Laxmi

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 5 Information Processing Additional Questions

Additional Questions and Answers

Exercise 5.1

Question 1.
Find the relationship between x and y if

Solution:
The relationship between x and y is y = -5x

Question 2.
Find the relationship between x and y if

Solution:
The relationship between x andy is y = 3x + 2

Exercise 5.2

Question 1.
Find the sum of the elements of Pascal’s triangle and find the relationship between the numbers obtained.
Solution:
Pascal’s triangle with sum of elements is given by

The numbers formed are 1, 2, 4, 8, 16, 32, 64, …………..
They can be written as 2⁰, 2¹, 2², 2³, 2⁴, 2⁵, 2⁶, …………..
So they are the powers of base 2
The relationship is given by 2^x-1 if x represents the row.

Exercise 5.3

Question 1.
If the following numbers are taken from Pascal’s triangle find the missing numbers.

1. 9, 1, _____ = 45, 1, 11
2. 1, 6, _____ = 1, 4, 15
3. 21, 8, 1 = _____, 6, 28

Solutions:

1. 55
2. 10
3. 1

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Tamilnadu Samacheer Kalvi 9th English Solutions Poem Chapter 7 The Stick-together families

The Stick-together families Warm Up:

At the heart of life lie the relationships you have with other people: with family, classmates and friends close-by and far away. All relationships are based on some commonly accepted values like respect, honesty, consideration and commitment. Think about all the important relationships in your life and complete the table given below.

The Stick-together families Textual Questions

A. Based on your understanding of the poem, answer the questions in a sentence or two.

1. “The gladdest people living are the wholesome folks who make A circle at the fireside that no power but death can break. ”

Question (a).
Who are the gladdest people living?
Answer:
The gladdest people are those who live together as wholesome folks.

Question (b).
Where do they gather?
Answer:
They gather at the fireside.

Question (c).
What can break their unity?
Answer:
Only death can break their unity.

2. “And the finest of conventions ever held beneath the sun Are the little family gatherings when the busy day is done. ”

Question (a).
When do they have their family gatherings?
Answer:
They have their family gatherings at the end of the busy day.

Question (b).
Where do they have their family conventions?
Answer:
The family conventions are held beneath the sun.

Question (c).
What does the poet mean by ‘finest conventions’?
Answer:
The finest conventions means a family get together.

3. “There are rich folk, there are poor folk, who imagine they are wise,
And they’re very quick to shatter all the little family ties. ”

Question (a).
What do the rich and poor folk imagine themselves to be?
Answer:
The rich and the poor folks imagine themselves to be wise.

Question (b).
What do they do to their families?
Answer:
They are quick to shatter their little family ties.

Question (c).
Whom does ‘they’ refer to?
Answer:
They refers to the rich and poor people who do not consider the little family ties as valuable and shatter them.

4. “There are some who seem to fancy that for gladness they must roam,
That for smiles that are the brightest they must wander far from home”

Question (a).
Why do they roam?
Answer:
They roam to attain gladness.

Question (b).
According to them, when do they get bright smiles?
Answer:
They get bright smiles when they wander far from home.

5. “But the gladdest sort of people, when the busy day is done,
Are the brothers and the sisters who together share their fun.”

Question (a).
Who are the gladdest people?
Answer:
The gladdest people are those brothers and sisters who share their fun.

Question (b).
When do they share their fun?
Answer:
They share their fun when the busy day is done.

Question (c).
What does ‘who’ refer to?
Answer:
Who refers to the brothers and sisters.

6. “It’s the stick-together family that wins the joys of earth,

That hears the sweetest music and that finds the finest mirth;”

Question (a).
Who wins the joys of the earth?
Answer:
The stick-together families wins the joys of earth.

Question (b).
How do they find their joy?
Answer:
They find their joys by being together and hearing the sweetest music

Question (c).
What does the poet mean by ‘stick-together family’?
Answer:
The stick-together family means those families who spend time together (Joint families) and share their fun and sorrows and can be only separated after death.

Additional Questions

1. “ The stick-together families are happier by far
Than the brothers and the sisters who take separate highways are.
The gladdest people living are the wholesome folks who make
A circle at the fireside that no power but death can break.
And the finest of conventions ever held beneath the sun
Are the little family gatherings when the busyday is done. ”

Question (a).
Give the rhyme scheme of the above lines.
Answer:
‘aabbcc’ is the rhyme scheme of the above lines.

Question (b).
Pick out the rhyming words.
Answer:
‘far and are’, ‘make and break’ and ‘sun and done’ are the rhyming words.

Question (c).
What does the poet mean by ‘day is done’?
Answer:
The poet means that the day has come to an end.

Question (d).
What can break the circle at the fireside?
Answer:
Death can break a circle at the fireside.

Question (e).
What do the stick-together families do when the busy day is done?
Answer:
The stick-together families have little family gatherings when the busy day is done.

Question (f).
Mention the figure of speech in the last line.
Answer:
The figure of speech in the last line is alliteration, (day is done)

2. “There are rich folk, there are poor folk, who imagine they are wise,
And they’re very quick to shatter all the little family ties.”

Question (a).
What does the poet mean by the term, ‘family ties’?
Answer:
The poet by the term ‘family ties’ means the family bonds.

Question (b).
Pick out the rhyming words.
Answer:
The rhyming words are wise and ties.

Question (c).
Who shatters the family ties?
Answer:
The rich and the poor folk who think they are intelligent are quick to shatter the family ties.

3. “Each goes searching after pleasure in his own selected way,
Each with strangers likes to wander, and with strangers likes to play.”

Question (a).
Identify the figure of speech in the first line.
Answer:
Alliteration is the figure of speech in the first line.

