Studies Solutions Chapter 7 Emergence of Kingdoms and Republics

Question 7.
“ The Mahajanapadas developed on the river banks.” Do you agree or not? Justify your answer.
Answer:
All the Mahajanapadas developed on the river banks. I agree with this statement. Archeologists found hundreds of sites in the Ganges valley. As this plain receives very high rainfall, it is very fertile. These rivers bring silt from the Himalayas and flow throughout the year. Transportation is also easy from these places. So Mahajanapadas developed on the river banks. If we observe all the Mahajanapadas, all the janapadas emerged around the river banks of Ganga, Yamuna, Narmada and Godavari.

Question 8.
How do you appreciate the works of craftsmen in the times of Mahajanapadas?
Answer:
During the period of Mahajanapadas, craftsmen played a key role in the development of Mahajanapadas. Availability of iron facilitated craft production also. Blacksmiths made necessary tools for agriculture like ploughshares, sickles, axes, arrows etc., With the use of iron ploughshare productivity improved. Potters made pots for cooking and storing grains. Carpenters made carts and with the help of these carts transportation was made easy. Weavers weaved cloth which was exported to other places and it helped the economy of Mahajanapadas. Potters made special type of pottery known as painted grey ware, which became famous in those days.
In this way craftsmen participated in the development of Mahajanapadas.

Question 9.
What were the taxes collected by the rulers of the Mahajanapadas?
Answer:
The taxes collected by the Mahajanapadas were :

  1. 1/6th of the total agricultural produce as a tax on crops.
  2. Craftsmen had to pay taxes in the form of labor.
  3. Taxes on the sale and purchase of goods and services for trade.
  4. Taxes on herders in the form of animals or animal products and taxes on hunters and gatherers in the nature of their collection from forests.

AP Board 6th Class Social Studies Solutions Chapter 7 Emergence of Kingdoms and Republics

Question 10.
How are present-day elections different from the way in which rulers were chosen in Janapadas?
Answer:
Choosing of rulers in ‘janapadas’ – Men were chosen ‘rajas’ by performing big sacrifices. The ‘Ashwamedha was one such ritual that was used to identify a ‘raja’. The ‘raja’ chosen by this sacrifice was considered very powerful.
Electing rulers today – Today, we have a democratic system of government. Each citizen has a right to cast his vote and to form the government through his elected representative.

Question 11.
What is similar in the way crops were grown in the Mahajanapadas and how they are grown today?
Answer:
The crops that were grown in the Mahajanapadas were wheat, barley, peas, and lentils. These crops are grown in the same way as those were grown in the ancient days.
In the time of Mahajanapads, they planted paddy saplings instead of grains.
Even today the same system was followed by the farmers.

Question 12.
How can you appreciate the role of natural resources in the emergence of Magadha as a powerful kingdom?
Answer:
The rivers made the land very fertile and the Grihapatis could irrigate their lands easily. The rivers were also used for transporting goods and armies. Parts of the Magadha were forested. Elephants were captured from there and trained for fighting in the armies. Wood from the forests was used for building fortresses and palaces and chariots. In the southern parts of Magadha, there were iron ore deposits that could be used for making weapons, etc.
All this enabled Magadha to emerge as a very powerful kingdom. The kingdom extended from the northwest part of India to Odisha.
Thus Magadha used the natural wealth of the region to build a powerful kingdom.

Question 13.
Locate the 16 Mahajanapadas and their capitals in the following India outline map.
Answer:
AP Board 6th Class Social Studies Solutions Chapter 7 Emergence of Kingdoms and Republics 1

AP Board 6th Class Social Studies Solutions Chapter 7 Emergence of Kingdoms and Republics

Question 14.
Solve the puzzle with new terms you have learned in this lesson. Take the support of your teacher.
AP Board 6th Class Social Studies Solutions Chapter 7 Emergence of Kingdoms and Republics 2
Down ↓

  1. The mahajanapada located on the extreme eastern side
  2. This is the capital city of Avanti
  3. Farmers used this metal to make plough shares.
  4. The Mahajanapada located on the banks of river Godavari
  5. The slaves employed at Gahapatis
  6. Kusinara is the capital of this kingdom
  7. The capital city of Surasena Cross

Cross →

  1. The powerful kingdom of Mahajanapadas
  2. another name for Varanasi
  3. Kaushambi is the capital of this Mahajanapada
  4. Mahabharata tells us about the battle among the kings of this mahajanapada
  5. 1 /6th of farm produce collected as tax from farmers
  6. These condemned caste systems and the use of yagnas
  7. Vajji has this type of government.

Answer:
AP Board 6th Class Social Studies Solutions Chapter 7 Emergence of Kingdoms and Republics 4

AP Board 6th Class Social Studies Solutions Chapter 7 Emergence of Kingdoms and Republics

Project Work

Collect the information about 16 Mahajanapadas, and the state, in which they were located. Prepare a table as given below. Refer to India Political map given.
AP Board 6th Class Social Studies Solutions Chapter 7 Emergence of Kingdoms and Republics 3
Answer:

S.No.MahajanapadaCapitalModern Location / State
1.AngaChampaWest Bengal
2.MagadhaGirivraja/RajagrihaGaya & Patna / Bihar
3.KasiBanaras/VaranasiUttar Pradesh
4.VatsaKausambiAround Allahabad / UP
5.KosalaShravastiModern Avadh region / Eastern UP
6.SurasenaMathuraWestern UP region
7.KuruIndraprastaMeerut & Southeastern Haryana
8.MatsyaViratnagarAround Jaipur
9.ChediSothiratiBundelkhand region
10.AvantiUjjaini/MahismatiAround Malwa MP
11.GandharaTaxilaRawalpindi/Pakistan
12.KambojaPoonchaKashmir & Hindukush
13.AsmakaPratisthan / PothanTelangana & Maharashtra
14.VajjiVaishaliBihar
15.MallaKusiharaDeoria & UP
16.PanchalaAhichatra/KampliyaWestern UP

6th Class Social Studies 7th Lesson Emergence of Kingdoms and Republics InText Questions and Answers

Let’s Do

(Textbook Page No. 75)

Question 1.
Look at the physical map of India and identify the plains through which the rivers Ganga and Yamuna flow
Answer:
AP Board 6th Class Social Studies Solutions Chapter 7 Emergence of Kingdoms and Republics 5

AP Board 6th Class Social Studies Solutions Chapter 7 Emergence of Kingdoms and Republics

Question 2.
Identify the modern cities of India like Delhi, Allahabad, Varanasi, Lucknow and Patna.
Answer:
AP Board 6th Class Social Studies Solutions Chapter 7 Emergence of Kingdoms and Republics 6

Question 3.
Do you think this area will be similar to your villages? Give your reasons.
Answer:
I don’t think that our villages are similar to that of Delhi, Allahabad, Varanasi, Lucknow, and Patna. They are well developed from the time of Mahajanapadas itself. Because of high fertility lands and transportation facilities these areas developed from that time. Now they became big cities and Delhi being our national capital. So, we can’t compare our villages to the cities mentioned above.

(Textbook Page No. 77)

Question 4.
The important Janapadas of those times are shown on the map.
Look at the map and fill in the blanks.
1. The …………. Janapada was settled on both the banks of the Yamuna.
2. The Panchala was settled on both the banks of the river …………..
3. The ………….. Janapada was situated on the western side of the Surasena.
4. The ………….. Janapada was on the extreme north.
5. The ………….. Janapada was situated on the banks of the river Godavari.
6. The Gandhara was situated on the banks of the river …………..
Answer:
1) Kuru
2) Ganga
3) Matsya
4) Kambhoja
5) Asmaka
6) Kubha

AP Board 6th Class Social Studies Solutions Chapter 7 Emergence of Kingdoms and Republics

Think and Respond

(Textbook Page No. 75)

Question 1.
Find out the names of a few Janas (tribes) who initially settled down in the Indo – Gangetic plain,
Answer:

  1. Magadha dynasty
  2. Imperial Kanauj
  3. Mughal Empire
  4. Maratha Empire are some of the janapadas who initially settled down in the Indo – Gangetic plains.

