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Students can Download Computer Science Chapter 1 Function Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Computer Science Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Computer Science Solutions Chapter 1 Function

Samacheer Kalvi 12th Computer Science Function Text Book Back Questions and Answers

PART – 1
I. Choose The Best Answer

Question 1.
The small sections of code that are used to perform a particular task is called ……………………….
(a) Subroutines
(b) Files
(c) Pseudo code
(d) Modules
Answer:
(a) Subroutines

Question 2.
Which of the following is a unit of code that is often defined within a greater code structure?
(a) Subroutines
(b) Function
(c) Files
(d) Modules
Answer:
(b) Function

Question 3.
Which of the following is a distinct syntactic block?
(a) Subroutines
(b) Function
(c) Definition
(d) Modules
Answer:
(c) Definition

Question 4.
The variables in a function definition are called as ……………………….
(a) Subroutines
(b) Function
(c) Definition
(d) Parameters
Answer:
(d) Parameters

Question 5.
The values which are passed to a function definition are called ……………………….
(a) Arguments
(b) Subroutines
(c) Function
(d) Definition
Answer:
(a) Arguments

Question 6.
Which of the following are mandatory to write the type annotations in the function definition?
(a) Curly braces
(b) Parentheses
(c) Square brackets
(d) Indentations
Answer:
(b) Parentheses

Question 7.
Which of the following defines what an object can do?
(a) Operating System
(b) Compiler
(c) Interface
(d) Interpreter
Answer:
(c) Interface

Question 8.
Which of the following carries out the instructions defined in the interface?
(a) Operating System
(b) Compiler
(c) Implementation
(d) Interpreter
Answer:
(c) Implementation

Question 9.
The functions which will give exact result when same arguments are passed are called ……………………….
(a) Impure functions
(b) Partial Functions
(c) Dynamic Functions
(d) Pure functions
Answer:
(d) Pure functions

Question 10.
The functions which cause side effects to the arguments passed are called ……………………….
(a) Impure functions
(b) Partial Functions
(c) Dynamic Functions
(d) Pure functions
Answer:
(a) Impure functions

PART – II
II. Answer The Following Questions

Question 1.
What is a subroutine?
Answer:
Subroutines are the basic building blocks of computer programs. Subroutines are small sections of code that are used to perform a particular task that can be used repeatedly. In Programming languages these subroutines are called as Functions.

Question 2.
Define Function with respect to Programming language?
Answer:
A function is a unit of code that is often defined within a greater code structure. Specifically, a function contains a set of code that works on many kinds of inputs, like variants, expressions and produces a concrete output.

Question 3.
Write the inference you get from X: = (78)?
Answer:
Value 78 being bound to the name X.

Question 4.
Differentiate interface and implementation?
Answer:
Interface:
Interface just defines what an object can do, but won’t actually do it.

Implementation:
Implementation carries out the instructions defined in the interface.

Question 5.
Which of the following is a normal function definition and which is recursive function definition?
Answer:
(I) Let Recursive sum x y:
return x + y

(II) let disp:
print ‘welcome’

(III) let Recursive sum num:
if (num! = 0) then return num + sum (num – 1) else
return num

1. Recursive function
2. Normal function
3. Recursive function

PART – III
III. Answer The Following Questions

Question 1.
Mention the characteristics of Interface?
Answer:
Characteristics of interface:

1. The class template specifies the interfaces to enable an object to be created and operated properly.
2. An object’s attributes and behaviour is controlled by sending functions to the object.

Question 2.
Why strlen is called pure function?
Answer:
strlen (s) is called each time and strlen needs to iterate over the whole of ‘s’. If the compiler is smart enough to work out that strlen is a pure function and that ‘s’ is not updated in the lbop, then it can remove the redundant extra calls to strlen and make the loop to execute only one time. This function reads external memory but does not change it, and the value returned derives from the external memory accessed.

Question 3.
What is the side effect of impure function. Give example?
Answer:
Impure Function:

• The return value of the impure functions does not solely depend on its arguments passed. Hence, if you call the impure functions with the same set of arguments, you might get the different return values. For example, random( ), Date( ).
• They may modify the arguments which are passed to them.

Question 4.
Differentiate pure and impure function?
Answer:
Pure Function:

1. The return value of the pure functions solely depends on its arguments passed.
2. If you call the pure functions with the same set of arguments, you will always get the same return values.
3. They do not have any side effects.
4. They do not modify the arguments which are passed to them.

Impure Function:

1. The return value of the impure functions does not solely depend on its arguments passed.
2. If you call the impure functions with the same set of arguments, you might get the different return values. For example, random( ), Date( ).
3. They have side effects.
4. They may modify the arguments which are passed to them.

Question 5.
What happens if you modify a variable outside the function? Give an example?
Answer:
When a function depends on variables or functions outside of its definition block, you can never be sure that the function will behave the same every time it’s called.
For example let y: = 0
(int) inc (int) x
y: = y + x;
return (y)
In the above example the value of y get changed inside the function defintion due to which the result will change each time. The side effect of the inc ( ) function is it is changing the data ‘ of the external visible variable ‘y’.

PART – IV
IV. Answer The Following Questions

Question 1.
What are called Parameters and write a note on?
Answer:

1. Parameter without Type
2. Parameter with Type Parameters (and arguments)

Parameters are the variables in a function definition and arguments are the values which are passed to a function definition.

(I) Parameter without Type
Let us see an example of a function definition:
(requires: b> = 0)
(returns: a to the power of b)
let rec pow a b: =
if b = 0 then 1
else a * pow a (b – 1)
In the above function definition variable ‘b’ is the parameter and the value which is passed to the variable ‘b’ is the argument. The precondition (requires) and postcondition (returns) of the function is given. Note we have not mentioned any types: (data types). Some language compiler solves this type (data type) inference problem algorithmically, but some require the type to be mentioned.

In the above function definition if expression can return 1 in the then branch, by the typing rule the entire if expression has type int. Since the if expression has type ‘int ’, the function’s return type also be ‘inf. ‘b ’is compared to 0 with the equality operator, so ‘b ’is also a type of ‘int. Since a is multiplied with another expression using the * operator, ‘a’ must be an int.

(II) Parameter with Type
Now let us write the same function definition with types for some reason:
(requires: b > 0)
(returns: a to the power of b)
let rec pow (a: int) (b: int): int : =
if b = 0 then 1
else a * pow b (a – 1)
When we write the type annotations for ‘a ’ and ‘b ’ the parentheses are mandatory. Generally we can leave out these annotations, because it’s simpler to let the compiler infer them. There are times we may want to explicitly write down types. This is useful on times when you get a type error from the compiler that doesn’t make sense. Explicitly annotating the types can help with debugging such an error message.

The syntax to define functions is close to the mathematical usage: the definition is introduced by the keyword let, followed by the name of the function and its arguments; then the formula that computes the image of the argument is written after an = sign. If you want to define a recursive function: use “let rec ” instead of “let
Syntax: The syntax for function definitions:

let rec fnal a2 … an : = k
Here the fn is a variable indicating an identifier being used as a function name. The names ‘al ’ to ‘an ’ are variables indicating the identifiers used as parameters. The keyword ‘rec ’ is required if fn ’ is to be a recursive function; otherwise it may be omitted.
For example: let us see an example to check whether the entered number is even or odd.
(requires: x> = 0)
let rec even x : = x = 0 || odd (x – 1)
return ‘even’
(requires: x> = 0)
let odd x : =
x< >0 && even (x – 1)
return ‘odd’
The syntax for function types:
x → y
x₁ → x₂ → y
x₁ → … → x_n → y
The ‘x’ and ‘y’ are variables indicating types. The type x → y is the type of a function that gets an input of type ‘x’ and returns an output of type ‘y’. Whereas x₁ → x₂ → y is a type of a function that takes two inputs, the first input is of type ‘x1 ’ and the second input of type ‘x2’, and returns an output of type ‘y’. Likewise x1 → … → xn → y has type ‘x’ as input of n arguments and ‘y’ type as output.

