Prasanna

Students can Download Maths Chapter 5 Geometry Ex 5.6 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 5.6

Miscellaneous Practice Problems

Question 1.
Find the value of x if ∠AOB is a right angle.
Solution:
Given that ∠AOB = 90°

∠AOB = ∠AOC + ∠COB = 90° (Adjacent angles)
3x + 2x = 90°
5x = 90°

Question 2.
In the given figure, find the value of x.
Solution:
Since ∠BOC and ∠AOC are linear pair, their sum = 180°
2x + 23 + 3x – 48 = 180°
5x – 25 = 180°
5x – 25 + 25 = 180° + 25

Question 3.
Find the value of x, y and z.
Solution:

∠DOB and ∠BOC are linear pair
∴ ∠DOB + ∠BOC = 180°
x + 3x + 40 = 180°
4x + 40 = 180°
4x + 40 – 40 = 180° – 40°
4x = 140°

Also ∠BOD and ∠AOC are vertically opposite angles.
∴ ∠BOD = ∠AOC
x = z + 10
35° = z + 10
z + 10 – 10 = 35 – 10
z = 25°
Again ∠AOD and ∠AOC are linear pair.
∴ ∠AOD + ∠AOC = 180°
y + 30 + z + 10 = 180°
y + 30 + 25 + 10 = 180°
y + 65 = 180°
y + 65 – 65 = 180° – 65
y = 115°
∴ x = 35°,
y = 115°,
z = 25°

Question 4.
Two angles are in the ratio 11 : 25. If they are linear pair, find the angles.
Solution:
Given two angles are in the ratio 11 : 25.
Let the angles be 11x and 25x.
They are also linear pair
∴ 11x + 25x = 180°.

∴ The angles 11x = 11 × 5° = 55° and 25x = 25 × 5 = 125°.
∴ The angles are 55° and 125°.

Question 5.
Using the figure, answer the following questions and justify your answer.
(i) Is ∠1 adjacent to ∠2?
(ii) Is ∠AOB adjacent to ∠BOE?
(iii) Does ∠BOC and ∠BOD form a linear pair?
(iv) Are the angles ∠COD and ∠BOD supplementary.
(v) Is ∠3 vertically opposite to ∠1 ?

Solution:
(i) Yes, ∠1 is adjacent to ∠2.
Because they both have the common vertex ‘O’ and the common arm OA . Also their interiors do not overlap.
(ii) No, ∠AOB and ∠BOB are not adjacent angles because they have overlapping interiors.
(iii) No, ∠BOC and ∠BOD does not form a linear pair.
Because ∠BOC itself a straight angle, so the sum of ∠BOC and ∠BOD exceed 180°.
(iv) Yes, the angles ∠COD and ∠BOD are supplementary ∠COD + ∠BOD = 180°, [∵ linear pair of angles]
∴ ∠COD and ∠BOD are supplementary.
(v) No. ∠3 and ∠1 are not formed by intersecting lines. So they are not vertically opposite angles.

Question 6.
In the figure POQ, ROS and TOU are straight lines. Find the x°.

Solution:
Given TOU is a straight line.
∴ The sum of all angles formed at a point on a straight line is 180°
∠TOR + ∠ROP + ∠POV + ∠VOU = 180°.
36° + 47°+ 45° + x° = 180°.
128° + x° = 180°
128° + x° – 128° = 180° – 128°
x = 52°

Question 7.
In the figure AB is parallel to DC. Find the value of ∠1 and ∠2. Justify your answer.

Solution:
Given AB || DC
AB and CD are parallel lines Taking CE as transversal we have.
∠1 = 30°, [∵ alternate interior angles]
Taking DE as transversal
∠2 = 80°.[∵ alternate interior angles]
∠1 = 30° and ∠2 = 80°
Justification:
CDE is a triangle

Question 8.
In the figure AB is parallel to CD. Find x, y and z.

Solution:
Given AB || CD
∴ Z = 42 (∵ Alternate interior angles)
Also y = 42° [vertically opposite angles]
Also x° + 63° + z° = 180°
x° + 63° + 42° = 180°
x° + 105° = 180°
x°+ 105° – 105° = 180° – 105°
x° = 75°
∴ x = 75°;
y = 42°;
z = 42°

Question 9.
Draw two parallel lines and a transversal. Mark two alternate interior angles G and H. If they are supplementary, what is the measure of each angle?
Solution:
l and m are parallel lines and n is the transversal.
∠G and ∠H are alternate interior angles.
∠G = ∠H …… (1)
Given ∠G and ∠G are Suplementary

Question 10.
A plumber must install pipe 2 parallel to pipe 1. He knows that ∠1 is 53. What is the measure of ∠2?

Solution:
Given ∠1 = 53°

Clearly ∠1 and ∠2 are interior angles on the same side of the transversal and so they are supplementary.
∠1 + ∠2 = 180°
53° + ∠2 = 180°
53° + ∠2 – 53° = 180° – 53°
∠2 = 127°

Challenge Problems

Question 11.
Find the value of y.

Solution:
Cleary POQ is a straight line”
Sum of all angles formed at a point on a straight line is 180°
∴ ∠POT + ∠TOS + ∠SOR + ∠ROQ = 180°
60° + (3y – 20°) + y° + (y + 10°) = 180°
60° + 3y – 20° + y° + y° + 10° = 180°
5y + 50° = 180°
5y + 50° – 50° = 180° – 50
5y = 130°
\(y=\frac{130^{\circ}}{5}\)
y = 26°

Question 12.
Find the value of z.