Question (b).
Write the alliterated words in the first line.
Answer:
The alliterated words are searching and selected in the first line.

4. “There are some who seem to fancy that for gladness they must roam,
That for smiles that are the brightest they must wander far from home.
That the strange friend is the true friend, and they travel far astray
They waste their lives in striving for a joy that’s far away,
But the gladdest sort of people, when the busy day is done,
Are the brothers and the sisters who together share their fun. ”

Question (a).
Mention the rhyme scheme?
Answer:
‘aabbcc’ is the rhyme scheme.

Question (b).
Give the rhyming words for away and home.
Answer:
The rhyming words for away and home are astray and roam respectively.

Question (c).
What is meant by ‘astray’?
Answer:
Being lost in this world due to bad ways is the meaning of astray.

Question (d).
How do they waste their lives?
Answer:
They waste their lives in striving for a joy that cannot be easily got.

5. “And, O weary, wandering brother, if contentment you would win,
Come you back unto the fireside and be comrade with your kin.”

Question (a).
What are the alliterated words in the first line.
Answer:
The alliterated words are weary, wandering.

Question (b).
Who is a comrade?
Answer:
A comrade is a friend.

Question (c).
What should the weary wandering brother do?
Answer:
The weary wandering brother should come back to the fireside and be friends with the kith and kin.

Question (d).
What do you mean by ‘kin’?
Answer:
Kin means relatives.

B. Based on the understanding of the poem, fill in the blanks using the words and phrases given below to make a meaningful summary of the poem.

The poet brings out the difference in the attitudes of children living in joint family and nuclear family. The (1) …………………., are the happiest of all. Where as the (2) …………………… of nuclear families take (3) ………………………. The gladdest people are the children from (4) …………………….. who circle near the fireside. No power other than death can break them. The (5) …………………. imagine themselves to be wise and in the process they (6) ……………………. ties. Each of them goes searching for pleasure in their own selected way. They harvest only (7) …………. and find empty joy. But the wisest among them are the children of the stick-together families. When the busy day is done, they together (8) ……………… The stick-together family wins (9) ………………. The old house shelters all the (10) …………………… The poet invites wandering brothers to come and join the stick-together families in their fireside and have fun.
Answers

1. stick-together families
2. brothers and sisters
3. separate ways
4. joint family
5. rich and the poor folk
6. shatter their family
7. bitterness
8. share their fun
9. the joy of earth
10. charm of life

C. Answer the following questions in about 80-100 words.

Question 1.
The stick-together families are the happiest of all. Explain.
Answer:
The stick-together families are the happiest of all. There is so much fun living together in joint families. There is a lot of excitement in joint families. The gladdest people living are the good folks who create a circle for themselves. They share their joys and sorrows. The finest of all large formal gatherings is the little family gatherings. The wisest children are in the stick-together families. Some travel far away to seek joy but the happiest kind of people, are the ones who share their fun in stick-together families. It is sure that the stick-together family wins the joys of the earth.

Question 2.
Bring out the difference between the children of the joint family and nuclear family.
Answer:
The joint families are definitely happier than the members in the nuclear family who take separate ways. The joint family meets at the end of the day while the nuclear family meets occasionally. Members of the nuclear family go searching after pleasure in their own selected way but the joint family loves to travel together. People in nuclear families waste their lives in searching for a joy that’s far away. But the happiest kind of people in joint families share their fun together. The children in the stick-together families wins the joys of the earth whereas children in joint families travel far away seeking pleasure in vain.

Additional Questions:

Question 1.
What do those who think wise of themselves do?
Answer:
The rich and poor folk think wise of themselves and quickly destroy all the little family bonds. They do not like to mingle with one another and prefer to travel and seek pleasure. Every person goes searching after pleasure in his/her own selected way with strangers who likes to wander and play. On the contrary, they reap unpleasantness and find an empty joy. However, the children who are the wisest are surely those who live in the stick-together kind in the joint families and not those who are in nuclear families who think they are wise

Question 2.
Why do some people roam and what happens as they journey along?
Answer:
There are some who seem to like the thought of roaming around to seek happiness. They strongly feel that the brightest smile can be seen only when they travel far away from home. Such people also think that a strange friend is the best friend who will be true. However, when they travel far away and get lost in their path, they do waste their lives in searching for a joy that’s far away. Therefore, they really lack in happiness and lose the joy searching for gladness.

Appreciate The Poem

D. Answer the following

Question 1.
There are rich folk, there are poor folk, who imagine they are wise,…
Pick out the words in alliteration.
Answer:
‘There’ and ‘they’ are the words that alliterate.

Question 2.
Mention the rhyme scheme of the poem.
Answer:
The rhyme scheme of the poem is ‘aabbcc’ for all the stanzas.

Listening Activity

E. Listen to the passage and fill in the blanks with appropriate answer.

(For listening to the passage refer to our website www.fullcircleeducation.in) Family is where we all belong to and from where our identity comes from. A person is valued based on his family and upbringing. Family is a bond, a long lasting relationship that holds a bond with each other. There are many values that one has to learn to get the family bonding in the right manner. Bonding does not happen overnight. It forms with every second, every minute that you spend with your loved ones. The understanding, the acceptance, the belonging and the security all enclosed together is how a family bond is formed.

A close family bond is like a safe harbour, where we feel secure and where we trust that we have someone always there to whom we could turn to when we need them the most. It is through a family that we learn the values of love, trust, hope, belief, cultures, morals, traditions and every little matter that concerns to us. A strong foundation for any individual comes from being with a supportive family.