Question 2.
What do you mean by Janapada? How is it different from Mahajanapadas?
Answer:
People of different tribes settled down to practice agriculture in many parts of the valley. These tribes were called Jana and the place they settled was called Janapada. Many groups of such villages and towns are called Mahajanapadas.

Question 3.
How is paddy grown today? (Textbook Page No. 78)
Answer:

  1. Land is to be prepared and levelled.
  2. Planting on time to be done.
  3. Fertilization to be done.
  4. Field is to be watered.
  5. Pests is to be controlled.
  6. Harvest on time.
  7. Storing safely.
  8. Milling efficiently.

AP Board 6th Class Social Studies Solutions Chapter 7 Emergence of Kingdoms and Republics

Question 4.
Describe the relationship between Grihapatis and Craftsmen of the village. (Textbook Page No. 79)
Answer:
In most villages, there Were craft persons like blacksmiths who made tools necessary for agriculture (like ploughshares, sickles, axes, arrows, etc), potters who made pots for cooking and storing grains, carpenters who made carts, ploughs, furniture etc., and weavers who wove cloth for the villagers.
Probably the Grihapatis gave them grains in return for their products. These craft products were necessary for agriculture, but the Grihapatis may not have had the time or skill to make them.

(Textbook Page No. 80)

Question 5.
The people of the towns would haMe needed grain, milk, meat etc. How do you think they got them if most townspeople were not doing any farming?
Answer:
Even though the townspeople were riot doing, any farming they get milk and meat by purchasing them from nearby villages and shops situated in the town.

Question 6.
Have you ever seen a fort on TV or have you ever been there? Why were there big walls around the fort?
Answer:
I have seen Red fort. I had been to Delhi and I have seen the Red Fort there. There are big walls around the fort. To protect the enemy kings by not entering them into the fort big walls were constructed around the fort.

AP Board 6th Class Social Studies Solutions Chapter 7 Emergence of Kingdoms and Republics

Question 7.
What were they made of? How do you think the kings managed to get the wealth needed for all this?
Answer:
The Red fort got its name from the stone used to construct it. The entire fort is made up of red sandstone. Maybe it was constructed with the tax amount collected and conquering other kingdoms.

Question 8.
Why did the king of Mahajanapadas need armies?
Answer:
The kings of Mahajanapadas were afraid of attacks from other kings and enemies. They needed protection. So the king of Mahajanapada needed armies.

(Textbook Page No. 81)

Question 9.
In which form the hunter-gatherers paid taxes to their kings?
Answer:
Hunters and gatherers paid taxes to the Raja from what they got from the forest like hides, woods, and honey etc.

Question 10.

  1. If everyone was forced to give away a part of their earnings as tax, how did it affect their lives?
    Answer:
    If everyone was forced to give away a part of their earnings, they will lose the earning for that part which was given as tax.
  2. Why do you think they agreed to pay the taxes? Do you think they benefited in any way from the new arrangements?
    Answer:
    They will get encouragement, and support from the king – besides protection.
  3. What is bhaga? Does the government of our times take the produce of farmers in , a similar way?
    Answer:
    The Grihapatis had to divide their crops into six parts and one part of them is to be given to king as tax. This was called bhaga.
    The government of our times does not collect produce from the farmers. The government collects tax from the farmers as per the quantity sold.

AP Board 6th Class Social Studies Solutions Chapter 7 Emergence of Kingdoms and Republics

(Textbook Page No. 82)

Question 11.

  1. Why were the kings of Mahajanapadas keen to increase craft production and trade?
    Answer:
    The kings of Mahajanapadas were k^en to increase craft production and trade because they would get more taxes. The wealth of the kingdom will increase.
  2. How did the headmen of the villages benefit from the imposition of taxes by the kings?
    Answer:
    The kings of Mahajanapadas wanted the village headmen to collect the taxes on their behalf. This might have helped the headmen to increase their power in the villages.

(Textbook Page No. 83)

Question 12.
Write a couple of lines on each of the natural resources of Magadha and how it could have been used by the kings.
Answer:

  1. Magadha kingdom was spread on both sides of the river Ganga. The river made the land fertile and the Grihapatis could irrigate the land easily and produce was high.
  2. The river was also used for transporting goods and armies.
  3. Elephants were captured from the forests that spread over in the kingdom and trained for fighting in the wars.
  4. In southern parts of Magadha, there were iron ore deposits that were used for making weapons etc.

Question 13.
Compare the gana form of government in Vajji with the present-day republic. Who was not allowed to participate in the assembly of Vajji Mahajanapada?
Answer:
Vajji had gana form of government which was nearly equal to the present-day form of government. Gana was ruled by a group of leaders instead of a single ruler. In the present republic type of government, we will elect local loaders and they will discuss our problems in the assemblies through discussion and debate.
Women, slaves and wage earners are not allowed to participate in the assembly of vajji mahajanapadas.

AP Board 6th Class Social Studies Solutions Chapter 7 Emergence of Kingdoms and Republics

Explore

Question 1.
A famous religious epic tells us about many of these Janapadas. Find out about it. (Textbook Page No. 77)
Answer:
The Mahabharat.

Do you know

Question 1.
Make a list of the Mahajanapadas and the cities which were situated in the Ganges valley. (Textbook Page No. 76)
Answer:

  1. Kasi – Banaras
  2. Kosala – Shravasti
  3. Anga – Champa
  4. Magadha – Girivraja or Rajagriha
  5. Vajji or Vriji – Vaishali
  6. Malya – Kushinagar
  7. Vatsa – Kausambi
  8. Kuru – Indraprasta/Hastinapur
  9. Pachala – Ahichhtra
  10. Surasena – Mathura

AP Board 7th Class Hindi Solutions 5th Lesson ईमानदारी का फल

SCERT AP Board 7th Class Hindi Solutions 5th Lesson ईमानदारी का फल Textbook Questions and Answers.

AP State Syllabus 7th Class Hindi 5th Lesson Questions and Answers ईमानदारी का फल

7th Class Hindi 5th Lesson ईमानदारी का फल Textbook Questions and Answers

सोचिए-बोलिए
AP Board 7th Class Hindi Solutions 5th Lesson ईमानदारी का फल 1

प्रश्न 1.
चित्र में क्या – क्या दिखाई दे रहे हैं? (చిత్రంలో ఏమేమి కనిపించుచున్నవి?)
उत्तर:
चित्र में कुछ पेड़ – पौधे, फूल, एक झोपडी, एक घर, घास, पर्वत, जड.पी. हाई स्कूल, एक आदमी और एक बालक दिखाई दे रहे हैं। बडे – बडे पत्थर भी दिखाई दे रहे हैं।
(చిత్రంలో కొన్ని చెట్లు, మొక్కలు, పూలు, ఒక గుడిసె, ఒక ఇల్లు, గడ్డి, పర్వతములు, జెడ్.పి స్కూల్, ఒక మనిషి మరియు ఒక బాలుడు కనిపించుచున్నారు.)

AP Board 7th Class Hindi Solutions 5th Lesson ईमानदारी का फल

प्रश्न 2.
लड़का क्या कर रहा है? (బాలుడు ఏమి చేయుచుచున్నాడు?)
उत्तर:
लड़का एक आदमी को रुपयों का बंडल दे रहा है और वह आदमी उसे ले रहा है।
(బాలుడు ఒక మనిషికి రూపాయల కట్టను ఇచ్చుచున్నాడు. ఆ మనిషి దానిని తీసుకొనుచున్నాడు.)

Intext Questions & Answers

प्रश्न 1.
जंगल में क्या – क्या मिलते हैं? (అడవిలో ఏమేమి లభించును?)
उत्तर:
जंगल में तरह – तरह के पेड़, फल, फूल, पत्ते,वनस्पति, खनिज आदि मिलते हैं।
(అడవిలో రకరకాల చెట్లు, పండ్లు, పూలు, ఆకులు, వనస్పతి, ఖనిజ పదార్థములు మున్నగునవి లభించును.)