Question 2.
Identify in the following program
Answer:
let rec gcd a b : =
if b < > 0 then gcd b (a mod b) else return a
(I) Name of the function
gcd

(II) Identify the statement which tells it is a recursive function
let rec

(III) Name of the argument variable
a, b

(IV) Statement which invoke the function recursively
gcd b(a mod b) [when b < > 0]

(V) Statement which terminates the recursion
return a (when b becomes 0).

Question 3.
Explain with example Pure and impure functions?
Answer:
Pure functions:
Pure functions are functions which will give exact result when the same arguments are passed. For example the mathematical function sin (0) always results 0. This means that every time you call the function with the same arguments, you will always get the same result. A function can be a pure function provided it should not have any external variable which will alter the behaviour of that variable.
Let us see an example
let square x
return: x * x
The above function square is a pure function because it will not give different results for same input. There are various theoretical advantages of having pure functions. One advantage is that if a function is pure, then if it is called several times with the same arguments, the compiler only needs to actually call the function once. Let’s see an example let i: = 0;
if i < strlen (s) then – Do something which doesn’t affect s ++ i If it is compiled, strlen (s) is called each time and strlen needs to iterate over the whole of ‘s’.

If the compiler is smart enough to work out that strlen is a pure function and that ‘s’ is not updated in the loop, then it can remove the redundant extra calls to strlen and make the #loop to execute only one time. From these what we can understand, strlen is a pure function because the function takes one variable as a parameter, and accesses it to find its length.

This function reads external memory but does not change it, and the value returned derives from the external memory accessed. Impure functions: The variables used inside the function may cause side effects through the functions which are not passed with any arguments.

In such cases the function is called impure function. When a function depends on variables or functions outside of its definition block, you can never be sure that the function will behave the same every time it’s called. For example the mathematical function random Q will give different outputs for the same function call, let Random number let a : = random( ) if a > 10 then
return: a
else
return: 10
Flere the function Random is impure as it is not sure what will be the result when we call the function.

Question 4.
Explain with an example interface and implementation?
Answer:
Interface Vs Implementation:
An interface is a set of action that an object can do. For example when you press a light switch, the light goes on, you may not have cared how it splashed the light. In Object Oriented Programming language, an Interface is a description of all functions that a class must have in order to be a new interface.

In our example, anything that “ACTSLIKE” a light, should have function defnitions like turn on ( ) and a turn off ( ). The purpose of interfaces is to allow the computer to enforce the properties of the class of TYPE T (whatever the interface is) must have functions called X, Y, Z, etc.

A class declaration combines the external interface (its local state) with an implementation of that interface (the code that carries out the behaviour). An object is an instance created from the class. The interface defines an object’s visibility to the outside world.

The difference between interface and implementation is:

Interface:
Interface just defines what an object can do, but won’t actually do it. Implementation carries out the instructions defined in the interface.

Implementation:
Implementation carries out the instructions defined in the interface.
In object oriented programs classes are the interface and how the object is processed and executed is the implementation.

Characteristics of interface

1. The class template specifies the interfaces to enable an object to be created and operated properly.
2. An object’s attributes and behaviour is controlled by sending functions to the object.

For example, let’s take the example of increasing a car’s speed.

The person who drives the car doesn’t care about the internal working. To increase the speed of the car he just presses the accelerator to get the desired behaviour. Here the accelerator is the interface between the driver (the calling / invoking object) and the engine (the called object). In this case, the function call would be Speed (70): This is the interface.

Internally, the engine of the car is doing all the things. It’s where fuel, air, pressure, and electricity come together to create the power to move the vehicle. All of these actions are separated from the driver, who just wants to go faster.

Let us see a simple example, consider the following implementation of a function that finds the minimum of its three arguments:
let min 3 x y z : =
if x < y then
if x < z then x else z
else
if y < z then y else z

Practice Programs
Question 1.
Write algorithmic function definition to find the minimum among 3 numbers?
Answer:
let min 3 x y z : =
if x < y then
if x < z then x else z
else
if y < z then y else z

Question 2.
Write algorithmic recursive function definition to find the sum of n natural numbers?
Answer:
let rec sum num:
if (num! = 0) then return num + sum (num – 1)
else
return num

Samacheer kalvi 12th Computer Science Function Additional Questions and Answers

PART – 1
I. Choose The Best Answer

Question 1.
……………………… are expressed using statements of a programming language.
(a) Algorithm
(b) Procedure
(c) Specification
(d) Abstraction
Answer:
(a) Algorithm

Question 2.
……………………… are the basic building blocks of a computer programs.
(a) Code
(b) Subroutines
(c) Modules
(d) Variables
Answer:
(b) Subroutines

Question 3.
In programming languages, subroutines are called as …………………………..
(a) Functions
(b) Task
(c) Modules
(d) Code
Answer:
(a) Functions

Question 4.
Find the correct statement from the following.
(a) a : = (24) has an expression
(b) (24) is an expression
Answer:
(a) a : = (24) has an expression

Question 5.
……………………… binds values to names.
(a) Algorithms
(b) Variables
(c) Interface
(d) Definitions
Answer:
(d) Definitions

Question 6.
Identify the statement which is wrong.
(a) Definitions are expressions
(b) Definitions are distinct syntactic blocks.
(c) Definitions can have expressions, nested inside them.
Answer:
(a) Definitions are expressions

Question 7.
The name of the function in let rec pow ab : = is …………………………
(a) Let
(b) Rec
(c) Pow
(d) a b
Answer:
(c) Pow

Question 8.
In function definition pre condition is given by ……………………….
(a) Needed
(b) Let
(c) Returns
(d) Requires
Answer:
(d) Requires

Question 9.
In function definition post condition is given by …………………………
(a) Needed
(b) Let
(c) Returns
(d) Requires
Answer:
(c) Returns

Question 10.
In b = 0, = is ……………………….. operator
(a) Assignment
(b) Equality
(c) Logical
(d) Not equal
Answer:
(b) Equality

Question 11.
The formula should be written after ……………………….. sign
(a) +
(b) –
(c) =
(d) ++
Answer:
(c) =

Question 12.
To define a recursive function, …………………………. is used.
(a) Let
(b) Let r
(c) Let rfn
(d) Let rec
Answer:
(d) Let rec

Question 13.
Find which is false.
(a) All function definitions are static
(b) All function definitions are dynamic
Answer:
(b) All function definitions are dynamic

Question 14.
A ……………………….. combines the external interface with an implementation of that interface.
Answer:
class declaration

Question 15.
An …………………………. is an instance created from the class.
(a) Object
(b) Functions
(c) Subroutines
(d) Definitions
Answer:
(a) Object

Question 16.
Find the statement which is not true.
(a) The interface defines an objects visibility to the outside world
(b) Interface defines what an object can do.
(c) In object oriented programs, objects are interfaces
Answer:
(c) In object oriented programs, objects are interfaces

Question 17.
An ………………………… attributes and behaviour is controlled by sending functions to the object.
Answer:
Objects

Question 18.
The class template specifies the ………………………. to enable an object to be created and operated properly.
Answer:
Interfaces

Question 19.
The accelerator is the …………………………… between the driver and the engine.
(a) Interface
(b) Object
(c) Instruction
(d) Code
Answer:
(a) Interface

Question 20.
sin (0) = 0 is an example for ………………………. function.
(a) Impure
(b) Pure
(c) Interface
(d) Instruction
Answer:
(b) Pure

Question 21.
Find the impure function from the following.
(a) Sin (0)
(b) Square x
(c) Strlen (s)
(d) None of these
Answer:
(d) None of these

Question 22.
The function random ( ) is an example for …………………….. functions.
Answer:
Impure

Question 23.
Why is the function random( ) is a impure function?
(a) It gives different outputs for same function call
(b) It gives different outputs when 0 is given
(c) It will not give different output
Answer:
(a) It gives different outputs for same function call

Question 24.
Which function definition, doesn’t modify the arguments passed to them?
(a) Pure function
(b) Impure function
(c) Object
(d) Interface
Answer:
(a) Pure function

Question 25.
How many parameters are defined in the function let rec gcd a b : =
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(c) 2

Question 26.
In the function definition, the keyword let is followed by …………………………
(a) Function name
(b) Arguments
(c) Parameters
(d) Implementations
Answer:
(a) Function name

