Category: Class 10

Students can go through AP SSC 10th Class Maths Notes Chapter 2 Sets to understand and remember the concepts easily.

AP State Syllabus SSC 10th Class Maths Notes Chapter 2 Sets

→ Set theory is comparatively a new concept in mathematics.

→ This theory was developed by George Cantor.

→ A well defined collection of objects or ideas is known as “set”.

→ Well defined means that:
i) All the objects in the set should have a common feature or property.
ii) It should be possible to decide whether any given object belongs to the set or not.
Example or comparision of well defined and not well defined collections:

Not well defined collections	Well defined collections
i) A family of rich persons	i) A family of persons having more than one crore rupees
ii) A group of tall students	ii) A group of students, with height 160 cm or more
iii) A group of numbers	iii) A group of even natural numbers less than 15

→ Some more examples of well defined collections:
i) Vowels of English alphabets, namely a, e, i, o, u.
ii) Odd natural numbers less than 11, namely 1, 3, 5, 7, 9.
iii) The roots of the equation x² – 3x + 2 = 0, i.e., 1 and 2.

→ Objects, elements and members of a set are synonymous words.

→ Sets are usually denoted by the capital letters like A, B, C, X, Y, Z, etc.

→ An object belonging to a set is known as a member/element/individual of the set.

→ The elements of a set are represented by small case letters,
i.e., a, b, c, , x, y, z, etc.

→ If ‘b’ is an element of a set A, then we say that ‘b’ belongs to A.

→ The word ‘belongs to’ is denoted by the Greek symbol ‘∈’.

→ Thus, in a notation form, ‘b’ belongs to A is written as b ∈ A and ‘c’ does not belong to ‘A’ is written as c ∉ A.

→ Representation of sets: Sets are generally represented by the following two methods.
i) Roster (or) Tabular form
ii) Rule method (or) Set builder form.

→ Roster (or) Tabular form: In this form, all elements of the set are written, separated by commas, within curly brackets.
Example:
i) The set of all natural numbers less than 5 is represented as N = {l,2,3,4}
ii) The set of all letters in the word “JANUARY” is represented as B = {A, J, N, R, U, Y}
Note:
a) In a set notation, order is not important.
b) The elements of a set are generally not repeated in a particular set.

→ Set builder form (or) Rule method: In this method, a set is described by using a representative and stating the property (or) properties which the elements of the set satisfy, through the representative.
Example:
i) Set of all natural numbers less than 5.
A = {x : x ∈ N, x < 5}
ii) Set of vowels of the English alphabet.
V = {x : x is a vowel in the English alphabet)
Note: It may be observed that we describe the set by using a symbol (x or y or z etc.) for elements of the set.

→ Types of set:

→ Empty set (or) Null set (or) Void set: A set, which does not contain any element is called an empty set (or) a null set (or) a void set.

→ Empty set is denoted by ∅ (or) { }

Example :
A = (x : x is a natural number smaller than 1}
B = {x : x² – 2 = 0 and x is a rational number}
C = (x : x is a man living on the moon}
Note: ∅ and { 0 } are two different sets. {0} is a set containing the single element ‘0’ while { } is a null set.

→ Singleton set: A set consisting of a single element is called a singleton set.
Examples:
{ 0 }, {- 7} are singleton sets.

→ Finite set: A set which is possible to count the number of elements of that set is called finite set.
Example – 1 : The set {3, 4, 5, 6} is a finite set, because it contains a definite number of elements i.e., only 4 elements.
Example – 2 : The set of days in a week is a finite set.

Example – 3 : An empty set, which does not contain any element (no element) is also a finite set.

→ Infinite set: A set whose elements cannot be listed, that type of set is called infinite set.

Example : i) B = {x : x is an even number}
ii) J = {x : x is a multiple of 7}
iii) The set of all points in a plane. s|s A set is infinite if it is not finite.

→ Equal sets: Two sets are said to be equal, if they have exactly the same elements.
For example: The set A and B are having same elements i.e., watch, ring, flower are said to be equal sets.

→ Cardinal number: The number of elements in a set is called the cardinal number of the set.
Example: Consider the finite set A = {1, 2, 4}
Number of elements in set ‘A’ is 3.
It is represented by n(A) = 3

→ Universal set: A set which consists of all the sets under consideration (or) discussion is called the universal set. (or) A set containing all objects or elements and of which all other sets or subsets.
It is usually denoted by ∪ (or) μ.

The universal set is usually represented by rectangles.
Example:
i) The set of real numbers is universal set for number theory.
Here ‘R’ is a universal set.
ii) If we want to study various groups of people of our state, universal set is the set of all people in Andhra Pradesh.

→ Subset: If every element of first set (A) is also an element of second set (B), then first set (A) is said to be a subset of second set (B).

→ It is represented as A ⊂ B.
Example :
Set A = {2, 4, 6, 8} is a subset of .
Set B = {1,2, 3, 4,5, 6, 7, 8}

→ Empty set is a subset of every set.

→ Every set is a subset of itself.

→ Consider ‘A’ and ‘B’ are two sets, if A ⊂ B and B ⊂ A ⇔ A = B.

→ A set doesn’t change if one or more elements of the set are repeated.

→ If A ⊂ B, B ⊂ C ⇒ A ⊂ C.

→ Venn Diagrams: Venn-Euler diagram or simply Venn diagram is a way of representing the relationships between sets.

→ These diagrams consist of rectangles and closed curves usually circles.
Example: Consider that U = {1, 2, 3, ……, 10} is
the universal set of which, A = {2, 4, 6, 8, 10} is a subset.
Then the Venn diagram is as:

→ Basic operations on sets: In sets, we define the operations of union, intersection and difference of sets.

→ Union of sets: The union of two or more sets is the set of all those elements which are either individual (or) both in common.

→ In symbolic form, union of two sets A and B is written as A ∪ B and usually read as “A union B”.

→ Set builder form of A ∪ B is A ∪ B = (x : x ∈ A or x ∈ B}

→ The union of the sets can be represented by a Venn diagram as shown (shaded portion).

→ It is evident from the definition that A ⊆ A ∪ B; B ⊆ A ∪ B.

→ Roster form of union of sets : Let A = {a, e, i, o, u} and B = (a, i, u} then A ∪ B = {a, e, i, o, u} ∪ { a, i, u} = {a, e, i, o, u}

→ Intersection of sets: The intersection of two sets A and B is the set of all those elements which belong to both A and B.

→ We denote intersection by A ∩ B.

→ We read A ∩ B as “A intersection B”.

→ Symbolically, we write A ∩ B = (x : x ∈ A and x ∈ B}

→ The intersection of A and B can be illustrated in the Venn diagram as shown in the shaded portion in the adjacent figure.

→ The intersection of A and B can be illustrated in the Roster form:
If A = {5, 6, 7, 8} and B = {7, 8, 9, 10} then A ∩ B = {7, 8}

→ Disjoint set: Consider A and B are two finite sets and if there are no common element in both A and B. Such set is known as disjoint set (or A ∩ B = ∅).
(or)
Two sets (finite) are said to be disjoint sets if they have no common elements. That is if the intersection of two sets is a null set they are disjoint sets.

→ The disjoint sets can be represented by means of the Venn diagrams as shown in the adjacent figure.

→ Difference of sets: The difference of sets A and B is the set of elements which belong to ‘A’ but do not belong to ‘B’.

→ We denote the difference of A and B by A – B or simply “A minus B”.

→ Set builder form of A – B is (x : x ∈ A and ∉ B}

→ Venn-diagram of A – B is

→ Venn-diagram of B – A is

→ A – B ≠ B – A

→ Fundamental theorem on sets:
If A and B are any two sets then n (A ∪ B) = n (A) + n (B) – n (A ∩ B) where
n (A ∪ B) = number of elements in the set (A ∪ B), also called cardinal number of A ∪ B
n (A) = number of elements in the set A also called cardinal number of A
n (B) = number of elements in the set B also called cardinal number of B

Students can go through AP SSC 10th Class Maths Notes Chapter 1 Real Numbers to understand and remember the concepts easily.

AP State Syllabus SSC 10th Class Maths Notes Chapter 1 Real Numbers

→ “God made the integers. All else is the work of man” …… Leopold Kronecker

→ Euclid’s division lemma: Given positive integers a, b there exists unique pair of integers q and r satisfying
a = bq + r; 0 ≤ r < b
This result was first published / recorded in book VII of Euclid’s “The Elements”.

→ Euclid’s division algorithm is a technique to compute the Highest Common Factor (H.C.F) of two given numbers.
E.g: HCF of 80 and 130
130 = 80 × 1 + 50 80 = 50 × 1 + 30
50 = 30 × 1 + 20 30 = 20 × 1 + 10
20 = 10 × 2 + 0 and H.C.F = 10

→ Euclid’s division algorithm can also be extended to all integers.

→ Numbers which can be expressed in the form p/q, where q ≠ 0 and ‘p and q’ are integers are called rational numbers; represented by Q.
Q = { \(\frac{p}{q}\) ; q ≠ 0; p, q ∈ Z} .

→ Every rational number can be expressed either as a terminating decimal or as a non-terminating recurring decimal.

→ Numbers which can’t be expressed in p/q form are called irrational numbers represented by S. You may notice that the first letter of surds is ‘S’.
Eg: √2, √3, √5, …….. etc,

→ The combined set of rationals and irrationals is called the set of Real numbers; represented by R.
R = Q ∪ S.

→ Diagramatic representation of the number system:

where N = the set of natural numbers; W = the set of whole numbers;
Z = the set of integers; Q = the set of rational numbers;
S = the set of irrational numbers ; R = the set of real numbers.

→ Fundamental Theorem of Arithmetic: Every composite number can be expressed as a product of primes uniquely, (i.e.,) if x is a composite number, then
x = \(p_{1}^{l} \cdot p_{2}^{m} \cdot p_{3}^{n}\) …… where p₁, p₂, p₃, ….. are prime numbers and l, m, n, …… are natural numbers.
Eg: 420 = 2 × 210 = 2 × 2 × 105 = 2 × 2 × 3 × 35 = 2 × 2 × 3 × 5 × 7
i. e. 420 = 2² × 3¹ × 5¹ × 7¹ and the factorisation on the R.H.S is unique.
Note: R.H.S is called exponential form of 420.

→  To find the H.C.F. of two or more numbers:
Step (i): Express given numbers in their exponential form.
Step (ii): Take the common bases.
Step (iii): Assign the respective smallest exponent from their exponential forms.
Step (iv): Take the product of the above.
Eg: H.C.E of 60 and 75 is
Step (i) 60 = 2² × 3 × 5 ; 75 = 3 × 5²
Step (ii) 3^O × 5^O [taking common bases]
Step (iii) 3¹ × 5¹ [∵ smallest exponent among 3¹ and 3¹ is 1]
Step (iv) 3 × 5 = 15 [smallest exponent among 5¹ and 5² is 1]
∴ H.C.F = 15
(i.e.) The highest common factor of the given set of numbers is the product of the com¬mon bases with the respective least exponents.

→ To find the L.C.M. of two or more numbers:
Step – 1: Express the given numbers in their exponential forms.
Step – 2: Take every base.
Step – 3: Assign the respective greatest exponent to each base.
Step – 4: Take the product of the above.
Eg: L.C.M. of 60 and 75 is
Step – 1: 60 = 2² × 3 × 5 ; 75 = 3 × 5²
Step – 2: 2^O × 3^O × 5^O
Step – 3: 2² × 3¹ × 5²
Step – 4: 4 × 3 × 25 = 300
L.C.M = 300

→ We may notice that the product of any two numbers N₁ and N₂ is equal to the product of their L.C.M. (L) and H.C.F. (H).
i.e., N₁ . N₂ = L.H

→ Let x = p/q be a rational number. If the numerator p is divided by the denominator q, we get the decimal form of x. The decimal form of x may or may not be terminating, i.e., every rational number can be expressed either as a terminating decimal or a non-terminating decimal. This gives us the following theorems.
Theorem – 1: Let ‘x’ be a rational number when expressed in decimal form, terminates, then x can be expressed in the form p/q where p, q are co-primes and the prime factorization of q is of the form 2ⁿ × 5^m, where n and m are non-negative integers.
Theorem – 2: Let x = p/q be a rational number, where q is of the form 2ⁿ × 5^m then x has a decimal expansion that terminates.
Theorem – 3: Let x = p/q be a rational number, where p, q are co-primes and the prime factorization of q is not of the form 2ⁿ . 5^m (n, m ∈ Z⁺) then x has a decimal expansion which is non-terminating recurring decimal.
Theorem – 4: Let ‘p’ be a prime number. If p divides a² then p divides a, where ‘a’ is a positive integer.

→ If a is a non-zero rational number and b is any irrational number, then (a + b), (a – b), a/b and ab are all irrational numbers.

→ Properties of Real Numbers:
If a, b and c are any three real numbers we may notice that

• a + b is also a real number – closure property w.r.t. addition
  a.b is also a real number – closure property w.r.t. multiplication
• a + b = b + a – commutative property w.r.t. addition
  a . b = b . a – commutative property w.r.t. multiplication
• (a + b) + c = a + (b + c) – associative law w.r.t.
  addition (a.b).c = a.(b.c) – associative law w.r.t. multiplication
• a + 0 = 0 + a = a, where ‘0’ is the additive identity,
  a × 1 = 1 × a = a, where 1 is the multiplicative identity,
• a + (-a) = (-a) + a = 0 where (a) and (-a) are additive inverse of each other.
  a × \(\frac{1}{a}\) = \(\frac{1}{a}\) × a = 1 where a and \(\frac{1}{a}\) are multiplicative inverse of each other.

→ If aⁿ = x, where a and x are positive integers and a ≠ 1, then we define log_ax = n read as logarithm of x to the base a is equal to n.
Eg.: 2⁴ = 16 ⇒ log₂16 = 4

→  log_ax + log_ay = log_axy

→ log_aa = 1

→ log_ax – log_ay = log_a\(\frac{x}{y}\)

→ log_a1 = 0

→ log_ax^m = m log_ax

→ In general, the bases in the logarithms are 10 (or) e, where e’ is approximated to 2.718.

→ If p is a prime number and p divides a² then p divides ‘a’ also.

AP State Syllabus SSC 10th Class Maths Solutions 14th Lesson Statistics InText Questions

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 14 Statistics InText Questions and Answers.

10th Class Maths 14th Lesson Statistics InText Questions and Answers

Think & Discuss

(Page No. 327)

Question 1.
The mean value can be calculated from both ungrouped and grouped data. Which one do you think is more accurate? Why?
Answer:
Mean calculated from ungrouped data is more accurate than, mean calculated from the grouped data. Since its calculation takes all the observations in the data into consideration.

Question 2.
When it is more convenient to use grouped data for analysis?
Answer:
Grouped data is convenient when the values f_i and x_i are low.

(Page No. 331)

Question 3.
Is the result obtained by all the three methods the same?
Answer:
Yes. Mean obtained by all the three methods are equal.

Question 4.
If x_i and f_i are sufficiently small, then which method is an appropriate choice?
Answer:
Direct method.

Question 5.
If x_i and f_i are numerically large numbers, then which methods are appropriate to use?
Answer:
Assumed – Mean method and Step – Deviation method.