प्रश्न 2.
परियों के बारे में तुम क्या जानते हो? (దేవదూతల గురించి నీకు ఏమి తెలియును?)
उत्तर:
कहा जाता है कि परियाँ बहुत सुंदर होती हैं। अगर किसी पर महरबान हो जाए तो अपनी जादू की छड़ी से उसकी कोई कामना पूरी कर देती हैं। यों तो परियाँ कई प्रकार की होती हैं – जैसे जंगल परी, पर्वत परी, जल परी आदि।
(దేవతలు చాలా అందంగా ఉంటారని అంటూ ఉంటారు. ఎవరి మీదైనా ప్రేమ, ఇష్టం కల్గినట్లయితే అవి తమ మాయాజాలంతో వారి ఇష్టాన్ని తీరుస్తాయి. సాధారణంగా వీరు చాలా రకాలుగా ఉంటారు. ఉదా : వన దేవత, పర్వత మరియు జల దేవత.)

Improve Your Learning

सुनिए-बोलिए

प्रश्न 1.
नदी से क्या – क्या लाभ हैं? (నదుల వలన ఏమేమి లాభములు కలవు?)
उत्तर:
नदी से हमें कई लाभ मिलते हैं जैसे पीने का पानी मिलता है। नदियों के पानी से सिंचाई का प्रबंध किया जाता है। नदी यातायात का साधन भी हैं। नदियों के पानी को इकट्टा करके बिजली की उत्पत्ति की जाती है।
(నదుల నుండి మనకు అనేక లాభములు కలుగును ఉదాహరాణకి త్రాగునీరు లభించును. నదీ నీటిలో నీటి తడుపు ఏర్పాట్లు చేయబడును. నదులు ప్రయాణ మార్గాలు కూడా. నదీ నీటిని పోగు చేసి విద్యుత్ ఉత్పత్తి చేయవచ్చు.)

प्रश्न 2.
कुछ पेड़ों के नाम बताइए| (కొన్ని చెట్ల పేర్లను తెలపండి.)
उत्तर:
बरगद, पीपल, इमली, नीम, नींबू, जामून, नारियल और ताड़ आदि कुछ पेडों के नाम हैं।
(మర్రిచెట్టు, రావిచెట్టు, చింతచెట్టు, వేపచెట్టు, నిమ్మచెట్టు, నేరేడు చెట్టు, కొబ్బరి చెట్టు మరియు తాటిచెట్టు అనునవి కొన్ని చెట్ల పేర్లు)

AP Board 7th Class Hindi Solutions 5th Lesson ईमानदारी का फल

प्रश्न 3.
ईमानदारी का फल कैसा होता है? (నిజాయితీ ఫలితము ఎలా ఉంటుంది?)
उत्तर:
ईमानदारी का फल मीठा होता है। (నిజాయితీ యొక్క ఫలితము తీయగా ఉంటుంది.)

पढिए

अ) जोड़ी बनाइए।
AP Board 7th Class Hindi Solutions 5th Lesson ईमानदारी का फल 2
उत्तर:
AP Board 7th Class Hindi Solutions 5th Lesson ईमानदारी का फल 3

आ) पाठ में नीचे दिये गये वाक्यों के सही क्रम को पहचानकर उनकी क्रम संख्या कोष्ठक में लिखिए।

1. उसने सोने की कुल्हाड़ी दिखाई। [ 3 ]
2. एक लकड़हारा था। [ 1 ]
3. हाँ, यही मेरी कुल्हाड़ी है। [ 5 ]
4. लकड़हारा रोने लगा। [ 2 ]
5. यह मेरी नहीं है। [ 4 ]

इ) सही वर्तनी वाले शब्दों पर गोला “O” बनाइए।
AP Board 7th Class Hindi Solutions 5th Lesson ईमानदारी का फल 4a

ई) नीचे दिये गये वाक्यों में चित्रों से संबंधित शब्दों पर “O” गोला बनाइए।
AP Board 7th Class Hindi Solutions 5th Lesson ईमानदारी का फल 5

लिखिए

अ) नीचे दिये गये प्रश्नों के उत्तर छोटे – छोटे वाक्यों में लिखिए।
క్రింది ఇవ్వబడిన ప్రశ్నలకు సమాధానములు చిన్న – చిన్న వాక్యములలో ఇవ్వండి.)

1. लकड़हारा क्या करता था? (Sev Senyas Sansats?)
उत्तर:
लकड़हारा जंगल में लकडियाँ काटकर उन्हें बेचता था।
(కట్టెలు నరుకువాడు అడవిలో కట్టెలు నరికి వాటిని అమ్ముకునేవాడు.)

2. नदी की देवी क्यों खुश हुई? (568 vos 305220060?)
उत्तर:
लकडहारे की ईमानदारी को देखकर नदी की देवी खुश हुई।
(కట్టెలు నరుకువాని నిజాయితీని చూసి నదీ దేవత సంతోషించినది.)

AP Board 7th Class Hindi Solutions 5th Lesson ईमानदारी का फल

आ) नीचे दिये गये प्रश्न का उत्तर पाँच – छह वाक्यों में लिखिए।
క్రింది ఇవ్వబడిన ప్రశ్నకు సమాధానము 5 -6 వాక్యములలో వ్రాయండి.)

“ईमानदारी का फल” कहानी का सारांश लिखिए। (“Resran 2080” s av5g dowon provis.)
उत्तर:
एक लकड़हारे की कुल्हाडी नदी में गिर गई। नदी की देवी ने उसे सोने और चाँदी की कुल्हाड़ियाँ दिखाई। लेकिन उसने ईमानदारी से अपनी कुल्हाड़ी ही माँगी। इससे उसे तीनों कुल्हाड़ियाँ ईनाम के रूप में मिलीं। इससे यह शिक्षा मिलती है कि ईमानदारी का फल मीठा होता है।
(ఒక కట్టెలు నరుకువాని గొడ్డలి నదిలో పడిపోయినది. నదీ దేవి అతనికి బంగారపు మరియు వెండి గొడ్డళ్ళు చూపినది. కానీ అతడు నిజాయితీతో తన గొడ్డలినే అడిగెను. అందువలన అతనికి బహుమతిగా మూడు గొడ్డళ్ళు లభించినవి. దీని నుండి మనకు నిజాయితీ ఫలితము తీయగా ఉంటుందన్న పాఠం లభించుచున్నది.)

इ) उचित शब्दों से खाली जगह भरिए।

1. वह ………. में लकड़ियाँ काटता था। (जंगल / मैदान)
उत्तर:
जंगल

2. नदी में ……….. जा गिरी। (कुल्हाड़ी / लकड़ी)
उत्तर:
कुल्हाड़ी

3. नदी की देवी ……….. आयी। (अंदर / बाहर)
उत्तर:
बाहर

4. देवी ने उसको ………. कुल्हाड़ियाँ दे दी। (एक | तीनों)
उत्तर:
तीनों

5. ईमानदारी का फल ……… होता है। (मीठा/ कडुआ)
उत्तर:
मीठा

ई) संकेतों के आधार पर शब्द लिखिए।
AP Board 7th Class Hindi Solutions 5th Lesson ईमानदारी का फल 6
उत्तर:
नाम
धाम
काम
आम