Question 27.
Find the correct statement from the following function definitions. let rec p on a b : =
(a) data type of the parameters are given
(b) data type of the parameters are not mentioned
Answer:
(b) data type of the parameters are not mentioned

Question 28.
If a function is not a recursive one, then ……………………………. is used
(a) abc
(b) gcd
(c) let
(d) let rec
Answer:
(c) let

Question 29.
Find the name of the function,
let rec even x : =
(a) Let
(b) Rec
(c) Even
(d) x
Answer:
(c) Even

Question 30.
Match the following function definitions with their terms.
let rec odd xy : =

1. Keyword – (i) Xy
2. Recursion – (ii) Odd
3. Function name – (iii) Rec
4. Parameters – (iv) let

(a) 1 – (iv) 2 – (iii) 3 – (ii) 4 – (i)
(b) 1 – (i) 2 – (ii) 3 – (iii) 4 – (iv)
(c) 1 – (iv) 2 – (i) 3 – (ii) 4 – (iii)
(d) 1 – (i) 2 – (iv) 3 – (ii) 4 – (iii)
Answer:
(a) 1 – (iv) 2 – (iii) 3 – (ii) 4 – (i)

Question 31.
In object oriented programming language, an is a description of all functions that a class must have
(a) Object
(b) Class
(c) Interface
(d) Code
Answer:
(c) Interface

Question 32.
The ……………………… defines an object’s visibility to the outside world.
(a) Object
(b) Interface
(c) Pure function
(d) Impure function
Answer:
(b) Interface

Question 33.
Find the correct statement.
(i) Evaluation of pure function causes side effects to its output.
(ii) Evaluation of Impure function causes side effects to its output.
Answer:
(ii) Evaluation of Impure function causes side effects to its output.

PART – II
II. Answer The Following Questions

Question 1.
What are the two types of parameter passing?
Answer:

1. Parameter without type
2. Parameter with type

Question 2.
In the function definition
let rec pow a b : = Is it recursive function. If so Explain. Why?
Answer:
Yes it is a recursive function. It is given in the function definition as rec which indicates recursive function.

Question 3.
Write the syntax for the function definitions?
Answer:
let rec f_n a₁ a₂ … a_n : = k
f_n : Function name
a₁ … a_n – variable
rec: recursion

Question 4.
Define recursive functions: How will you define it?
Answer:
A function definition which calls itself is called recursive functions. It is given by let rec.

PART – III
III. Answer The Following Questions

Question 1.
Write note on Definitions?
Answer:
Definitions bind values to names, Definitions are not expressions, Definitions are distinct syntactic blocks. Definitions can have expressions nested inside them, and vice – versa.

Question 2.
Write the pre condition and post condition for the function?
Answer:
(requires: b > 0)
– (returns: a to the power of b) let rec pow(a : int) (b : int): int: =

1. Pre condition : b > 0
2. Post condition : a to the power of b.

Question 3.
Give function definition for the Chameleons of Chromeland problem?
Answer:
let rec monochromatize abc : =
if a > 0 then
a, b, c : = a – 1, b – 1, c + 2
else
a: = 0, b: = 0, c: = a + b + c
return c

Question 4.
Give the flow chart for Chameleons of Chromeland problem?
Answer:
The algorithm is depicted in the flowchart as below:

Question 5.
Give the example function definition for parameter with type?
Answer:
Parameter with Type:
Now let us write the same function definition with types for some reason:
(requires: b> 0)
(returns: a to the power of b ) let rec pow (a: int) (b: int): int : =
if b = 0 then 1
else a * pow b (a – 1)

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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.13

Choose the correct or most suitable answer from given four alternatives.
Question 1.
If \(\int f(x) d x\) = g(x) + c, then \(\int f(x) g^{\prime}(x) d x\)

Solution:
(a)

Question 2.
If , then the value of k is ……………
(a) log 3
(b) -log 3
(c) \(-\frac{1}{\log ^{3}}\)
(d) \(\frac{1}{\log 3}\)
Solution:
(c)

Question 3.
If \(\int f^{\prime}(x) e^{x^{3}} d x\) = (x – 1)e^x2, then f(x) is …………………

Solution:
(d)

Question 4.
The gradient (slope) of a curve at any point (x, y) is \(\frac{x^{2}-4}{x^{2}}\). If the curve passes through the point(2, 7), then the equation of the curve is ………….
(a) y = x + \(\frac{4}{x}\) + 3
(b) y = x + \(\frac{4}{x}\) + 4
(c) y = x² + 3x + 4
(d) y = x² – 3x + 6
Solution:

Question 5.

(a) cot (xe^x) + c
(b) sec (xe^x) + c
(c) tan (xe^x) + c
(d) cos (xe^x) + c
Solution:
(c)

Question 6.
\(\int \frac{\sqrt{\tan x}}{\sin 2 x} d x\) is ……………..

Solution:
(a)

Question 7.
\(\int \sin ^{3} x d x\) is …………….

Solution:
(c)
Hint: sin³x = \(\frac{1}{4}\) (3 sin x – sin 3x)

Question 8.

Solution:
(b)

Question 9.

(a) tan^-1 (sin x) + c
(b) 2 sin^-1 (tan x) + c
(c) tan^-1 (cos x) + c
(d) sin^-1 (tan x) + c
Solution:
(d)

= sin^-1 (t) + c
= sin^-1 (tan x) + c

Question 10.

(a) x² + c
(b) 2x² + c
(c) \(\frac{x^{2}}{2}\) + c
(d) \(-\frac{x^{2}}{2}\) + c
Solution:
(c)

Question 11.
\(\int 2^{3 x+5} d x\) is ……………

Solution:
(d)

Question 12.

Solution:
(b)

Question 13.

Solution:
(d)

Question 14.
\(\int \frac{x^{2}+\cos ^{2} x}{x^{2}+1}\) cosec²xdx is …………….
(a) cot x + sin^-1 x + c
(b) -cot x + tan^-1 x + c
(c) -tan x + cot^-1 x + c
(d) -cot x – tan^-1 x + c
Solution:
(d)

Question 15.
\(\int x^{2} \cos x d x\) is ……………
(a) x² sin x + 2x cos x – 2 sin x + c
(b) x² sin x – 2x cos x – 2 sin x + c
(c) -x² sin x + 2x cos x + 2 sin x + c
(d) -x² sin x – 2x cos x + 2 sin x + c
Solution:
(a)
Hint: \(\int x^{2} \cos x d x\)
By Bernoullis formula dv = cosxdx
u = x² v = sinx
u’ = 2x v₁ = -cos x
u” = 2 v₂ = -sinx
= uv – u’v₁ + u”v₂
= x²sin x + 2x cos x – 2 sin x + c

Question 16.

Solution:
(b)

Question 17.
\(\int \frac{d x}{e^{x}-1}\) is …………….
(a) log |e^x| – log |e^x – 1| + c
(b) log |e^x| + log |e^x – 1| + c
(c) log |e^x – 1| – log |e^x| + c
(d) log |e^x + 1| – log |e^x| + c
Solution:
(c)

Question 18.

Solution:
(b)
We know that

Question 19.

Solution:
(d)

Question 20.

Solution:
(a)
We know that

Question 21.

Solution:
(c)
Hint:

By Bernoullis formula,

Question 22.

Solution:
(d)

Question 23.

Solution:
(c)

Question 24.

Solution:
(a)

Question 25.