Do these

(Page No. 334)

Question 1.
Find the mode of the following data.
a) 5, 6, 9, 10, 6, 12, 3, 6, 11, 10, 4, 6, 7.
Answer:
Mode = 6 (most repeated value of the data).

b) 20, 3, 7, 13, 3, 4, 6, 7, 19, 15, 7, 18, 3.
Answer:
Mode = 3,7.

c) 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6.
Answer:
Mode = 2, 3, 4, 5, 6.

Question 2.
Is the mode always at the centre of the data?
Answer:
No. Mode may not beat the centre always.

Question 3.
Does the mode change, if another observation is added to the data in Example? Comment.
Answer:
In the example the observations are 0, 1, 2, 2, 2, 3, 3, 4, 5, 6.
Here mode = 2.
If we add ‘2’ then mode doesn’t change.
It we add ‘3’ then the mode will be ‘2’ and ‘3’.
Even if we add other then 3 the mode will not be changed.
So we cannot decide whether the mode changes or not. It depends on the situation.

Question 4.
If the maximum value of an observation in the data in Example 4 is changed to 8, would the mode of the data be affected? Comment.
Answer:
If the maximum value is altered to 8, the mode remains the same. Mode doesn’t consider the values but consider their frequencies only.

Think & Discuss

(Page No. 336)

Question 1.
It depends upon the demand of the situation whether we are interested in finding the average marks obtained by the students or the marks obtained by most of the students.
a) What do we find in the first situation?
Answer:
We find A.M.

b) What do we find in the second situation?
Answer:
We find the mode.

Question 2.
Can mode be calculated for grouped data with unequal class sizes?
Answer:
Yes. Mode can be calculated for grouped data with unequal class sizes.

AP State Syllabus SSC 10th Class Maths Solutions 13th Lesson Probability InText Questions

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

10th Class Maths 13th Lesson Probability InText Questions and Answers

Do This

(Page No. 307)

Outcomes of which of the following experiments are equally likely ?
Question 1.
Getting a digit 1, 2, 3, 4, 5 or 6 when a die is rolled.
Answer:
Equally likely.

Question 2.
Picking a different colour ball from a bag of 5 red balls, 4 blue balls and 1 black ball.
Note: Picking two different colour balls …………..
i.e., picking a red or blue or black ball from a …………
Answer:
Not equally likely.

Question 3.
Winning in a game of carrom.
Answer:
Equally likely.

Question 4.
Units place of a two digit number selected may be 0, 1, 2, 3, 4, 5, 6, 7, 8 or 9.
Answer:
Equally likely.

Question 5.
Picking a different colour ball from a bag of 10 red balls, 10 blue balls and 10 black balls.
Answer:
Equally likely.

Question 6.
a) Raining on a particular day of July.
Answer:
Not equally likely.

b) Are the outcomes of every experiment equally likely?
Answer:
Outcomes of all experiments need not necessarily be equally likely.

c) Give examples of 5 experiments that have equally likely outcomes and five more examples that do not have equally likely outcomes.
Answer:
Equally likely events:

1. Getting an even or odd number when a die is rolled.
2. Getting tail or head when a coin is tossed.
3. Getting an even or odd number when a card is drawn at random from a pack of cards numbered from 1 to 10.
4. Drawing a green or black ball from a bag containing 8 green balls and 8 black balls.
5. Selecting a boy or girl from a class of 20 boys and 20 girls.
6. Drawing a red or black card from a deck of cards.

Events which are not equally likely:

1. Getting a prime or composite number when a die is thrown.
2. Getting an even or odd number when a card is drawn at random from a pack of cards numbered from 1 to 5.
3. Getting a number which is a multiple of 3 or not a multiple of 3 from numbers 1, 2, …… 10.
4. Getting a number less than 5 or greater than 5.
5. Drawing a white ball or green ball from a bag containing 5 green balls and 8 white balls.

Question 7.
Think of 5 situations with equally likely events and find the sample space.    (Page No. 309)
Answer:
a) Tossing a coin: Getting a tail or head when a coin is tossed.
Sample space = {T, H}.
b) Getting an even or odd number when a die is rolled.
Sample space = (1, 2, 3, 4, 5, 6}.
c) Winning a game of shuttle.
Sample space = (win, loss}.
d) Picking a black or blue ball from a bag containing 3 blue balls and 3 blackballs = {blue, black}.
e) Drawing a blue coloured card or black coloured card from a deck of cards = {black, red}.

Question 8.
i) Is getting a head complementary to getting a tail? Give reasons.   (Page No. 311)
Answer:
Number of outcomes favourable to head = 1
Probability of getting a head = \(\frac{1}{2}\) [P(E)]
Number of outcomes not favourable to head = 1
Probability of not getting a head = \(\frac{1}{2}\) [P(\(\overline{\mathrm{E}}\))]
Now P(E) + P(\(\overline{\mathrm{E}}\)) = \(\frac{1}{2}\) + \(\frac{1}{2}\) = 1
∴ Getting a head is complementary to getting a tail.

ii) In case of a die is getting a 1 comple-mentary to events getting 2, 3, 4, 5, 6? Give reasons for your answer.
Answer:
Yes. Complementary events.
∵ Probability of getting 1 = \(\frac{1}{6}\) [P(E)]
Probability of getting 2, 3, 4, 5, 6 = P(E) = P(\(\overline{\mathrm{E}}\)) = \(\frac{5}{6}\)
P(E) + P(\(\overline{\mathrm{E}}\)) = \(\frac{1}{6}\) + \(\frac{5}{6}\) = \(\frac{6}{6}\) = 1

iii) Write of five new pair of events that are complementary.
Answer:

1. When a dice is thrown, getting an even number is complementary to getting an odd number.
2. Drawing a red card from a deck of cards is complementary to getting a black card.
3. Getting an even number is complementary to getting an odd number from numbers 1, 2, ….. 8.
4. Getting a Sunday is complementary to getting any day other than Sunday in a week.
5. Winning a running race is complementary to loosing it.

Try This

Question 1.
A child has a dice whose six faces show the letters A, B, C, D, E and F. The dice is thrown once. What is the probability of getting (i) A? (ii) D?     (Page No. 312)
Answer:
Total number of outcomes (A, B, C, D, E and F) = 6.
i) Number of favourable outcomes to A = 1
Probability of getting A =
P(A) = \(\frac{\text { No.of favourable outcomesto } \mathrm{A}}{\text { No.of all possible outcomes }}\) = \(\frac{1}{6}\)

ii) No. of outcomes favourable to D = 1
Probability of getting D
= \(\frac{\text { No.of outcomes favourble to } \mathrm{D}}{\text { All possible outcomes }}\) = \(\frac{1}{6}\)

Question 2.
Which of the following cannot be the probability of an event?     (Page No. 312)
(a) 2.3
(b) -1.5
(c) 15%
(d) 0.7
Answer:
a) 2.3 – Not possible
b) -1.5 – Not possible
c) 15% – May be the probability
d) 0.7 – May be the probability

Question 3.
You have a single deck of well shuffled cards. Then, what is the probability that the card drawn will be a queen?     (Page No. 313)
Answer:
Number of all possible outcomes = 4 × 13 = 1 × 52 = 52
Number of outcomes favourable to Queen = 4 [♥ Q, ♦ Q, ♠ Q, ♣ Q]
∴ Probability P(E) = \(\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}\)
= \(\frac{4}{52}\) = \(\frac{1}{13}\)

Question 4.
What is the probability that it is a face card?     (Page No. 314)
Answer:
Face cards are J, Q, K.
∴ Number of outcomes favourable to face card = 4 × 3 = 12
No. of all possible outcomes = 52
P(E) = \(\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}\)
= \(\frac{12}{52}\) = \(\frac{3}{13}\)

Question 5.
What is the probability that it is a spade?       (Page No. 314)
Answer:
Number of spade cards = 13
Total number of cards = 52
Probability
= \(\frac{\text { Number of outcomes favourable to spades }}{\text { Number of all outcomes }}\)
= \(\frac{13}{52}\) = \(\frac{1}{4}\)

Question 6.
What is the probability that is the face card of spades?       (Page No. 314)
Answer:
Number of outcomes favourable to face cards of spades = (K, Q, J) = 3
Number of all outcomes = 52
P(E) = \(\frac{3}{52}\)

Question 7.
What is the probability it is not a face card?       (Page No. 314)
Answer:
Probability of a face card = \(\frac{12}{52}\) from (1)
∴ Probability that the card is not a face card

(or)
Number of favourable outcomes = 4 × 10 = 40
Number of all outcomes = 52
∴ Probability
= \(\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}\)
= \(\frac{40}{52}\) = \(\frac{10}{13}\)

Think & Discuss

(Page No. 312)

Question 1.
Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of any game?
Answer:
Probability of getting a head is \(\frac{1}{2}\) and of a tail is \(\frac{1}{2}\) are equal.
Hence tossing a coin is a fair way.

Question 2.
Can \(\frac{7}{2}\) be the probability of an event? Explain.
Answer:
\(\frac{7}{2}\) can’t be the probability of any event.
Since probability of any event should lie between 0 and 1.

Question 3.
Which of the following arguments are correct and which are not correct? Give reasons.
i) If two coins are tossed simultaneously, there are three possible outcomes – two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is \(\frac{1}{3}\).
Answer:
False.
Reason:
All possible outcomes are 4
HH, HT, TH, TT
Thus, probability of two heads = \(\frac{1}{4}\)
Probability of two tails = \(\frac{1}{4}\)
Probability of one each = \(\frac{2}{4}\) = \(\frac{1}{2}\).

ii) If a dice is thrown, there are two possible outcomes – an odd number or an even number. Therefore, the probability of getting an odd number is \(\frac{1}{2}\).
Answer:
True.
Reason:
All possible outcomes = (1, 2, 3, 4, 5, 6) = 6
Outcomes favourable to an odd number (1, 3, 5) = 3
Outcomes favourable to an even number = (2, 4, 6) = 3
∴ Probability (odd number)
= \(\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}\)
= \(\frac{3}{6}\) = \(\frac{1}{2}\).

AP State Syllabus SSC 10th Class Maths Solutions 12th Lesson Applications of Trigonometry InText Questions

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 12 Applications of Trigonometry InText Questions and Answers.

10th Class Maths 12th Lesson Applications of Trigonometry InText Questions and Answers

Do This

(Page No. 297)

Question 1.
Draw diagram for the following situations:
i) A person is flying a kite at an angle of elevation a and the length of thread from his hand to kite is ‘l’.
ii) A person observes two banks of a river at angles of depression θ₁ and θ₂ (θ₁ < θ₂) from the top of a tree of height ‘h’ which is at a side of the river. The width of the river is ‘d’.
Answer:

In the figure
‘D’ is the position of person
CD is the height of the tree
AB is the width of the river
Angles of depression are θ₁ and θ₂.

Think & Discuss

(Page No. 297)

Question 1.
You are observing top of your school building at an angle of elevation a from a point which is at d meter distance from foot of the building. Which trigonometric ratio would you like to consider to find the height of the building?
Answer:

The trigonometric ratio which connects d and h is tan α.

Question 2.
A ladder of length x meter is leaning against a wall making angle θ with the ground. Which trigonometric ratio would you like to consider to find the height of the point on the wall at which the ladder is touching?
Answer:

We use ‘sin θ’ as it is the ratio between the side opp. to θ and hypotenuse.

AP State Syllabus SSC 10th Class Maths Solutions 11th Lesson Trigonometry InText Questions

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

10th Class Maths 11th Lesson Trigonometry InText Questions and Answers

Do This

Identify “Hypotenuse”, “Opposite side” and “Adjacent side” for the given angles in the given triangles.  (Page No. 271)

Question 1.
For angle R
Answer:

In the △PQR
Opposite side = PQ
Adjacent side = QR
Hypotenuse = PR

Question 2.
i) For angle X
ii) For angle Y        (Page No. 271)
Answer:

In the △XYZ,
i) For angle X
Opposite side = YZ
Adjacent side = XZ
Hypotenuse = XY
ii) For angle Y
Opposite side = XZ
Adjacent side = YZ
Hypotenuse = XY

Question 3.
Find (i) sin C (ii) cos C and (iii) tan C in the given triangle.    (Page No. 274)

Answer:
By Pythagoras theorem
AC² = AB² + BC²
13² = AB² + 5²
AB² = 169 – 25
AB² = √144
⇒ AB = √144 = 12

Question 4.
In a triangle XYZ, ∠Y is right angle, XZ = 17 cm and YZ = 15 cm, then find (i) sin X (ii) cos Z (iii) tan X.    (Page No. 274)
Answer:
Given △XYZ, ∠Y is right angle.

By Pythagoras theorem
XZ² = YZ² + XY²
17² = 15² + XY²
XY² = 17² – 15² = 289 – 225
XY² = 64
XY = √64 = 8

Question 5.
In a triangle PQR with right angle at Q, the value of ∠P is x, PQ = 7 cm and QR = 24 cm, then find sin x and cos x.    (Page No. 274)
Answer:
Given right angled triangle is PQR with right angle at Q. The value of ∠P is x.

By Pythagoras theorem
PR² = PQ² + QR²
PR² = 7² + 24²
PR² = 49 + 576
PR² = 625
PR² = √625 = 25
sin x = \(\frac{QR}{PR}\) = \(\frac{24}{25}\)
cos x = \(\frac{PQ}{PR}\) = \(\frac{7}{25}\)

Try This

Question 1.
Write lengths of “Hypotenuse”, “Opposite side” and “Adjacent side” for the given angles in the given triangles.
1. For angle C
2. For angle A          (Page No. 271)
Answer:

By Pythagoras theorem
AC² = AB² + BC²
(5)² = AB² + 4²
25 = AB² + 16
AB² = 25 – 16
AB² = 9
AB = √9 = 3
For angle C:
Opposite side = AB = 3 cm
Adjacent side = BC = 4 cm
Hypotenuse = AC = 5 cm
For angle A:
Opposite side = BC = 4 cm
Adjacent side = AB = 3 cm
Hypotenuse = AC = 5 cm

Question 2.
In a right angle triangle ABC, right angle is at C. BC + CA = 23 cm and BC – CA = 7 cm, then find sin A and tan B.    (Page No. 274)
Answer:
In a right angle triangle ABC, right angle is at C.

BC = \(\frac{30}{2}\) = 15
BC = 15
Substituting BC = 15 in equation (1)
BC + CA = 23
CA = 23 – BC = 23 – 15
CA = 8
By Pythagoras theorem
AB² = AC² + BC²
= 8² + 15²
= 64 + 225 = 289
= √289 = 17
sin A = \(\frac{BC}{AB}\) = \(\frac{15}{17}\)
cos B = \(\frac{AC}{BC}\) = \(\frac{8}{15}\)

Question 3.
What will be the ratios of sides for sec A and cot A?    (Page No. 275)
Answer:
sec A = \(\frac{\text { Hypotenuse }}{\text { Adjacent side of the angle } \mathrm{A}}\)
cot A = \(\frac{\text { Adjacent side of the angle } A}{\text { Opposite side of the angle } A}\)

Think & Discuss

Question 1.
Discuss with your friends
i) sin x = \(\frac{4}{3}\) does exist for some value of angle x?
ii) The value of sin A and cos A is always less than 1. Why?
iii) tan A is product of tan and A.        (Page No. 274)
Answer:
i) The value of sin 0 always lies between 0 and 1. Here sin x = \(\frac{4}{3}\) which is greater than 1. So, it does not exist.
ii)

We observe in above sin A, cos A, hypotenuse is in denominator which is greater than other two sides
∴ sin A = \(\frac{\text { Opposite side }}{\text { Hypotenuse }}\) = \(\frac{Nr}{Dr}\)
here denominator is more than numerator. Hence its value will be less than 1.
cos A = \(\frac{\text { Adjacent side }}{\text { Hypotenuse }}\)
here also adjacent side is always less than hypotenuse. Hence its value is also less than or equal to 1.

iii) The symbol tan A is used as an abbreviation for “the tan of the angle A”.
tan A is not the product of “tan” and A. “tan” separated from A’ has no meaning.