Family is one among the greatest gifts that we get from God. To have parents, who support us, teach us values in life, and gives us a strong foundation in character, teach us the importance of love and being loved, trust to be there for one another and many other morals that could be obtained only from a family. A gift not only with lovable parents, but siblings who care and love us beyond themselves. We cannot buy or demand all these things in life, as we are being given to understand their importance.

To be part of a happy family, one should always thank God for the blessing we have in lives, as having a family who cares and loves us is the greatest blessing that any person could get in life.

1. A person is valued based on his …………………. .
2. ………………….. does not happen overnight.
3. A close family bond is like a ……………………… .
4. A strong foundation for any individual comes from being with a …………………… .
5. A gift not only with …………….. but …………………….. who care and love us beyond themselves.

Answers

1. family and upbringing
2. Bonding
3. safe harbour
4. supportive family
5. lovable parents, siblings

Speaking:

F. “The building actually rests on the well laid out foundation and hence is strong and still.” How can this be related to a family? Discuss with your partner and share your views in the class.

The building actually rests on the well laid out foundation. Therefore it is strong and still. If the foundation is strong the building is strong. Likewise, if we lay a strong foundation for the children, then they will grow strong and still. For such a foundation to be strong, as parents and teachers, we must nurture them with virtues like prayer, faith, courage and other values.

You need a really solid foundation of friends and family to keep you where you need ‘ to be. If the mason is able and his materials are good, the house built will be strong and still. Just like the firm foundation with good material, the parents and teachers too should build a child on a strong foundation with good values like discipline, responsible citizenship, empathy, sympathy, sacrifice and so on.

It is surely the family which provides a child with roots, much- needed structure, and unconditional love. Families also provide their children with a happy home – a place where a child is always safe and welcome.

Writing:

G. Write a four-line poem with rhyming words describing your family.

I live in a lovely family off our Ladies two with Gentlemen two Missing days with Grandpa too Hope to hear the same from you.

The Stick-together families About The Poet:

Edgar Albert Guest born on 20th August 1881, was an English-born American poet, popular in the first half of the 20th century, commonly known as the People’s Poet due to his inspirational and optimistic view of everyday life. Elis first poem appeared on 11th December 1898. For 40 years, Guest was widely read throughout North America. Guest has penned around 11,000 poems that has appeared in 300 newspapers and collected in more than 20 books. Even today it occasionally appears in periodicals such as Reader’s Digest.

The Stick-together families Summary:

The poet envisions the joint families who are much happier than the brothers and sisters who take their separate ways. The gladdest people are those who circle around a fireside that no power but death can break. The finest conventions beneath the sun is the family gatherings after a busy day. Rich or poor folks, who think they are wise, break their family ties and wander away searching for joy in strangers and material things. But in fact, the stick together families alone enjoy the true joys of the earth, enjoy the sweetest music and finds mirth. An old home shelters the charm that life can give, and a happiest spot to live. Hence the poet asks people who go their separate ways to come together and comrade with their kin.

The Stick-together families Glossary:
Textual:
astray – away from the correct path or correct way of doing something
comrade – a friend
conventions – a large formal meeting of people who have a similar interest
mirth – laughter, humour or happiness
shatter – to break suddenly into very small pieces

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 5 Information Processing Intext Questions

Activity (Text book Page No. 91)

Question 1.
Observe the pattern given below. Continue the pattern for three more steps.

Let, ‘x’ be the number of steps and ‘y’ be the number of match sticks. Tabulate the values of ‘x’ and ‘y’ and verify the relationship y = 7x + 5.
Solution:
Three more patterns are

From the table y = 7x + 5 is verified.

Try These (Text book Page No. 92)

Question 1.
In the given figure, let Y denote the number of steps and y denote its area. Find the relationship between x and y by tabulation.

Solution:
Let x denote the number of steps and y denote the area.
In the first shape let x = 1 and the area be 1 cm²
when x = 2 : Area = 22 = 4 cm²
whenx = 3 : Area = 32 = 9 cm² and so on.
Tabulating the values of x and y

From the table:
x = 1 ⇒ y = 1²
x = 2 ⇒ y = 2²
x = 4 ⇒ y = 4²
x = 5 ⇒ y = 5²
Hence the relationship between x and y is y = x²

Question 2.
In the figure, let x denotes the number of steps and y denotes the number of matchsticks used. Find the relationship between Y and y by tabulation.

Solution:
Let x denote the number of steps and y denote the number of matchsticks used.
In step 1, x = 1 ⇒ y = number of mathsticks used is 1
In step 2, x = 2 ⇒ y = number of mathsticks used is 4
In step 3, x = 3 ⇒ y = number of mathsticks used is 7 and so on.
The values of v andy are tabulated as

x = 1 ⇒ y = 1 = 3(1) – 2
x = 2 ⇒ y = 4 = 3(2) – 2
x = 3 ⇒ y = 7 = 3(3) – 2
x = 4 ⇒ y = 10 = 3(4) – 2
From the table y = 3x – 2

Question 3.
Observe the table given below.

Find the relationship between x and y. What will be the value of y, when x = 8.
Solution:
Soil When x = – 2 y = 2 (-2) = -4
When x = -1 y = 2 (-1) = -2
When x = 0 y = 2 (0) = 0
When x =1 y = 2(1) = 2
When x = 2 y = 2 (2) = 4
When x = 8 y = 2 (8) = 16.
Also y = 2x is the relation between x and y.

Activity (Text book Page No. 93 & 94)

Question 1.
Complete the following Pascal’s Triangle by observing the number pattern.

Solution:
Pascal’s triangle is given by

Question 2.
Observe the above completed and r find the sequence that you see in it and complete them.