Solution:
(d)
Hint: Let I = \(\int e^{\sqrt{x}} d x\)
t = \(\sqrt{x}\)

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Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 1 Numbers Ex 1.2

Question 1.
Fill in the blanks:
(i) If a number has 5 or 6 digits in it then, its square root will have………digits.
(ii) The value of 180 lies between integers………and……….
(iii) \(\sqrt{10}\) × \(\sqrt{6}\) × \(\sqrt{15}\) =……………
(iv) \(\frac{\sqrt{300}{\sqrt{192}}\) =…………….
(v) \(\sqrt{65.61}\) =…………….
Solution:
(i) 3
(ii) 13, 14
(iii) 30
(iv) \(\frac{5}{4}\)
(v) 8.1

Question 2.
Estimate the value of the following square roots to the nearest whole number:
(i) \(\sqrt{440}\)
(ii) \(\sqrt{800}\)
(iii) \(\sqrt{1020}\)
Solution:
(i) We have 20² = 400
21² = 441
∴ \(\sqrt{440}\)  \(\widetilde { – } \) 21

(ii) We have 28² = 784
29² = 841
∴ \(\sqrt{800}\) \(\widetilde { – } \) 28

(iii) We have 31² = 961
32² = 1024
∴ \(\sqrt{1020}\) \(\widetilde { – } \) = 32

Question 3.
Find the least number that must be added to 1300 so as to get a perfect square. Also find the square root of the perfect square.
Solution:
We work out the process of finding square root by long division method.
The given number is 1300

So we have 36² < 1300 < 37²
Also 1300 is (469 – 400) = 69 less than 37². So if we add 69 to 1300 it will be perfect square. Hence the required, least number is 69 and the perfect square number is 1300 + 69 = 1369

∴ \(\sqrt{1369}\) = 37

Question 4.
Find the least number that must be subtracted to 6412 so as to get a perfect square. Also find the square root of the perfect square.
Solution:
Let us work out the process of finding the square root of 6412 by long division method.

The remainder in the last step is 12. Is if 12 be subtracted from the given number the remainder will be zero and the new number will be a perfect square.
∴ The required number is 12.
The square number is 6412 – 12 = 6400
Also \(\sqrt{6400}\) = 80

Question 5.
Find the square root by long division method.
(i) 17956
(ii) 11025
(iii) 6889
(iv) 1764
(v) 418609
Solution:

Question 6.
Find the square root of the following decimal numbers:
(i) 2.89
(ii) 1.96
(iii) 67.24
(iv) 31.36
(v) 2.0164
(vi) 13.9876
Solution:

Question 7.
Find the square root of each of the following fractions:
(i) \(\frac{144}{225}\)
(ii) 7\(\frac{18}{49}\)
(iii) 6\(\frac{1}{4}\)
(iv) 4\(\frac{25}{36}\)
Solution:

Question 8.
Say True or False:
(i) \(\frac{\sqrt{32}}{\sqrt{8}}=2\)
(ii) \(\sqrt{\frac{625}{1024}}=\frac{25}{32}\)
(iii) \(\sqrt{28}{7}=2\sqrt{7}\)
(iv) \(\sqrt{225}{64}=\sqrt{289}\)
(v) \(\sqrt{1 \frac{400}{441}}=1 \frac{20}{21}\)
Solution:
(i) true
(ii) true
(iii) false
(iv) false
(v) false

Objective Type Questions

Question 9.
\(\sqrt{48}\) is approximately equal to
(a) 5
(b) 6
(c) 7
(d) 8
Solution:
(c) 7
Hint:
\(\sqrt{49}\) = 7

Question 10.
\(\sqrt{128}\) – \(\sqrt{98}\) + \(\sqrt{18}\) =
(a) \(\sqrt{2}\)
(b) \(\sqrt{8}\)
(c) \(\sqrt{48}\)
(d) \(\sqrt{32}\)
Solution:
(d) \(\sqrt{32}\)
Hint:
\(\sqrt{128}-\sqrt{98}+\sqrt{18}=8 \sqrt{2}-7 \sqrt{2}+3 \sqrt{2}=4 \sqrt{2}=\sqrt{32}\)

Question 11.
\(\sqrt{22+\sqrt{7+\sqrt{4}}}=\)
(a) \(\sqrt{25}\)
(b) \(\sqrt{33}\)
(c) \(\sqrt{31}\)
(d) \(\sqrt{29}\)
Solution:
(a) \(\sqrt{25}\)
Hint:
\(\sqrt{22+\sqrt{7+\sqrt{4}}}=\sqrt{22+\sqrt{7+2}}=\sqrt{22+3}=\sqrt{25}\)

Question 12.
The number of digits in the square root of 123454321 is
(a) 4
(b) 5
(c) 6
(d) 7
Solution:
(b) 5
Hint:
\(=\frac{n+1}{2}=\frac{10}{2}=5\)

Students can Download Maths Chapter 1 Numbers Ex 1.1 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 1 Numbers Ex 1.1

Question 1.
Fill in the blanks:
i. The ones digit in the square of 77 is …………
ii. The number of non-square numbers between 24² and 25² is ……………….
iii. If a number ends with 5, its square ends with …………
iv. A square number will not end with numbers ………….
v. The number of perfect square numbers between 300 and 500 is …………
Solution:
i. 9
ii. 48
iii. 5
iv. 2, 3, 7, 8
v. 5

Question 2.
Say True or False:
i. When a square number ends in 6, its square root will have 6 in the unit’s place.
ii. A square number will not have odd number of zeros at the end.
iii. The number of zeros in the square of 961000 is 9.
iv. (7, 24, 25) is a Pythagorean triplet.
v. The square root of 221 is 21.
Solution:
i. True
ii. True
iii. False
iv. True
v. False

Question 3.
What will be the ones digit in the squares of the following numbers?
(i) 36
(ii) 252
(iii) 543
Solution:
(i) If a number ends with 6, its square ends with 6.
∴ Ones’ digit in the square of 36 is 6.

(ii) If a number ends with 2, its square ends with 4.
∴ Ones’ digit in the square of 252 is 4

(iii) If a number ends with 3, its square ends with 9.
∴ Ones’ digit in the square of 543 is 9.

Question 4.
Study the given numbers and justify why each of them obviously cannot be a perfect square.
(i) 1000
(ii) 34567
(iii) 408
Solution:
We know that the numbers end with odd number of zeros, 7 and 8 not perfect squares.
∴ 1000, 34567 and 408 cannot be perfect squares.

Question 5.
Find the sum without actually adding the following odd numbers:
(i) 1 + 3 + 5 + 7 +……..+ 35
(ii) The first 99 odd natural numbers.
Solution:
1 + 3 + 5 + 7 +……..+ 35.
Here there are 18 odd numbers from 1 to 35.
Sum of first n consecutive odd natural numbers = n²
∴ Sum of first 18 consecutive odd natural numbers = 18² = 18 × 18 = 324

(ii) The first 99 odd natural numbers.
Sum of first n consecutive natural numbers = n²
∴ Sum of 99 odd natural numbers = 99² = 99 × 99 = 9801

Question 6.
Express
(i) 15² and
(ii) 19²
as the sum of two consecutive positive integers.
Solution:

Question 7.
Write
(i) 10² and
(ii) 11²
as the sum of consecutive odd natural numbers.
Solution:
(i) 10²
10² = 100 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
(ii) 11²
11² = 121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

Question 8.
Find a Pythagorean triplet whose
(i) largest member is 65
(ii) smallest member is 10
Solution:
(i) Largest number is 65
Given largest number is 65.
We know that 2m, m² – 1, m² + 1 form a Pythagorean triplet.
let m² + 1 = 65
m² = 65 – 1
m² = 64
m² = 8 × 8
m = 8
∴ 2m = 2 × 8 = 16
m² – 1 = 64 – 1 = 63
∴ The required Pythagorean triplet is (16, 63, 65)

(ii) Smallest number is 10
We know that (2m, m² – 1, m² + 1) form a Pythagorean triplet.
Given smallest number is 10
let 2m = 10
m = \(\frac{10}{2}\)
m = 5
m² + 1 = 5² + 1 = 25 + 1 = 26
m² – 1 = 5² – 1 = 25 – 1 = 24
∴ The required Pythagorean triplet is (10, 24, 26)

Question 9.
Find the square root of the following by repeated subtraction method.
(i) 144
(ii) 256
(iii) 784
Solution:
(i) 144
Step 1: 144 – 1 = 143
Step 2: 143 – 3 = 140
Step 3: 140 – 5 = 135
Step 4: 135 – 7 = 128
Step 5: 128 – 9 = 119
Step 6: 119 – 11 = 108
Step 7: 108 – 13 = 95
Step 8: 95 – 15 = 80
Step 9: 80 – 17 = 63
Step 10: 63 – 19 = 44
Step 11: 44 – 21 = 23
Step 12: 23 – 23 = 0.
We get 0 in the 12th step.
∴ 144 is a perfect square and ⇒ \(\sqrt{144}\) = 12.