Solution:
The sum of angles at a point is 360°.
∴ ∠QOP + ∠PON + ∠NOM + ∠MOQ = 360°
3z + (2z – 5) + (z + 10) + (4z – 25) = 360°
3z + 2z + z + 4z — 5 +10 — 25 = 360°
10z – 20° = 360°
10z – 20° + 20 = 360°+ 2
10z = 380°
\(z=\frac{380^{\circ}}{10}\)
z = 38°

Question 13.
Find the value of x and y if RS is parallel to PQ.

Solution:
Given RS || PQ
Considering the transversal RU, we have y = 25° (corresponding angles)
Considering ST as transversal

Question 14.
Two parallel lines are cut by a transversal. For each pair of interior angles on the same side of the transversal, if one angle exceeds the twice of the other angle by 48°. Find the angles.
Solution:
Let the two parallel lines be m and n and l be the transversal
Let one of the interior angles on the same side of the transversal be x°
Then the other will be 2x + 48.
We know that they are supplementary.

Question 15.
In the figure, the lines GH and IJ are parallel. If ∠1 = 108° and ∠2 = 123°, find the value of x, y and z.
Solution:
Given GH || IZ
∠1 = 108°
∠2 = 123°
∠1 + ∠KGH = 180 [linear pair]
108° + ∠KGH = 180°

108° + ∠KGH – 108° = 180° – 108°
∠KGH = 72°
∠KGH = x° (corresponding angles if KG is a transversal)
∴ x° = 72°
Similarly
∠2 + ∠GHK = 180° (∵ linear pair)
123° + ∠GHK = 180°
123° + ∠GHK – 123° = 180° – 123°
∠GHK = 57°
Again ∠GHK = y° (corresponding angles if KH is a transversal)
y = 57°
x° +y° + z° = 180° (sum of three angles of a triangle is 180°)
72° + 57° + z° = 180°
129° + z° = 180°
129° + z° – 129° = 180° – 129°
z = 51°
x = 72°,
y = 57°,
z = 51°

Question 16.
In the parking lot shown, the lines that mark the width of each space are parallel. If
∠1 = (2x – 3y)°; ∠2 = (x + 39)° find x° and y°.

Solution:
From the picture
∠2 + 65° = 180° [Sum of interior angles on the same side of a transversal]
x + 39° + 65° = 180°
x + 104° = 180°
x + 104° – 104° = 180° – 104°
x = 76°
Also from the picture
∠1 = 65° [alternate exterior angles]
2x – 3y = 65°
2 (76) – 3y = 65°
152° – 3y = 65°
152° – 3y – 152° = 65 – 152°
-3y = -87

Question 17.
Draw two parallel lines and a transversal. Mark two corresponding angles A and B. If ∠A = 4x, and ∠B = 3x + 7, find the value of x. Explain.
Solution:
Let m and n are two parallel lines and l is the transversal.
A and B are corresponding angles.
We know that corresponding angles are equals,

Question 18.
In the figure AB in parallel to CD. Find x°, y° and z°.
Solution:

Given AB||CD
Then AD and BC are transversals.
x = 48°, alternate interior angles; AD is transversal y = 60°, alternate interior angles; BC is transversal
∠AEB + 48° + y° = 180°, (sum of angles of a triangle is 180°)

Question 19.
Two parallel lines are cut by transversal. If one angle of a pair of corresponding angles can be represented by 42° less than three times the other. Find the corresponding angles.
Solution:
We know that the corresponding angles are equal.
Let one of the corresponding angles be x.
Then the other will be 3x – 42°.

Question 20.
In the given figure, ∠8 = 107°, what is these sum of the angles ∠2 and ∠4.
Solution:
Given ∠8 = 107°
∠2 = 107°
[∵ ∠8 and ∠2 are alternate exterior angles, ∵ ∠8 = ∠2]

Students can Download Maths Chapter 5 Geometry Ex 5.5 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 5.5

Question 1.
Construct the following angles using ruler and compass only.
(i) 60°
(ii) 120°
(iii) 30°
(iv) 90°
(v) 45°
(vi) 150°
(vii) 135°
Solution:
(i) 60°
Construction :
Step 1: Drawn a line and marked a point ‘A’ on it.
Step 2: With A as center drawn an arc of convenient radius to meet the line at a point B.

Step 3: With the same radius and B as center drawn an arc to cut the previous arc at C.
Step 4: Joined AC. The ∠ABC is the required angle with the measure 60°.

(ii) 120°
Construction :

We know that there are two 60° angles in 120°.
∴ We can construct two 60° angles consecutively construct 120°
Step 1: Drawn a line and marked a point ‘A’ on it.
Step 2: With ‘A’ as center, drawn an arc of convenient radius to the line at a point B.
Step 3: With the same radius and B as center, drawn an arc to cut the previous arc at C.
Step 4: With the same radius and C as center, drawn an arc to cut the arc drawn in step 2 at D.
Step 5: Joined AD. Then ∠BAD is the required angle with measure 120°.

(iii) 30°
Constructions :

Since 30° is half of 60°, we can construct 30° by bisecting the angle 60°.
Step 1: Drawn a line and marked a point A on it.
Step 2: With A as center drawn an arc of convenient radius to the line to meet at a point B.
Step 3: With the same radius and B as center drawn an arc to cut the previous arc at C.
Step 4: Joined AC to get ∠BAC = 60°
Step 5: With B as center drawn an arc of convenient radius in the interior of ∠BAC
Step 6: With the same radius and C as center drawn an arc to cut the previous arc at D.
Step 7: Joined AD.
∴ ∠BAD is the required angle of measure 30°.

