Class 7

Samacheer Kalvi 7th English Book Answers Solutions Guide

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Samacheer Kalvi 7th Science Book Answers Solutions Guide

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Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 5 Geometry Additional Questions

Students can Download Maths Chapter 5 Geometry Additional 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 1 Chapter 5 Geometry Additional Questions

Exercise 5.1

Question 1.
Can two adjacent angles be supplementary?
Solution:
Yes, In the figure
Samacheer Kalvi 7th Maths Term 1 Chapter 5 Geometry Ex 5.1 101
∠AOB and ∠BOC are adjacent angles.
Also ∠AOB + ∠BOC = 180°
∴ ∠AOB and ∠BOC are supplementary

Question 2.
Can two obtuse angles form a linear pair?
Solution:
No, the sum of the measures of two obtuse angles is more than 180°.

Question 3.
Can two right angles form a linear pair?
Solution:
Yes, because the sum of two right angles is 180° and form a linear pair.

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Question 4.
Find x, y and z from the figure.
Samacheer Kalvi 7th Maths Term 1 Chapter 5 Geometry Ex 5.1 102
Solution:
x = 55° vertically opposite angles
y + 55° = 180°
y = 180°- 55°
y = 125°

Execise 5.2

Question 1.
Can two lines intersect in more than one point ?
Solution:
No, two lines cannot intersect in more than one point.

Question 2.
In the figure EF parallel to GH
Solution:
∠EAB = 60° and ∠ACD = 105°
Determine (i) ∠CAF and
(ii) ∠BAC
Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 5 Geometry Additional Questions 50
Solution:
(i) Since EF || GH and AC is a transversal
⇒ ∠CAF + ∠ACH = 180°
⇒ ∠CAF + 105° = 180° .
= 75°
(ii) ∴ EF || GH and AC is transversal.
∴ ∠EAC = ∠ACH [ ∵ Alternate interior angles]
⇒ ∠BAC = 105°
⇒ ∠BAC + ∠BAB = 105°
⇒ ∠BAC + 60° = 105°
⇒ ∠BAC = 105° – 60°
= 45°
∴ ∠CAF = 75° and ∠BAC = 45°.

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Question 3.
In the given figure, the arms of two angles are parallel. If ∠ABC = 70°, then find
(i) ∠DGC
(ii) ∠DEF
Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 5 Geometry Additional Questions 60
Solution:
We have AB||ED and BC || EF
(i) BC is transversal
∠DGC = ∠ABC [corresponding angles]
But ∠ABC = 70°
∠DGC = 70°

(ii) ED is a transversal to BC||EF
∴ ∠DEF = ∠DGC [corresponding]
∠DGC = 70°
∠DEF = 70°

Exercise 5.6

Question 1.
In the following figure, show that CD || EF
Solution:
Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 5 Geometry Additional Questions 74
∠BAD = ∠BAE + ∠EAD
= 40°+ 30° = 70°.
and ∠CDA = 70°
∠BAD = ∠CDA
But they form a pair of alternate angles
⇒ AB || CD
Also ∠BAE + ∠AEF = 40° + 140° = 180°
But they form a pair of interior opposite angles.
⇒ AB || EF
From (1) and (2), we get
AB || CD || EF
⇒ CD || EF
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Question 2.
In the adjoining figure, the lines \(\overleftrightarrow { AB } \) and \(\overleftrightarrow { CD } \) intersect at ‘O’. If ∠COB = 50°, find the measures of the other three angles.
Solution:
∠COB = 50°
∠AOD = 50° (vertically opposite angles)
Now ∠AOC and ∠COB form a linear pair,
Thus ∠AOC + ∠COB = 180°
⇒ ∠AOC + 50° = 180°.
∠AOC = 180° – 50° = 130°
Also ∠AOC and ∠BOD are vertically opposite angles.
∴ ∠BOD = ∠AOC = 130°
Thus the three angles are
∠AOD = 50°
∠AOC =130°
∠BOD = 130°

Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 1 Number System Ex 1.5

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Question 1.
One night in Kashmir, the temperature is -5°C. Next day the temperature is 9°C. What is the increase in temperature?
Solution:
Temperature in the first day = -5°C
Temperature in the next day = 9°C
∴ Increase in temperature = 9°C – (-5°C)
= 9°C + (+5°C) = 14°C