(ii) 256
Step 1: 256 – 1 = 255
Step 2: 255 – 3 = 252
Step 3: 252 – 5 = 247
Step 4: 247 – 7 = 240
Step 5: 240 – 9 = 231
Step 6: 231 – 11 = 220
Step 7: 220 – 13 = 207
Step 8: 207 – 15 = 192
Step 9: 192 – 17 = 175
Step 10: 175 – 19 = 156
Step 11: 156 – 21 = 135
Step 12: 135 – 23 = 112
Step 13: 112 – 25 = 87
Step 14: 87 – 27 = 60
Step 15: 60 – 29 = 31
Step 16: 31 – 31 = 0
We have subtracted odd numbers starting from 1 repeatedly from 256. We get 0 in the 8th step.
∴ 256 is a perfect square and \(\sqrt{256}\) = 16

(iii) 784
Step 1: 784 – 1 = 783
Step 2: 783 – 3 = 780
Step 3: 780 – 5 = 775
Step 4: 775 – 7 = 768
Step 5: 768 – 9 = 759
Step 6: 759 – 11 = 748
Step 7: 748 – 13 = 735
Step 8: 735 – 15 = 720
Step 9: 720 – 17 = 703
Step 10: 703 – 19 = 684
Step 11: 684 – 21 = 663
Step 12: 663 – 23 = 640
Step 13: 640 – 25 = 615
Step 14: 615 – 27 = 588
Step 15: 588 – 29 = 559
Step 16: 559 – 31 = 528
Step 17: 528 – 33 = 495
Step 18: 495 – 35 = 460
Step 19: 460 – 37 = 423
Step 20: 423 – 39 = 384
Step 21: 384 – 41 = 343
Step 22: 343 – 43 300
Step 23: 300 – 45 = 255
Step 24: 255 – 47 = 208
Step 25: 208 – 49 = 159
Step 26: 159 – 51 = 108
Step 27: 108 – 53 = 55
Step 28: 55 – 55 = 0
We have subtracted odd numbers starting from 1 repeatedly from 784, we get zero in the 28th step.
∴ 784 is a perfect square. \(\sqrt{784}\) = 28

Question 10.
Find the square root by prime factorisation method.
(i) 1156
(ii) 4761
(iii) 9025
Solution:
(i) 1156

1156 = 2 × 2 × 17 × 17
1156 = 2² × 17²
1156 = (2 × 17)²
∴ \(\sqrt{1156}\) = \((\sqrt{2×17})^{2}\) = 2 × 17 = 34
∴ \(\sqrt{1156}\) = 34

(ii) 4761

4761 = 3 × 3 × 23 × 23
4761 = 3² × 23²
4761 = (3 × 23)²
\(\sqrt{4761}\) = \((\sqrt{3 × 23})^{2}\)
\(\sqrt{4761}\) = 3 × 23
\(\sqrt{4761}\) = 69

(iii) 9025

9025 = 5 × 5 × 19 × 19
9025 = 5² × 19²
9025 = (5 × 19)²
\(\sqrt{925}\) = \((\sqrt{5 × 19})^{2}\) = 5 × 19 = 95

Question 11.
Examine if each of the following is a perfect square:
(i) 725
(ii) 190
(iii) 841
(iv) 1089
Solution:
(i) 725

725 = 5 × 5 × 29 = 5² × 29
Here the second prime factor 29 does not have a pair.
Hence 725 is not a perfect square number.

(ii) 190

190 = 2 × 5 × 19
Here the factors 2, 5 and 9 does not have pairs.
Hence 190 is not a perfect square number.

(iii) 841
841 = 29 × 29
Hence 841 is a perfect square

(vi) 1089
1089 = 3 × 3 × 11 × 11 = 33 × 33

Hence 1089 is a perfect square

Question 12.
Find the least number by which 1800 should be multiplied so that it becomes a perfect square. Also, find the square root of the perfect square thus obtained.
Solution:

We find 1800 = 2 × 2 × 3 × 3 × 5 × 5 × 2
= 2² × 3² × 5² × 2
Here the last factor 2 has no pair. So if we multiply 1800 by 2, then the number becomes a perfect square.
∴ 1800 × 2 = 3600 is the required perfect square number.
∴ 3600 = 1800 × 2
3600 = 2² × 3² × 5² × 2 × 2
3600 = 2² × 3² × 5² × 2²
= (2 × 3 × 5 × 2)²
\(\sqrt{3600}\) = \((\sqrt{2 × 3 × 5 × 2})^{2}\) = 2 × 3 × 5 × 2 = 60
∴ \(\sqrt{3600}\) = 60.

Question 13.
Find the smallest number by which 10985 should be divided so that the quotient is a perfect square.
Solution:
We find 10985 = 5 × 13 × 13 × 13
= 5 × 13 × 13²
Here the prime factors 5 and 13 do not have pairs.
∴ We can divide 10985 by 65 (5 × 13) to get a perfect square
∴ When we divide 10985 by 65 we get quotient 169.
169 = 13 × 13
\(\sqrt{169}\) = 13

Question 14.
Is 2352 a perfect square? If not, find the smallest number by which 2352 must be multiplied so that the product is a perfect square. Find the square root of the new number.
Solution:

We find 2352 = 2 × 2 × 2 × 2 × 3 × 7 × 7
2352 = 2² × 2² × 3 × 7²
Here the factor 3 has no pair.
∴ 2352 is not a perfect square.
We have to multiply 2352 by 3 so that the product is a perfect square.
Img 13
∴ 2352 × 3 = 2² × 2² × 7² × 3 × 3
7056 = 2² × 2² × 7² × 3²
7056 = (2 × 2 × 7 × 3)²
∴ \(\sqrt{7056}\) = \((\sqrt{2 × 2 × 7 × 3})^{2}\)
= 2 × 2 × 7 × 3 = 84
\(\sqrt{7056}\) = 84

Question 15.
Find the least square number which is divisible by each of the numbers 8, 12 and 15.
Solution:
The least number divisible by each of the numbers 8, 12, 15 is their L.C.M

∴ LCM of 8, 12, 15 is (4 × 3 × 2 × 5) = 120
Resolving 120 into prime factors
We get 120 = 2 × 2 × 2 × 3 × 5
Grouping into pairs of equal factors

120 = (2 x 2) x 2 x 3 x 5
∴ The factors 2, 3 and 5 had no pairs.
Thus to make 120 a perfect square, we must multiply it by (2 x 3 x 5) = 30.
∴ The least square number divisible by 8,12 and 15 is 120 x 30 = 3600

Students can Download Maths Chapter 4 Statistics Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 4 Statistics Intext Questions

Exercise 4.1
Try These (Text book Page no. 77)

Question 1.
Arrange the given data in ascending and descending order:
9, 34, 4, 13, 42, 10, 25, 7, 31, 4, 40
Solution:
Ascending order: 4, 4, 7, 9, 10, 13, 25, 31, 34, 40, 42.
Descending order : 42, 40, 34, 31, 25, 13, 10, 9, 7, 4, 4

Question 2.
Find the range of the given data : 53, 42, 61, 9, 39, 63, 14, 20, 06, 26, 31, 4, 57
Solution:
Ascending order of the given data:
4, 6, 9, 14, 20, 26, 31, 39, 42, 53, 57, 61, 63
Here largest value = 63
Smallest value = 4
∴ Range = Largest value – smallest value = 63 – 4 = 59

Think (Text book Page no. 79)

How will you change the given series as continuous series
15 – 25
28 – 38
41 – 51
54 – 64
Solution:
Given series
15 – 25
28 – 38
41 – 51
54 – 64
Difference in the gap = 28 – 25 = 3
Here half of the gap = \(\frac{1}{2}\)(3) = 1.5
∴ 1.5 is the adjustment factor. So we subtract 1.5 from the lower limit and add 1.5 to the upper limit to make it as a continuous series.