(iv) 90°
Construction :

Step 1: Drawn a line and marked a point ‘A’ on it.
Step 2: With ‘A’ as center, drawn an arc of convenient radius to the line at a point B.
Step 3: With the same radius and B as center drawn an arc to cut the previous arc at ‘C’.
Step 4: With the same radius and C as center, drawn an arc to cut the arc drawn in step 2 at D.
Step 5: Joined AD. ∠BAD = 120°.
Step 6: With C as center, drawn an arc of convenient radius in the interior of ∠CAD.
Step 7: With the same radius and D as center, drawn an arc to cut the arc at E.
Step 8: Joined AF ∠BAE = 90°.

(v) 45°
Construction :
Step 1: Drawn a line and marked a point A on it

Step 2: With A as center, drawn an arc of convenient radius to the line at a point B.
Step 3: With the same radius and B as center drawn an arc to cut the previous arc at C.
Step 4: With the same radius and C as center, drawn an arc to cut the arc drawn in step 2 at D.
Step 5: Joined AD. ∠BAD = 120°.
Step 6: With G as center and any convenient radius drawn an arc in the interior of ∠GAB
Step 7: With the same radius and B as center drawn an arc to cut the arc at F.
Step 8: Joined AF. ∠BAF = 45°

(vi) 150°
Construction :

Since 150° = 60° + 60° + 30°; we construct as follows
Step 1: Drawn a line and marked a point A on it.
Step 2: With ‘A’ as center, drawn a full arc of convenient radius to the line at a point B and at E the other end.
Step 3: With the same radius and B as center, drawn an arc to cut the previous arc at C.
Step 4: With the same radius and C as center drawn an arc to cut the already drawn arc at D.
Step 5: With D as center, drawn an arc of convenient radius in the interior of ∠DAE
Step 6: With E as center and with the same radius drawn an arc to cut the previous arc at F.
Step 7: Joined AF, ∠FAB = 150°.

(vii) 135°
Construction :

Step 1: Drawn a line and marked a point A on it.
Step 2: With ‘A’ as center, drawn an arc of convenient radius to the line at a point B.
Step 3: With the same radius and B as center drawn an arc to cut the previous arc at C.
Step 4: With the same radius and C as center, drawn an arc to cut the arc at D.
Step 5: With C and D as centers drawn arcs of convenient (same) radius in the interior of ∠CAD. Marked the point of intersection as E.
Step 6: Joined AE, through G. ∠BAE = 90°.
Step 7: Drawn angle bisector to ∠GAH through F.
Now ∠BAF = 135°.

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 6 Information Processing Intext Questions

Try These (Textbook Page No. 138)

Question 1.
Now drop the condition that each digit must be used exactly once. List the numbers that are possible now and find the numbers that were not listed above.
Solution:
Here repetition of digits allowed.
Fixing 9 for a thousand places we have

Fixing 6 for thousand places we have

Fixing 5 for thousands place

Fixing 3 for thousands place, we get

Thus we have 4 × 4 × 4 × 4 = 256 numbers.
(ii) The underlined 24 numbers were listed as the numbers with non-repeated digits. Remaining numbers are not listed in the previous list.

Question 2.
Mother had a lot of wooden pieces in different shapes with her. She gave 6 triangles to Kannagi and 6 circles to Madhan and asked them to create different figures using them. They tried and got some interesting figures. Can you create figures like them that are nice and interesting?

Solution:

Try These (Textbook Page No. 139)

Question 1.
Form a group with two of your friends and try this. All three of you together have to draw a scene. First, one of your friends should draw one part, next the other friend has to continue it, and finally you have to complete it. No discussion or any other communication is allowed. Finally, each person tells what (s)he actually intended to draw the full picture.
Solution:
Activity to be done by the students themselves.

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 6 Information Processing Additional Questions

Question 1.
In a meeting, breakfast is served with coffee, Tea and Milk. Breakfast items are Idly, Dosai and Pongal. Each one can have a drink and breakfast. How many such combinations can be there?
Solution:

So they have 9 combinations such as

1. Coffee and Idly
2. Coffee and Dosai
3. Coffee and Pongal
4. Milk and Idly
5. Milk and Dosai
6. Milk and Pongal
7. Tea and Idly
8. Tea and Dosai
9. Tea and Pongal.

Question 2.
Place the numbers from 1 to 12 exactly once in the circles so that the sum of numbers in each of the 5 lines of the star is 24.

Solution:
By trial and error, we arrange the numbers to get the sum 24.

Question 3.
Place the numbers from 1 to 12 in the circles without repetition to get a sum of 28, 29, 30 and 31.

Solution:

Question 4.
Find the minimum number of straight lines used in forming the figure.

Solution:

The straight lines may be labelled as
Vertical lines = 5
Horizontal lines = 3
Standing lines = 6
Total of 14 lines are needed.

Question 5.
Find the dots in the 12th figure of the following:

Solution:

10th figure will have 10 × 3 = 30 dots.
And 12th figure will have 12 × 3 = 36 dots.

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.3

Question 1.
Find the missing values.