Question 2.
Is \(\frac{\sin \mathrm{A}}{\cos \mathrm{A}}\) equal to tan A?      (Page No. 275)
Answer:
Yes, \(\frac{\sin \mathrm{A}}{\cos \mathrm{A}}\) = tan A
Proof:

Question 3.
Is \(\frac{\cos \mathrm{A}}{\sin \mathrm{A}}\) equal to cot A?     (Page No. 275)
Answer:
Yes, \(\frac{\cos \mathrm{A}}{\sin \mathrm{A}}\) = tan A
Proof:

cot A = \(\frac{\text { Adjacent side }}{\text { Opposite side }}\)

Do this

Question 1.
Find cosec 60°, sec 60° and cot 60°.       (Page No. 279)
Answer:

Consider an equilateral triangle ABC. Since each angle is 60° in an equilateral triangle, we have ∠A = ∠B = ∠C = 60° and the sides of equilateral triangle is AB = BC = CA = 2a units.
Draw the perpendicular line AD from vertex A to BC as shown in the given figure.
Perpendicular AD acts as “angle bisector of angle A” and “bisector of the side BC” in the equilateral triangle ABC.
Therefore, ∠BAD = ∠CAD = 30°. Since point D divides the side BC into equal halves.
BD = \(\frac{1}{2}\) BC = \(\frac{2a}{2}\) = a units.
Consider right angle triangle ABD in the above given figure.
We have AB = 2a and BD = a
Then AD² = AB² – BD²
(By Pythagoras theorem)
= (2a)² – (a)² = 3a²
Therefore, AD = a√3
From definitions of trigonometric ratios,

Try this

Question 1.
Find sin 30°, cos 30°, tan 30°, cosec 30°, sec 30° and cot 30° by using the ratio concepts.     (Page No. 279)
Answer:

Consider an equilateral triangle ABC. Since each angle is 60° in an equilateral triangle, we have ∠A = ∠B = ∠C = 60° and the sides of equilateral triangle is AB = BC = CA = 2a units.
Draw the perpendicular line AD from vertex A to BC as shown in the given figure.
Perpendicular AD acts as “angle bisector of angle A” and “bisector of the side BC” in the equilateral triangle ABC.
Therefore, ∠BAD = ∠CAD = 30°. Since point D divides the side BC into equal halves.
BD = \(\frac{1}{2}\) BC = \(\frac{2a}{2}\) = a units.
Consider right angle triangle ABD in the above given figure.
We have AB = 2a and BD = a
Then AD² = AB² – BD²
(By Pythagoras theorem)
= (2a)² – (a)² = 3a²
Therefore, AD = \(\sqrt{3 a^{2}}\) = √3a
BD = a, AD = √3a and hypotenuse = AB = 2a and ∠DAB = 30°.
sin 30° = \(\frac{BD}{AB}\) = \(\frac{a}{2a}\) = \(\frac{1}{2}\)

Question 2.
Find the values for tan 90°, cosec 90°, sec 90° and cot 90°.     (Page No. 281)
Answer:
From the adjacent figure, the trigonometric ratios of ∠A, gets larger and larger in △ABC till it becomes 90°.

As ∠A get larger and larger, ∠C gets smaller and smaller. Therefore, the length of the side AB goes on decreasing. The point A gets closer to point B. Finally when ∠A is very close to 90°, ∠C becomes very close to 0° and the side AC almost coincides with side BC.

∴ AB = 0 and AC = BC = r
From trigonometric ratios
sin A = \(\frac{BC}{AC}\)
sin A = \(\frac{AB}{AC}\)
If A = 90°, then AB = 0 and AC = BC = r,

Think & Discuss

Question 1.
Discuss with your friend about the following conditions:
What can you say about cosec 0° = \(\frac{1}{\sin 0^{\circ}}\)? Is it defined? Why?    (Page No. 280)
Answer:
sin 0° = 0
cosec 0° = \(\frac{1}{\sin 0^{\circ}}\) = \(\frac{1}{0}\) = not defined.
It is not defined.
Reason:
Division by ‘0’ is not allowed, hence \(\frac{1}{0}\) is indeterminate.

Question 2.
What can you say about cot 0° = \(\frac{1}{\tan 0^{\circ}}\). Is it defined? Why?    (Page No. 281)
Is it defined? Why?
Answer:
tan 0° = 0
cot 0° = \(\frac{1}{\tan 0^{\circ}}\) = \(\frac{1}{0}\) = undefined.
Reason:
Division by ‘0’ is not allowed, hence \(\frac{1}{0}\) is indeterminate.

Question 3.
sec 0° = 1. Why?      (Page No. 281)
Answer:
sec 0° = \(\frac{1}{\cos 0^{\circ}}\) [∵ cos 0° = 1]
= \(\frac{1}{1}\) = 1

Question 4.
What can you say about the values of sin A and cos A, as the value of angle A increases from 0° to 90°? (Observe the above table)
i) If A ≥ B, then sin A ≥ sin B. Is it true?
ii) If A ≥ B, then cos A ≥ cos B. Is it true? Discuss.      (Page No. 282)
Answer:
i) Given statement
“If A ≥ B, then sin A ≥ sin B”
Yes, this statement is true.
Because, it is clear from the table below that the sin A increases as A increases.

ii) Given statement
“If A ≥ B, then cos A ≥ cos B”
No, this statement is not true.
Because, it is clear from the table below that cos A decreases as A increases.

Think & Discuss

Question 1.
For which value of acute angle
(i) \(\frac{\cos \theta}{1-\sin \theta}\) + \(\frac{\cos \theta}{1+\sin \theta}\) = 4 is true?
For which value of 0° ≤ θ ≤ 90°, above equation is not defined?    (Page No. 285)
Answer:


⇒ cos θ = cos 60° [from trigonometric ratios table]
⇒ θ = 60°
Given statement is true for the acute angle i.e., θ = 60°.

Question 2.
Check and discuss the above relations in the case of angles between 0° and 90°, whether they hold for these angles or not? So,          (Page No. 286)
i) sin (90° – A) = cos A
ii) cos (90° – A) = sin A
iii) tan (90° – A) = cot A and
iv) cot (90° – A) = tan A
v) sec (90° – A) = cosc A
vi) cosec (90° – A) = sec A
Answer:
Let A = 30°
i) sin (90° – A) = cos A
⇒ sin (90° – 30°) = cos 30°
⇒ sin 60° = cos 30° = \(\frac{\sqrt{3}}{2}\)

ii) cos (90° – A) = sin A
⇒ cos (90° – 30°) = sin 30°
⇒ cos 60° = sin 30° = \(\frac{1}{2}\)

iii) tan (90° – A) = cot A
⇒ tan (90° – 30°) = cot 30°
⇒ tan 60° = cot 30° = √3

iv) cot (90° – A) = tan A
⇒ cot (90° – 30°) = tan 30°
⇒ cot 60° = tan 30° = \(\frac{1}{\sqrt{3}}\)

v) sec (90° – A) = cosec A
⇒ sec (90° – 30°) = cosec 30°
⇒ sec 60° = cosec 30° = 2

vi) cosec (90° – A) = sec A
⇒ cosec (90° – 30°) = sec 30°
⇒ cosec 60° = sec 30° = \(\frac{2}{\sqrt{3}}\)

So, the above relations hold for all the angles between 0° and 90°.

Do this

(Page No. 290)

Question 1.
i) If sin A = \(\frac{15}{17}\) then find cos A.
Answer:
Given sin A = \(\frac{15}{17}\)
cos A = \(\sqrt{1-\sin ^{2} A}\) [From Identity -I]

ii) If tan x = \(\frac{5}{12}\), then find sec x. (AS j)
Answer:
Given tan x = \(\frac{5}{12}\)
We know that sec² x – tan² x = 1
sec² x = 1 + tan² x

iii) If cosec θ = \(\frac{25}{7}\), then find cot θ.
Answer:
Given cosec θ = \(\frac{25}{7}\)
We know that cosec² θ – cot² θ = 1
cot² θ = cosec² θ – 1

Try This

(Page No. 290)

Question 1.
Evaluate the following and justify your answer.

i) \(\frac{\sin ^{2} 15^{\circ}+\sin ^{2} 75^{\circ}}{\cos ^{2} 36^{\circ}+\cos ^{2} 54^{\circ}}\)
Answer:
Given \(\frac{\sin ^{2} 15^{\circ}+\sin ^{2} 75^{\circ}}{\cos ^{2} 36^{\circ}+\cos ^{2} 54^{\circ}}\)

[∵ sin (90° – θ) = cos θ; cos (90° – θ) = sin θ]
= \(\frac{1}{1}\) = 1 [By sin² θ + cos² θ = 1]

ii) sin 5° cos 85° + cos 5° sin 85°
Answer:
Given sin 5° cos 85° + cos 5° sin 85°
= sin 5° . cos (90° – 5°) + cos 5° . sin (90° – 5°)
= sin 5° . sin 5° + cos 5° . cos 5°
[∵ sin (90° – θ) = cos θ; cos (90° – θ) = sin θ]
= sin² 5° + cos² 5°
= 1 [∵ sin² θ + cos² θ = 1]

iii) sec 16° cosec 74° – cot 74° tan 16°.
Answer:
Given sec 16° cosec 74° – cot 74° tan 16°
= sec 16° . cosec (90° – 16°) – cot (90° – 16°) . tan 16°
= sec 16°. sec 16° – tan 16° . tan 16° [∵ cosec (90° – θ) = sec θ; cot (90° – θ) — tan θ]
= sec² 16° – tan² 16°
= 1 [∵ sec² θ – tan² θ = 1]

Think & Discuss

(Page No. 290)

Question 1.
Are these identities true for 0° ≤ A ≤ 90°? If not for which values of A they are true?
i) sec² A – tan² A = 1
Answer:
Given identity is sec² A – tan² A = 1
Let A = 0°
L.H.S. = sec² 0° – tan² 0°
= 1 – 0 = 1 = R.H.S.
Let A = 90°
tan A and sec A are not defined.
So it is true.
∴ For all given values of ‘A’ such that 0° ≤ A ≤ 90° this trigonometric identity is true.

ii) cosec² A – cot² A = 1
Answer:
Given identity is cosec² A – cot² A = 1
Let A = 0°
cosec A and cot A are not defined for A = 0°.
Therefore identity is true for A = 0°.
Let A = 90°
cosec A = cosec 90° = 1
cot A = cot 90° = 0
L.H.S. = l² – 0² = 1 – 0 = 1 = R.H.S.
∴ This identity is true for all values of A, such that 0° ≤ A ≤ 90°.

AP State Syllabus SSC 10th Class Maths Solutions 10th Lesson Mensuration InText Questions

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

10th Class Maths 10th Lesson Mensuration InText Questions and Answers

Try this

(Page No. 245)

Question 1.
Consider the following situations. In each find out whether you need volume or area and why?
i) Quantity of water inside a bottle.
ii) Canvas needed for making a tent.
iii) Number of bags inside the lorry.
iv) Gas filled in a cylinder.
v) Number of match sticks that can be put in the match box.
Answer:
i) Volume: 3-d shape
ii) Area: L.S.A. / T.S.A.
iii) Volume: 3-d shape
iv) Volume: 3-d shape
v) Volume: 3-d shape

Question 2.
Write 5 more such examples and ask your friends to choose what they need?
Answer:
Student’s Activity.

Question 3.
Break the pictures in the previous figure into solids of known shapes.    (Page No. 246)
Answer:

Question 4.
Think of 5 more objects around you that can be seen as a combination of shapes. Name the shapes that combined to make them.    (Page No. 246)
Answer:
Student’s Activity.

Try this

Question 1.
Use known solid shapes and make as many objects (by combining more than two) as possible that you come across in your daily life.
[Hint: Use clay, or balls, pipes, paper cones, boxes like cube, cuboid etc]      (Page No. 252)
Answer:
Student’s Activity.

Think & Discuss

Question 1.
A sphere is inscribed in a cylinder. Is the surface of the sphere equal to the curved surface of the cylinder? If yes, explain how.      (Page No. 252)

Answer:
Yes, the surface area of the sphere is equal to the curved surface area of the cylinder.
Let the radius of the “cylinder be ‘r’ and its height ‘h’.
Then radius of sphere = r.
Then curved surface area of cylinder = 2πrh = 2πr (r + r)
[∵ height = diameter of the sphere = diameter of the cylinder = 2r]
= 2πr (2r) = 4πr²
And surface area of the sphere = 4πr²
∴ C.S.A. of cylinder = Surface area of sphere.

Try This

(Page No. 257)

Question 1.
If the diameter of the cross – section of a wire is decreased by 5%, by what percentage should the length be increased so that the volume remains the same ?
Answer:
Let the radius of wire = r
(if diameter = 2r)
and length of wire = h = l (say)
then volume of wire = πr²l₁ = v₁ ….(1)
Now the diameter after decreasing 5%

then radius of wire after decreasing
= \(\frac{190r}{100}\) × \(\frac{1}{2}\) = \(\frac{95r}{100}\)
and let the length of wire = (l₂)

So it length is increased by 24.22%. Its volume remains same.

Question 2.
Surface area of a sphere and cube are equal. Then find the ratio of their volumes.
Answer:
Let a cube with side ‘a’.
Then its surface area = 6a²
By problem, surface area of the sphere = 4πr²
Surface area of the cube = 6a²

Do This

(Page No. 263)

Question 1.
A copper rod of diameter 1 cm. and length 8 cm; is drawn into a wire of length 18 m of uniform thickness. Find the thickness of the wire.
Answer:
Volume of the copper rod (cylinder)
= πr²h
= \(\frac{22}{7}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) × 8
= \(\frac{44}{7}\) m²
If V is the radius of the wire, then its volume = πr²h
∴ The volume of rod is equal to the volume of the wire. We have

∴ Thickness = d = 2 × 0.03 ≃ 0.06 cm.

Question 2.
Pravali house has a water tank in the shape of a cylinder on the roof. This is filled by pumping water from a sump (an underground tank) which is in the shape of a cuboid. The sump has dimensions 1.57 m × 1.44 m × 9.5 cm. The water tank has radius 60 cm. and height 95 cm. Find the height of the water left in the sump after the water tank has been completely filled with water from the sump which had been full of water. Compare the capacity of the tank with that of the sump, (π = 3.14)
Answer:
Volume of the water in the sump = 1.57 × 1.44 × 0.95 [∵ V = lbh]
= 2.14776 m³ = 2147760 cm³
Volume of the tank on the roof = πr²h
= 3.14 × 60 × 60 × 95
= 1073880 cm³
∴ Volume of the water left in the sump after filling the tank
= 2147760 – 1073880 = 1073880 cm³
Let the height of the water in the tank be h.
∴ 157 × l44h = 1073880
h = \(\frac{1073880}{157 \times 144}\) = 47.5 cm.
∴ Ratio of the volume of the sump and tank = 2147760 : 1073880 = 2 : 1
∴ Sump can hold two times the water that can be hold in the tank.