Think (Text book Page no. 80)

If we want to represent the given data by 5 classes, then how shall we find the interval?
Solution:
We can find the class size by the formula
Number of class intervals = \(\frac{Range}{Class size}\)

Try These (Text book Page no. 82)

Question 1.
Prepare a frequency table for the data : 3, 4, 2, 4, 5, 6, 1, 3, 2, 1, 5, 3, 6, 2, 1, 3, 2, 4
Solution:
Ascending order of the given data.
1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 5, 5, 6, 6
The distribution table:

∴ Frequency Table:

Question 2.
Prepare a grouped frequency table for the data :
10, 9, 3, 29, 17, 34, 23, 20, 39, 42, 5, 12, 19, 47, 18, 19, 27, 7, 13, 40, 38, 24, 34, 15, 40
Largest value = 47
Smallest value = 3
Range = Largest value – Smallest value = 47 – 3 = 44
Suppose we take class size as 10, then Number of class intervals possible
= \(\frac{Range}{Class size}\) = \(\frac{44}{10}\) = 4.4
\(\tilde { – } \) 5

Exercise 4.2
Think (Text book Page no. 94)

When joining two adjacent midpoints w ithout using a ruler, can you get a polygon?
Solution:
No, because it may be curved lines and they are not considered as polygons.

Students can Download Maths Chapter 3 Geometry Additional Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 3 Geometry Additional Questions

Question 1.
In ΔPQR, PS is a median. If QS = 7 cm find the length of QR?
Solution:
PS is the median ⇒ S is the midpoint of QR

Given
QS = 7 cm
∴ SR = 7 cm
QR = QS + SR
= 7 + 7 = 14cm

Question 2.
In ΔABC, G is the centroid. If AD = 6 cm, BC = 4 cm and BE = 9 cm find the perimeter of ΔBDG.
Solution:
In ΔABC, G is the centroid.

If AD = 6 cm
⇒ GD = \(\frac{1}{3}\) of AD = \(\frac{1}{3}\) (6)
BE = 9 cm
⇒ BG = \(\frac{2}{3}\) of BE = \(\frac{2}{3}\) (9) = 6cm
Also D is the midpoint of BC ⇒ BD
= \(\frac{1}{1}\) of BC = \(\frac{1}{2}\) (4) = 2cm
∴ Perimeter of ΔBDG = BD + GD + BG = 2 + 2 + 6 = 10 cm

Question 3.
Construct a rhombus FISH with FS = 8 cm and ∠F = 80°
Solution:
Given FS = 8 cm and ∠F = 80°

Steps :
(i) Drawn a line segment FS = 7 cm.
(ii) At F, made ∠SFX = ∠SFY = 40° on either side of FS.
(iii) At S, made ∠FSP = ∠FSQ = 40° on either side of FS
(iv) Let FX and SP cut at H and FY and SQ cut at I.
(v) FISH is the required rhombus

Calculation of Area:
Area of the rhombus FISH = \(\frac{1}{2}\) × d₁ × d₂ sq.units = \(\frac{1}{2}\) × 5.9 × 7 cm² = 20.65 cm²

Students can Download Maths Chapter 2 Life Mathematics Additional Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 2 Life Mathematics Additional Questions

Question 1.
120 men had food for 200 days. After 5 days 30 men left the camp. How long will the remaining food last.
Solution:
Since 30 men left after 5 days, the remaining food is sufficient for 120 men for 195 days. Suppose the remaining food lasts for x days for the remaining 90 men.
We have

More men means less days the food lasts
∴ It is inverse proportion
120 : 90 = x : 195
Product of extremes = Product of means
120 × 195 = 90 × x
x = \(\frac{120×195}{90}\)
x = 90
x = 260.
∴ Remaining food last for 260 days.

Question 2.
15 men earn Rs 900 in 5 days, how much will 20 men earn in 7 days?
Solution:
In one day 15 men earn Rs 900
In one day 15 men earn \(\frac{900}{5}\) = Rs 180
In one day 1 men earn \(\frac{180}{5}\) = Rs 12
∴ 1 men earn in 7 days = 12 × 7 = Rs 84
∴ 20 men earn in 7 days = 84 × 20 = 1680

Question 3.
A and B together can do a piece of work in 10 days, B and C can do the same work together in 12 days, A and C can do together in 15 days. How long will it take to complete the work working three of them altogether?
Solution:
(A + B)’s 1 day’s work = \(\frac{1}{10}\)……….(1)
(B + C)’s 1 day’s work = \(\frac{1}{12}\)……….(2)
(A + C)’s 1 day’s work = \(\frac{1}{15}\)……….(3)
(l) + (2) + (3) ⇒
[A + B + B + C + A + C]’s 1 day work = \(\frac{1}{10}\) + \(\frac{1}{12}\) + \(\frac{1}{15}\)
(2A + 2B + 2C)’s 1 day work = \(\frac{6 + 5 + 4}{60}\)
2(A + B + C)’s 1 day work = \(\frac{15}{60}\)
(A + B + C)’s 1 day’s work = \(\frac{1}{4 × 2}\) = \(\frac{1}{8}\)
∴ A + B + C work together to finish the work in 8 days.

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Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 3 Geometry Ex 3.3

Question 1.
Construct the following rhombuses with the given measurements and also find their area.
(i) FACE, FA = 6 cm and FC = 8 cm
Solution:
Given FA = 6 cm and FC = 8cm

Steps :
(i) Drawn a line segment FA = 6 cm.
(ii) With F and A as centres, drawn arcs of radii 8 cm and 6 cm respectively and let them cut at C.
(iii) Joined FC and AC.
(iv) With F and C as centres, drawn arcs of radius 6 cm each and let them cut at E. Joined FE and EC.
(v) FACE is the required rhombus.

Calculation of Area :
Area of the rhombus = \(\frac{1}{2}\) × d₁ × d₂ sq.units = \(\frac{1}{2}\) × 8 × 9 sq.units = 36 cm²

(ii) RACE, RA = 5.5 cm and AE = 7 cm
Solution:
Given RA = 5.5 cm and AE = 7 cm

Steps :
(i) Drawn a line segment RA = 5.5 cm.
(ii) With R and A as centres, drawn arcs of radii 5.5 cm and 7 cm respectively and let them cut at E.
(iii) Joined RE and AE.
(iv) With E and A as centres, drawn arcs of radius 5.5 cm each and let them cut at C.
(v) Joined AC and EC.
(vi) RACE is the required rhombus.

Calculation of Area :
Area of the rhombus = \(\frac{1}{2}\) × d₁ × d₂ sq.units = \(\frac{1}{2}\) × 7 × 8.5 cm² = 29.75 cm²

(iii) CAKE, CA = 5 cm and ∠A = 65°
Solution:
Given CA = 5 cm and ∠A = 65°

(i) Drawn a line segment CA = 5 cm.
(ii) At A on AC, made ∠CAX = 65°
(iii) With A as centre, drawn arc of radius 5 cm. Let it cut AX at K.
(iv) With K and C as centres, drawn arcs of radius 5 cm each and let them cut at E. Joined KE and CE.
(v) CAKE is the required rhombus.

Calculation of Area :
Area of the rhombus = \(\frac{1}{2}\) × d₁ × d₂ sq.units = \(\frac{1}{2}\) × 5.4 × 8.5 cm² = 22.95 cm²

(iv) MAKE, MA= 6.4 cm and ∠M = 80°
Solution:
Given MA = 6.4 cm and ∠M = 80°

Steps :
(i) Drawn a line segment MA = 6.4 cm.
(ii) At M on MA, made ∠AMX = 80°
(iii) With M as centres, drawn arc of radius 6.4 cm. Let it cut MX at E.
(iv) With E and A as centres, drawn arcs of radius 6.4 cm each and let them cut at K.
(v) Joined EK and AK.
(vi) MAKE is the required rhombus.

Calculation of Area :
Area of the rhombus = \(\frac{1}{2}\) × d₁ × d₂ sq.units = \(\frac{1}{2}\) × 8.2 × 9.8 cm² = 40.18 cm²

(v) LUCK, LC = 7.8 cm and UK = 6 cm
Solution:
Given LC = 7.8 cm and UK = 6 cm

Steps :
(i) Drawn a line segment LC = 7.8 cm.
(ii) Drawn the perpendicular bisector XY to LC. Let it cut LC at ‘O’
(iii) With O as centres, drawn arc of radius 3 cm on either side of O which cut OX at K and OY at U.
(iv) Joined LU, UC, CK and LK.
(v) LUCK is the required rhombus.