(i) 1,2, 3, 4, 5, 6, 7.
(ii) 1, 3, _____, _____, _____, _____.
(iii) 1, _____, _____, _____, _____,
(iv) _____, _____, _____, _____.
Solution:
(i) 1,2, 3, 4, 5, 6, 7.
(ii) 1,3,6,10,15,21.
(iii) 1,4,10,20,35.
(iv) 1,5,15,35.

Question 3.
Observe the sequence of numbers obtained in the 3rd and 4th slanting rows of Pascal’s Triangle and find the difference between the consecutive numbers and complete the table given below.

Solution:

Try These (Text book Page No. 96)

Question 1.
Observe the pattern of numbers given in the slanting rows earlier and complete the Pascal’s Triangle.

Solution:

Question 2.
Complete the given Pascal’s Triangle. Find the common property of the numbers filled by you. Can you relate this pattern with the pattern discussed in situation 2. Discuss.

Solution:

Common Properties:
The numbers filled by me are even numbers. Also they make triangular shape.
Yes, this pattern and the pattern given in situation 2 have the same properties.

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 5 Information Processing Ex 5.3

Miscellaneous Practice Problems

Question 1.
Choose the correct relationship between x and y for the given table.

(i) y = x + 4
(ii) y = x + 5
(iii) y = x + 6
(iv) y = x + 7
Answer:
(iii) y = x + 6

Question 2.
Find the triangular numbers from the Pascal’s Triangle and colour them.

Solution:
Triangular numbers are numbers the objects of which can be arranged in the form of equilateral triangle.
Example : 1, 3, 6, 10, 15,…
From Pascal’s Triangle, the triangular numbers are

Question 3.
Write the first five numbers in the third slanting row of the Pascal’s Triangle and find their squares. What do you infer?
Solution:

Numbers in the 3rd slanding row are 1, 3, 6, 10, 15, 21,….
The squares are 1², 3², 6², 10². 15², 21²,…. = 1, 9, 36, 100, 225, 441,…

From the above table we can conclude that the squares of the triangular numbers are the sum of cubes of natural numbers.

Challenge Problems

Question 4.
Tabulate and find the relationship between the variables (x and y) for the following patterns.

Solution:
(i) Let the number of steps be x and the number of shapes be y.
Tabulating the values of x and y

From the table
x = 1 ⇒ y = 1 = 12
x = 2 ⇒ y = 4 = 22
x = 3 ⇒ y = 9 = 32
x = 4 ⇒ y = 16 = 42
Hence the relationship between x and y is y = x2.

(ii) Let the number of steps be x and the number of shapes be y.
Tabulating the values of x and y

From the table x = 1 ⇒ y = 1 = 1
x = 2 ⇒ y = 2 + 1 = 3
x = 3 ⇒ y = 3 + 2 = 5
x = 4 ⇒ y = 4 + 3 = 7
x = 5 ⇒ y = 5 + 4 = 9
Hence the relationship between x and y is y = 2x-1.

Question 5.
Verify whether the following hexogonal shapes form a part of the Pascal’s Triangle.

Solution:
In Pascal’s Triangle product of the 3 alternate numbers given around the hexagon is equal to the product of remaining three numbers.

1 × 13 × 66 = 11 × 1 × 78 = 858
∴ It form a part of Pascal’s Triangle.

5 × 21 × 20 = 10 × 6 × 35 = 2100
∴ It form a part of Pascal’s Triangle

8 × 45 × 84 = 28 × 9 × 120 = 30240
∴ It form a part of Pascal’s Triangle

56 × 210 × 126 = 70 × 84 × 252 = 1481760
∴ It form a part of Pascal’s Triangle

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 5 Information Processing Ex 5.2

Question 1.
Complete the Pascal’s Triangle.

Solution:

Question 2.
The following hexagonal shapes are taken from Pascal’s Triangle. Fill in the missing numbers.

Solution:

Question 3.
Complete the Pascal’s Triangle by taking the numbers 1,2,6,20 as line of symmetry.

Solution:
Corresponding numbers are equal about the line of symmetry.

Objective Type Questions

Question 1.
The elements along the sixth row of the Pascal’s Triangle is
(i) 1,5,10,5,1
(ii) 1,5,5,1
(iii) 1,5,5,10,5,5,1
(iv) 1,5,10,10,5,1
Answer:
(iv) 1,5,10,10,5,1

Question 2.
The difference between the consecutive terms of the fifth slanting row containing four elements of a Pascal’s Triangle is
(i) 3,6,10,…
(ii) 4,10,20,…
(iii) 1,4,10,…
(iv) 1,3,6,…
Answer:
(ii) 4,10,20,…

Question 3.
What is the sum of the elements of ninth row in the Pascal’s Triangle?
(i) 128
(ii) 254
(iii) 256
(iv) 126
Answer:
(iii) 256

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 5 Information Processing Ex 5.1

Question 1.
Match the given patterns of shapes with the appropriate number pattern and its generalization.


Solution:
(i) (d)
(ii) (a)
(iii) (c)
(iv) (c)
(v) (b)

Objective Type Questions

Question 2.
Identify the correct relationship between x andy from the given table.

(i) y = 4x
(ii) y = x + 4
(iii) y = 4
(iv) y = 4 × 4
Answer:
(i) y = 4x

Question 3.
Identify the correct relationship between x and y from the given table.

(i) y = -2x
(ii) y = +2x
(iii) y = +3x
(iv) y = -3x
Answer:
(iv) y = -3x

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 4 Geometry Additional Questions

Additional Questions and Answers

Exercise 4.1

Question 1.
“The sum of any two angles of a triangle is always greater than the third angle”. Is this statement true. Justify your answer.
Solution:
No, the sum of any two angles of a triangle is not always greater than the third angle. In an isosceles right angled triangles, the angle will be 90°, 45°, 45°.
Here sum of two angles 45° + 45° = 90°.