AP Board 7th Class Hindi Solutions 5th Lesson ईमानदारी का फल

उ) वर्ण विच्छेद कीजिए।

1. जंगल : ज् + अं + ग् + अ + ल् + अ

2. कुल्हाड़ी : ………………………….
उत्तर:
क् + उ + ल् + इ + आ + ड् + ई

3. आधार : ………………………………
उत्तर:
आ + ध् + आ + र् + अ

4. प्रणाम : ………………………………..
उत्तर:
प् + र् + अ + ण् + आ + म् + अ

5. कहानी : ………………………………….
उत्तर:
क् + अ + ह् + आ + न् + ई

भाषांश

अ) अंत्याक्षरी विधि के अनुसार नीचे दिये गये शब्दों के आधार पर चार शब्द बनाइए।
उदा : लकड़हारा – राम – मगर – रमेश – शरत
AP Board 7th Class Hindi Solutions 5th Lesson ईमानदारी का फल 7
उत्तर:
AP Board 7th Class Hindi Solutions 5th Lesson ईमानदारी का फल 8

आ) पर्यायवाची शब्द लिखिए।

1. जल : पानी, नीर

2. जंगल : …………
उत्तर:
कानन, वन

3. लकड़ी : ………………..
उत्तर:
डंडा, काठ

4. मीठा : ………………
उत्तर:
मधुर, रसीला

5. खुश : …………………
उत्तर:
आनंद, संतोष

AP Board 7th Class Hindi Solutions 5th Lesson ईमानदारी का फल

इ) विलोम शब्द लिखिए।

1. बाहर × अंदर
2. आना × जाना
3. गिरना × उठना
4. ईमान × बेईमान
5. अपना × पराया

सृजनात्मकता

अ) चित्र को देखकर छोटे – छोटे शब्द लिखिए।
AP Board 7th Class Hindi Solutions 5th Lesson ईमानदारी का फल 9
1. लड़का
उत्तर:
2. आशीश
3. चरण
4. रास्ता
5. अंधा

आ) परियोजना कार्य

कुछ नीति वाक्यों का संकलन कीजिए। (కొన్ని నీతి వాక్యాములను సంగ్రహించండి.)
उत्तर:
नीति वाक्य :

  1. पाप से घृणा करो, पापी से नहीं। – महात्मा गाँधी
  2. ईश्वर के हाथ देने के लिए खुले हैं, लेने के लिए तुम्हें प्रयत्न करना होगा : गुरु नानक देव।
  3. प्रार्थना करनेवाले होठों से सहायता करनेवाले हाथ कहीं अच्छे हैं – मदर तेरेसा
  4. मानव की सेवा ही माधव की सेवा है। – मदर तेरेसा।
  5. मूर्ख अपने घर में पूजा जाता, मुखिया अपने गाँव में परंतु विद्वान सर्वत्र पूजा जाता है। – चाणक्य
  6. रास्ते कभी खत्म नहीं होते, अकसर लोग हिम्मत हार जाते हैं।
  7. मीठा जबान, अच्छी आदतें, अच्छा व्यवहार और अच्छे लोग हमेशा सम्मानित होते हैं।
  8. कड़ी मेहनत ही आपको अच्छी किस्मत दे सकती है।
  9. कम बोलना एक अच्छी आदत है।
  10. घमंड पतन का लक्षण है।

AP Board 7th Class Hindi Solutions 5th Lesson ईमानदारी का फल

इ) अनुवाद कीजिए।

1. लकड़हारा लकड़ियाँ काटता है। …………………………………………
उत्तर:
लकड़हारा लकड़ियाँ काटता है। కట్టెలు నరుకువాడు కట్టెలు నరుకుతాడు.

2. जंगल में अनेक जीव – जंतु रहते हैं। …………………………………………
उत्तर:
जंगल में अनेक जीव – जंतु रहते हैं। అడవిలో అనేక జంతు జీవములు ఉండును.

3. नदी में मछलियाँ रहती हैं। …………………………………………
उत्तर:
नदी में मछलियाँ रहती हैं। నదిలో చేపలు ఉండును.

4. सोना बहुत महँगा है। …………………………………………
उत्तर:
सोना बहुत महँगा है। బంగారము చాలా పిరుము.

5. बड़ों को प्रणाम करना चाहिए। …………………………………………
उत्तर:
बड़ों को प्रणाम करना चाहिए। పెద్దలకు నమస్కరించవలెను.

व्याकरणांश

अ) परिभाषा :

किसी व्यक्ति, स्थान, गुण या वस्तु के नाम को “संज्ञा” कहते हैं।
(ఎవరైనా ఒక వ్యక్తి, స్థానము, గుణము లేదా వస్తువు పేర్లను నామవాచకము (సంజ్ఞ) అని అందురు.)
उदाहरण : मोहन, लड़का, दिल्ली, पानी, मेज़, धर्म आदि।

आ) निम्न लिखित वाक्यों में संज्ञा शब्दों को पहचानकर लिखिए।

1. सब धर्म अच्छे होते हैं।
उत्तर:
धर्म

2. मैदान में घोड़ा है।
उत्तर:
मैदान, घोडा

3. दीपा गाती है।
उत्तर:
दीपा

4. वे विजयवाडा में रहते हैं।
उत्तर:
विजयवाडा

5. ‘हिंदी” सरल भाषा है।
उत्तर:
हिंदी, भाषा

AP Board 7th Class Hindi Solutions 5th Lesson ईमानदारी का फल

इ) नीचे दिये गये वाक्यों में संज्ञा शब्दों को रेखांकित कीजिए।

1. वह जंगल में लकड़ियाँ काटता था।
उत्तर:
जंगल, लकड़ियाँ

2. नदी की देवी जल से बाहर आयी।
उत्तर:
नदी, देवी, जल

3. यह मेरी कुल्हाड़ी नहीं है।
उत्तर:
कुल्हाड़ी

4. एक लकड़हारा वन में गया।
उत्तर:
लकडहारा, वन

5. देवी को उस पर दया आयी।
उत्तर:
देवी, दया

अध्यापकों के लिए सूचना : ఉపాధ్యాయులకు సూచన :

ईमानदारी से संबंधित कुछ घटनाओं का वर्णन कीजिए।
उत्तर:
एक घटना :
एक बार मेरा दोस्त दिल्ली गया। उसके पास एक लाख रुपये थे। जिन्हें वह अपने पिता के इलाज के लिए लेकर जा रहा था। वह रेलगाडी में अपना पर्स खो दिया। अस्पताल पहुंचने पर उसने अपनी जेब देखी तो उसमें पर्स नहीं था। उसके चेहरे की हवाएँ उडने लगी।

इतने में एक व्यक्ति वहाँ पहुँचा। उसने मेरे दोस्त का पर्स लौटाते हुए कहा कि उसे यह पर्स रेलगाड़ी में पड़ा मिला था। मेरे दोस्त के यह पूछने पर कहा कि उसे यहाँ का पता किसने दिया तो उसने बताया था कि उसके पर्स में अस्पताल का नाम तथा डॉक्टर द्वारा लिखी गयी दवाइयों की पर्ची को पढ़कर वह यहाँ तक पहुंचा है। मेरा दोस्त उस अनजान व्यक्ति की ईमानदारी को आज भी याद करता है।

एक और घटना :
यह घटना तेलंगाणा की राजधानी हैदराबाद की है। बीते बुधवार को बाईस वर्षीय गोपाल अपने खाते में मौजूद कुल एक लाख रुपये में से हजार रुपये निकालने स्टेट बैंक आफ़ हैदराबाद के एक ए.टी.एम.पर गया। पैसे निकालने के क्रम में मशीन के किनारे का एक दरवाज़ा खुल गया और उसमें से सारे नोट बाहर गिर गए। वहाँ न कोई सुरक्षा गार्ड तैनात था न ही सी.सी.टी.वी कैमरा लगा था। यानी चुपचाप सुरक्षित तरीके ढेर सारी रकम ले जाने के लिए गोपाल के सामने पूरा मौका था। लेकिन उसके भीतर पल भर के लिए भी ऐसा ख्याल नहीं आया, और उसने बाहर खडे हुए अपने साथियों से सुरक्षा गार्ड को खोजने के लिए कहा। गार्ड के न मिलने पर पुलिस को फोन करके सारी स्थिति बताई। इतने सारे रुपये देखकर न गोपाल के भीतर लालच नहीं आया, बल्कि उसके बेरोजगार दोस्त भी उसे कुछ गलत करने के लिए उकसाने के बजाय उसकी ईमानदारी में सहभागी बने। यह पुलीस और सबसे ज्यादा उस बैंक के लिए भी राहत की बात थी। अंदाज़ा लगाया जा सकता है कि इन युवकों ने इससे उलट व्यवहार किया होता तो बैंक और उसके कर्मचारी की लापरवाही का अंजाम क्या हो सकता था। हैदराबाद पुलीस ने उचित ही गोपाल को सम्मानित करने का फैसला किया है।