Think & Discuss

(Page No. 262)

Question 1.
Which barrel shown in the below figure can hold more water? Discuss with your friends.

Answer:
r₁ = \(\frac{1}{2}\) = 0.5 cm; h₁ = 4 cm
Volume of the 1st barrel = πr²h
= \(\frac{22}{7}\) × 0.5 × 0.5 × 4 = 3.142 cm³
r₂ = \(\frac{4}{2}\) = 2 cm
∴ h = 1 cm
Volume of the 2nd barrel
V = πr²h = \(\frac{22}{7}\) × 2 × 2 × 1 = 12.57 cm³
Hence the volume of the 2nd barrel is more than the first barrel.

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

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

10th Class Maths 9th Lesson Tangents and Secants to a Circle InText Questions and Answers

Do this

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

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

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

We can draw only two tangents from an exterior point.

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

Answer:
P and M are the tangents to the given circles.

Question 4.
Draw a circle and a secant PQ of the circle on a paper as shown below. Draw various lines parallel to the secant on both sides of it. What happens to the length of chord coming closer and closer to the centre of the circle?      (Page No. 227)

Answer:

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

Question 5.
What is the longest chord?        (Page No. 227)
Answer:
Diameter is the longest of all chords.

Question 6.
How many tangents can you draw to a circle, which are parallel to each other?    (Page No. 227)
Answer:
Only one tangent can be drawn parallel to a given tangent.
To a circle, we can draw infinitely many pairs of parallel tangents.

Try this

(Page No. 228)

How can you prove the converse of the above theorem.
Question 1.
“If a line in the plane of a circle is perpendicular to the radius at its end point on the circle, then the line is tangent to the circle”.
Answer:
Given: Circle with centre ‘O’, a point A on the circle and the line AT perpendicular to OA.
R.T.P: AT is a tangent to the circle at A.
Construction:

Suppose AT is not a tangent then AT produced either way if necessary, will meet the circle again. Let it do so at P, join OP.
Proof: Since OA = OP (radii)
∴ ∠OAP = ∠OPA But ∠OPA = 90°
∴ Two angles of a triangle are right angles which is impossible.
∴ Our supposition is false.
∴ Hence AT is a tangent.

We can find some more results using the above theorem.
i) Since there can be only one perpendicular OP at the point P, it follows that one and only one tangent can be drawn to a circle at a given point on the circumference.
ii) Since there can be only one perpendicular to XY at the point P, it follows that the perpendicular to a tangent at its point of contact passes through the centre.

Question 2.
How can you draw the tangent to a circle at a given point when the centre of the circle is not known?    (Page No. 229)
Answer:

Steps of Construction:

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

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

Hint: Draw equal angles ∠QPX and ∠PRQ. Explain the construction.
Answer:

Steps of construction:

1. Draw any two chords AB and AC in the given circle.
2. Draw the perpendicular bisectors to AB and AC, they meet at the centre of the circle.
3. Tet O be the centre, join OP.
4. Draw a perpendicular to OP at P and extend it on either sides which forms a tangent to the circle at ‘P’.

Try this

Question 1.
Use Pythagoras theorem and write proof of above theorem “the lengths of tangents drawn from an external point to a circle are equal.”      (Page No. 231)
Answer:
Given: Two tangents PA and PB to a circle with centre O, from an exterior point P.
R.T.P: PA = PB

Proof: In △OAP; ∠OAP = 90°
∴ AP² = OP² – OA²
[∵ Square of the hypotenuse is equal to the sum of squares on the other two sides – Pythagoras theorem]
[∵ OA = OB, radii of the same circle]
= BP² [∵ In AOBP; OB² + BP² = OP²
⇒ BP² – OP² – OB²]
⇒ AP² – BP²
⇒ PA – PB Hence proved.

Question 2.
Draw a pair of radii OA and OB such that ∠BOA = 120°. Draw the bisector of ∠BOA and draw lines perpendiculars to OA and OB at A and B. These lines meet on the bisector of ∠BOA at a point which is the external point and the perpendicular lines are the required tangents. Construct and justify.    (Page No. 235)
Answer:

Justification:
OA ⊥ PA
OB ⊥ PB
Also in △OAP, △OBP
OA = OB
∠OAP = ∠OBP
OP – OP
∴ △OAP ≅ △OBP
∴ PA = PB. [Q.E.D.]

Do this

Question 1.
Shankar made the following pictures also with washbasin.

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

Question 2.
Make some more pictures and think of the shapes they can be divided into different parts.      (Page No. 237)
Answer:

Question 3.
Find the area of sector, whose radius is 7 cm. with the given angles.    (Page No. 239)
i) 60° ii) 30° iii) 72° iv) 90° v) 120°
Answer:

Question 4.
The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 10 minutes.    (Page No. 239)
Answer:
Angle made by minute hand in 1 m = \(\frac{360^{\circ}}{60}\) = 6°
Angle made by minute hand in 10m = 10 × 6 = 60°
The area swept by minute hand is in the shape of a sector with radius r = 14 cm and angle x = 60°

Area swept by the minute hand in 10 minutes = 102.66 cm².

Try this

Question 1.
How can you find the area of major segment using area of minor segment?    (Page No. 239)
Answer:

Area of the major segment = Area of the circle – Area of the minor segment.

AP State Syllabus SSC 10th Class Maths Solutions 8th Lesson Similar Triangles InText Questions

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

10th Class Maths 8th Lesson Similar Triangles InText Questions and Answers

Do these

(Page No. 194)

Question 1.
Fill in the blanks with similar/not similar.
i) All squares are ………. (similar)
ii) All equilateral triangles are ………. (similar)
iii) All isosceles triangles are ………. (similar)
iv) Two polygons with same number of sides are ………, if their corresponding angles are equal and corresponding sides are equal. (similar)
v) Reduced and enlarged photographs of an object are ………. (similar)
vi) Rhombus and squares are ……… to each other. (not similar)

Question 2.
Write True / False for the following statements.
i) Any two similar figures are congruent.
Answer:
False
ii) Any two congruent figures are similar.
Answer:
True
iii) Two polygons are similar if their corresponding angles are equal.
Answer:
False

Question 3.
Give two different examples of pair of
i) Similar figures
ii) Non-similar figures
Answer:
i) Similar figures:
a) Any two circles
b) Any two squares
c) Any two equilateral triangles
ii) Non-similar figures:
a) A square and a rhombus
b) A square and a rectangle

Question 4.
What value(s) of x will make DE || AB, in the given figure?  (Page No. 200)
AD = 8x + 9, CD = x + 3,
BE = 3x + 4, CE = x.

Answer:
Given : In AABC, DE // AB AD = 8x + 9, CD = x + 3,
BE = 3x + 4, CE = x
By Basic proportionality theorem,
If DE // AB then we should have
\(\frac{\mathrm{CD}}{\mathrm{DA}}\) = \(\frac{\mathrm{CE}}{\mathrm{EB}}\)
\(\frac{x+3}{8x+9}\) = \(\frac{x}{3x+4}\)
⇒ (x + 3) (3x + 4) = x (8x + 9)
⇒ x (3x + 4) + 3 (3x + 4) – 8x² + 9x
⇒ 3x² + 4x + 9x + 12 = 8x² + 9x
⇒ 8x² + 9x – 3x² – 4x – 9x -12 = 0
⇒ 5x² – 4x – 12 = 0
⇒ 5x² – 10x + 6x – 12 = 0
⇒ 5x (x – 2) + 6 (x – 2) = 0
⇒ (5x + 6) (x – 2) = 0
⇒ 5x + 6 = 0 or x – 2 = 0
⇒ x = \(\frac{-6}{5}\) or x = 2;
x cannot be negative.
∴ The value x = 2 will make DE // AB.

Question 5.
In △ABC, DE || BC. AD = x, DB = x – 2, AE = x + 2 and EC = x – 1. Find the value of x.    (Page No. 200)

Answer:
Given: In △ABC, DE // BC
∴ By Basic proportionality theorem, we have
\(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{\mathrm{AE}}{\mathrm{EC}}\)
\(\frac{\mathrm{x}}{\mathrm{x}-2}\) = \(\frac{x+2}{x-1}\)
⇒ x (x – 1) = (x + 2) (x – 2)
⇒ x² – x = x² – 4
⇒ -x = -4
∴ x = 4

Try This

Question 1.
E and F are points on the sides PQ and PR respectively of △PQR. For each of the following, state whether EF || QR or not?      (Page No. 197)
i) PE = 3.9 cm, EQ = 3 cm,
PF = 3.6 cm and FR = 2.4 cm.
Answer:

Here
\(\frac{\mathrm{PE}}{\mathrm{EQ}}\) = \(\frac{3.9}{3}\) = \(\frac{1.3}{1}\)
\(\frac{\mathrm{PF}}{\mathrm{FR}}\) = \(\frac{3.6}{2.4}\) = \(\frac{3}{2}\)
\(\frac{\mathrm{PE}}{\mathrm{EQ}}\) ≠ \(\frac{\mathrm{PF}}{\mathrm{FR}}\)
Hence, EF is not parallel to QR.

ii) PE = 4 cm, QE = 4.5 cm,
Answer:
Here
\(\frac{\mathrm{PE}}{\mathrm{EQ}}\) = \(\frac{4}{4.5}\) = \(\frac{8}{9}\)
\(\frac{\mathrm{PF}}{\mathrm{RF}}\) = \(\frac{8}{9}\)
\(\frac{\mathrm{PE}}{\mathrm{EQ}}\) = \(\frac{\mathrm{PF}}{\mathrm{RF}}\)
∴ EF // QR
Hence, EF is parallel to QR.

iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 1.8 cm and PF = 3.6 cm.
Answer:

Given: PQ = 1.28 cm, PE = 1.8 cm
⇒ EQ = PE – PQ = 1.8 – 1.28
⇒ EQ = 0.52 cm
Also, PR = 2.56 cm, PE = 3.6 cm, FR = PF – PR = 3.6 cm – 2.56 cm
FR = 1.04 cm

∴ EF // QR (By converse of Basic proportionality theorem)
Hence, EF is parallel to QR.

Question 2.
In the following figures DE || BC.    (Page No. 198)
i) Find EC.

Answer:
\(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{AE}{EC}\)
⇒ \(\frac{1.5}{3}\) = \(\frac{1}{EC}\)
∴ EC = \(\frac{3}{1.5}\) = 2 cm

ii) Find AD.

Answer:

Think & Discuss

Question 1.
Can you give some more examples from your daily life where scale factor is used?    (Page No. 192)
Answer:
Scale factor is used in drawing maps, designing machines and in sculpture, etc.

Question 2.
Can you say that a square and a rhombus are similar? Discuss with your friends.Write why the conditions are not sufficient.       (Page No. 193)
Answer:
A square □ ABCD and a rhombus ▱ PQRS are not similar.
Though the ratio of their corresponding sides are equal, the corresponding angles are not equal.

but ∠A ≠ ∠P; ∠B ≠ ∠Q
∠C ≠ ∠R; ∠D ≠ ∠S

Try This

(Page No. 207)

Question 1.
Are the triangles similar ? If so, name the criterion of similarity. Write the similarity relation in symbolic form.
i)

!! ∠G = ∠I alt. int. angles for the ∠F = ∠K parallel lines GF // KI
Answer:
∠FHG = ∠IHK (Vertically opp. angle)
∴ ∠GHF and ∠IKH are similar by AAA similarity rule.
△GHF ~ △IKH.

ii)

Answer:
\(\frac{PQ}{LM}\) = \(\frac{6}{3}\) = 2;
\(\frac{QR}{MN}\) = \(\frac{10}{4}\) = 2.5;
\(\frac{PQ}{LM}\) ≠ \(\frac{QR}{MN}\)
△PQR and △LMN are not similar.
△PQR ~ △LMN

iii)

Answer:
∠A = ∠A (Common)
\(\frac{AB}{AX}\) = \(\frac{5}{3}\); \(\frac{AC}{AY}\) = \(\frac{5}{3}\)
∴ △ABC and △AXY are similar by SAS similarity condition.
△ABC ~ △AXY.

iv)

Answer:

∴ △ABC and △APJ are not similar.

v)

Answer:
∠A = ∠B = 90°
∠AOQ = ∠POB (∵ Vertically opposite angles)
∠Q = ∠P (alternate interior angles)
∴ △AOQ and △BOP are similar by AAA criterion.
△AOQ ~ △BOP.

vi)

Answer:
△ABC and △QPR are similar by AAA similarity condition.
△ABC ~ △QPR.

vii)

Answer:
∠A = ∠P
\(\frac{AB}{PQ}\) = \(\frac{2}{5}\); \(\frac{AC}{PR}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
∴ \(\frac{AB}{PQ}\) ≠ \(\frac{AC}{PR}\)
Hence not similar.

viii)

Answer:
\(\frac{AB}{PQ}\) = \(\frac{6}{2.5}\); \(\frac{AC}{PR}\) = \(\frac{10}{5}\)
∴ △ABC and △PQR are not similar.

Question 2.
Explain why the triangles are similar and then find the value of x.
i)

Answer:
Given: In △PQR and △LTS
∠Q = ∠T; ∠R = ∠S = 90°
∴ ∠P = ∠T
(by angle sum property of triangles)
Hence, △PQR ~ △LTS [∵ AAA]

ii)

Answer:
Given: In △ABC and △PQC
∠B = ∠Q
[∵ ∠PQC = 180°- 110° = 70° – linear pair of angles]
∠C = ∠C [∵ Common]
∠A = ∠P [∵ Angle Sum property of triangles]
△ABC ~ △PQC by AAA similarity condition.
Then the ratio of their corresponding sides are equal.

iii)

Answer:
Given: In △ABC and △ECD
∠A = ∠E
∠ACB = ∠ECD [∵ Vertically opposite]
∴ ∠B = ∠D [∵ angle sum property]
∴ △ABC ~ △EDC

iv)

Answer:
Given: In △RAB and △RST
∠R = ∠R (common)
∠A = ∠S and ∠B = ∠T
[∵ Pair of corresponding angles for AB // ST]
∴ △RAB ~ △RST [∵ AAA similarity]

v)

Answer:
Given: In △PQR and △PMN
∠P = ∠P [∵ Common]
∠Q = ∠M [∵ Pair of corresponding angles for MN // QR]
∠R = ∠N
△PQR ~ △PMN [∵ AAA similarity]

[From the figure, PR = 4 + x]
⇒ 3 × 4 = 4 + x
⇒ x = 12 – 4 = 8

vi)

Answer:
Given: In △XYZ and △XBA,
∠X = ∠X [∵ Common]
∠B = ∠Y [∵ Pair of corresponding ∠A = ∠Z angles for AB // ZY]
∴ △XYZ ~ △XBA [∵ AAA similarity]

[From the figure, XZ = 7.5 + x]
\(\frac{3}{2}\) = \(\frac{7.5+\mathrm{x}}{\mathrm{x}}\)
3x = 15 + 2x;
3x – 2x = 15
x = 15

vii)

Answer:
Given: With the given conditions, we can’t find the value of x.
Note: If it is given that ∠A = ∠E then
we can say that △ABC ~ △EDC by AAA rule

viii)

Answer:
In △ABC and △BEC
∠ABC = ∠CEB (given)
∠C = ∠C (Common angle)
∴ △ABC ~ △BEC
(A.A. Criterion similarity)

Think & Discuss

(Page No. 203)

Question 1.
Discuss with your friends that in what way similarity of triangles is dif¬ferent from similarity of other polygons?
Answer:
In two triangles if the corresponding angles are equal then they are similar, whereas in two polygons if the corre-sponding angles are equal, they may not be similar, i.e., In triangles,
(Pairs of corresponding angles are equal) ⇔ (Ratio of corresponding sides are equal).
But this is not so with respect to polygons.