Question 2.
An atom can contain protons which have a positive charge (+) and electrons which have a negative charge (-). When an electron and a proton pair up, they become neutral (0) and cancel the charge at. Now determine the net charge:
(i) 5 electrons and 3 protons → -5 + 3 = -2 that is 2 electrons \(\ominus\ominus\)
(ii) 6 protons and 6 electrons →
(iii) 9 protons and 12 electrons →
(iv) 4 protons and 8 electrons →
(v) 7 protons and 6 electrons →
Solution:
(ii) 6 protons and 6 electrons → (+6) + (-6) = 0
(iii) 9 protons and 12 electrons → (+9) + (-12) = 9-12 = -3 ⇒ 3 electrons \(\ominus\ominus\ominus\)
(iv) 4 protons and 8 electrons → (+4) + (-8) = +4 – 8 = -4 ⇒ 4 electrons \(\ominus\ominus \ominus\ominus\)
(v) 7 protons and 6 electrons → (+7) + (-6) = +1 = 1 proton \(\oplus\)

Question 3.
Scientists use the Kelvin scale (K) as an alternative temperature scale to degrees Celsius (°C) by the relation T°C = (T + 273)K. Convert the following to Kelvin:
(i) -275°C
(ii) 45°C
(iii) -400°C
(iv) -273°C
Solution:
(i) -275°C = (-275 + 273)K = -2K
(ii) 45°C = (45 + 273)K = 318 K
(iii) -400°C = (-400 + 273)K = -127 K
(iv) -273°C = (-273 + 273) K = 0K

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Question 4.
Find the amount that is left in the student’s bank account, if he has made the following transaction in a month. His initial balance is ₹ 690.
(i) Deposit (+) of ₹ 485
(ii) Withdrawal (-) of ₹ 500
(iii) Withdrawal (-) of ₹ 350
(iv) Deposit (+) of ₹ 89
(v) If another ₹ 300 was withdrawn, what would the balance be?
Solution:
(i) Initial balance of student’s account = ₹ 690
Deposited amount = ₹ 485 (+)
∴ Amount left in the account = ₹ 690 + ₹ 485 = ₹ 1175

(ii) Balance in the account = ₹ 1175
Amount withdrawn = ₹ 500 (-)
Amount left = ₹ 1175 – ₹ 500 = ₹ 675

(iii) Balance in the account = ₹ 675
Amount withdrawn = ₹ 350 (-)
Amount left = ₹ 675 – ₹ 350 = ₹ 325

(iv) Balance in the account = ₹ 325
Amount deposited = ₹ 89(+)
Amount left = ₹ 325 + ₹ 89 = ₹ 414

(v) Balance in the account = ₹ 414
Amount withdrawn = ₹ 300 (-)
Amount left = ₹ 414 – ₹ 300 = ₹ 114

Question 5.
A poet Tamizh Nambi lost 35 pages of his ‘lyrics’ when his file had got wet in the rain. Use integers, to determine the following.
(i) If Tamil Nambi wrote 5 pages per day, how many day’s work did he lose?
(ii) If four pages contained 1800 characters, (letters) how many characters were lost?
(iii) If Tamil Nambi is paid ₹ 250 for each page produced, how much money did he lose?
(iv) If Kavimaan helps Tamizh Nambi and they are able to produce 7 pages per day, how many days will it take to recreate the work lost?
(v) Tamizh Nambi pays Kavimann ₹ 100 per page for his help. How much money does Kavimaan receive?
Solution:
Total pages lost – 35
One day work = 5 page 35
35 pages = \(\frac{35}{5}\) = 7 days work
∴ 7 day’s work he lost.