Solution:
(i) Given Height h = 10 m ; Parallel sides a = 12 m ; b = 20 m
Area of the Trapezium = \(\frac{1}{2}\)h(a + b) sq. units = \(\frac{1}{2}\) × 10 × (12 + 20)m²
= (5 × 32)m² = 160 m²

(ii) Given the parallel sides a = 13 cm ; 6 = 28 cm
Area of the trapezium = 492 sq. cm
\(\frac{1}{2}\)h(a + b) = 492
\(\frac{1}{2}\) × h × (13 + 28) = 492
h × 41 = 492 × 2
h = \(\frac { 492\times 2 }{ 41 } \)
h = 24 cm

(iii) Given height ‘h’ = 19 m; Parallel sides b = 16 m
Area of the trapezium = 323 sq. m
\(\frac{1}{2}\)h(a + b) = 323
\(\frac{1}{2}\) × h × (a + 16) = 323
a + 16 = \(\frac { 323\times 2 }{ 19 } \) = 34
a = 34 – 16 = 18 m
a = 18 m

(iv) Given the height h- 16 cm ; Parallel sides a = 15 cm
Area of the trapezium = 360 sq. cm
\(\frac{1}{2}\) × h × (a + b) = 360
\(\frac{1}{2}\) × 16 × (15 + 6) = 360
15 + b = \(\frac{360}{8}\) =45
b = 45 – 15 = 30
b = 30 cm
Tabulating the results we get

Question 2.
Find the area of a trapezium whose parallel sides are 24 cm and 20 cm and the distance between them is 15 cm.
Solution:
Given the parallel sides a = 24 cm; b = 20 cm

Distance between a and b is ‘h’ = 15 cm
Area of the trapezium = \(\frac{1}{2}\) × h × (a + b) sq. units
= \(\frac{1}{2}\) × 15 × (24 + 20) cm²
= \(\frac{1}{2}\) × 15 × 44 = 330 cm²
Area of the trapezium = 330 cm²

Question 3.
The area of a trapezium is 1586 sq. cm. The distance between its parallel sides is 26 cm. If one of the parallel sides is 84 cm then find the other side.
Solution:
Given one parallel side = 84 cm. Let the other parallel side be ‘b’ cm.
Distance between a and b is h = 26 cm.
Area of the trapezium = 1586 sq. cm

∴ The other parallel side = 38 cm.

Question 4.
The area of a trapezium is 1080 sq. cm. If the lengths of its parallel sides are 55.6 cm and 34.4 cm. Find the distance between them.
Solution:
Length of the parallel sides a = 55.6 cm ; b = 34.4 cm
Area of the trapezium = 1080 sq. cm

Distance between parallel sides = 24 cm.

Question 5.
The area of a trapezium is 180 sq. cm and its height is 9 cm. If one of the parallel sides is longer than the other by 6 cm. Find the length of the parallel sides.
Solution:
Let one of the parallel side be ‘a’ cm. Given one parallel sides is longer than the other by 6 cm.
i.e. b = a + 6 cm Also given height ‘h’ = 9 cm
Area of trapezium = 180 sq. cm

2a + 6 = 40
2a = 40 – 6 = 34
a = \(\frac{34}{2}\) = 17 cm
b = a + 6= 17 + 6 = 23 cm
∴ The parallel sides are a = 17 cm and b = 23 cm

Question 6.
The sunshade of a window is in the form of isoceles trapezium whose parallel sides are 81 cm and 64 cm and the distance between them is 6 cm. Find the cost of painting the surface at the rate of ₹ 2 per sq. cm.
Solution:
Given the parallel sides a = 81 cm ; b = 64 cm
Distance between ‘a’ and ‘b’ is height h = 6 cm
Area of the trapezium = \(\frac{1}{2}\) × h(a + b) sq. units
= \(\frac{1}{2}\) × 6³ × (81 + 64) = 3 × 145 cm² = 435 cm²
Cost of painting 1 cm² = ₹ 2
Cost of painting 435 cm² = ₹ 435 × 2 = ₹ 870
Cost of painting = ₹ 870 .

Question 7.
A window is in the form of trapezium whose parallel sides are 105 cm and 50 cm respectively and the distance between the parallel sides is 60 cm. Find the cost of the glass used to cover the window at the rate of ₹ 15 per 100 sq. cm.
Solution:
Given the parallel sides a = 105 cm ; b = 50 cm ; Height = 60 cm
Area of the trapezium = \(\frac{1}{2}\) × h × (a + b)sq. units = \(\frac{1}{2}\) × 60 × (105 + 50) cm²
= 30 × 155 cm² = 4650 cm²
For 100 cm² cost of glass used = ₹15
∴ For 4650 cm² cost of glass = ₹ \(\frac{4650}{100}\) × 15 = ₹ 697.50
Cost of the glass used = ₹ 697.50

Objective Type Questions

Question 8.
The area of the trapezium, if the parallel sides are measuring 8 cm and 10 cm and the height 5 cm is
(i) 45 sq. cm
(ii) 40 sq. cm
(iii) 18 sq. cm
(iv) 50 sq. cm
Solution:
(i) 45 sq. cm
Hint: \(\frac{1}{2}\) × h × (a + b) = \(\frac{1}{2}\) × 5 × (10 + 8) = 45

Question 9.
In a trapezium if the sum of the parallel sides is 10 m and the area is 140 sq.m, then the height is
(i) 7 cm
(ii) 40 cm
(iii) 14 cm
(iv) 28 cm
Solution:
(iv) 28 cm
Hint: Area = \(\frac{1}{2}\) × h × (a + b) = 140 = \(\frac{1}{2}\) × h × 10 ⇒ h = 28