Do This

Question 1.
In △ACB, ∠C = 90° and CD ⊥ AB. Prove that \(\frac{\mathrm{BC}^{2}}{\mathrm{AC}^{2}}\) = \(\frac{BD}{AD}\).    (Page No. 218)

Answer:
Proof: △ADC and △CDB are similar.

[Ratio of areas of similar triangles is equal to the ratio of squares of their corresponding sides.]
From (1) and (2),
\(\frac{BD}{AD}\) = \(\frac{\mathrm{BC}^{2}}{\mathrm{AC}^{2}}\) (Q.E.D.)

Question 2.
A ladder 15 m long reaches a window which is 9 m above the ground on one side of a street. Keeping its foot at the same point, the ladder is turned to other side of the street to reach a window 12 m high. Find the width of the street.    (Page No. 218)
Answer:

Let A and D be the windows on the either sides of the street.
From Pythagoras theorem,
AC² = AB² + BC²
15² = 9² + BC²
BC² = 225 – 81
BC = √144 = 12 ….. (1)
Also, CD² = DE² + CE²
15² = 12² + CE²
CE² = 225 – 144
CE = √81 = 9
∴ BE = BC + CE = 12 + 9 = 21
Width of the street = 21 m.

Question 3.
In the given figure if AD ⊥ BC, prove that AB² + CD² = BD² + AC².    (Page No. 219)

Answer:
Given: In △ABC, AD ⊥ BC.
R.T.P: AB² + CD² = BD² + AC²
Proof: △ABD is a right angled triangle
AB² – BD² = AD² ……. (1)
△ACD is a right angle triangle
AC² – CD² = AD²
From (1) and (2)
AB² – BD² = AC² – CD²
AB² + CD² = BD² + AC²

Think & Discuss

Question 1.
For a right angled triangle with integer sides atleast one of its measurements must be an even number. Why? Discuss this with your friends and teachers.    (Page No. 215)
Answer:

Let l, m, n are integer sides of a right
angled triangle.
then l² – m² + n²
⇒ n = l² – m² = (l + m) (l – m)
Now
Case I: Both l, m are even the (l + m) is even then (l + m) (l – m) is also even. So ‘n’ is even. Here all are even.
Case II: Both l, m are odd then (l + m) and (l – m) become even. Then the product of even numbers is even so ‘n’ is even.
Here only ‘n’ is even.
Case III: If we consider l is even, m is’ odd then ‘n’ will be odd. So here T is even. We observe in all above three cases at least one of l, m, n is even; Hence proved.

AP State Syllabus SSC 10th Class Maths Solutions 7th Lesson Coordinate Geometry InText Questions

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

10th Class Maths 7th Lesson Coordinate Geometry InText Questions and Answers

Do these

Question 1.
From the figure write coordinates of the points A, B, C, D, E, F, G, H.          (Page No. 159)
Answer:
Given: Knight is at the origin (0, 0).

Therefore, A (- 1, 2), B (1, 2), C (2, 1), D (2, – 1), E (1, – 2), F (-1, -2), G (-2, -1) and H (-2, 1).

Question 2.
Find the distance covered by the Knight in each of its 8 moves i.e., find the distance of A, B, C, D, E, F, G and H from the origin.    (Page No. 159)
Answer:
Origin (0, 0).
Points A, B, C, D, E, F, G and H.
Distance of any point P(x, y) from the

Question 3.
What is the distance between two points H and C? And also find the distance between two points A and B.        (Page No. 159)
Answer:
Given: H (- 2, 1), C (2, 1), A (- 1, 2), B (1, 2).
Distance between any two points P(x₁, y₁) and Q(x₂, y₂) is
\(\overline{\mathrm{PQ}}\) = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
∴ Distance between H and C is HC
= \(\sqrt{[2-(-2)]^{2}+(1-1)^{2}}\)
= \(\sqrt{4^{2}+(0)^{2}}\)
= \(\sqrt{16}\)
= 4 units
Distance between A and B is
AB = = \(\sqrt{[1-(-1)]^{2}+(2-2)^{2}}\)
= \(\sqrt{2^{2}+0^{2}}\)
= \(\sqrt{4+0}\)
= 2 units
[!! H, C are points on a line parallel to X – axis.
∴ HC = |x₂ – x₁| = |2 – (- 2)| = 4 Similarly,
AB = |x₂ – x₁| = |-1-1| = 2 ]

Question 4.
Where do these following points lie (-4, 0), (2, 0), (6, 0), (-8, 0) on coordinate plane?    (Page No. 160)
Answer:
Given points are (- 4, 0), (2, 0), (6, 0), (- 8, 0).
All these points have their y-coordinates = 0
∴ These points lie on X-axis.

Question 5.
What is the distance between points (- 4,0) and (6, 0) on coordinate plane?    (Page No. 160)
Answer:
Given points = (- 4, 0) and (6, 0).
These two points lie on the X – axis.
∴ Distance between them = |x₂ – x₁| = |16 – (-4)| = 16 + 41 = 10 units.

Question 6.
Find the distance between the following pairs of points:    (Page No. 162)
i) (3, 8), (6, 8).
Answer:
Given points = A (3, 8) and B (6, 8)
These two points lie on the line parallel to X – axis.
Distance between A (3, 8) and B (6, 8) = |x₂ – x₁|
= |6 – 3| = 3 units.

ii) (-4, -3), (-8, -3).
Answer:
Given points = A (- 4, – 3) and B (- 8, – 3)
These two points lie on the line parallel to X – axis.
∴ Distance between A (- 4, – 3) and B (- 8, – 3) = |x₂ – x₁|
= |-8 – (-4)|
= |-8 + 4|
= |-4| = 4 units.

iii) (3, 4), (3, 8).
Answer:
Given points = A (3, 4) and B (3, 8) These two points lie on the line parallel to Y – axis.
∴ Distance between A (3, 4) and B (3, 8)= |y₂ – y₁|
= |8 – 4| = 4 units.

iv) (-5, -8), (-5, -12).
Answer:
Given points = A (-5, -8) and B (-5, -12)
These two points lie on the line parallel to Y – axis.
∴ Distance between A (-5, -8) and B (-5, -12) = |y₂ – y₁|
= |-12 – (-8)|
= |-12 + 8|
= |-4| = 4 units.

Question 7.
Find the distance between the following points.    (Page No. 162)
i) A = (2, 0) and B(0, 4)
Answer:
Given points = A (2, 0) and B (0, 4)

ii) P(0, 5) and Q(12, 0).
Answer:
Given points = P (0, 5) and Q (12, 0)

Question 8.
Find the distance between the following pair of points.    (Page No. 164)
i) (7, 8) and (- 2, 3)
Answer:
Given points = (7, 8) and (- 2, 3)
Distance formula

ii) (- 8, 6) and (2, 0)
Answer:
Given points = (- 8, 6) and (2, 0).
Distance formula

Try these

Question 1.
Where do these following points lie – (0, -3), (0, -8), (0, 6), (0, 4) on coordinate plane?    (Page No. 161)
Answer:
As the x – coordinate of all these points is zero, all points lie on Y – axis.

Question 2.
What is the distance between (0, -3), (0, -8) and justify that the distance between two points on Y – axis is |y₂ – y₁| on coordinate plane?    (Page No. 161)
Answer:
As the given two points lie on Y-axis, distance between them is
|y₂ – y₁| = |-3 + 8| = |5| = 5 units.
Let (0, y₁) and (0, y₂) be any two points on Y-axis, then distance between them

[∵ distance can’t be negative]

Question 3.
Find the distance between points ‘O’(origin) and ‘A’ (7, 4).    (Page No. 162)
Answer:
Given: Origin and a point (7, 4).
Distance of a point (x, y) from the origin is \(\sqrt{x^{2}+y^{2}}\)
= \(\sqrt{7^{2}+4^{2}}\) = \(\sqrt{49+16}\) = √65 units.

Question 4.
Find the distance between A(1, -3) and B(-4, 4) and rounded to two decimals.      (Page No. 164)
Answer:

Think & Discuss

Question 1.
How will you find the distance between two points in which x or y coordinates are same but not zero?  (Page No. 161)
Answer:
Let the points be A(2, 3), B(2, 5).
Here the x-coordinates are same, then the distance between the points A and B is |y₂ – y₁| = |5 – 3| = 2 units.
If the points are P (4, 3), Q (- 8, 3),
here the y-coordinates are same. In such a case, the distance is given by |x₂ – x₁| = |-8-4| = |-12|
= 12 units.
i.e.,

Question 2.
Ramu says the distance of a point P(x, y) from the origin O(0, 0) is \(\sqrt{x^{2}+y^{2}}\). Do you agree with Ramu or not? Why?    (Page No. 163)
Answer:
Yes. The distance between O(0, 0) and
P(x, y) is \(\sqrt{(x-0)^{2}+(y-0)^{2}}\)
= \(\sqrt{x^{2}+y^{2}}\)

Question 3.
Ramu also writes the distance formula as AB = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\). Why?    (Page No. 163)
Answer:
(x₁ – x₂)² is same as (x₂ – x₁)²
and (y₁ – y₂)² is same as (y₂ – y₁)²

Question 4.
Sridhar calculated the distance between T(5, 2) and R(-4, -1) to the nearest tenth is 9.5 units. Now you find the distance between P (4, 1) and Q (-5, -2). Do you get the same answer that Sridhar got? Why?    (Page No. 164)
Answer:
Given points (4, 1), (-5, -2)

We got the same answer.
∵ (x)² = (-x)²
(x₁ – x₂)² = (x₂ – x₁)² or
The given points (4, 1) and (-4, -1) are images to each other and (5, 2), (-5, -2) are also images to each other.

Do these

Question 1.
Find the point which divides the line segment joining the points (3, 5) and (8, 10) internally in the ratio 2 : 3.  (Page No. 171)
Answer:
Let P (x, y) be the required point then, P (x, y) =

∴ P (x, y) = (5, 7)

Question 2.
Find the midpoint of the line segment joining the points (2, 7) and (12, -7).  (Page No. 171)
Answer:
Midpoint of the line joining the points (x₁, y₁) and (x₂, y₂) is

Question 3.
Find the centroid of the triangle whose vertices are (-4, 6), (2, -2) and (2, 5) respectively.    (Page No. 173)
Answer:
Given points: (- 4, 6), (2, – 2) and (2,-5).
The coordinates of the centroid


∴ the centroid is (0, 3)

Question 4.
Find the trisectional points of line joining (2, 6) and (-4, 8).  (Page No. 175)
Answer:
A (2, 6) and B (- 4, 8) be the given points.
Let P, Q divide the line joining of \(\overline{\mathrm{AB}}\) in the ratio 1 : 2 and 2 : 1.
Section formula (x, y)

For P(x, y) the ratio is 1 : 2.
P(x,y).

For Q (x, y) the ratio is 2 : 1.
Q (x, y)

∴ The points of trisection are \(\left(0, \frac{20}{3}\right)\) and \(\left(-2, \frac{22}{3}\right)\).

Question 5.
Find the trisectional points of line joining (-3, -5) and (-6, -8).  (Page No. 175)
Answer:
Given: A (- 3, – 5) and B (- 6, 8).
Let P and Q be the points of trisection of \(\overline{\mathrm{AB}}\), then P divides \(\overline{\mathrm{AB}}\) in the ratio 1 : 2.

Q (x, y) divides \(\overline{\mathrm{AB}}\) in the ratio 2 : 1
Q (x, y)
∴ The points of trisection are P (-4, -6) and Q (-5, -7).

Try these

(Page No. 172)

Let A(4,2), B(6, 5) and C(l, 4) be the vertices of △ABC.

Question 1.
The median from A meets BC at D. Find the coordinates of the point D.
Answer:
D is the midpoint of BC

Question 2.
Find the coordinates of the point P on AD such that AP : PD = 2 : 1.      (Page No. 172)
Answer:
P is a point on AD which divides AD in the ratio 2 : 1
∴ P(x,y)

Midpoint of the line joining the points (x₁, y₁) and (x₂, y₂) is

Question 3.
Find the points which divide the line segment BE in the ratio 2 : 1 and also that divide the line segment CF in the ratio 2 : 1.    (Page No. 172)
Answer:
Given: B (6, 5), E\(\left(\frac{5}{2}, 3\right)\)
Let it be P(x, y) =

Similarly, let P divide C(1, 4) and F\(\left(5, \frac{7}{2}\right)\) in the ratio 2:1.

Question 4.
What do you observe? Justify the point that divides each median in the ratio 2 : 1 is the centriod of a triangle.    (Page No. 172)
Answer:
From the above problems, we conclude that the point ‘P’ divides each median in the ratio 2:1.
i.e., the three medians are concurrent at P, which is called the centroid.
A centriod divides each median in the ratio 2 : 1.

Question 5.
The points (2, 3) (x, y) (3, -2) are the vertices of a triangle. If the centroid of this triangle is origin then find (x, y).   (Page No. 173)
Answer:
Given vertices of triangle are (2, 3) (x, y) (3, -2)
now formula for centroid

∴ 5 + x = 0 and 1 + y = 0
⇒ x = -5 and y = -1
∴ (x, y) = (-5, -1)

Think & Discuss

(Page No. 174)

Question 1.
The line joining points A(6, 9) and B(-6, -9) are given.
a. In which ratio does origin divide \(\overline{\mathrm{AB}}\)? And what it is called for \(\overline{\mathrm{AB}}\)?
Answer:
Given : A (6, 9), B (-6, -9)
Let origin O(0, 0) divides \(\overline{\mathrm{AB}}\) in the ratio k : 1 internally.
[By section formula]

The ratio is 1 : 1.
Here the origin bisects \(\overline{\mathrm{AB}}\).
∴ Origin is called the midpoint of \(\overline{\mathrm{AB}}\).

b. In which ratio does the point P(2, 3) divide \(\overline{\mathrm{AB}}\)?
Answer:
Given: A (6, 9), B (-6, -9) and P (2, 3) divide \(\overline{\mathrm{AB}}\) internally in the ratio say k : 1. [By section formula]
Then P (2, 3) =

⇒ 2k + 2 = -6k + 6 and 3k + 3 = -9k + 9
⇒ 8k = 6 – 2 and 3k + 9k = 9 – 3
⇒ k = \(\frac{4}{8}\) and 12k = 6
⇒ k = \(\frac{1}{2}\) and k = \(\frac{6}{12}\)
⇒ k = \(\frac{1}{2}\)
∴ The ratio (k : 1) = \(\left(\frac{1}{2}: 1\right)\) = 2 : 1

c. In which ratio does the point Q(-2, -3) divide \(\overline{\mathrm{AB}}\)?
Answer:
Let Q divide \(\overline{\mathrm{AB}}\) in the ratio say k : 1 internally, then

[By section formula]

⇒ 6k + 6 = -2k-2 and -9k + 9 = -3k-3
⇒ -6k + 2k = -2-6 and -9k + 3k = -3-9
⇒ -4k = -8 and -6k = -12
⇒ k = \(\frac{-8}{-4}\) = 2 and k = \(\frac{-12}{-6}\) = 2
∴ The ratio is k : 1 = 2 : 1

d. Into how many equal parts is \(\overline{\mathrm{AB}}\) divided by P and Q?
Answer:
Since P, Q divide \(\overline{\mathrm{AB}}\) in the ratio 1 : 2 and 2 : 1, \(\overline{\mathrm{AB}}\) is divided into three equal parts by P and Q.

e. What do we call P and Q for \(\overline{\mathrm{AB}}\)?
Answer:
P and Q are the points of trisection of \(\overline{\mathrm{AB}}\).