Question 2.
The three angles of a triangle are in the ratio 1:2:1. Find all the angles of the triangle. Classify the triangle in two different ways.
Solution:
Let the angles of the triangle be x, 2x, x.
Using the angle sum property, we have
x + 2x + x = 180°
4x = 180°
x = \(\frac{180^{\circ}}{4}\)
x = 45°
2x = 2 × 45° = 90°
Thus the three angles of the triangle are 45°, 90°, 45°.
Its two angles are equal. It is an isoscales triangle. Its one angle is 90°.
∴ It is a right angled triangle.

Question 3.
Find the values of the unknown x and y in the following figures

Solution:
(i) Since angles y and 120° form a linear pair.
y + 120° = 180°
y = 180° – 120°
y = 60°
Now using the angle sum property of a triangle, we have
x + y + 50° = 180°
x + 60° + 50° = 180°
x + 110° = 180°
x = 180° – 110°= 70°
x = 70°
y = 60

(ii) Using the angle sum property of triangle, we have
50° + 60° + y = 180°
110° + y = 180°
y = 180° – 110°
y = 70°
Again x and y form a linear pair
∴ x + y = 180°
x + 70° = 180°
x = 180° – 70°= 110°
∴ x = 110°; y = 70°

Question 4.
Two angles of a triangle are 30° and 80°. Find the third angle.
Solution:
Let the third angle be x.
Using the angle sum property of a triangle we have,
30° + 80° + x = 180°
x + 110° = 180°
x = 180° – 110° = 70°
Third angle = 70°.

Exercise 4.2

Question 1.
In an isoscleles ∆ABC, AB = AC. Show that angles opposite to the equal sides are equal.
Solution:

Given: ∆ABC in which \(\overline{A B}\) = \(\overline{A C}\).
To Prove: ∠B = ∠K.
Construction: Draw AD ⊥ BC.
Proof: In right ∆ADB and right ∆ADC.
we have side AD = side AD (common)
AB = AC (Hypoteneous) (given)
∆ADB = ADC (RHS criterion]
∴ Their corresponding parts are equal. ∠B = ∠C.

Question 2.
ABC is an isosceles triangle having side \(\overline{A B}\) = side \(\overline{A C}\). If AD is perpendicular to BC, prove that D is the mid-point of \(\overline{B C}\).
Solution:
In ∆ABD and ∆ACD, we have
∠ADB = ∠ADC [∵ AD ⊥ BC]
Side \(\overline{A D}\) = Side \(\overline{A D}\) [Common]
Side \(\overline{A B}\) = Side \(\overline{A C}\) [Common]
Using RHS congruency, we get
∆ABD ≅ ∆ACDc
Their corresponding parts are equal

∴ BD = CD
∴ O is the mid point of BC.

Question 3.
In the figure PL ⊥ OB and PM ⊥ OA such that PL = PM.
Prove that ∆PLO ≅ ∆PMO.
Solution:

In ∆PLO and ∆PMO, we have
∠PLO = ∠PMO = 90° [Given]
\(\overline{O P}\) = \(\overline{O P}\) [Hypotenuse]
PL = PM
Using RHS congruency, we get
∆PLO ≅ ∆PMO

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 2 Chapter 3 Geometry Intext Questions

Exercise 3.1
Try This (Text book Page No. 60)

Question 1.
How will you prove ∆ ABC ~ AD AC?
Proof:
In ∆ ABC & ∆ DAC, ∠C is common and ∠BAC = ∠ADC = 90°
Therefore ∆ ABC ~ AD AC (AA similarity)

Try This (Text book Page No. 61)

Question 1.
Check whether the following are Pythagorean triplets

1. 57, 176, 185
2. 264, 265, 23
3. 8, 41, 40

Solution:
1. For Pythagorean triplet, the sum of the squares of 2 sides is equivalent to square of 3rd side (hypotenuse)
Let us check of
57² + 176² whether = 185² or not
57² = 3249
176² = 30976
185² = 34225
57² + 176² = 3249 + 30976 = 34225 = 185²
∴ 57, 176 & 185 are Pythagorean triplet.

2. 23, 264, 265
23² = 529
264² = 69696
23² + 264² = 70225
265² = 70225
∴ 23² + 264² = 265²
∴ Pythagorean triplet

3. 8, 41, 40
8² = 64
40² = 1600
8² + 40² = 1664
41² = 1681
8² + 40² ≠ 41² = 1
∴ They are not Pythagorean triplet

Activity – 1 (Text book Page No. 62)

We can construct sets of Pythagorean triplets as follows. Let m and n be any two positive integers (m > n):
(a, b, c) is a Pythagorean triple if a = m² – n², b = 2mn and c = m² + n² (Think, why?) Complete the table.

Solution:

Activity – 2 (Text book Page No. 65)

Question 1.
Find all integer-sided right angled triangles with hypotenuse 85
Solution:
(x + y)² – 2xy = 85²

Pythagorean triplets with hypotenuse 85.

Exercise 3.3
Try These (Text book Page No. 68)

Question 1.
The area of the trepezium is ……..
Answer:
\(\frac{1}{2}\) x h x (a + b) sq. units!

Question 2.
The distance between the parallel sides of a trapezium is called as ………
Answer:
its height

Question 3.
If the height and parallel sides of a trapezium are 5 cm, 7 cm and 5 cm respectively, then its area is ……..
Answer:
30 sq cm
Hint:
= \(\frac{1}{2}\) x h x (a + b) sq. units
= \(\frac{1}{2}\) x 5 x (7 + 5)
= \(\frac{1}{2}\) x 5 x 12 = 30 sq.cm

Question 4.
In an isosceles trapezium, the non-parallel sides are ……….. in length.
Answer:
Equal.