(ii) Number of characters in four pages = 1800
Number of characters in one page = \(\frac{1800}{4}\) = 450
∴ Number of characters in 35 pages = 450 × 35 = 15,750 characters

(iii) Payment for one page = ₹ 250
∴ Payment for 35 pages = ₹ 250 × ₹ 35 = ₹ 8,750

(iv) Number of pages recreated a day = 7
∴ To recreate 35 pages day’s needed = \(\frac{35}{7}\) = 5 days

(v) Payment of Kavimaan = ₹ 100 per page
∴ for 35 pages payment = ₹ 100 × 35 = ₹ 3,500

Question 6.
Add 2 to me. Then multiply by 5 and subtract 10 and divide new by 4 and I will give you 15! Who am I?
Solution:
According to the problem {[(I + 2) × 5] – 10} ÷ 4 = 15
{[(I + 2) × 5] – 10} = 15 × 4 = 60
I + 2 = \(\frac{70}{5}\) = 14
(I + 2) × 5 = 60 + 10 = 70
I = 14 – 2 ; I = 12

Question 7.
Kamatchi, a fruit vendor sells 30 apples and 50 pomegranates. If she makes a profit of ? 8 per apple and loss ? 5 per pomegranate. What will be her overall profit or loss?
Solution:
Number of apples Kamatchi sold = 30
Profit per apple = ₹ 8(+)
∴ Profit for 30 apples = 30 × 8 = ₹ 240
Number of pomegranates sold 50
Loss per pomegranate = ₹ 5(-)
Loss on selling 50 pomegranates = 50 × (-5) = ₹ -250
Overall loss = -250 + 240 = ₹ -10
i.e. loss ₹ 10.

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.5

Question 8.
During a drought, the water level in a dam fell 3 inches per week for 6 consecutive weeks. What was the change in the water level in the dam at the end of this period?
Solution:
Water level fall per week = -3 inches
∴ Water level decrease for 6 weeks = 6 ₹ (-3) = 18 inches
∴ decrease of 18 inches of water level.

Question 9.
Buddha was born in 563 BC (BCE) and died in 483 BC (BCE). Was he alive in 500 BC (BCE)? and find his life time. (Source: Compton’s Encyclopedia)
Solution:
Years in BCC (BCE) are taken as negative integers.
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.5 2
Buddha was bom in -563
and died in -483
So he was alive in 500 BC (BCE)
Life time = -483 – (-563) = -483 + 563 = +80
Buddha’s life time = 80 years.

Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 1 Number System Ex 1.2

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Question 1.
Fill in the blanks
(i) -44 + ____ = -88
(ii) ___ – 75 = -45
(iii) ___ – (+50) = -80
Solution:
(i) -44
(ii) 30
(iii) -30

Question 2.
Say True or False.
(i) (-675) – (-400) = -1075
(ii) 15 – (-18) is the same as 15 + 18
(iii) (-45) – (-8) = (-8) – (-45)
Solution:
(i) False
(ii) True
(iii) False

Question 3.
Find the value of the following.
(i) -3 – (-4) using number line.
Solution:
We start at zero facing positive direction. Move 3 units backward to represent (-3). Then turn towards the negative side and move 4 units backwards. We reach+1.
Samacheer Kalvi 7th Maths Term 1 Chapter 1 Number System Ex 1.2 1
∴ (-3) – (-4) = +1

(ii) 7 – (-10) using number line
Solution:
Samacheer Kalvi 7th Maths Term 1 Chapter 1 Number System Ex 1.2 2
We start at zero facing positive direction. Move 7 units forward to represent (+7). Then turn towards the negative side and move 10 units backwards.
We reach +17
∴ 1 – (-10) = +17

(iii) 35 – (-64)
Solution:
35 – (-64) = 35+ (Additive inverse of-64) = 35 + (+64) = 99
∴ 35 – (-64) = 99

(iv) -200 – (+100)
Solution:
-200 – (+100) = -200 + (Additive inverse of+100) = -200 + (-100) = -300
-200 – (+100) = -300

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Question 4.
Kabilan was having 10 pencils with him. He gave 2 pencils to senthil and 3 to Karthick. Next day his father gave him 6 more pencils, from that he gave 8 to his sister. How many pencils are left with him?
Solution:
Total pencils Kabilan had = 10
No. of pencils given to Senthil = 2
No. of pencils given to Karthick = 3.
Now number of pencils left with Kabilan = 10 – 2 – 3 = 8 – 3 = 5
Number of pencils got from his father = 6
No. total pencils Kabilan had = 5 + 6 = 11
Number of pencils given to his sister = 8
Number of pencils left with Kabilan = 11 – 8 = 3