Do these

Question 1.
Find the area of the triangle whose vertices are
(5, 2) (3,-5) and (-5,-1).  (Page No. 180)
Answer:
Given: The vertices of the triangle are (5, 2), (3, -5) and (-5, -1).
Area of the triangle

Question 2.
(6, -6), (3, -7) and (3, 3).    (Page No. 180)
Answer:
Given: The vertices of a*triangle are (6, -6), (3, -7) and (3, 3).
Area of a triangle

Question 3.
Verify whether the following points are collinear or not.  (Page No. 182)
i) (1, -1), (4, 1), (-2, -3).
Answer:
Given: Three points (1, -1), (4, 1), (-2, -3).
Area of the triangle

As the area of the triangle is ‘O’, the three points are collinear.

ii) (1, -1), (2, 3), (2, 0).
Answer:
Given points are (1, -1), (2, 3), (2, 0).

Area of the triangle formed by the given three points is

△ ≠ 0.
Hence the points are not collinear.

iii) (1, -6), (3, -4), (4, -3).
Answer:
The given points are (1, -6), (3, -4), (4, -3).
Area of a triangle

Area of the triangle formed by the given three points is

As △ = 0, the points are collinear.

Question 4.
Find the area of the triangle whose lengths of sides are 15 m, 17 m, 21 m. (use Heron’s Formula)  (Page No. 183)
Answer:
Given: The sides of a triangle
a = 15 m; b = 17 m and c = 21 m.

Note: As the ‘height’ is not given we can’t verify the above answer by A = \(\frac{1}{2}\)bh

Question 5.
Find the area of the triangle formed by the points (0, 0), (4, 0), (4, 3) by using Heron’s formula.  (Page No. 183)
Answer:
The given points are O (0, 0), A (4, 0) and B (4, 3).
Then the sides

Heron’s formula

Verification:
If the sides are 3, 4 and 5 units, clearly the triangle is a right triangle.

Try these

Question 1.
Take a point A on X-axis and B on Y-axis and find area of the triangle AOB. Discuss with your friends what did they do.      (Page No. 178)
Answer:

[∴ Axes are perpendicular to each other]
Consider the points A(5, 0) and B(0, 6).
△AOB = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × 6 × 5 = 15 sq. units.
Area of a triangle A (x, 0), O (0, 0) and
B (0, y) is \(\frac{1}{2}\)xy.

Question 2.
Find the area of the square formed by (0, -1), (2, 1), (0, 3) and (-2, 1) taken in order are as vertices.    (Page No. 178)
Answer:

Let A (0, -1), B (2, 1), C (0, 3) and D (- 2, 1) are the vertices of a square.
Area of the square ABCD = side²
= AB²
But, AB = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
= \(\sqrt{(2-0)^{2}+(1+1)^{2}}\)
= \(\sqrt{2^{2}+2^{2}}\)
= \(\sqrt{4+4}\)
= √8
Area of square = √8 × √8
= 8 sq. units.

Think & Discuss

Question 1.
Let A(x₁, y₁), B(x₂, y₂), C(x₃, y₃). Then find the area of the following triangles in a plane. And discuss with your friends in groups about the area of that triangle.    (Page No. 178)

Answer:
i) Given △AOB where A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃)

But we know that the origin O is (0, 0) which is given as C.
\(\overline{\mathrm{CB}}\) = x₂ – x₃ = x₂ – 0 = x₂
\(\overline{\mathrm{AB}}\) = y₁ – y₂ = y₁ – 0 = y₁
∴ Area of △ABC = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\)x₂y₁

ii)

A(x₁, y₁) = (x₁, y₁)
B(x₂, y₂) = (0, y₂)
C(x₃, y₃) = (0, 0)
∴ Area of △ABC = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × y₂ × x₁
= \(\frac{1}{2}|\)x₁y₂| sq. units.

iii)

∴ AB = x₁ – x₂
BC = y₂ – y₃
A(x₁, y₁) = (-x₁, y₁)
B(x₂, y₂) = (x₂, y₂)
C(x₃, y₃) = (x₃, -y₃)
∴ Area of △ABC

iv)

AB = (y₂ – y₁)
BC = (x₁ – x₃)
A(x₁, y₁) = (x₁, y₁)
B(x₂, y₂) = (x₁, y₂)
C(x₃, y₃) = (x₃, y₂)
∴ Area of △ABC = \(\frac{1}{2}\) × base × height

Question 2.
Find the area of the triangle formed by the following points    (Page No. 181)
i) (2, 0), (1, 2), (1, 6)
Answer:
Take the third point as (-1, 6).

ii) (3, 1), (5, 0), (1, 2)
Answer:

iii) (-1.5, 3), (6, 2), (-3, 4)
Answer:
Take the second point as (6, -2)

Question 3.
What do you observe?          (Page No. 181)
Answer:
We observe that the area formed by above all triangles is zero.

Question 4.
Plot these points on three different graphs. What do you observe?      (Page No. 181)
Answer:

We observe that the points are collinear.

Question 5.
Can we have a triangle having area zero square units area?    (Page No. 181)
Answer:
No.

Question 6.
What does it mean?      (Page No. 181)
Answer:
If the area of the triangle formed by any three points is zero, it means the points are collinear.

Do these

Question 1.
Plot these points on the coordinate axis and join them:
Which gives a straight line? Which is not? Why?    (Page No. 185)
i) A(1, 2), B(-3, 4), C(7, -1)
Answer:
Points ABC gives a straight line.

ii) P(3, -5), Q(5, -1), R(2, 1), S(1, 2)
Answer:
Points PQRS doesn’t give a straight line.

Question 2.
Find the slope of \(\overleftrightarrow{\mathbf{A B}}\) with the given end points,      (Page No. 188)
i) A(4, -6), B(7, 2)
Answer:

ii) A(8, -4), B(-4, 8)
Answer:

iii) A(-2, -5), B(l, -7)
Answer:

Try these

(Page No. 188)

Question 1.
Find the slope of AB with the points lying on
i) A(2, 1), B(2, 6)
Answer:

Question 2.
A(-4, 2), B(-4, -2)
Answer:

Question 3.
A(-2, 8), B(-2, -2)
Answer:

Question 4.
Justify that the line AB line segment formed by given points is parallel to Y-axis. What can you say about their slope? Why?
Answer:
In the above problem, all points are of the form (K, y) where K is a fixed number and y is a variable.
∴ All lines in the above problem are parallel to Y – axis. Slope of lines parallel to y – axis are not defined.

Think & Discuss

Question 1.
Does y = x + 7 represent a straight line? Draw the line on the coordinate plane. At which point does this line intersect Y – axis?
How much angle does it make with X – axis? Discuss with your friends.        (Page No. 185)
Answer:
Yes. y = x + 7 represents a straight line.

Angle made by y = x + 7 with X-axis is 45°. (∵ (0, 7) and (- 7, 0) are equidistant from the origin and hence the triangle formed is right isosceles triangle.)

Question 2.
Find the slope AB with the points lying on A(3, 2), B(- 8, 2). When the line AB parallel to X-axis? Why? Think and discuss with your friends in groups.      (Page No. 188)
Answer:
Given : A (3, 2), B (-8, 2)

Yes. The line is parallel to X-axis as the points are of the form (x₁, K), (x₂, K)

AP State Board Syllabus AP SSC 10th Class Hindi Textbook Solutions उपवाचक Chapter 4 अनोखा उपाय Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Hindi Solutions उपवाचक Chapter 4 अनोखा उपाय

10th Class Hindi उपवाचक Chapter 4 अनोखा उपाय Textbook Questions and Answers

प्रश्न 1.
राजा कुमारवर्मा के राज्य में अकाल की स्थिति क्यों उत्पन्न हुई होगी?
उत्तर:
अकाल की स्थिति तो प्रकृति से संबंधित अंश है। अकाल की स्थिति का कारण वर्षा की कमी है। यदि वर्षा की कमी हो या वर्षा बराबर नहीं हो तो अकाल की स्थिति उत्पन्न होगी। इसी कारण राजा कुमारवर्मा के राज्य में अकाल की स्थिति उत्पन्न हुई होगी।

प्रश्न 2.
अकाल की समस्या के निवारण के लिए राजा ने क्या-क्या उपाय सोचे होंगे?
उत्तर:
राजा कुमार वर्मा ने अकाल की समस्या के निवारण के लिए कई बुद्धिमानों और विद्वानों को बुलवाया होगा। राजभंडार का अनाज प्रजा में बाँट दिया होगा। अडोस – पडोस के राज्यों से अनाज़ उधार लिये होंगे। सभी तरह से खुशहाल किसी राज्य के राजा से, वहाँ के शासन नियमों के पता प्राप्त करने भेंट किये होंगे।

प्रश्न 3.
राजा कुमारवर्मा की जगह पर यदि तुम होते तो अकाल की समस्या से कैसे जूझते?
उत्तर:
राजा कुमारवर्मा की जगह पर यदि मैं होता तो अकाल की समस्या से इस प्रकार जूडता :

• राज्य के धान्यागार में से धान लेकर सबको बाँट दूँ।
• अडोस – पडोस के राज्यों के राजाओं से धन – धान्य आदि उदार लेकर जनता में बाँट दूँ।
• नीतिशास्त्र, धर्मशास्त्र आदि को पढ़कर समस्या को हल करने का प्रयत्न करूँ।
• नीति कोविद, धर्मकोविद, ज्योतिष्य कोविदों को बुलाकर उनसे अकाल के कारणों के बारे में चर्चा करता।
• वेदों में लिखा गया है कि यज्ञ यागादि कार्य करने से अकाल की स्थिति दूर होगी। – इस कथन के अनुसार मैं यज्ञ यागादि कार्यक्रम करता।
• शास्त्र, सांकेतिक वैज्ञानिकता की सहायता से मेघ मदन कार्यक्रम करता।

अनोखा उपाय Summary in English

Once there was a king named Kumara Varma who ruled over the kingdom Harithanagar. He was a good administrator. During his reign, his kingdom flourished well and the people lived happily.

Once a severe famine broke out in the kingdom. The harvests in the kingdom dried up. The ponds, as well as the lakes, dried up. Even the two perennial rivers in the kingdom became small gutters.

There was no fodder for the cattle. Many formers started selling their cattle at cheaper prices. In such a situation the king distributed the grain in the treasury to the people. He borrowed the grain from the neighbouring kingdoms. Yet, he couldn’t find a solution.

“All the neighbouring kingdoms are green and prosperous. The people in those kingdoms are leading a comfortable and happy life. Why all this is happening in my kingdom? What is the reason for it? How can I solve this problem?” – These questions role in the King’s heart. He was deeply moved to see the plight of the kingdom.

He held a meeting with the learned, the scholars, and the geniuses in the kingdom and asked them to find out a solution for the problem.

They advised : “Your Magesty! please visit the best ever kingdom and meet the king there. Know the legislative policies of the administration followed in that kingdom. Take up the reforms in administrative methods existed here, based on them. Through this, the conditions of the kindgom will be set right.”

The King Kumara Varma liked this suggestion. Immediately he decided to meet Satya Sinh, the king of his neighbouring kingdom. He sent a message by his servants to Satya Sinh revealing that the people in his kingdom were facing difficulties due to server famine conditions. He wanted to visit thier kingdom to know the policies adopted there and to seek an advice to solve the problem.

Invited by him, Kumara Varma visited the kingdom of Satya Sinh and received a royal welcome. Kumara Varma was amazed to see that kingdom. The water bodies on the four sides and the rivers appeared to the full. The fields were green with harvests. The groves were full of flowers and fruits. He was overjoyed to see the pleasant atmosphere prevailed there.

He said to Satya Sinh, “Friend! Your kingdom is not less than a paradise. You are following good methods unknown to me. It is the reason why your people are living happily. I too want my people to live contentedly. Please advise me concerning good governance.”

Satya Sinh replied, “O king! Don’t implore me. A person who committed a mistake has no right to give guidance to others. I will relate an incident to you. Once I was wandering in the garden accompanied by my bodyguards. At the very moment, I had to meet my royal mother on an urgent matter. I ordered my bodyguards to stay there until I come. Later I forget about them. The following day it rained heavily. Suddenly I remembered them and went to the garden. They got drenched fully and were still waiting for me. As it was my fault, I have no right to give you advice. Hence, forgive me.” Kumara Varma was astounded to hear his words.

Then he met the royal mother and said that he needed a guiding hand in respect of good governance.

She said, “Son! To be frank, I too am faulty. Once my son presented his wife with a beautiful jewel. I was greedy for it. It’s not good on the part of a royal mother to have greed for jewels. I’m not deserved as I committed such a mistake”.

Kumara Varma later met royal preceptor and sought his advice regarding good governance.

Then the royal preceptor said, “O King ! Forgive me. I’m not deserved for it. Once a scholar came to our kingdom from a faraway country and wanted to visit our king. I had to tell the king that he was a great scholar, as there was no time to test his erudition. The king relied on my words and presented him with many gifts. Later I came to know that he was not such an erudite and he was a common scholar. Because of my sloth, I could not guide our king in a proper way. Hence, I’m not deserved”.

The king Kumara Varma sank in thoughts. He learnt a lesson from these three incidents that one should not commit even a small mistake. If one committed any mistake, one should rectify it. He followed this lesson. Within no time his kingdom achieved progress and the people lived happily ever after.

This story was extracted from ‘Bangaru Kundelu’ written by a renowned Telugu writer late Sri Ravuri Bhradwaja who won Jnanpith Award for the year 2012. Sri Syed Mateen Ahmed translated this story into Hindi.

अनोखा उपाय Summary in Telugu

చాలా సంవత్సరాల క్రితం నాటి మాట. ఒక రాజ్యం ఉండేది. దాని పేరు హరిత నగరం. హరిత నగరం రాజుగారు కుమారవర్మ. అతడు మంచి పరిపాలకుడు. కుమార వర్మ పరిపాలనా కాలంలో రాజ్యం పచ్చగా ఉంది. కానీ ఒకసారి రాజ్యం మొత్తం పంటలు ఎండిపోయాయి. చెరువులు – కుంటలు ఎండిపోయాయి. కేవలం రెండే రెండు జీవనదులు మిగిలిపోయాయి. అవి కూడా చిన్న – చిన్న కాలువల్లా తయారయ్యాయి.