Calculation of Area :
Area of the rhombus = \(\frac{1}{2}\) × d₁ × d₂ sq.units = \(\frac{1}{2}\) × 7.8 × 6 cm² = 23.4 cm²

(vi) DUCK, DC = 8 cm and UK = 6 cm
Solution:
Given DC = 8 cm and UK = 6 cm

Steps :
(i) Drawn a line segment DC = 8 cm.
(ii) Drawn the perpendicular bisector XY to DC. Let it cut DC at ‘O’
(iii) With O as centres, drawn arc of radius 3 cm on either side of O which cut OX at U and OYat K.
(iv) Joined DK, KC, CU and DU.
(v) DUCK is the required rhombus.

Calculation of Area :
Area of the rhombus = \(\frac{1}{2}\) × d₁ × d₂ sq.units = \(\frac{1}{2}\) × 8 × 6 cm² = 24 cm²

(vii) PARK, PR = 9 cm and ∠P = 70°
Solution:
Given PR = 9 cm and ∠P = 70°

Steps :
(i) Drawn a line segment PR = 9 cm.
(ii) At P, made ∠RPX = ∠RPY = 35° on either side of PR.
(iii) At R, made ∠PRQ = ∠PRS = 35° on either side of PR
(iv) Let PX and RQ cut at A and PY and RS at K.
(v) PARK is the required rhombus

Calculation of Area :
Area of the rhombus = \(\frac{1}{2}\) × d₁ × d₂ sq.units = \(\frac{1}{2}\) × 9 × 6.2 cm² = 27.9 cm²

(viii) MARK, AK =7.5 cm and ∠A = 80°
Solution:
Given AK = 7.5 cm and ∠A = 80°

(i) Drawn a line segment AK = 7.5 cm.
(ii) At A, made ∠KAX = ∠KAY = 40° on either side of AK.
(iii) At K, made ∠AKP = ∠AKQ = 40° on either side of AK
(iv) Let AX and KP cut at M and AY and KQ at R.
(v) MARK is the required rhombus

Calculation of Area :
Area of the rhombus = \(\frac{1}{2}\) × d₁ × d₂ sq.units = \(\frac{1}{2}\) × 7.5 × 6.4 cm² = 24 cm²

Question 2.
(i) Construct the following rectangles with the given measurements and also find their area.
(i) HAND, HA = 7 cm and AN = 4 cm
Solution:
Given HA = 7 cm and AN = 4 cm

Steps :
(i) Drawn a line segment HA = 7 cm.
(ii) At H, constructed HX ⊥ HA.
(iii) With H as centre, drawn an arc of radius 4 cm and let it cut at HX at D.
(iv) With A and D as centres, drawn arcs of radii 4 cm and 7 cm respectively and let them cut at N.
(v) Joined AN and DN.
(vi) HAND is the required rectangle.

Calculation of Area :
Area of the rectangle HAND = l × b sq.units = 7 × 4 cm² = 28 cm²

(ii) SAND, SA = 5.6 cm and SN = 4.4 cm
Solution:
Given SA = 5.6 cm and SN = 4.4 cm

Steps :
(i) Drawn a line segment SA = 5.6 cm.
(ii) At S, constructed SX ⊥ SA.
(iii) With S as centre, drawn an arc of radius 4.4 cm and let it cut at SX at D.
(iv) With A and D as centres, drawn arcs of radii 4.4 cm and 5.6 cm respectively and let them cut at N.
(v) Joined DN and AN.
(vi) SAND is the required rectangle.

Calculation of Area :
Area of the rectangle SAND = l × b sq.units = 5.6 × 4.4 cm² = 26.64 cm²

(iii) LAND, LA = 8 cm and AD = 10 cm
Solution:
Given LA = 8 cm and AD = 10 cm

Steps :
(i) Drawn a line segment LA = 8 cm.
(ii) At L, constructed LX ⊥ LA.
(iii) With A as centre, drawn an arc of radius 10 cm and let it cut at LX at D.
(iv) With A as centre and LD as radius drawn an arc. Also with D as centre and LA as radius drawn another arc. Let then cut at N.
(v) Joined DN and AN.
(vi) LAND is the required rectangle.

Calculation of Area :
Area of the rectangle LAND = l × b sq.units = 8 × 5.8 cm² = 46.4 cm²

(iv) BAND, BA = 7.2 cm and BN = 9.7 cm
Solution:
Given = 7.2 cm and BN = 9.7 cm

Steps :
(i) Drawn a line segment BA = 7.2 cm.
(ii) At A, constructed NA ⊥ AB.
(iii) With B as centre, drawn an arc of radius 9.7 cm and let it cut at AX at N.
(iv) With B as centre and AN as radius drawn an arc. Also with N as centre and BA as radius drawn another arc. Let then cut at D.
(v) Joined ND and BD.
(vi) BAND is the required rectangle.

Calculation of Area :
Area of the rectangle BAND = l × b sq.units = 7.2 × 6.7 cm² = 48.24 cm²

Question 3.
Construct the following squares with the given measurements and also find their area.
(i) EAST, EA = 6.5 cm
Solution:
Given side = 6.5 cm

Steps :
(i) Drawn a line segment EA = 6.5 cm.
(ii) At E, constructed EX ⊥ EA.
(iii) With E as centre, drawn an arc of radius 6.5 cm and let it cut EX at T.
(iv) With A and T as centre drawn an arc of radius 6.5 cm each and let them cut at S.
(v) Joined TS and AS.
(vi) EAST is the required square.

Calculation of Area :
Area of the square EAST = a² sq.units = 6.5 × 6.5 cm² = 42.25 cm²

(ii) WEST, ST = 6 cm
Solution:
Given side of the square = 6 cm

Steps :
(i) Drawn a line segment ST = 6 cm.
(ii) At S, constructed SX ⊥ ST.
(iii) With S as centre, drawn an arc of radius 6 cm and let it cut SX at E.
(iv) With E and T as centre drawn an arc of radius 6 cm each and let them cut at W.
(v) Joined TW and EW.
(vi) WEST is the required square.

Calculation of Area :
Area of the square WEST = a² sq.units = 6 × 6 cm² = 36 cm²

(iii) BEST, BS = 7.5 cm
Solution:
Given diagonal = 7.5 cm

Steps :
(i) Drawn a line segment BS = 7.5 cm.
(ii) Drawn the perpendicular bisector XY to BS. Let it bisect BS at O.
(iii) With O as centre, drawn an arc of radius 3.7 cm on either side of O which cut OX at T and OY at E
(iv) Joined BE, ES, ST and BT.
(v) BEST is the required square.

Calculation of Area :
Area of the square BEST = a² sq.units = 5.3 × 5.3 cm² = 28.09 cm²

(iv) REST, ET = 8 cm
Solution:
Given diagonal = 8 cm

Steps:
(i) Drawn a line segment ET = 8 cm.
(ii) Drawn the perpendicular bisector XY to ET. Let it bisect ET at O.
(iii) With O as centre, drawn an arc of radius 4 cm on either side of O which cut OX at R and OY at S
(iv) Joined ES, ST, TR and ER.
(v) REST is the required square.

Calculation of Area :
Area of the square REST = a² sq.units = 5.7 × 5.7 cm² = 32.49 cm²

Students can Download Maths Chapter 3 Geometry Ex 3.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 3 Geometry Ex 3.2

Miscellaneous and Practice Problems

Question 1.
Identify the centroid of ΔPQR.

Solution:
In ΔPQR, PT = TR ⇒ QT is a median from vertex Q.
QS = SR ⇒ PS is a median from vertex P.
QT and PS meet at W and therefore W is the centroid of ΔPQR.

Question 2.
Name the orthocentre of ΔPQR.

Solution:
∠P = 90°
This is a right triangle
∴ orthocentre = P [∴ In right triangle orthocentre is the vertex containing 90°]

Question 3.
In the given figure, A is the midpoint of YZ and G is the centroid of the triangle XYZ. If the length of GA is 3 cm, find XA.