Question 5.
A lift is on the ground floor. If it goes 5 floors down and then moves up to 10 floors from there, then in which floor will the lift be?
Solution:
Initially the lift will be in the ground floor representing ‘0’
It goes to 5 floors down ⇒ -5
Then it moves 10 floors up +10.
Now the lift will be = 0 – 5 + 10 = -5 + 10
= 5th floor (above the ground floor)

Question 6.
When Kala woke up, her body temperature was 102°F. She took medicine for fever. After 2 hours it was 2°F lower. What was her temperature then?
Solution:
Kala’s temperature initially = 102°F
After two hours the temperature decreased = -2°F
Now the final temperature = 102°F – 2°F = 100°F

Question 7.
What number should be added to (-17) to get -19?
Solution:
According to the problem = -17 + A number = -19
The number = -19 + 17 = -2
∴ -2 should be added to -17 to get -19

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Question 8.
A student was asked to subtract (-12) from -47. He got -30. Is he correct? Justify.
Solution:
Subtracting -12 from -47, we get
-47 – (-12) = -47 + (Additive inverse of-12)
= -47 + (+12) = -35
But the students answer is -30.
So he is not correct.

Objective Type Questions

Question 9.
(-5) – (-18)
(i) 23
(ii) -13
(iii) 13
(iv) -23
Solution:
(iii) 131

Question 10.
(-100) – 0 + 100 =
(i) 200
(ii) 0
(iii) 100
(iv)-200
Solution:
(ii) 0

Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 5 Geometry Ex 5.1

Students can Download Maths Chapter 5 Geometry Ex 5.1 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 1 Chapter 5 Geometry Ex 5.1

Question 1.
Name the pairs of adjacent angles.
Solution:
(i) ∠ABG and ∠GBC are adjacent angles.
(ii) ∠BCF and ∠FCD are adjacent angles.
(iii) ∠BCF and ∠FCE are adjacent angles.
(iv) ∠FCE and ∠ECD are adjacent angles.
Samacheer Kalvi 7th Maths Term 1 Chapter 5 Geometry Ex 5.1 1

Question 2.
Find the angle ∠JIL from the given figure.
Solution:
∠LIK and ∠KIJ are adjacent angles.
∴ ∠JIL = ∠LIK + ∠KIJ
= 38° + 27°
= 65°
∴ ∠JIL = 65°
Samacheer Kalvi 7th Maths Term 1 Chapter 5 Geometry Ex 5.1 2

Question 3.
Find the angles ∠GEH from the given figure.
Solution:
Samacheer Kalvi 7th Maths Term 1 Chapter 5 Geometry Ex 5.1 3
∠HEF = ∠HEG + ∠GEF
120° = ∠HEG + 34°
120°- 34° = ∠GEH + 34° – 34°
∠GEH = 86°

Question 4.
Given that AB is a straight line. Calculate the value of x° in the following cases.
Samacheer Kalvi 7th Maths Term 1 Chapter 5 Geometry Ex 5.1 5
Solution:
(i) Since the angles are linear pair ∠AOC + ∠BOC = 180°
72° + x° = 180°
72° + x° – 12° = 180°- 72°
x° = 108°

(ii) Since the angles are linear pair
∠AOC + ∠BOC = 180°
3x + 42° = 180°
Samacheer Kalvi 7th Maths Term 1 Chapter 5 Geometry Ex 5.1 6

(iii) Since the angles are linear pair
∠AOC + ∠BOC = 180°
4x° + 2x° = 180°
6x° = 180°
x° = 180°

Question 5.
One angle of a linear pair is a right angle. What can you say about the other angle?
Solution:
If the angle are linear pair, then their sum is 180°.
Given one angle is right angle ie 90°.
∴ The other angle = 180° -90° = 90°
∴ The other angle also a right angle

Question 6.
If the three angles at a point are in the ratio 1 : 4 : 7, find the value of each angle?
Solution:
We know that the sum of angles at a point is 360°.
Given the three angles are in the ratio 1:4:7.
Let the three angles be 1x, 4x, 7x.
Samacheer Kalvi 7th Maths Term 1 Chapter 5 Geometry Ex 5.1 7
∴ The three angles are 30°, 120° and 210°.