రాజ్యంలో పశువులకు మేత దొరకడం కూడా చాలా కష్టంగా ఉంది. చాలా మంది రైతులు తమ పశువులను చౌక ధరలకు అమ్మివేయడం ప్రారంభించారు. ఇలాంటి పరిస్థితుల్లో రాజభండాగారంలోని ధాన్యాన్ని రాజుగారు ప్రజలకు పంచి పెట్టడం ఆరంభించిరి. ఇరుగు – పొరుగు రాజ్యాల నుండి ధాన్యాన్ని అప్పుగా తీసుకోవడం ప్రారంభించిరి. కానీ రాజుగారికి భవిష్యత్తుపై దిగులు బాధిస్తోంది. రాజుగారు ఉత్పన్నమైన ఈ పరిస్థితులను గురించి గంభీరంగా ఆలోచించసాగిరి. కానీ దీనికి పరిష్కారం లభించలేదు. “ఇరుగు-పొరుగు రాజ్యాలన్నీ పచ్చగా ఉన్నాయి. అక్కడి ప్రజలంతా సుఖంగా ఉన్నారు. కానీ మన రాజ్యంలోనే ఇలా ఎందుకు జరుగుతున్నది? దీనికి కారణం ఏమిటి? ఈ సమస్యను ఎలా పరిష్కరించడం?” – అని రాజుగారి మనస్సులో ఎన్నో ప్రశ్నలు ఉదయిస్తున్నాయి.

రాజుగారు కుమారవర్మ ఈ సమస్యను పరిష్కరించుటకు, బుద్ధిమంతులను, విద్వాంసులను, పరిస్థితులను క్షుణ్ణంగా అర్థం చేసుకుని జవాబు చెప్పగల మేధావులను పిలిపించారు.

వారి చర్చలో కొంత మంది బుద్ధిమంతులు “ఓ మహారాజా, తప్పులు చాలా రకాలుగా ఉంటాయి. కొన్ని తప్పులు సరళంగానే గుర్తించబడతాయి, కొన్ని తప్పులను గుర్తించలేం! – అని చెప్పిరి. కొంత మంది “ఓ రాజా! కొన్ని తప్పుల (పొరపాట్లు) కు సంకేతాలు ఉంటాయి. మరికొన్ని తప్పులకు సంకేతాలు ఉండవు అని చెప్పిరి.

కొంత మంది విద్వాంసులు- ఓ ప్రభూ! కొన్ని తప్పులు (పొరపాట్లు) సంస్కరణల రూపంలో ఏర్పడతాయి. మరికొన్ని సంస్కరణలే తప్పులుగా మారతాయి. ఇలాంటిదే ఏదో తెలిసీ – తెలియని విషయం దాగి ఉండవచ్చు. అందువలననే ఈ రోజున మన రాజ్యంలో ఇలాంటి పరిస్థితి ఏర్పడినది” అని చెప్పిరి.

అప్పుడు రాజుగారు కుమారవర్మ “అలా అయితే మీరే చెప్పండి. ఇప్పుడు నన్ను ఏమి చేయమంటారు?” అని వారిని అడిగెను.

అందరూ తర్జన – భర్జన చేసి రాజుగారికి ఈ సలహా ఇచ్చి – “అన్ని విధాలా సుభిక్షితంగా ఉన్న రాజ్యంలోని రాజుగారిని కలవండి. అక్కడి పరిపాలనా నియమాలను తెలుసుకోండి. వాటి ఆధారంగా ఇక్కడి పరిపాలనా విధానాలలో మార్పులు (సంస్కరణలు) తీసుకురండి. దీని ద్వారా రాజ్యంలోని పరిస్థితులు చక్కబడతాయి”.

రాజుగారైన కుమారవర్మకు ఈ సలహా చాలా బాగా నచ్చింది. ఆయన వెంటనే తన పొరుగు రాజ్యపు రాజుగారైన సత్యసింహను కలవడానికి నిర్ణయించుకుని తన సేవకుల ద్వారా ఆ రాజుగారికి సందేశం ఈ విధంగా పంపెను“రాజాధిరాజు, మహారాజు అయిన సత్యసింహనకు సాదర నమస్కారములు. మా రాజ్యంలో కరువు కాటకాలతో ప్రజలు ఇబ్బందులు పడుతున్నారు. ఈ సమస్యను పరిష్కరించుటకు మీ ఉచిత సలహా మరియు మీ శాసన నియమాలను తెలుసుకొనుటకు మేము మీ రాజ్యం దర్శించాలని కోరుకుంటున్నాము. మీరు నా కోరికను మన్నిస్తారని ఆశిస్తున్నాను.”

దానికి ప్రత్యుత్తరంగా మహారాజు సత్యసింహ్ తన సందేశం ఈ విధంగా పంపెను. “మీరు – నేను ఇరుగు-పొరుగు రాజులం. ఏదేని ఒక సమస్య విషయంలో ఒకరినొకరు పరస్పరం సహాయం చేసుకోవడం మన కర్తవ్యం. మా రాజ్యానికి మీకు సాదర స్వాగతం. మీరు మా ఆత్మీయ ఆదరణీయ అతిథులు. అతిథి రూపంలో మిమ్ములను సత్కరించు సౌభాగ్యం మాకు కల్గుచున్నందులకు, మేము మీకు కృతజ్ఞులం.”

ఈ ప్రత్యుత్తరం చదవగానే రాజుగారైన కుమారవర్మకు తన రాజ్యానికి సంబంధించిన కరువు కాటకాల సమస్యకు పరిష్కారం కొంతవరకు దొరికినట్లు భావించిరి. అయినప్పటికీ రాజుగారు పొరుగున ఉన్న రాజ్యాన్ని సందర్శించి ఆ రాజును కలుచుట కోరుకొనెను.

చూస్తూ చూస్తుండగానే ఆ రోజు రానే వచ్చింది. రాజుగారైన కుమారవర్మకు పొరుగు రాజ్యం నుండి భవ్యమైన స్వాగతం లభించినది. రాజ్యాన్ని చూసి రాజుగారు ఆశ్చర్యచకితులైరి. నాలుగువైపులా జలాశయాలు నిండుగా ఉన్నవి. నదులన్నీ నిండుగా ఉన్నవి. కాలువలు ప్రవహిస్తూ ఉన్నవి. చల్లని గాలులు వీస్తూ ఉన్నవి. పంట పొలాలన్నీ పచ్చని పైరు పంటలతో నిండుగా ఉన్నాయి. పూల సుగంధం అంతటా వ్యాపిస్తోంది. తోటలన్నీ పండ్లు – పూలతో నిండుగా ఉన్నాయి. వీటన్నిటినీ చూడగానే రాజుగారి మనస్సు అవధులు లేని సంతోషంతో నిండిపోయినది.

మహారాజు కుమారవర్మ మహారాజుగారైన సత్యసింహను కలిసిరి. “మిత్రమా, మీ రాజ్యం ఏ స్వర్గానికి తక్కువ లేదు. నాకు తెలియని శాసన నియమాలను మీరు పాటించుచున్నట్లు ఉన్నారు అని నాకు అన్పించుచున్నది. అందువలననే మీ ప్రజలందరూ సుఖంగా ఉన్నారు. నేను కూడా మా దేశ ప్రజలను సుఖంగా చూడదలచుచున్నాను. దయచేసి మీరు నాకు సుపరిపాలన హితోపదేశం చేయండి”- అని కుమారవర్మ, సత్యసింహను కోరిరి.

మహారాజుగారైన సత్యసింహలవారు మొదట తిరస్కరించిరి. కానీ రాజుగారైన కుమారవర్మగారి ప్రార్థన మీదట ఆయన ఈ విధంగా బదులిచ్చిరి ” లేదు మహారాజా, నన్ను అభ్యర్థించవద్దు. నేను దోషిని. దోషి అయిన వానికి హితోపదేశం చేసే హక్కు లేదు. నేను మీకు ఒక సంఘటన వినిపిస్తాను. నేను ఒకసారి నా అంగరక్షకునితో ఇదే విధంగా తోటలో విహరిస్తూ ఉన్నాను. అప్పుడే నేను రాజమాతతో అత్యవసర విషయమై మాట్లాడుటకు వెళ్ళవలసి వచ్చినది. నేను తిరిగి వచ్చేవరకు అంగరక్షకులను అక్కడే నిలబడి ఉండమని ఆదేశించితిని. రాజమాతతో మాట్లాడుతూ- మాట్లాడుతుండగా రాత్రి అయిపోయినది. అక్కడే నేను భోజనం చేసి నిద్రించితిని. మరుసటి రోజు ఉదయం లేచి చూడగా బాగా వర్షం కురియుచున్నది. నేను తోటలోకి వెళ్ళి చూడగా అంగరక్షకులు అక్కడే నిలబడి తడచిపోతూ ఉండడం గమనించితిని.

నేను మాటల్లో అంగరక్షకులను మరచిపోయినంతగా నిమగ్నమైయున్నాను. వాళ్ళను వెళ్ళిపొమ్మని కూడా చెప్పలేనంతగా మాటల్లో మునిగిపోతిని. ఇది నేను చేసిన తప్పు. అందువలన అలా తప్పు చేసిన రాజుకు హితోపదేశం చేయు హక్కు లేదు. నన్ను క్షమించండి.”

రాజుగారైన కుమారవర్మ మహారాజు సత్యసింహ్ చెప్పిన విషయాన్ని సంఘటనను) పూర్తి ధ్యాసతో విన్నారు. రాజమాత గారే తనకు హితోపదేశం చేయగలరని అనుకుని ఆయన రాజమాత దర్శనం చేసుకొని వారిని హితోపదేశం చేయవలసినదిగా కోరిరి.

“కుమారా! నిజం చెప్పవలెనన్న నేను కూడా దోషినే. ఒకసారి నా కుమారుడు తన భార్యకు ఒక అందమైన నగ తయారుచేయించి ఇచ్చెను. నా మనస్సులో ఆ నగ పట్ల దురాశ కలిగినది. నేను నా కుమారుడిని లేదా నా కోడల్ని ఆ నగ ఇమ్మని అడిగినట్లయితే వారెప్పటికీ కాదనరు. ఒక రాజమాతకు నగల పట్ల వ్యా మోహం ఉండడం తప్పు. వేరొకరి వస్తువు పట్ల దురాశ కలగడం తప్పు. అలాంటి తప్పు చేసిన నేను నీకు హితోపదేశం చేసేంత యోగ్యురాలిని కాను”అని రాజుగారైన కుమారవర్మతో రాజమాత చెప్పినది. (రాజా సత్యసింహ గారి తల్లి.)

రాజుగారైన కుమారవర్మ ఆశ్చర్యచకితులైరి. తదుపరి రాజగురువును కలిసి తనకు హితోపదేశం చేయవలసినదిగా కోరిరి.

అప్పుడు రాజగురువుగారు ఇట్లు అనిరి – “మహారాజా నన్ను క్షమించండి. నేను దీనికి యోగ్యుడను కాను. ఒకసారి సుదూర దేశం నుండి ఒక పండితుడు విచ్చేసెను. ఆయన రాజ దర్శనాన్ని కోరెను. అతని పాండిత్యాన్ని పరీక్షించే సమయం లేకపోవడం వల్ల నేను రాజుగారతో అతడు చాలా గొప్ప పండితుడు అని చెప్పాను. రాజుగారు నా పై అంతులేని నమ్మకాన్ని ఉంచుతారు. ఆయన పండితునికి కుప్పలు – తెప్పలుగా బహుమతి ఇచ్చి పంపారు. ముందు ముందు నాకు ఆ పండితునికి అంత పాండిత్యం లేదని సాధారణ పండితుడు మాత్రమేనని తెలిసినది. నా సోమరితనం వల్ల నేను రాజుగారికి తగిన మార్గదర్శకత్వం చేయలేకపోయాను. అలాంటి తప్పు చేసిన నేను నీకు హితోపదేశం చేయగల యోగ్యుడను కాను.”

కుమారవర్మగారు చాలా ఆలోచనలో మునిగిరి. ఈ మూడు సంఘటనల ఆధారంగా తను చిన్న – చిన్న తప్పులు కూడా చేయరాదన్న గుణపాఠం నేర్చుకొనిరి. ఏదైనా తప్పు జరిగితే దానిని సరిదిద్దవలెనని అనుకొనెను. రాజుగారు ఈ గుణపాఠాన్ని పాటించిరి. కొద్దిరోజుల్లోనే తన రాజ్యం తిరిగి సస్యశ్యామలమైనది. సుఖసంతోషాలతో నిండిపోయినది.

(2012 సంవత్సరమునకు జ్ఞానపీఠ పురస్కారాన్ని పొందిన స్వర్గీయ శ్రీ రావూరి భరద్వాజ గారు తెలుగులోని ఒక గొప్ప ప్రసిద్ది చెందిన రచయిత. ప్రస్తుత ఈ కథ ఆయన రచించిన బంగారు కుందేలులో నుండి గ్రహించబడినది. దీనిని శ్రీ సయ్యద్ మతీన్ అహ్మద్ గారు హిందీలోనికి అనువదించిరి.)