Question 5.
To construct a trapezium, ………… measurements are enough.
Answer:
Four.

Question 6.
If the area and sum of the parallel sides are 60 cm2 and 12 cm, its height is ………..
Answer:
10 cm
Hint:
Area of the trapezium = \(\frac{1}{2}\) x h (a + b)
60 = \(\frac{1}{2}\) x h x (12)
h = \(\frac{60×2}{12}\) = 10 cm

Exercise 3.4
Think (Text book Page No. 74)

Question 1.
Can a rhombus, a square or a rectangle be called as a parallelogram? Justify your answer.
Solution:
Yes, a rhombus, a square or a rectangle can be called as parallelogram as the opposite sides are equal and parallel and diagonals bisect each other in this figures.

Try These (Text book Page No. 74)

Question 1.
In a parallelogram, the opposite sides are …….. and ……….
Answer:
equal, parallel

Question 2.
If ∠A of a parallelogram ABCD is 100° then, find ∠B, ∠C and ∠D .
Answer:

∠B = 80°
∠C = 100°
∠D = 80°

Question 3.
Diagonals of a parallelogram each other.
Answer:
Bisect

Question 4.
If the base and height of the parallelogram are 20 cm and 5 cm then, its area is ………
Answer:
100 sq.cm

Question 5.
Find the unknown values in the given parallelograms and write the property Used to find them.

Solution:

Students can Download Maths Chapter 4 Geometry Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 7th 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 7th Maths Solutions Term 2 Chapter 4 Geometry Intext Questions

Exercise 4.1

Try These (Text book Page No. 66)

Answer the following questions.

Question 1.
Triangle is formed by joining three ______ points.
Answer::
Non collinear

Question 2.
A triangle has ______ vertices and ______ sides.
Answer:
three, three

Question 3.
A point where two sides of a triangle meet is known as ______ of a triangle.
Answer:
vertese

Question 4.
Each angle of an equilateral triangle is of measure.
Answer:
same

Question 5.
A triangle has angle measurements of 29°, 65° and 86°. Then it is ______ triangle.
(i) an acute angled
(ii) a right angled
(iii) an obtuse angled
(iv) a scalene
Answer:
(i) an acute angled

Question 6.
A triangle has angle measurements of 30°, 30° and 120°. Then it is ______ triangle.
(i) an acute angled
(ii) scalene
(iii) obtuse angled
(iv) right angled
Answer:
obtuse angled

Question 7.
Which of the following can be the sides of a triangle?
(i) 5.9.14
(ii) 7,7,15
(iii) 1,2,4
(iv) 3, 6, 8
Answer:
(iv) 3, 6, 8
Solution:
(i) Here 5 + 9 = 14 = the measure of the third side.
In a triangle the sum of the measures of any two sides must be greater than the third side.
∴ 5, 9, 14 cannot be the sides of a triangle.

(ii) 7.7.15
Here sum of two sides 7 + 7 = 14 < the measures of the thrid side.
So 1,1, 15 cannot be the sides of a triangles.

(iii) 1,2,4
Here sum of two sides 1 + 2 = 3 < the measure of the third side.
∴ 1, 2, 4 cannot be the sides of a triangle.

(iv) 3, 6, 8
Sum of two sides 3 + 6 = 9 > the third side.
∴ 3, 6, 8 can be the sides of a triangle.

Question 8.
Ezhil wants to fence his triangular garden. If two of the sides measure 8 feet and 14 feet then the length of the third side is ______
(i) 11 ft
(ii) 6 ft
(iii) 5 ft
(iv) 22 ft
Answer:
(i) 11 ft

Question 9.
Can we have more than one right angle in a triangle?
Solution:
No, we cannot have more than one right angle in a triangle.
Because the sum of three angles of a triangle is 180°.
But if two angles are right angles then their sum itself become 180°.

Question 10.
How many obtuse angles are possible in a triangle?
Solution:
Only one.

Question 11.
In a right triangle, what will be the sum of other two angles?
Solution:
Sum of three angles of a triangle = 180°
If one angle is right angle (i.e. 90°) .
Sum of other two sides = 180° – 90° = 90°

Question 12.
Is it possible to form an isosceles right angled triangle? Explain.
Solution:
Yes, it is possible.
If one angle is right angle, then the other two angles will be 45° and 45°.

Exercise 4.2

Try These (Text book Page No. 76)

Question 1.
Measure and group the pair of congruent line segments.

Solution:
\(\overline{A B}\) = 3 cm
\(\overline{C D}\) = 4.8 cm
\(\overline{I J}\) = 4.8 cm
\(\overline{P Q}\) = 3 cm
\(\overline{R S}\) = 1.7 cm
\(\overline{X Y}\) = 1.7 cm
From the above measurement S, we can conclude that
(i) \(\overline{A B}\) ≅ \(\overline{P Q}\)
(ii) \(\overline{C D}\) ≅ \(\overline{I J}\)
(iii) \(\overline{R S}\) ≅ \(\overline{X Y}\)

Try These (Text book Page No. 77)

Question 1.
Find the pairs of congruent angles either by superposition method or by measuring them.

Solution:
From the given figures
∠ABC = 50°
∠EFG = 120°
∠HIJ = 120°
∠KLH = 90°
∠PON = 50°
∠RST = 90°
From the above measures, we can conclude that
(i) ∠ABC = ∠PON
(ii) ∠EFG = ∠HIJ
(iii) ∠KLH ≅ ∠RST

Try These (Text book Page No. 83)

Question 1.
If ∆ABC ≅ ∆XYZ then list the corresponding sides and corresponding angles.