Question 10.
When the non-parallel sides of a trapezium are equal then it is known as
(i) a square
(ii) a rectangle
(iii) an isoceles trapezium
(iv) a parallelogram
Solution:
(iii) an isoceles trapezium

Students can Download Maths Chapter 5 Geometry Ex 5.4 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 5.4

Question 1.
Construct the following angles using protractor and draw a bisector to each of the angle using ruler and compass.
(a) 60°
(b) 100°
(c) 90°
(d) 48°
(e) 110°.
Solution:
(a) 60°
Construction:

Step 1: Drawn the given angle ∠ABC with the measure 60° using protractor.
Step 2: With B as centre and convenient radius, drawn an arc to cut BA and BC. Marked the points of intersection as E on BA and F on BC.
Step 3: With the same radius and E as centre drawn an arc in the interior of ∠ABC and another arc of same measure with centre at F to cut the previous arc.
Step 4: Marked the point of intersection as G. Drawn a ray BX through G. BG is the required bisector of the given ∠ABC

(b) 100°

Construction :
Step 1: Drawn the given angle ∠ABC with the measure 100° c using protractor.
Step 2: With B as centre and convenient radius, drawn an arc to cut BA and BC. Marked the points of intersection as E on BA and F on BC.
Step 3: With the same radius and E as centre drawn an arc in the interior of ∠ABC and another arc of the same measure with centre at F to cut the previous arc.
Step 4: Marked the point of intersection at G. Drawn a ray BX through G.
BG is the required bisector of angle ∠ABC
∠ABG = ∠GBC = 50°

(c) 90°
Construction :

Step 1: Drawn the given angle ∠ABC with the measure 90° using protractor.
Step 2: With B as center and convenient radius, drawn an arc to cut BA and BC. Marked the points of intersection as E on BA and F on BC.
Step 3: With the same radius and E as center drawn an arc in the interior of ∠ABC and another arc of same measure with center at F to cut the previous arc.
Step 4: Mark the point of interaction as G. Drawn a ray BX through G. BG is the required bisector of the given angle ∠ABC
∠ABG = ∠GBC = 45°

(d) 48°

Construction :
Step 1: Drawn the given angle ∠ABC with the measure 48° using protractor.
Step 2: With B as center and convenient radius, drawn an arc to cut BA and to cut BA and BC. Marked the points of intersection as E on BA and F on BC.
Step 3: With the same radius and E as center drawn an arc in the interior of ∠ABC and another arc of the same measure with center at F to cut the previous arc.
Step 4: Marked the point of intersection as G. Drawn a ray BX through G. BG is the required bisector of the given angle ∠ABC
Now ∠ABC = ∠GBC = 24°

(e) 110°
Construction:

Step 1: Drawn the given angle ∠ABC with the measure 110° using protractor.
Step 2: With B as center and convenient radius, drawn an arc to cut BA and BC. Marked points of intersection as E on BA and F BC.
Step 3: With the same radius and E as center, drawn an arc in the interior of ∠ABC and another arc of same measure with center at F to cut the previous arc.
Step 4: Mark the point of intersection as G. Drawn a ray BX through G. BG is the
required bisector of the given angle ∠ABC
∠ABG = ∠GBC = 55°

Students can Download Maths Chapter 5 Geometry Ex 5.3 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 1 Chapter 5 Geometry Ex 5.3

Question 1.
Draw a line segment of given length and construct a perpendicular bisector to each line segment using scale and compass
(a) 8 cm
(b) 7 cm
(c) 5.6 cm
(d) 10.4 cm
(e) 58 cm
Solution:
(a) 8 cm
Construction :

Step 1: Drawn a line. Marked two points A and B on it so that AB = 8 cm
Step 2: Using compass with A as centre and radius more than half of the length of AB, drawn two arcs of the same length one above AB and one below AB
Step 3: With the same radius and B as centre drawn two arcs to cut the arcs drawn in step 2. Marked the points of intersection of the arcs as C and D.
step 4: Joined C and D, CD intersect AB. Marked the point of intersection as ‘O’.
CD is the required perpendicular bisector of AB.

(b) 7 cm
Construction :

step 1: Drawn a line and marked points A and B on it so that AB = 7 cm.
step 2: Using compass with A as centre and radius more than half of the length of AB drawn two arcs of same length one above AB and one below AB.
step 3: With the same radius and B as centre drawn two arcs to cut the already drawn arcs in step 2. Marked the intersection of the arcs as C and D
step 4: Joined C and D, CD is the required perpendicular bisector of AB.

(c) 5.6 cm.
Construction :

Step 1: Drawn a line and marked two points A and B on it so that AB = 5.6cm
Step 2: Using compass with A as centre and radius more than half of the length of AB, drawn two arcs of the same length, one above AB and one below AB
Step 3: With the same radius and B as centre drawn two arcs to cut the arcs drawn in step 2 and marked the points of intersection of the arcs as C and D
Step 4: Joined C and D. CD intersects AB. Marked the point of intersection as ‘O’CD is the required perpendicular bisector of AB.
Now ∠AOC = 90° AO = BO = 2.8 cm

(d) 10.4 cm
Construction :

Step 1: Drawn a line and marked two points A and B on it so that AB = 10.4 cm.
Step 2: Using compass with A as centre and radius more than half of the length of AB, drawn two arcs of same length one above AB and one below AB.
Step 3: With the same radius and B as centre drawn two arcs to cut the arcs drawn in step 2 and marked the points of intersection of the arcs as C and D.
Step 4: Joined C and D. CD intersects AB. Marked the points of intersection as O. CD is the required perpendicular bisector.
Now ∠AOC = 90° ; AO = BO = 5.2 cm

(e) 58 mm

Construction :

Step 1: Drawn a line. Marked two points A and B on it so that
AB = 5.8 cm = 58 mm.
Step 2: Using compass with A as centre and radius more than half of the length of AB, drawn two arcs of the same length one above AB and one below AB.
Step 3: With the same radius and B as centre drawn two arcs to cut the arcs of drawn in step 2. Marked the points of intersection of the arcs as C and D.
Step 4: Joined C and D. CD intersects AB. Marked the point of intersection as O. CD is the required perpendicular bisector. ∠AOC = 90°
AO = BO = 2.9 cm

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Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 6 Information Processing Ex 6.3

Question 1.
How many Triangles are there in each of the following figures?