AP State Board Syllabus AP SSC 10th Class Hindi Textbook Solutions उपवाचक Chapter 3 अपने स्कूल को एक उपहार Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Hindi Solutions उपवाचक Chapter 3 अपने स्कूल को एक उपहार

10th Class Hindi उपवाचक Chapter 3 अपने स्कूल को एक उपहार Textbook Questions and Answers

प्रश्न 1.
राजू को उसका पुराना स्कूल कैसा लगता था?
उत्तर:
राजू अपने पुराने स्कूल में वह एक मेधावी छात्र रहा था। पुराने स्कूल में उसके बहुत सारे मित्र थे और सभी अध्यापक भी उसे पसंद किया करते थे। जब भी कोई कठिनाई होती तो राजू सबसे पहले उसकी मदद के लिए पहुँच जाता। उसकी टाँगें बहुत दुर्बल थीं, इस कारण उसे खेलने में रुचि न थी। जब भी उनके स्कूल में मैच होता, राजू अपने साथियों को खेलते हुए देखता और जोर – शोर से उनका उत्साह बढ़ाता। राजू का विचार था कि यदि स्वर्ग में भी स्कूल हो तो वह भी उसके पुराने स्कूल से ज्यादा अच्छा तो नहीं हो सकता।

प्रश्न 2.
राजू के प्रति नये स्कूल के साथियों का व्यवहार कैसा था?
उत्तर:
राजू नये स्कूल में प्रेवश करते ही सभी बच्चे उसके टांगों की ओर संकेत करके हँस रहे थे, और उसका मजाक उडाते थे। कुछ ही क्षणों में पूरा मैदान और बरामदे कौतूहल से देखने और उसकी ओर इशारा करके हँसनेवालों से भर गये। स्कूल के अध्यापक भी पास से ऐसे गुजरे जैसे कुछ हो ही नहीं रहा है। अध्यापक के राजू से पूछने पर उसने बताया कि वह एक गाँव के स्कूल से आया है। यह सुनकर सभी छात्र बहुत हँसे। लेकिन मधुर स्वभाव वाले राजू को गुस्सा नहीं आया। अगले दिन, उससे और अगला दिन और फिर महीने का हर दिन उसके लिए ऐसा ही रहा। उसका कोई मित्र भी नहीं बन पाया था। आधी छुट्टी में जब बाकी सभी लडके खेलने जाते तो वह कक्षा में ही बैठा रहता। अब तक पूरा स्कूल जान गया था कि राजू एक ‘गंवार’ लडका है और उसे अपने गाँव के स्कूल का बडा ‘घमंड’ है।

प्रश्न 3.
राजू ने अपने स्कूल को किस तरह उपहार समर्पित किया?
उत्तर:
राजू बहुत मेधावी छात्र था। वह गाँव के स्कूल से शहर के स्कूल में पढ़ने के लिए आया था। वह जल्दी नहीं चल सकता था। सभी छात्र उसका बहुत मजाक उडाते थे | सभी छात्र जान गये थे कि राजू एक गंवार लडका है । आखिर राजू ने इस स्थिति से निबटने के लिए बड़ी चतुराई से एक योजना बनायी। इसलिए उसने घर पर अधिक रिश्रम किया। आखिर उसे नये स्कूल के समक्ष प्रमाणित करना था कि उसका पुराना गाँव का स्कूल कोई कम नहीं था। राजू अपनी पढ़ाई करता रहा। उसने परीक्षाएँ अच्छी तरह लिखीं। परिणाम निकलने के दिन पूरे आत्मविश्वास के साथ परीक्षा – फल देखने गया। फिर वह कक्षा में प्रथम आया था। प्रथम आने पर वह इतना प्रसन्न कभी नहीं हुआ था, क्योंकि अब उसने अपने स्कूल को सुंदर और समुचित उपहार समर्पित किया है।

अपने स्कूल को एक उपहार Summary in English

Raju was an intelligent student. He was a child prodigy. He would always come first in the class. He was fond of reading story books. His favourite subjects were Hindi and English. He would read books concerning science subject also. So he became well – versed in science and history. As to mathematics, he was a magician. He would answer the problems even before the teacher wrote on the blackboard.

In old school he had many friends. Teachers too liked him very much. If anybody else was in difficulties, he would go first to help him/her. In that school nobody cared for his physical disability. His legs were lean and weak. His knees were not strong enough to bear his body for much time. Hence, he would not play any games. But he would watch his friends playing games and encourage them. Inspired by him, they would play the games with much enthusiasm.

He thought about his old school throughout the night and he wished that his new school too would be like the old one. He was of the opinion that his old school was greater than the heaven. At the time of his leaving the old school, his friends sank in sorrow. Even his teachers and the headmaster requested his father to leave him there. As his father got transferred, he had to join another school.

The following day he woke up early and donned his uniform. He looked at himself in the mirror and thought that his new school too would be good. He didn’t take tiffin properly. His parents understood the situation and did not compel him. His father took him to school by his vehicle and saw him off smilingly.

Raju would walk slowly while limping. The fellow students and everyday in the new school looked at his legs and made fun of him. The teachers also didn’t show any concern towards him.

Soon after the first period started, the teacher made Raju sit on the least row. Asked by the teacher, he introduced himself that he had come from a village school. On hearing that, the students as well as the teacher laughed at him. As he was a boy with composure, he controlled his anger. He felt the need of being bold. He thought that everything would be alright later. But this went on in every period. He was humiliated by the teachers and fellow students.

Raju wanted to prove that village schools are not less than those of town. That evening his parents asked him about the new school. But he kept quiet and gave a smile in reply. He even didn’t like to tell a lie.

In the classroom he was not given any chance to answer the questions asked by the teachers He could make friends with nobody. While the other students kept on playing in a recess during lunch time, he remained quiet in the classroom. Everyone thought that he was proud of his village school.

Finally, Raju made a plan to come out of this situation. He gave up raising his hand when the teachers asked questions. As a result ,the students and the teachers didn’t concentrate on him.

Raju wanted to get through the annual examinations with flying colours. So he worked hard. He wanted to prove that his old school was not a fools’ paradise.

Everybody in the school expected that he would fail. They jeered at him why they saw him with books. But he continued his preparation boldly and ardently. The examinations were over within a week. After the examinations, he left for his village.

Later, he returned to the town one day before the results day. The following day he, confidently went to school along with his father to know result. He stood first in his class. His father felt very happy. Raju’s joy knew no bounds. He never felt so much happy like that, earlier. Thus, he presented a good and proper gift to his old school.

अपने स्कूल को एक उपहार Summary in Telugu

తన పాత పాఠశాలలో అతడు ఒక మేధావి విద్యార్థి. గత 5 సంవత్సరాలుగా తను తన తరగతిలో అందరి కంటే ముందుండేవాడు. అతనికి కథల పుస్తకాలు చదవడం అంటే చాలా ఇష్టం. అందుకని అతనికి ఆంగ్ల భాష, హిందీ భాష బాగా ఇష్టం. అతనికి సాధారణ జ్ఞానానికి సంబంధించిన పుస్తకాలు కూడా చాలా ఇష్టం. అందువలన అతని విజ్ఞానశాస్త్రం మరియు చరిత్రకు సంబంధించిన విజ్ఞానం వికసించినది. అతడు లెక్కల విషయం (గణితశాస్త్రం) లో అయితే గణిత మాంత్రికుడే. ఉపాధ్యాయుడు నల్లబల్లపై ప్రశ్నలు వ్రాయడానికి పూర్వమే అతడు వాటికి సమాధానం చెప్పడానికి ధైర్యంతో చేతులెత్తేవాడు.

తనకు పాత పాఠశాలలో చాలా మంది మిత్రులు ఉన్నారు. ఉపాధ్యాయులు కూడా ఆ పిల్లవాణ్ణి ఇష్టపడేవారు. అతడు అందర్నీ సంతోషంగా కలుస్తూ “హలో” అని పలకరించేవాడు. ఎప్పుడయినా ఎవరైనా కష్టాలలో ఉంటే రాజు అందరికంటే ముందు తాను సహాయం చేయడానికి బయలుదేరేవాడు. తన పాత పాఠశాలలో ఎవ్వరూ కూడా తన బలహీనతను పట్టించుకునేవారు కాదు. అతని కాళ్ళు చాలా బలహీనంగా సన్నగా ఉండేవి. అతని మోకాళ్ళలో శక్తి లేదు. అధిక . సమయం తన కాళ్ళు తన శరీర భారాన్ని మోసేవికావు. అందువలన అతడు ఎక్కువ సమయం తన కాళ్ళపై నిలబడగలిగేవాడు కాదు. అందువలనే అతడు ఆటలు కూడా ఆడేవాడు కాదు. ఎప్పుడైనా తన పాఠశాలలో ఏవైనా ఆటల పోటీలు ఉంటే రాజు తన స్నేహితుల ఆటను చూస్తూ వారిని ఉత్సాహంగా ఉత్తేజపరిచేవాడు. తన స్నేహితులు ఆట ఓడిపోయేటట్లు – ఉంటే రాజు ప్రేరణతో వారిలో ఆశ సంచరించేది అంతే ఇంకేముంది నూతన స్పూర్తితో ఆట ఆడేవారు.

రాజు రాత్రంతా తన పాత పాఠశాలను గురించి ఆలోచిస్తూ తన నూతన పాఠశాల కూడా తన పాత పాఠశాల అంత మంచిదిగా ఉండాలని ప్రార్థించాడు. ఒకవేళ స్వర్గంలో కూడా పాఠశాల ఉండి ఉంటే అది కూడా తన పాత పాఠశాల కంటే మంచిగా ఉండదని రాజు అభిప్రాయం. తను పాఠశాల విడిచి వచ్చేటప్పుడు ఎంతమంది స్నేహితులు ఏడ్చారు? తన స్నేహితులు, ఉపాధ్యాయులు, అక్కడ దాకా ఎందుకు ప్రధానోపాధ్యాయులు కూడా తన తండ్రిగారితో అతడిని అక్కడే వదిలి వెళ్ళమని ప్రార్థించారు. కానీ వారెవరి మాటను పట్టించుకోలేదు. వాళ్ళ నాన్నగారు బదిలీ అయ్యిి. తన ఏకైక కుమారుణ్ణి అక్కడే వదిలి వెళ్ళాలనే విషయాన్ని వాళ్ళు ఆలోచించనే లేదు.

మరుసటి రోజు ప్రాతఃకాలాన్నే రాజు త్వరగా లేచి తన యూనిఫారంను ధరించాడు. అద్దంలో తనను తాను చూచుకుని సమదుస్తులైతే బాగానే ఉన్నాయి. పాఠశాల కూడా బాగానే ఉండి ఉంటుంది అని అనుకున్నాడు. అతడేమి టిఫిన్ కూడా సరిగా తినలేదు. అమ్మ-నాన్నలు అర్థం చేసుకుని అతనిపై ఎటువంటి వత్తిడి చేయలేదు. నాన్నగారు తనను బడి (పాఠశాల) గేటు వరకు తన బండిపై తీసుకువెళ్ళి వదిలి నవ్వుతూ వీడ్కోలు చెప్పారు.

రాజు నెమ్మది – నెమ్మదిగా (ఎందుకంటే అతడు వేగంగా నడవలేడు.) కుంటుకుంటూ నడుస్తున్నాడు. అతని మధురమైన నవ్వును గుర్రుగా చూస్తూ, అతని కాళ్ళవైపు సైగచేసి నవ్వుతూ అతనిని అవహేళన చేస్తున్నారు. అక్కడి జనం.

కొద్ది క్షణాలలోనే మైదానం అంతా, వరండా అంతా కుతూహలంతో తననే చూస్తూ అతనివైపు సైగ చేస్తూ నవ్వుకునే వారి సంఖ్య పెరిగిపోయింది. పాఠశాల ఉపాధ్యాయులు కూడా అక్కడేమీ జరగటం లేదన్నట్లు తన ముందుగానే వెళ్ళిపోసాగిరి.

మొదటి పీరియడ్ ప్రారంభం కాగానే ఉపాధ్యాయుడు రాజూని అందరికంటే వెనుక కూర్చోబెట్టెను. రాజూని తనను పరిచయం చేసుకొనమని చెప్పగా తను ఓ గ్రామ పాఠశాల నుండి వచ్చినట్లు చెప్పెను. అది విన్న ఉపాధ్యాయునితో సహా పిల్లలందరూ నవ్విరి. మధుర స్వభావం కల రాజు ఇంతకు ముందెన్నడూ కోపాన్ని ప్రదర్శించలేదు. అతడు స్వయంగా తనకు తాను ధైర్యంగా ఉండవలసిన అవసరాన్ని గుర్తించాడు. త్వరగానే అన్నీ సర్దుకుంటాయి అని అనుకున్నాడు. కాని పూర్తిగా రోజంతా ప్రతి పీరియడ్ లో ఇలాగే జరిగింది. ప్రతిసారి ఒక కొత్త ఉపాధ్యాయుడు రావడం, రాజును ప్రతిసారి అవమానించడం జరిగింది. రాజును, తన పాఠశాలను ఎగతాళి చేయడంలో వారు విజయాన్ని సాధించారా?

లేదు, రాజు వేరే మట్టితో తయారుచేయబడినవాడు. అతడు గ్రామ పాఠశాలలు పట్టణ పాఠశాలల కంటే తక్కువేమీ కాదని నిరూపించదలచాడు. ఆ రోజు సాయంత్రం రాజు తన అమ్మ-నాన్నలకు ఏమీ చెప్పలేదు. వాళ్ళు జిజ్ఞాసపూర్వకంగా అడుగుతున్న ప్రశ్నలకు జవాబులేమీ చెప్పకుండా నవ్వుతూ ఉండిపోయాడు. ఎందుకంటే తనకు అబద్దం చెప్పడం కూడా ‘ : ఇష్టం లేదు.

మరుసటి రోజు, దాని మరుసటి రోజు, తర్వాత నెలలోని ప్రతిరోజూ అతనికి ఇలానే ఉంది. అతడు తరగతి గదిలో చెయ్యి ఎత్తినా తనను ప్రశ్నలకు జవాబులు చెప్పే అవకాశం ఇవ్వలేదు. అతనికి ఎవ్వరూ స్నేహితులు కూడా ఏర్పడలేదు. భోజన విరామ సమయంలో మిగిలిన పిల్లలందరూ ఆడుకుంటుంటే తను మాత్రం తరగతి గదిలోనే కూర్చునేవాడు. ఇప్పటి వరకూ అందరికీ తెలిసింది ఏమిటంటే రాజు ఒక గ్రామీణ బాలుడని, అతనికి తన గ్రామ పాఠశాల అంటే చాలా గర్వమని.

ఆఖరికి రాజు ఈ స్థితి నుండి బయటపడడానికి తెలివిగా ఒక ప్రణాళిక సిద్ధం చేసుకున్నాడు. తరగతి గదిలో ఉపాధ్యాయులు ప్రశ్నలు అడిగినపుడు చేతులు ఎత్తడం మానేశాడు. తత్ఫలితంగా విద్యార్థులు, ఉపాధ్యాయులు అతనిపై దృష్టి పెట్టడం మానేసిరి. ఒక నెల తర్వాత వార్షిక పరీక్షలు ఉన్నవని రాజుకి తెలుసు. అందువలన రాజు ఇంటివద్ద ఎక్కువ పరిశ్రమ చేసెను. అతనికి తన కొత్త పాఠశాల ముందు తన పాత పాఠశాల (గ్రామ పాఠశాల) మూర్చుల స్వర్గం కాదని నిరూపించదలచాడు.

అందరూ రాజు తప్పుతాడని అనుకున్నారు. సమయం గడిచేకొలది విద్యార్థులందరూ చదువులో మునిగిపోయిరి. రాజు చేతిలో పుస్తకాలను చూసి వారు హేళనగా నవ్వుకొనసాగిరి. రాజు ఇప్పుడు చాలా ధైర్యంగా తన మనస్సులోనే నవ్వుకుంటూ చదువుకుంటూ ఉండెను. ఒక వారంలోనే పరీక్షలు అయిపోయినవి. రాజు రెండు వారాలు సెలవుపై తన
గ్రామం తిరిగి వెళ్ళెను. ఆ తరువాత పరీక్షా ఫలితాలు వస్తాయి.

పరీక్షా ఫలితాలు వెలువడే ముందురోజు రాజు తిరిగివచ్చెను. మరుసటి రోజు ఆత్మవిశ్వాసంతో తన తండ్రిగారితో పరీక్షా ఫలితాలను చూడటానికి వెళ్ళెను. ఒకసారి మరలా రాజు తన తరగతిలో ప్రథమస్థానం పొందెను. తన నాన్నగారు సంతోషించిరి. కానీ రాజు సంతోషానికి అవధులు లేవు. ప్రథమస్థానం పొందినప్పుడు అతడెప్పుడూ ఇంత ఆనందం . పొందలేదు. అతడు తన పాత పాఠశాలకు ఒక అందమైన సముచితమైన మంచి బహుమతిని సమర్పించెను.