Solution:
If ∆ABC ≅ ∆XYZ
\(\overline{A B}\) ≅ \(\overline{X Y}\) – \(\overline{B C}\) ≅ \(\overline{Y Z}\)
\(\overline{A C}\) ≅ \(\overline{X Z}\)
And also
∠A ≅ ∠X – ∠B ≅ ∠Y
∠C ≅ ∠Z

Question 2.
Given triangles are congruent. Identify the corresponding parts and write the congruent statement.

Solution:
Given the set of triangles are congruent. Also we observe from the triangles that the corresponding sides.
\(\overline{A B}\) = \(\overline{A C}\)
\(\overline{B C}\) = \(\overline{Y Z}\)
\(\overline{A C}\) = \(\overline{X Z}\)
Here three sides of ∆ABC are equal to the corresponding sides of ∆XYZ.
This criterion of congruency is side – side – side.

Question 3.
Mention the conditions needed to conclude the congruency of the triangles with reference to the above said criterions. Give reasons for your answer.

Solution:
(i) In ∆ABC and ∆XYZ
if \(\overline{A B}\) = \(\overline{X Y}\)
\(\overline{B C}\) = \(\overline{Y Z}\)
\(\overline{A C}\) = \(\overline{X Z}\)
then ∆ABC ≅ ∆XYZ. By the Side – Side -Side Congruency Criterion.

then ∆AB ≅ ∆XYZ.
By Side – Angle – Side Criterion.

then ∆ABC ≅ ∆XYZ.
By Angle – Side – Angle Congruency Critirion.

then by RHS criterion.
∆ABC ≅ ∆XYZ

Students can Download Maths Chapter 4 Geometry Ex 4.3 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th 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 7th Maths Solutions Term 2 Chapter 4 Geometry Ex 4.3

Miscellaneous Practice Problems

Question 1.
In an isoscales triangle one angle is 76°. If the other two angles are equal, find them.
Solution:
In an isoscales triangle, angle opposite to equal sides are equal. Let the equal angles be x° and x°.
In a triangle the sum of the three angles is 180°.
x° + x° + 76° = 180°
x° (1 + 1) = 180° – 76° = 104°
2x = 104°
x = \(\frac{104^{\circ}}{2}\) = 52°
x = 52°
∴ Other two angles are 52° and 52°.

Question 2.
If two angles of a triangle are 46° each, how can you classify the triangle?
Solution:
Given two angles of the triangle are same and is equal to 46°. If two angles are equal the sides opposite to equal angles are equal. Therefore it will be an isoscales triangle.

Question 3.
If an angle of a triangle is equal to the sum of the other two angles, find the type of the triangle.
Solution:
Let ∠B is the greater angle then by the given condition ∠B = ∠A + ∠C.
Sum of three angle of a triangle = 180°.

∠A + ∠B + ∠C = 180°.
∠A + (∠A + ∠C) + ∠C) = 180°.
2∠A + 2∠C = 180°
2(∠A + ∠C) = 180°
∠A + ∠C = \(\frac{180^{\circ}}{2}\)
∠B = 90°
∴ One of the angle of the triangle = 90°
It will be a right angled triangle.

Question 4.
If the exterior angle of a triangle is 140° and its interior opposite angles are equal, find all the interior angles of the triangle.
Solution:
Given the exterior angle = 140°
Interior opposite angle are equal.

Let one of the interior opposite angle be x.
Then x + x = 140°.
[∵ Exterior angle = sum of interior opposite angles]
2x = 140°
x = \(\frac{140^{\circ}}{2}\) = 70°
x = 70°
Interior opposite angle = 70°, 70°.
Sum of the three angles of a triangle = 180°.
70° + 70° + Third angle = 180°
140° + Third angle = 180°
Third angle = 180° – 140° = 40°
∴ Interior angle are 40°, 70°, 70°.

Question 5.
In ∆JKL, if ∠J = 60° and ∠K = 40°, then find the value of exterior angle formed by extending the side KL.
Solution:
When extending the side KL, the exterior angle formed in equal
to the sum of the interior opposite angles.

∠JLX = ∠LJK + ∠LKJ
= 60°+ 40° =100°
Exterior angle formed = 100°

Question 6.
Find the value of ‘x’ in the given figure.

Solution:
Given ∠DCB = 1000 and ∠DBA = 128°
In the given figure
∠CBD + ∠DBA = 180°
∠CBD + 128° = 180°
∠CBD = 52°
Now exterior angle x = Sum of interior opposite angles.
x = ∠DCB + ∠CBD = 100° + 52° = 152°
x = 152°

Question 7.
If ∆MNO ≅ ∆DEF, ∠M = 60° and ∠E = 45° then find the value of ∠O.
Solution:
Given ∆MNO ≅ ∆DEF

∴ Corresponding parts of conqruent triangle are congruent.
∠M = ∠D = 60° [given ∠M = 60°]
∠N = ∠E = 45° [given ∠E = 45°]
∠O = ∠F
In triangle MNO, sum of the three angle – 180°.
∠M + ∠N + ∠O = 180°
60° + 45° + ∠O = 180°
105° + ∠O = 180°
∠O = 180° – 105° = 75°
Value of ∠O = 75°

Question 8.
In the given figure ray AZ bisects ∠BAD and ∠DCB, prove that
(i) ∆BAC ≅ ∆DAC
(ii) AB = AD

Solution:
(i) In ∆BAC and ∆DAC
∠BAC = ∠DAC [Given \(\overline{A Z}\) bisects ∠BAD]
∠BCA = ∠DCA[\(\overline{A Z}\) bisects ∠DCB]
AC = AC [∵ common side]
∴ Here AC is the included side of the angles. By ASA criterior, ∆BAC ≅ ∆DAC.