पाठ का सारांश

एक लकड़हारे की कुल्हाड़ी नदी में गिर गई। नदी की देवी ने उसे सोने और चाँदी की कुल्हाड़ियाँ दिखाई। लेकिन उसने ईमानदारी से अपनी कुल्हाड़ी ही माँगी। इससे उसे तीनों कुल्हाड़ियाँ ईनाम के रूप में मिलीं। इससे यह शिक्षा मिलती है कि ईमानदारी का फल मीठा होता है।

పాఠ్య సారాంశం

ఒక కట్టెలు నరుకువాని గొడ్డలి నదిలో పడిపోయినది. నదీ దేవి అతనికి బంగారపు మరియు వెండి గొడ్డళ్ళు చూపినది. కానీ అతడు నిజాయితీతో తన గొడ్డలినే అడిగెను. అందువలన అతనికి బహుమతిగా మూడు గొడ్డళ్ళు లభించినవి. దీని నుండి మనకు నిజాయితీ ఫలితము తీయగా ఉంటుందన్న పాఠం లభించుచున్నది.

Summary

A woodcutter’s axe fell into a river. The river goddess showed him golden and silver axes. But he asked the goddess honestly for his own axe. Because of this, he was rewarded with three axes. From this, we learn a moral that the result of honesty would be sweet.

व्याकरणांश (వ్యాకరణాంశాలు)

लिंग बदलिए (లింగములను మార్చండి)

लड़का – लड़की
देवी – देव
भगवान – भगवती
चाचा – चाची
पुत्र – पुत्री
पति – पत्नी
आचार्य – आचार्या
सुनार – सुनारिन
श्रीमान – श्रीमती
अध्यापक – अध्यापिका
पोता – पोती
ग्वाला – ग्वालिन
अध्यक्ष- अध्यक्षा
दूल्हा – दुल्हन
साला – साली

वचन बदलिए (వచనములను మార్చండి)

फल – फल
सेवा – सेवाएँ
गुण – गुण
मार्ग – मार्ग
सीख – सीख
नदी – नदियाँ
लकडी – लकड़ियाँ
कुल्हाडी – कुल्हाड़ियाँ
खुशी – खुशियाँ
ईनाम – ईनाम

विलोम शब्द (వ్యతిరేక పదములు)

दया × निर्दया
विनय × अविनय
उपकार × अपकार
परोपकार × अपकार
नैतिक × अनैतिक
गुण × अगुण
ईमानदार × बेईमान
सही × गलत
जीवन × मरण
महान × अल्प/तुच्छ
बेचना × खरीदना
रोना × हँसना
आधार × निराधार
अंदर × बाहर
खुश × दुःख
मीठा × कडुवा
खुशी × दुःखी

AP Board 7th Class Hindi Solutions 5th Lesson ईमानदारी का फल

शब्दार्थ (అర్థాలు) (Meanings)

जंगल = वन, అడవి, forest
मिठा = मधुर, తీయని, sweet
लकड़हारा = लकड़ियाँ काटनेवाला, కట్టెలు నరుకువాడు, wood cutter
नदी = सरिता, నది, river
ईमानदारी = सच्चाई, నిజాయితీ, honesty
सीख = नीति, నీతి, morality
चाँदी = रजत, వెండి, silver
सोना = कनक, బంగారం, gold
किनारा = तट, ఒడ్డు, bank
दिन = दिवस, రోజు, day
जल = पानी, నీరు, water
दया = कृपा, దయ, kindness
खुश = संतोष, సంతోషము, happiness
प्रणाम = नमस्कार, నమస్కారం, salutation

श्रुत लेख : శ్రుతలేఖనము : Dietations

अध्यापक या अध्यापिका निम्न लिखित शब्दों को श्रुत लेख के रूप में लिखवायें। छात्र अपनी-अपनी नोट पुस्तकों में लिखेंगे। अध्यापक या अध्यापिका इन्हें जाँचे।
ఉపాధ్యాయుడు లేదా ఉపాధ్యాయిని క్రింద వ్రాయబడిన శబ్దములను శ్రుతలేఖనంగా డిక్లేట్ చేయును. విద్యార్థులు వారి వారి నోట్ పుస్తకాలలో వ్రాసెదరు. ఉపాధ్యాయుడు లేదా ఉపాధ్యాయిని వాటిని దిద్దెదరు.

AP SSC 10th Class Maths Solutions Chapter 7 Coordinate Geometry Optional Exercise

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

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

10th Class Maths 7th Lesson Coordinate Geometry Optional Exercise Textbook Questions and Answers

Question 1.
Centre of the circle Q is on the Y-axis. And the circle passes through the points (0, 7) and (0, -1). Circle intersects the positive X-axis at (p, 0). What is the value of ‘p’ ?
Answer:
AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise 20
Given points A, B are on y – axis.
(∵ their x – coordinate is zero) then they will be the end points (0, 7) and (0, -1).
Centre (Q) = \(\left(\frac{0+0}{2}, \frac{7-1}{2}\right)\) = (0, 3)
then radius = AQ = BQ
= \(\sqrt{(0-0)^{2}+(3-(-1))^{2}}\)
= \(\sqrt{0^{2}+4^{2}}\) = √16 = 4
now P(p, 0) is also another point on circle then \(\overline{\mathrm{QP}}\) is also radius
∴ \(\overline{\mathrm{QP}}\) = \(\sqrt{(0-p)^{2}+(3-0)^{2}}\) = 4
⇒ \(\sqrt{p^{2}+3^{2}}\) = 4
⇒ p2 = 42 – 32 = 16 – 9 = 7
∴ P = ± √7 .

AP SSC 10th Class Maths Solutions Chapter 7 Coordinate Geometry Optional Exercise

Question 2.
A triangle ABC is formed by the points A(2, 3), B(-2, -3), C(4, -3). What is the point of intersection of side BC and angular bisector of angle A?
Answer:
Given: △ABC, where A (2, 3), B (- 2, – 3), C (4, – 3).
Let AD be the bisector to ∠A meeting BC at D.
AP SSC 10th Class Maths Solutions Chapter 7 Coordinate Geometry Optional Exercise 2
Then BD : DC = AB : AC
[∵ The bisector of vertical angle of triangle divides the base in the ratio of other two sides.]
AP SSC 10th Class Maths Solutions Chapter 7 Coordinate Geometry Optional Exercise 3
Now D is a point which divides \(\overline{\mathrm{BC}}\) in the ratio √l3 : √10 internally section formula (x, y) =
AP SSC 10th Class Maths Solutions Chapter 7 Coordinate Geometry Optional Exercise 4
(By rationalising the denominator of the x-coordinate).