Solution:
(i) The single portions A, B, C, D, E, F, G and H are 8 triangles.
Taking the combination of 2 E & F, F & G & H and E & 11 are 4 triangles.
There are 8 + 4 = 12 triangles.

(ii) The smaller portions A, B, C, D, E, F, G and H are 8 triangles.
Combining 2 at a time B & C, D & F, F & G, A & H are 4 triangles.
Taking 4 at a time
Combination of B, C, D and E; D, E, F and G; F, G, H and A; H, A, B and C are 4 triangles.
Total triangles = 8 + 4 + 4 = 16.

(iii) Every part A, B, C, D, E, F, G, H, I, J, K, L, M, N, O and P are 16 triangles.
Combining two at a time. A & B, B & C, C & D, D & A, J & K, L & M, N & O, P & I are 8 triangles.
Taking the combination of 4, we have I, J, E & A; K, L, F and B, M, N, G and C, P, O, H and D are 4 triangles.
Taking 8 parts together we have QRS, RST, STQ, and TRQ are 4 triangles.
Total triangles = 16 + 8 + 4 + 4 = 32 triangles.

(iv)

Taking smaller triangles we have AFB, BFG, BGC, CGH, HCD, DHI, DIE, EIJ, EJAand AJF ⇒ 10 triangles.
Taking the combination of 2.
CGD, CID, DJE, DHE, EIA, EFA, AGB, AJB, BHC and BFC are 10 triangles.
Combining 3 at a time we have CED, DAE, EBA, ACB, BDC, CEF, DBJ, GDA, EBH, ACI ⇒ 10 triangles.
Other triangles are BCE, CAD, DBE, ADB and ACE are 5 triangles.
Total 10 + 10 + 10 + 5 = 35 triangles.

Question 2.
Find the number of dots in the tenth figure of the following pattern.

Solution:
(i) The number of dots given are 1, 3, 6, 10, 15,…

1st number = 1
2nd number = 1 + 2 = 3
3rd number = 3 + 3 = 6
4th number = 6 + 4 = 10
5th number = 10 + 5 = 15
6th number will be 15 + 6 = 21
7th number will be 21 + 7 = 28
8th number will be 28 + 8 = 36
9th number will be 36 + 9 = 45
10th number will be 45 + 10 = 55
Number of dots in the 10th figure = 55
(ii) The number of dots given are 1, 4, 9, 11, 16, 25,…
The no. of dots are in the pattern 1 × 1, 2 × 2, 3 × 3, 4 × 4, 5 × 5, …… 10 × 10
The number of dots in the 10th figure = 100.

Question 3.

(i) Draw the next pattern.
(ii) Prepare a table for the number of dots used for each pattern.
(iii) Explain the pattern.
(iv) Find the number of dots in the 25th pattern.
Solution:

First number is 2
2nd number is 2 + 3 = 5
3rd number is 5 + 4 = 9
4th number is 9 + 5 = 14
5th number is 14 + 6 = 20
(iv) Number of dots in the 25th pattern is

The number of dots in the 25th pattern = 350.

Question 4.
Count the number of squares in each of the following figures?

Solution:

Smaller squares 4 × 4 = 16 (As numbered).
As a whole bigger = 4
Total squares = 20

As we see in the figure in the middle 1, 2, 3, and 4 are 4 small squares.
Also, we have 9 and 10 = 2 big squares.
Outer squares 5, 6, 7, & 8 = 4.
Total = 4 + 4 + 1 + 1 = 10 squares.

Question 5.
How many circles are there in the following figure?

Solution:

There are 7 Circles.

Question 6.
Find the minimum number of straight lines used in forming the following figures.

Solution:

There are 5 horizontal lines and 5 vertical lines.
Total of 10 lines minimum needed.

There are 12 straight lines used which is minimum.

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 1 Number System Additional Questions

Additional Questions and Answers

Exercise 1.1

Question 1.
When Malar woke up her temperature was 102°F. Two hours later it was 3° lower, what was her temperature then?
Solution:
Initially Malar’s temperature = 102°F
After two hours it lowered 3° ⇒ -3°F
Here present temperature = 102° + (-3°) = 99°F

Question 2.
An elevator is on the twentieth floor. It goes down 11 floors and then up 5 floors.
What floor is the elevator on now?
Solution:
Present location of the elevator = 20th floor
If it goes down 11 floor ⇒ (-11)
= 20 +(-11) = 9th floor
If it goes up 5 floor ⇒ 9 + 5
= 14th floor

Question 3.
16 + ___ = 16. The property expressed here is ____
Solution:
16 + 0 = 16.
0 is the additive identity on integers.