Solution:
Given A is the midpoint of YZ.
∴ ZA = AY
G is the centroid of ΔXYZ centroid divides each median in a ratio 2 : 1 ⇒ XG : GA = 2 : 1
\(\frac{XG}{GA}\) = \(\frac{2}{1}\)
\(\frac{XG}{3}\) = \(\frac{2}{1}\)
XG = 2 × 3
XG = 6 cm
XA = XG + GA
= 6 + 3 ⇒ XA = 9 cm

Challenging Problems

Question 4.
Find the length of an altitude on the hypotenuse of a right angled triangle of legs of length 15 feet and 20 feet.
Solution:

Since ∠B = 90°
Using pythagoras theorem
AC² = AB² + BC² = 20² + 15²
= 400 + 225
AC² = 62⁵
AC² = 25⁵
AC = 25
Area of a triangle ΔABC = \(\frac{1}{2}\) × 15 × 20 = 150 feet²
Again Area of ΔABC = \(\frac{1}{2}\) × AC × BD
150 = \(\frac{1}{2}\) × 25 × BD
BD = \(\frac{2 × 150}{25}\) = \(\frac{300}{25}\)
BD = 12 feet
∴ Length of the altitude on the hypotenuse of the right angled triangle is 12 feet.

Question 5.
If I is the incentre of ΔXYZ, ∠IYZ = 30° and ∠IZY = 40°, find ∠YXZ.

Solution:
Since I is the incentre of ΔXYZ
∠IYZ = 30° ⇒ ∠IYX = 30°
∠IZY = 40° ⇒ ∠IZX = 40°
∴ ∠XYZ = ∠XYI + ∠IYZ = 30° + 30°
∠XYZ = 60°
lll ly ∠XYZ = ∠XZI + ∠IZY = 40° + 40°
∠XYZ = 80°
By angle sum property of a triangle
∠XZY + ∠XYZ + ∠YXZ = 180°
80° + 60° + ∠YXZ = 180°
140° + ∠YXZ = 180°
∠YXZ = 180° – 140°
∠YXZ = 40°

Question 6.
In ΔDEF, DN, EO, FM are medians and point P is the centroid. Find the following.
(i) If DE = 44, then DM = ?
(ii) If PD = 12, then PN = ?
(iii) If DO = 8, then FD = ?
(iv) If OE = 36, then EP = ?

Solution:
Given DN, EO, FM are medians.
∴ FN = EN
DO = FO
EM = DM
(i) If DE = 44, then
DM = \(\frac{44}{2}\) = 22
DM = 22

(ii) If PD = 12, PN = ?
\(\frac{PD}{PN}\) = \(\frac{2}{1}\)
\(\frac{12 }{PN}\) = \(\frac{2}{1}\) ⇒ PN = \(\frac{12}{2}\) = 6
PN = 6

(iii) If DO = 8, then
FD = DO + OF = 8 + 8
FD = 16

(iv) If OE = 36,
then \(\frac{EP}{PO}\) = \(\frac{2}{1}\)
\(\frac{EP}{2}\) = PO
OE = OP + PE
36 = \(\frac{PE}{2}\) + PE
36 = \(\frac{PE}{2}\) + \(\frac{2PE}{2}\)
36 = \(\frac{3PE}{2}\)
PE = \(\frac{36 × 2}{3}\)
PE = 24

Students can Download Maths Chapter 3 Geometry Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 3 Geometry Intext Questions

Exercise 3.1
Think (Text book Page no. 53)

Question 1.
In any acute angled triangle, all three altitudes are inside the triangle. Where will be the orthocentre? In the interior of the triangle or in its exterior?

Solution:
Interior of the triangle.

Question 2.
In any right angled triangle, the altitude perpendicular to the hypotenuse is inside the triangle; the other two altitudes are the legs of the triangle. Can you identify the orthocentre in this case?

Solution:
Vertex containing 90°

Question 3.
In any obtuse angled triangle, the altitude connected to the obtuse vertex is inside the triangle, and the two altitudes connected to the acute vertices are outside the triangle. Can you identify the orthocentre in this case?

Solution:
Exterior of the triangle.

Try These (Text book Page no. 56)

Identify the type of segment required in each triangle:
(median, altitude, perpendicular bisector, angle bisector)

(i) AD = ……….
(ii) l1 = ………..
(iii) BD = …………
(iv) CD = …………
Solution:
(i) AD = Altitude
(ii) l₁ = Perpendicular bisector
(iii) BD = Median
(iv) CD = Angular bisector

Exercise 3.3
Activity 1. (Text book Page no. 60)

Question 1.
A pair of identical 30°-60°-90° set-squares are needed for this activity. Place them as shown in the figure.

1. What is the shape we get? It is a parallelogram.
2. Are the opposite sides parallel?
3. Are the opposite sides equal?
4. Are the diagonals equal?
5. Can you get this shape by using any other pair of identical set-squares?

Solution:

1. It is a parallelogram.
2. Yes
3. Yes
4. no
5. yes

Question 2.
We need a pair of 30°-60°-90° set- squares for this activity. Place them as shown in the figure.

(i) What is the shape we get?
(ii) Is it a parallelogram?
It is a quadrilateral; infact it is a rectangle. (How?)
(iii) What can We say about its lengths of sides, angles and diagonals? Discuss and list them out.
Solution:
(i) Rectangle
(ii) Yes, Opposite sides are equal. All angles = 90°
(iii) Opposite sides are equal.
All angles are equal and are = 90°.
Diagonals are equal

Question 3.
Repeat the above activity, this time with a pair of 45°-450-90° set-squares.

(i) How does the figure change now? Is it a parallelogram? It becomes a square! (How did it happen?)
(ii) What can we say about its lengths of sides, angles and diagonals? Discuss and list them out.
(iii) How does it differ from the list we prepared for the rectangle?
Solution:
(i) All sides are equal
(ii) All sides are equal
All angles = 90°
Diagonals equal
(iii) All sides are equal.
Diagonals bisects each other.

Question 4.
We again use four identical 30°-60°-90° set- squares for this activity.
Note carefully how they are placed touching one another.

(i) Do we get a parallelogram now?
(ii) What can we say about its lengths of sides, angles and diagonals?
(iii) What is special about their diagonals?
Solution:
(i) Yes
(ii) All sides equal.
(iii) Diagonals bisects perpendicularly.

Try These (Text book Page no. 62)

Question 1.
Say True or False:
(a) A square is a special rectangle.
(b) A square is a parallelogram.
(c) A square is a special rhombus.
(d) A rectangle is a parallelogram
Solution:
(a) True
(b) True
(c) True
(d) True

Question 2.
Name the quadrilaterals
(a) Which have diagonals bisecting each other.
(b) In which the diagonals are perpendicular bisectors of each other.
(c) Which have diagonals of different lengths.
(d) Which have equal diagonals.
(e) Which have parallel opposite sides.
(f) In which opposite angles are equal.
Solution:
(a) Square, rectangle, parallelogram, rhombus.
(b) Rhombus and square.
(c) Parallelogram and Rhombus
(d) Rectangle, square.
(e) Square, Rectangle, Rhombus, parallelogram.
(f) Square, rectangle, rhombus, parallelogram

Question 3.
Two sticks are placed on a ruled sheet as shown. What figure is formed if the four corners of the sticks are joined?
(a)
Two unequal sticks. Placed such that their midpoints coincide.
Solution:
Parallelogram

(b)
Two equal sticks. Placed such that their midpoints coincide.
Solution:
Rectangle

(c)
Two unequal sticks. Placed intersecting at mid points perpendicularly.
Solution:
Rhombus

(d)
Two equal sticks. Placed intersecting at mid points perpendicularly.
Solution:
Square

(e)
Two unequal sticks. Tops are not on the same ruling. Bottoms on the same ruling. Not cutting at the mid point of either.
Solution:
Quadrilateral

(f)
Two unequal sticks. Tops on the same ruling. Bottoms on the same ruling. Not necessarily cutting at the mid point of either.
Solution:
Trapezium