AP SSC 10th Class Maths Solutions Chapter 7 Coordinate Geometry Optional Exercise

Question 3.
The side BC of an equilateral △ABC is parallel to X – axis. Find the slopes of line along sides BC, CA and AB.
Answer:
AP SSC 10th Class Maths Solutions Chapter 7 Coordinate Geometry Optional Exercise 5
Given: △ABC; BC // X – axis.
Slope of BC = tan θ
where θ is the angle made of BC with + ve X – axis.
= tan ‘0’ = ‘0’
Slope of AB = tan 60° = √3
[∵ Each angle of an equilateral triangle is 60°]
Slope of AC = tan 60° = √3

Question 4.
A right triangle has sides ‘a’ and ‘b’ where a > b. If the right angle is bisected then find the distance between ortho centres of the smaller triangles using coordinate geometry.
Answer:
Given : △ABC; ∠B = 90°
AP SSC 10th Class Maths Solutions Chapter 7 Coordinate Geometry Optional Exercise 6

AP SSC 10th Class Maths Solutions Chapter 7 Coordinate Geometry Optional Exercise

Question 5.
Find the centroid of the triangle formed by the line 2x + 3y – 6 = 0 with the coordinate axes.
Answer:
Given : △AOB formed by the line 2x + 3y – 6 = 0 with axes.
Let A lie on X – axis and B on Y – axis.
∴ Y – coordinate of A = 0
X – coordinate of B = 0
∴ A (3, 0), B (0, 2)
AP SSC 10th Class Maths Solutions Chapter 7 Coordinate Geometry Optional Exercise 7

AP Board 7th Class Maths Solutions Chapter 5 Triangles Ex 5.1

SCERT AP 7th Class Maths Solutions Pdf Chapter 5 Triangles Ex 5.1 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 5th Lesson Triangles Exercise 5.1

Question 1.
Classify the following triangles based on the length of their sides.
AP Board 7th Class Maths Solutions Chapter 5 Triangles Ex 5.1 1
Answer:
In ∆ABC,
AB ≠ BC ≠ AC
So, ∆ABC is scalene traingle.

AP Board 7th Class Maths Solutions Chapter 5 Triangles Ex 5.1 2
Answer:
In ∆DEF,
DE = EF = FD = 2 cm
So, ∆DEF is an equilateral triangle.

AP Board 7th Class Maths Solutions Chapter 5 Triangles Ex 5.1 3
Answer:
In ∆XYZ,
XZ = YZ = 3.5 cm
So, ∆XYZ is an isosceles triangle.

AP Board 7th Class Maths Solutions Chapter 5 Triangles Ex 5.1 4
Answer:
Isosceles triangle (∵ Two sides are equal)

AP Board 7th Class Maths Solutions Chapter 5 Triangles Ex 5.1 5
Answer:
Equilateral triangle (∵ Three sides are equal)

AP Board 7th Class Maths Solutions Chapter 5 Triangles Ex 5.1 6
Answer:
Scalene triangle (∵ NO two sides are equal).

AP Board 7th Class Maths Solutions Chapter 5 Triangles Exercise 5.1

Question 2.
Classify the following triangles based on the measure of angles.
AP Board 7th Class Maths Solutions Chapter 5 Triangles Ex 5.1 7
Answer:
In ∆TUV, ∠T = 90°
So, ∆TUV is Right-angled triangle.

AP Board 7th Class Maths Solutions Chapter 5 Triangles Ex 5.1 8
Answer:
In ∆QRS, ∠Q = 108°
So, ∆QRS is an Obtuse angled triangle.

AP Board 7th Class Maths Solutions Chapter 5 Triangles Ex 5.1 9
Answer:
In ∆IJK, all are acute angles.
So, ∆IJK is an acute angled triangle.

Question 3.
Classify the following triangles based on their sides and also on their angles.
AP Board 7th Class Maths Solutions Chapter 5 Triangles Ex 5.1 10
Answer:
In ∆IJK, IJ ≠ KJ ≠ IK
So, ∆IJK is a scalene triangle.

AP Board 7th Class Maths Solutions Chapter 5 Triangles Ex 5.1 11
Answer:
In ∆LMN, ∠M = 90
So, ∆LMN is a right triangle.

AP Board 7th Class Maths Solutions Chapter 5 Triangles Exercise 5.1

AP Board 7th Class Maths Solutions Chapter 5 Triangles Ex 5.1 12
Answer:
In ∆TUV, TU = UV = TV = 4.5 cm
So, ∆TUV is an equilateral triangle.

Inter 2nd Year Maths 2B System of Circles Important Questions

Students get through Maths 2B Important Questions Inter 2nd Year Maths 2B System of Circles Important Questions which are most likely to be asked in the exam.

Intermediate 2nd Year Maths 2B System of Circles Important Questions

Question 1.
x2 + y2 + 4x + 8 = 0, x2 + y2 – 16y + k = 0 [A.P. & T.S. Mar. 16]
Solution:
g1 = 2; f1 = 0; c1 = 8
g2 = 0; f2 = – 4; c2 = k
Two circles are said to be orthogonal
2g1g2 + 2f1f2 = c1 + c2
2(2) (0) + 2(0) (-8) = 8 + k
0 + 0 = 8 + k
⇒ k = -8

Inter 2nd Year Maths 2B System of Circles Important Questions

Question 2.
(x – a)2 + (y – b)2 = c2, (x – b)2 + (y – a)2 = c2 (a ≠ b) [A.P. Mar. 15]
Solution:
(x2 + y2 – 2xa – 2yb – c2)
– (x2 + y2 – 2xb – 2ya – c2) = 0
– 2x (a – b) – 2y(b – a) = 0
(or) x – y = 0

Question 3.
Find the angle between the circles given by the equations. [T.S. Mar.17]
i) x2 + y2 – 12x – 6y + 41 = 0, x2 + y2 + 4x + 6y – 59 = 0.
Solution:
C1 = (6, 3)
r1 =(36 + 9 – 41)1/2
r1 = 2

C2 = (-2, -3)
r2 =(4 + 9 + 59)1/2
r2 = (72)1/2 = 6\(\sqrt{2}\)

C1C2 = d = \(\sqrt{(6+2)^{2}+(3+3)^{2}}\)
= \(\sqrt{64+36}\) = 10
cos θ = \(\frac{d^{2}-r_{1}^{2}-r_{2}^{2}}{2 r_{1} r_{2}}\)
= \(\frac{100-4-72}{2 \times 2 \cdot \sqrt{72}}=\frac{24}{4 \times 6 \sqrt{2}}=\frac{1}{\sqrt{2}}\)
θ = 45°

Inter 2nd Year Maths 2B System of Circles Important Questions

Question 4.
Find the equation of the circle which cuts the circles x2 + y2 – 4x – 6y + 11 = 0 and x2 + y2 – 10x – 4y + 21 = 0 orthogonally and has the diameter along the straight line 2x + 3y = 7. [A.P. Mar. 16; May 07]
Solution:
Let circle be x2 + y2 + 2gx + 2fy + c = 0 ……………….. (i)
Orthogonal to circle
2g (-2) +2f(-3) = 11 + c ……………………. (ii)
2g (-5) + 2f(-2) = 21 + c ……………………. (iii)
Subtracting it we get
-6g + 2f = 10 ……………………… (iv)
Circles centre is on 2x + 3y = 7
∴ -2g – 3f = 7 ……………………. (v)
Solving (iv) and (y)
f = -1, g = -2, c = 3
Equation of circle be x2 + y2 – 4x – 2y + 3 = 0

Question 5.
Show that the angle between the circles x2 + y2 = a2, x2 + y2 = ax + ay is \(\frac{3\pi}{4}\) [Mar. 14]
Solution:
Equations of the circles are
S ≡ x2 + y2 – a2 = 0
S’ ≡ x2 + y2 – ax – ay = 0
C1 (0, 0), C2 (\(\frac{a}{2}\), \(\frac{a}{2}\))
∴ C1C22 = (0 – \(\frac{a}{2}\))2 + (0 – \(\frac{a}{2}\))2
Inter 2nd Year Maths 2B System of Circles Important Questions 1