(ii) By (i) ∆BAC ≅ ∆DAC
BA = DA [By CPCTC]
i.e., AB = AD

Question 9.
In the given figure FG = FI and H is midpoint of GI, prove that ∆FGH ≅ ∆FHI
Solution:
In ∆FGH and ∆FHI
Given FG = HI
Also, GH = HI [∵ H is the midpoint of GI]

FH = FH [Common]
∴ By S.SS congruency criteria, ∆FGH ≅ ∆FIH. Hence proved.

Question 10.
Using the given figure, prove that the triangles are congruent. Can you conclude that AC is parallel to DE.
Solution:
In ∆ABC and ∆EBD,

AB = EB
BC = BD
∠ABC = ∠EBD [∵ Vertically opposite angles]
By SAS congruency criteria. ∆ABC ≅ ∆EBD.
We know that corresponding parts of congruent triangles are congruent.
∴ ∠BCA ≅ ∠BDE
and ∠BAC ≅ ∠BED
∠BCA ≅ ∠BDE means that alternate interior angles are equal if CD is the transversal to lines AC and DE.
Similarly, if AE is the transversal to AC and DE, we have ∠BAC ≅ ∠BED
Again interior opposite angles are equal.
We can conclude that AC is parallel to DE.

Challenge Problems

Question 11.
In given figure BD = BC, find the value of x.

Solution:
Given that BD = BC
∆BDC is on isoscales triangle.
In isoscales triangle, angles opposite to equal sides are equal.
∠BDC = ∠BCD ……(1)
Also ∠BCD + ∠BCX = 180° [∵ Liner Pair]
∠BCD + 115° = 180°
∠BCD = 180° – 115°
∠BCD = 65° [By (1)]
In ∆ADB
∠BAD + ∠ADB = ∠BDC
[∵ BDC is the exterior angle and ∠BAD and ∠ABD are interior opposite angles]
35° + x = 65°
x = 65° – 35°
x = 30°

Question 12.
In the given figure find the value of x.

Solution:
For ∆LNM, ∠LMK is the exterior angle at M.
Exterior angle = sum of opposite interior angles
∠LMK = ∠MLN + ∠LNM = 26° + 30° = 56°
∠JMK = 56° [∵ ∠LMK = ∠JMK]
x is the exterior angle at J for ∆JKM.
∴ x = ∠JKM + ∠KMJ [∵ Sum of interior opposite angles]
x = 58° + 56° [∵ ∠JMK = 56°]
x = 114°

Question 13.
In the given figure find the values of x and y.

Solution:
In ∆BCA, ∠BAX = 62° is the exterior angle at A.
Exterior angle = sum of interior opposite angles.
∠ABC + ∠ACB = ∠BAX
28°+ x = 62°
x = 62° – 28° = 34°
Also ∠BAC + ∠BAX = 180° [∵ Linear pair]
y + 62° = 180°
y = 180° – 62° = 118°
x = 34°
y = 118°

Question 14.
In ∆DEF, ∠F = 48°, ∠E = 68° and bisector of ∠D meets FE at G. Find ∠FGD.

Solution:
Given ∠F = 48°
∠E = 68°
In ∆DEF,
∠D + ∠F + ∠E = 180° [By angle sum property]
∠D + 68° + 68° = 180°
∠D + 116° = 180°
∠D = 180° – 116° = 64°
Since DG is the angular bisector of ∠D.
∠FDG = ∠GDE
Also ∠FDG + ∠GDE = ∠D
2 ∠FDG = 64°
2 ∠FDG = 64°
∠FDG = \(\frac{64^{\circ}}{2}\) = 32°
∠FDG = 32°
In ∆FDG,
∠FDG + ∠GFD = 180° [By angle sum property of triangles]
32° + ∠FDG + 48° = 180°
∠FDG + 80° = 180°
∠FDG = 180° – 80°
∠FDG = 100°

Question 15.
In the figure find the value of x.

Solution:
Exterior angle is equal to the sum of opposite interior angles.
in ∆TSP ∠TSP + ∠SPT = ∠UTP
75° + ∠SPT = 105°
∠SPT = 105° – 75°
∠SPT = 30° ……(1)
∠SPT + ∠TPR + ∠RPQ = 180° [∵ Sum of angles at a point on a line is 180°]
30° + 90° + ∠RPQ = 180°
120° + ∠RPQ = 180°
∠RPQ = 180° – 120°
∠RPQ = 60° …… (2)
∠VRQ + ∠QRP = 180° [∵ linear pair]
145° + ∠QRP = 180°
∠QRP = 180° – 145°
∠QRP = 35°
Now in ∆ PQR
∠QRP + ∠RPQ = x [∵ x in the exterior angle]
35° + 60° = x
95° = x

Question 16.
From the given figure find the value of y.

Solution:
From the figure,
∠ACB = ∠XCY [Vertically opposite angles]
∠ACB = 48° …(1)
In ∆ABC, ∠CBD is the exterior angle at B.
Exterior angle = Sum of interior opposite angles.
∠CBD = ∠BAC + ∠ACB
∠CBE + ∠EBD = 57° + 48°
65° + ∠EBD = 105°
∠EBD = 105° + 65° = 40° ……… (2)
In ∆EBD, y is the exterior angle at D.
y = ∠EBD + ∠BED
[∵ Exterior angle = Sum of opposite interior angles]
y = 40° + 97° [∵ From (2)]
y = 137°