Exercise 1.2

Question 1.
Roman civilization began in 509 BC and ended in 476 AD. How long did Romans civilization last.
Solution:
From the start of common era no. of years upto 476
AD = 476
From 509 BC to start of common era years = 509
Total years Roman civilization last = 476 + 509 = 985 years.

Question 2.
A submarine was situated 450 feet below sea level. If it descends 300 feet. What is its new position?
Solution:
Position of submarine = -450 ft.
Again it descends 300 feet ⇒ -300 feet
∴ New position = —450 + (-300) = -750 ft.
∴ It was 750 feet below sea level.

Question 3.
In January the high temperature recorded was 90°F and the low temperature was -2°F. Find the difference between the high and the low temperatures?
Solution:
The high temperature recorded = 90°F
The low temperature recorded = -2°F
Difference = 90°F – (-2°F)
= 90°F + (Additive inverse of -2°F)
= 90°F + (+2°F) = 92°F

Exercise 1.3

Question 1.
Ani is scuba diving. She descends 5 feet below sea level. She descends the same distance 4 more times. What is Anis final elevation?
Solution:
Ani descends 5 feet below sea level once she descends 4 more times
∴ She descends (5 × 4) + 5 feet in total = 20 + 5 = 25 feet below sea level

Question 2.
The price of a plant reduced ₹ 6 per week for 7 weeks. By how much did the price of the plant change over the 7 weeks?
Solution:
The price of plant reduced in a week = ₹ 6
∴ The price reduced in 7 weeks = 7 × 6 = 42

Question 3.
The product of three integers is -3. Determine all of the possible values for the three factors?
Solution:
Product of three integers = -3
Possible factors are (1 × -1 × 3), (-1 × -1 × -3), (1 × 1 × -3)

Exercise 1.6

Question 1.
Simplify the following using suitable properties.
(a) (-1650) × (-2) + (-1650) × (-98)
(b) (9150 × 405) – (8150 × 405)
Solution:
(a) (-1650) × (-2) + (-1650) × (-98)
= 1650 [(-1) × (-2) + (-1) × -98] = 1650 (2 + 98) [Distributive property]
= 1650 × 100= 1,65,000

(b) (9150 × 405) – (8150 × 405)
= 405 (9150 – 8150) = 405 × 1000
= 4,05,000

Question 2.
Which is greater: (9 + 7) × 1000 or 9 + 7 × 1000?
Solution:
(9 + 7) × 1000 = 16 × 1000 = 16,000
9 + 7 × 1000 = 9 + 7000 = 7,009
16,000 > 7009
∴ (9 + 7) × 1000 > [9 + 7 × 1000]

Question 3.
Simiplify: 80 ÷ [240 ÷ (-24)] + 7
Solution:
We have
80 ÷ [240 ÷ (-24)] + 7
= 80 ÷ \(\left[\frac{240}{-24}\right]\) + 7
= 80 ÷ (-10) + 7 = \(-\left[\frac{80}{10}\right]\) + 7 = (-8) + 7 = -1

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 1 Number System Ex 1.6

Miscellaneous Practice Problems

Question 1.
What should be added to -1 to get 10?
Solution:
(-1) + a number = 10
∴ The number = 10 + 1 = 11

Question 2.

Solution:

Question 3.
Substract 94860 from (-86945)
Solution:
-86945 – (94860) = -86945 + (Additive inverse of 94860)
= -86945 + (-94860) = -1,81,805

Question 4.
Find the value of (-25) + 60 + (-95) + (-385)
Solution:
(-25) + 60 + (-95) + (-385) = 35 + (-95) + (-385) = -60 + (-385) = -445

Question 5.
Find the sum of (-9999) (-2001) and (-5999).
Solution:
(-9999) + (-2001) + (-5999) = -12,000 + (-5999) = -17,999

Question 6.
Find the product of (-30) × (-70) × 15.
(-30) × (-70) × 15 = (+2100) × 15 = 31,500

Question 7.
Divide-72 by 8.
Solution:
\(\frac{-72}{8}\) = -9

Question 8.
Find two pairs of integers whose product is +15.
Solution:
(i) (+3) × (+5)
(ii) (-3) × (-5)

Question 9.
Check the following for equality.
(i) (11 + 7) + 10 and 11 + (7 + 10)
(ii) (8 – 13) × 7 and 8 – (13 × 7)
(iii) [(-6) – (+8)] × (-4) and (-6) – [8 × (-4)]
(iv) 3 × [(-4) + (-10)] and [3 × (-4) + 3 × (-10)]
Solution:
(i) LHS = (11 + 7) + 10 = 18 + 10 = 28
RHS = 11 + (7 + 10)
= 11 + (17) = 28
LHS = RHS
∴ (11 + 7) + 10 = 11 + (7 + 10)

(ii) LHS = (8 – 13) × 7 = -5 × 7 = -35
RHS = 8 – (13 × 7) = 8 – 91 = -83
LHS ≠ RHS
∴ (8 – 13) × 7 ≠ 8 – (13 × 7)

(iii) LHS = [(-6) – (+8)] × (-4) = [(-6) + (-8)] × (-4) = (-14) × (-4) = +56
RHS = (-6) – [8 × (-4)] = -6 – (-32)
= -6 + (+32) = +26
LHS ≠ RHS
∴ [(-6) – (+8)] × (-4) ≠ (-6) – [8 × (-4)]

(iv) LHS = 3 × [(-4) + (-10)] = 3 × (-14) = -42
RHS = [3 × (-4) + 3 × (-10)] = (-12) + (-30) = -42
LHS = RHS
3 × [(-4) + (-10)] = [3 × (-4) + 3 × (-10)]