Question 6.
If x + y = 3 ¡s the equation of the chord AB of the circle x2 + y2 – 2x + 4y – 8 = 0, find the equation of the circle having \(\overline{\mathrm{AB}}\) as diameter. [A.P. Mar. 15]
Solution:
Required equation of circle passing through
intersection S = 0 and L = 0 is S + λL = 0
(x2 + y2 – 2x + 4y – 8) + λ(x + y – 3) = 0
(x2 + y2 + x(-2 + λ)+ y(4 + λ) – 8 – 3λ = 0 ………………. (i)
x2 + y2 + 2gx + 2fy + c = 0 ……………………. (ii)
Comparing (i) and (ii) we get
g = \(\frac{(-2+\lambda)}{2}\), f = \(\frac{(4+\lambda)}{2}\)
Centre lies on x + y = 3
∴ \(-\left(\frac{-2+\lambda}{2}\right)-\left(\frac{4+\lambda}{2}\right)\) = 3
2 – λ – 4 – λ = 6
-2λ = 8 ⇒ λ = – 4
Required equation of circle be
(x2 + y2 – 2x + 4y – 8) – 4(x + y – 3) = 0
x2 + y2 – 6x + 4 = 0

Inter 2nd Year Maths 2B System of Circles Important Questions

Question 7.
If two circles x2 + y2 + 2gx + 2fy = 0 and x2 + y2 + 2g’x + 2f’y = 0 touch each other then show that f’g = fg’. [T.S. Mar. 16]
Solution:
C1 = (-g, -f)
r1 = \(\sqrt{g^{2}+f^{2}}\)
C2 = (-g-1, -f-1)
r2 = \(\sqrt{g^{\prime 2}+r^{\prime 2}}\)
C1C2 = r1 + r2
(C1C2)2 = (r1r2)2
(g’ – g)2 + (f’ – f)2 = g2 + f2 + g’2 + f’2 + \(2 \sqrt{g^{2}+f^{2}} \sqrt{g^{2}+f^{1^{2}}}\)
-2(gg’ + ff’) = 2{g2g’2 + g2f’2 + f2g’2}1/2
Squaring again
(gg’ + ff’)2 = g2g’2 + f2f’2 + g2f’2 + g’2f2
g2g’2 + f2f’2 + 2gg’ff’ = g2g’2 + f2f’2 + g2f’2 + g’2f’2
2gg’ff’ = g2f’2 + f2g’2
⇒ g2f’2 + g’2f’2 – 2gg’ff’ = 0
(or) (gf’ – fg’)2 = 0 (or) gf’ = fg’

Question 8.
Show that the circles
S ≡ x2 + y2 – 2x – 4y – 20 = 0 …………………… (1)
and S ≡ x2 + y2 + 6x + 2y – 90 = 0 …………………(2)
touch each other internally. Find their point of contact and the equation of common tangent. [T.S. Mar. 15]
Solution:
Let C1, C2 be the centres and r1, r2 be the radii of the given circles (1) and (2). Then
C1 = (1, 2); C2 = (-3, -1); r1 = 5; r2 = 10.
C1C2 = distance between the centres = 5
|r1 – r2| = |5 – 10| = 5 = C1C2
∴ The given two circles touch internally. In
this case, the common tangent is nothing but the radical axis. Therefore its equation is S – S’ = 0.
i.e., 4x + 3y – 35 = 0
Now we find the point of contact. The point of contact divides \(\overline{\mathrm{C}_{1} \mathrm{C}_{2}}\) in the ratio 5 : 10
i.e., 1: 2 (externally)
∴ Point of contact
= \(\left(\frac{(1)(-3)-2(1)}{1-2}, \frac{(1)(-1)-2(2)}{1-2}\right)\)
= (5, 5)

Inter 2nd Year Maths 2B System of Circles Important Questions

Question 9.
Find the angle between the circles x2 + y2 + 4x – 14y + 28 = 0 and x2 + y2 + 4x – 5 = 0
Solution:
Equations of the given circles are .
x2 + y2 + 4x – 14y + 28 = 0
x2 + y2 + 4x – 5 = 0
Centres are C1 (-2, 7), C2 (-2, 0)
C1C2 = \(\sqrt{(-2+2)^{2}+(7-0)^{2}}\)
= \(\sqrt{0+49}\) = 7
r1 = \(\sqrt{4+49-28}\) = \(\sqrt{25}\) = 5
r2 = \(\sqrt{4+5}\) = \(\sqrt{9}\) = 3
If θ is the angle between the given circles,
then cos θ = \(\frac{d^{2}-r_{1}^{2}-r_{2}^{2}}{2 r_{1} r_{2}}\)
cos θ = \(\frac{49-25-9}{2(5)(3)}\) = \(\frac{15}{2.5 .3}\) = \(\frac{1}{2}\) = cos 60°
Angle between the circles = θ = \(\frac{\pi}{3}\)

Question 10.
If the angle between the circles x2 + y2 – 12x – 6y + 41 = 0 and x2 + y2 + kx + 6y – 59 = 0 is 45° find k.
Solution:
Suppose θ is the angle between the circles
x2 + y2 – 12x – 6y + 41 = 0
and x2 + y2 + kx + 6y – 59 = 0
g1 = -6, f1 = -3, c1 = 41,
g2 = \(\frac{k}{2}\), f2 = 3, c2 = -59
Inter 2nd Year Maths 2B System of Circles Important Questions 2
\(\frac{1}{\sqrt{2}}=\frac{6 k}{4 \cdot \sqrt{\frac{k^{2}}{4}+68}}\)
Squaring and cross – multiplying
4(\(\frac{k^{2}}{4}\) + 68) = 18k2
\(\frac{2\left[k^{2}+272\right]}{4}\) = 9k2
k2 + 272 = 18 k2
17k2 = 272
k2 = \(\frac{272}{17}\)
k2 = 16
k = ±4.

Question 11.
Find the equation of the circle which passes through (1, 1) and cuts orthogonally each of the circles.
x2 + y2 – 8x – 2y + 16 = 0 and …………….. (1)
x2 + y2 – 4x – 4y – 1 = 0. ………………. (2)
Solution:
Let the equation of the required circle be
x2 + y2 + 2gx + 2fy + c = 0 ……………. (3)
Then the circle (3) is orthogonal to (1) and (2).
∴ By applying the condition of orthogonality give in x2 + y2 + 2gx + 2fy + c = 0 we get
2g(-4) + 2f(-1) = c + 16 and …………………. (4)
2g (-2) + 2f(-2) = c – 1 ……………….. (5).
Given that the circle (3) is passing through (1, 1)
∴ 12 + 12 + 2g(1) + 2f(1) + c = 0
2g + 2f + c + 2 = 0 ………………….. (6)
Solving (4), (5) and (6) for g, f and c, we get
g = –\(\frac{7}{3}\), f = \(\frac{23}{6}\), c = -5
Thus the equation of the required circle is
3(x2 + y2) – 14x + 23y – 15 = 0.

Inter 2nd Year Maths 2B System of Circles Important Questions

Question 12.
Find the equation of the circle which is orthogonal to each of the following three circles.
x2 + y2 + 2x + 17y + 4 = 0 ………………… (1)
x2 + y2 + 7x + 6y + 11 = 0 ………………. (2)
and x2 + y2 – x + 22y + 3 = 0 ………………… (3)
Solution:
Let the equation of the required circle be
x2 + y2 + 2gx + 2fy + c = 0 ……………… (4)
Since this circle is orthogonal to (1), (2) and (3). by applying the condition of orthogonality given in x2 + y2 + 2gx + 2fy + c = 0
we have
2(g)(1) + 2(f) (\(\frac{17}{2}\)) = c + 4 ……………….. (5)
2(g) (\(\frac{7}{2}\)) + 2(f)(3) = c + 11 ………………. (6)
and 2(g) (-\(\frac{1}{2}\)) + 2(f)(11) = c + 3 ………………….. (7)
Solving (5), (6) and (7) for g, f, c we get
g = -3, f = -2 and c = -44
Thus the equation of the required circle is
x2 + y2 – 6x – 4y – 44 = 0.