Question 7.
Three are six angles at a point. One of them is 45° and the other five angles are all equal. What is the measure of all the five angles.
Solution:
We know that the sum of angles at a point is 360°.
One angle = 45°
Let the equal angles be x° each
∴ x° + x° + x° + x° + x° + 45° = 360°
5x° + 45° – 45° = 360° – 45°
Samacheer Kalvi 7th Maths Term 1 Chapter 5 Geometry Ex 5.1 50
5x° = 315°
Samacheer Kalvi 7th Maths Term 1 Chapter 5 Geometry Ex 5.1 51
∴ Measure of all 5 equal angles = 63°.

Question 8.
In the given figure, identify
(i) Any two pairs of adjacent angles.
(ii) Two pairs of vertically opposite angles.
Solution:
(i) (a) ∠PQT and ∠TOS are adjacent angles.
(b) ∠PQU and ∠UQR are adjacent angles.
(ii) (a) ∠PQT and ∠RQU are vertically opposite angles.
(b) ∠TQR and ∠PQU are vertically opposite angles.

Question 9.
The angles at a point are x°, 2x°, 3x°, 4x° and 5x°. Find the value of the largest angle?
Sum of angles at a point = 360°
∴ x° + 2x° + 3x° + 4x° + 5x° = 360°
15x° = 360°
x° = \(\frac{360^{\circ}}{15}\)
x° = 24°.
∴ The largest angle = 5x°
= 5 × 24° = 120°
The largest angle is 120°

Question 10.
From the given figure, find the missing angle.
Solution:
Lines \(\overleftrightarrow { RP }\) and \(\overleftrightarrow { SQ }\) are interesting at ‘O’
Samacheer Kalvi 7th Maths Term 1 Chapter 5 Geometry Ex 5.1 53
∴ Vertically opposite angles are equal.
∴ x = 105°
∴ Missing angle = 105°

Question 11.
Find the angles x° and y° in the figure shown.
Samacheer Kalvi 7th Maths Term 1 Chapter 5 Geometry Ex 5.1 54
Solution:
Consider the line m.
x° and 3x° are linear pair of angles
∴ x° + 3x° = 180°
Samacheer Kalvi 7th Maths Term 1 Chapter 5 Geometry Ex 5.1 56
Samacheer Kalvi 7th Maths Term 1 Chapter 5 Geometry Ex 5.1 55
x° = 45°
Also lines l and m intersects.
Vertically opposite angles are equal.
ie 3x° = y°
3 × 45° = y°
y = 135°
x° = 15° and
y° = 135°

Question 12.
Using the figure, answer the following questions.
Samacheer Kalvi 7th Maths Term 1 Chapter 5 Geometry Ex 5.1 66
(i) What is the measure of angle x°?
(ii) What is the measure of angle y°?
Solution:
From the figure x° and 125° are vertically opposite angles. So they are equal ie
ie x° = 125°
Also y° and 125° are linear pair of angles.
∴ y° + 125° = 180°
y° + 125° – 125° = 180°- 125°
y° = 55°
x° = 125°,
y° = 55°

Question 13.
Adjective angles have
(i) No common interior, no common arm, no common vertex.
(ii) One common vertex, one common arm, common interior
(iii) One common arm, one common vertex, no common interior.
(iv) One common arm, no common vertex, no common interior.
Solution:
(iii) one common arm, one common vertex, no common interior

Question 14.
In the given figure the angles ∠1 and ∠2 are
(i) Opposite angles
(ii) Adjacent angles
(iii) Linear angles
(iv) Supplementary angles
Samacheer Kalvi 7th Maths Term 1 Chapter 5 Geometry Ex 5.1 100
Solution:
(iii) Linear pair

Question 15.
Vertically opposite angles are
(i) not equal in measure
(ii) Complementary
(iii) supplementary
(iv) equal in measure
Solution:
(iv) equal in measure

Question 16.
The sum of all angles at a point is
(i) 360°
(ii) 180°
(iii) 90°
(iv) 0°
Solution:
(i) 360°

Question 17.
The measure of ∠BOC is
(i) 90°
(ii) 180°
(iii) 80°
(iv) 100°
Samacheer Kalvi 7th Maths Term 1 Chapter 5 Geometry Ex 5.1 58
Solution:
(iii) 80°