Question 10.
Kalaivani had ₹ 5000 in her bank account on 01.01.2018. She deposited ₹ 2000 in January and withdrew ₹ 700 in February. What was Kalaivani’s bank balance on 01.04.2018, if she deposited ₹ 1000 and withdraw ₹ 500 in March.
Solution:
Initial bank balance = ₹ 5000 ; Total deposits: January : ₹ 2000 ; March : ₹ 1000
Total deposits upto March = ₹ 5000 + ₹ 2000 + ₹ 1000 = ₹ 8000
Amount withdrawn: February : ₹ 700 (-)
March : ₹ 500 (-)
∴ Total amount withdrawn = (-700) + (-500) ₹ -1200
Net bank balance = ₹ 8000 – ₹ 1200 = ₹ 6800

Question 11.
The price of an item x increases by ₹ 10 every year and an item y decreases by ₹ 15 every year. If in 2018, the price of x is ₹ 50 andy is ₹ 90, then which item will be costlier in the year 2020?
Solution:
Amount increases for x every year = ₹ 10.
Price ofx in 2018 = ₹ 50 ; Price of x in 2019 = ₹ 50 + ₹ 10 = ₹ 60
Price of x in 2020 = ₹ 60 + ₹ 10 = ₹ 70 Amount decreases for y per year = ₹ 15
Price of y in 2018 = ₹ 90
Price of y in 2019 = ₹ 90 – ₹ 15 = ₹ 75
Price of y in 2020 = ₹ 75 – ₹ 15 = ₹ 60
Here 70 > 60. Item x will costlier in year 2020.

Question 12.
Match the statements in Column A and Column B.

Solution:
1. – d
2. – a
3. – e
4. – c
5. – b

Challenge Problems

Question 13.
Say True or False.
(i) The sum of a positive integer and a negative integer is always a positive integer.
(ii) The sum of two integers can never be zero
(iii) The product of two negative integers is a positive integer.
(iv) The quotient of two integers having opposite sign is a negative integer.
(v) The smallest negative integer is -1.
Solution:
(i) False
(ii) False
(iii) True
(iv) True
(v) False

Question 14.
An integer divided by 7 gives a result -3. What is that integer?
Solution:
According to the problem \(\frac{\text { An integer }}{7}\) = -3
∴ The integer = -3 × 7
The required integer = -21.

Question 15.
Replace the question mark with suitable integer in the equation.

Solution:

Question 16.
Can you give 10 pairs of single digit integers whose sum is zero?
Solution:
1 + (-1) + 2 + (-2) + 3 + (-3) + 4 + (-4) + 5 + (-5) = 0

Question 17.
If P = -15 and Q = 5 find (P – Q) – (P + Q).
Solution:
Given P = 15 ; Q = 5

Question 18.
If the letters in the English alphabets A to M represent the number from 1 to 13 respectively and N represents 0 and the letters O to Z correspond from -1 to -12, find the sum of integers for the names given below. For example,
MATH → Sum → 13 + 1 – 6 + 8 = 16
(i) YOUR NAME
(ii) SUCCESS
Solution:
Given

(i) My name LEENA → 12 + 5 + 5 + 0 + 1 = 23
(ii) SUCCESS → (-5) + (-7) + 3 + 3 + 5 + (-5) + (-5)
= -12 + 6 + 5 + (-10) = -6 + 5 + (-10) = (-1) + (-10)
= -11

Question 19.
From a water tank 100 litres of water is used every day. After 10 days there is 2000 litres of water in the tank. How much water was there in the tank before 10 days?
Solution:
Water used for one day = 100 litres.
Water used for 10 days = 100 × 10 = 1000 litres.
After 10 days water left in the tank = 2000 litres
Initially amount of water will be = 2000 + 1000 = 3000 litres

Question 20.
A dog is climbing down into a well to drink water. In each jump it goes down 4 steps. The water level is in 20th step. How many jumps does the dog take to reach the water level?
Solution:
The water in the well is at 20th step.
For each jump the dog goes low 4 steps. 5
∴ Number of jumps the dog to reach the water = \(\frac{20}{4}\) = 5 jumps

Question 21.
Kannan has a fruit shop. He sells 1 dozen banana at a loss of ? 2 each because it may get rotten next day. What is his loss?
Solution:
1 dozen = 12 bananas
For 1 banana loss = ₹ 2
For 12 bananas loss = ₹ 2 × 12 = ₹ 24

Question 22.
A submarine was situated at 650 feet below the sea level. If it descends 200 feet, what is its new position?
Solution:
Position of submarine = 650 feet below sea level = -650 feet
Again the depth it descends = 200 feet below = – 200 feet
∴ Position of submarine = (-650) + (-200) = -850 feet
The submarine will be 850 feet below the sea level.

Question 23.
In a magic square given below each row, column and diagonal should have the same sum. Find the values of x, y, and z.

Solution:
Column total = Row total = diagonal total
∴ 1 + y + (-6) = (-10) + (-3) + 4
y + (- 5) = -13 + 4
y = -9 + 5
y = -4
So 1 + (-10) + x = y + (-3) + (-2)
-9 + x = (-4) + (-3) + (-2)
-9 + x = -9
x = -9 + 9
x = 0
Now x + (-2) + z = (-10) + (-3) + 4
0 + (-2) + z = (-13) + 4
-2 + z = -9
z = -9 + 2 = -7
z = -7
∴ x = 0, y = -4, z = -7