Question 13.
If the straight line is represented by
x cos α + y sin α = p ………………….. (1)
intersects the circle
x2 + y2 = a2 ……………… (2)
at the points A and B, then show that the equation of the circle with \(\overline{\mathrm{AB}}\) as diameter is(x2 + y2 – a2) – 2p(x cos α + y sin α – p) = 0.
Solution:
The equation of the circle passing through the points A and B is S ≡ x2 + y2 + 2gx + 2fy + c = 0
(x2 + y2 – a2) + λ(x cos α + y sin α – p) = 0 ……………… (3)
The centre of this circle is
\(\left(-\frac{\lambda \cos \alpha}{2},-\frac{\lambda \sin \alpha}{2}\right)\)
If the circle given by (3) has \(\overline{\mathrm{AB}}\) as diameter then the centre of it must lie on (1)
∴ \(-\frac{\lambda \cos \alpha}{2}\) (cos α) – \(\frac{\lambda \sin \alpha}{2}\) (sin α) = p
i.e., –\(\frac{\lambda}{2}\) (cos2 α + sin2 α) p
i.e., λ = -2p
Hence the equation of the required circle is
(x2 + y2 – a2) – 2p(x cos α + y sin α – p) = 0.

Inter 2nd Year Maths 2B System of Circles Important Questions

Question 14.
Find the equation of the circle passing through the points of intersection of the circles.
x2 + y2 – 8x – 6y + 21 = 0 ……………….. (1)
x2 + y2 – 2x – 15 = 0 ……………….. (2)
and (1, 2).
Solution:
The equation of circle passing through the points of intersection of (1) arid (2) is
(x2 + y2 – 8x – 6y + 21) + λ (x2 + y2 – 2x – 15) = 0 ……………….. (3)
If it passes through (1, 2). we obtain
(1 + 4 – 8 – 12 + 21) + λ(1 + 4 – 2 – 15) = 0
i.e., 6 + λ(-12) = 0
i.e., λ = \(\frac{1}{2}\)
Hence the equation of the required circle is
(x2 + y2 – 8x – 6y + 21) + \(\frac{1}{2}\) (x2 + y2 – 2x – 15) = 0
i.e., 3(x2 + y2) – 18x – 12y + 27 = 0.

Question 15.
Let us find the equation the radical axis of the circles S ≡ x2 + y2 – 5x + 6y + 12 = 0 and S’ ≡ x2 + y2 + 6x – 4y – 14 = 0
Solution:
The given equations of circles are in general form. Therefore their radical axis is (S – S’ = 0)
i.e., 11x – 10y – 26 = 0

Question 16.
Let us find the equation of the radical axis of the circles
2x2 + 2y2 + 3x + 6y – 5 = 0 ………………….. (1)
and 3x2 + 3y2 – 7x + 8y – 11 = 0 ……………….. (2)
Solution:
Hence the given equations are not in general form, we get :
x2 + y2 + \(\frac{3}{2}\)x + 3y – \(\frac{5}{2}\) = 0 and
x2 + y2 – \(\frac{7}{3}\)x + \(\frac{8}{3}\) y – \(\frac{11}{3}\) = 0
Now the radical axis equation of given circles is
(\(\frac{3}{2}\) + \(\frac{7}{3}\))x + (3 – \(\frac{8}{3}\))y + (-\(\frac{5}{2}\) + \(\frac{11}{3}\)) = 0
i.e., 23x + 2y + 7 = 0

Inter 2nd Year Maths 2B System of Circles Important Questions

Question 17.
Let us find the radical centre of the circles
x2 + y2 – 2x + 6y = 0 ………………… (1)
x2 + y2 – 4x – 2y + 6 = 0 …………………. (2)
and x2 + y2 – 12x + 2y + 3 = 0 ………………. (3)
Solution:
The radical axis of (1) and (2) and (3)
x + 4y – 3 = 0 ………………… (4)
8x – 4y + 3 = 0 …………………. (5)
10x + 4y – 3 = 0 ……………… (6)
Solving (4) and (5) for the point of intersection we get (0, \(\frac{3}{4}\)) which is the required radical centre. Observe that the co-ordinates of this point satisfies (6) also.

Question 18.
Find the equation and length of the common chord of the two circles
S ≡ x2 + y2 + 3x + 5y + 4 = 0
and S’ ≡ x2 + y2 + 5x + 3y + 4 = 0
Solution:
Equations of the given circles are
S ≡ x2 + y2 + 3x + 5y + 4 = 0 ………………… (1)
S’ ≡ x2 + y2 + 5x + 3y + 4 ………………. (2)
Equations of the common chord is S – S’ = 0
-2x + 2y = 0
L ≡ x – y = 0 …………….. (3)
Inter 2nd Year Maths 2B System of Circles Important Questions 3

Question 19.
Show that the circles
S ≡ x2 + y2 – 2x – 4y – 20 = 0 ……………….. (1)
and S’ ≡ x2 + y2 + 6x + 2y – 90 = 0 ……………. (2)
touch each other internally. Find their point of contact and the equation of common tangent. [T.S. Mar. 15]
Solution:
Let C1, C2 be the centres and r1, r2 be the radii of the given circles (1) and (2). Then
C1 = (1, 2); C2 = (-3, -1); r1 = 5; r2 = 10
C1C2 = distance between the centres = 5
|r1 – r2| = |5 – 10| = 5 = C1C2
∴ The given two circles touch internally. In this case, the.common tangent is nothing but the radical axis. Therefore its equation is
S – S’ = 0
i.e., 4x + 3y – 35 = 0
Now we find the point of contact. The point of contact divides in the ratio 5 : 10 i.e., 1 : 2 (externally)
∴ Point, of contact
= \(\left(\frac{(1)(-3)-2(1)}{1-2}, \frac{(1)(-1)-2(2)}{1-2}\right)\)
= (5, 5).

Inter 2nd Year Maths 2B System of Circles Important Questions

Question 20.
Find the equation of the circle whose diameter is the common chord of the circles
S ≡ x2 + y2 + 2x + 3y + 1 = 0 ……………….. (1)
and S’ ≡ x2 + y2 + 4x + 3y + 2 = 0 ……………….. (2)
Solution:
Here the common chord is the radical axis of (1) and (2). The equation of the radical axis is S – S’ = 0.
i.e., 2x + 1 = 0 …………………… (3)
The equation of any circle passing through
the points of intersection of (1) and (3) is (S + λL = 0)
(x2 + y2 + 2x + 3y + 1) + λ(2x + 1) = 0
x2 + y2 + 2(λ + 1)x + 3y + (1 + λ) = 0 ……………………. (4)
The centre of this circle is (-(λ + 1), \(\frac{3}{2}\)).
For the circle (4), 2x + 1 = 0 is one chord. This chord will be a diameter of the circle (4) if the centre of (4) lies on (3).
∴ 2{-(λ + 1)} + 1 = 0
⇒ λ = –\(\frac{1}{2}\)
Thus equation of the circle whose diameter is the common chord (1) and (2)
(put λ = \(\frac{1}{2}\) in equation (4))
2(x2 + y2) + 2x + 6y + 1 = 0

Question 21.
Let us find the equation of a circle which cuts each of the following circles orthogonally
S’ ≡ x2 + y2 + 3x + 2y + 1 = 0 ………………… (1)
S” ≡ x2 + y2 – x + 6y + 5 = 0 …………….. (2)
and S” ≡ x2 + y2 + 5x – 8y + 15 = 0 ……………………. (3)
Solution:
The centre of the required circle is radical centre of (1), (2) and (3) and the radius is the length of the tangent from this point to any one of the given three circles. First we shall find the radical centre. For, the radical axis of (1) and (2) is
x – y = 1 ………………… (4)
and the radical axis of (2) and (3) is
3x – 7y = -5 …………………… (5)
The point of intersection (3, 2) of (4) and (5) is the radical centre of the circles (1), (2) and (3). The length of tangent from (3. 2) to the circle (1)
= \(\sqrt{3^{2}+2^{2}+3(3)+2(2)+1}=3 \sqrt{3}\)
Thus the required circle is
(x – 3)2 + (y – 2)2 = (3\(\sqrt{3}\))2
x2 + y2 – 6x – 4y – 14 = 0.