Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 5 Information Processing Intext Questions

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Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 5 Information Processing Intext Questions

Question 1.
Check whether the Tree diagrams are equal or not. Solution:
(i) Their algebraic expressions are a × (b – c) and (a × b) – (a × c)
∴ distributive property of multiplication over subtraction
∴ They are equal
(ii) Their algebraic expressions are a × (b – c) and (a × b) – c
Both are not equal [By BODMAS rule]

Question 2.
Check whether the following algebraic expressions are equal or not by using Tree diagrams.
(i) (x + y) + z and x + (y + z)
(ii) (p × q) × r and p × (q × r)
(iii) a – (b – c) and (a – b) – c
Solution:
(i) (x + y) + z and x + (y + z)
The tree diagrams are Subtraction is not associative and the expressions are not equal.
(ii) (p × q) × r and p × (q × r)
The tree diagram is Multiplication is associative
∴ Both expressions are equal.
(iii) a – (b – c) and (a – b) – c
Their tree diagrams are Subtraction is not associative
∴ Both expressions are not equal.

Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 5 Information Processing Additional Questions

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Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 5 Information Processing Additional Questions

Question 1.
Draw tree diagrams for the following questions:
(i) The number of books sold in a book fair is as follows. First day 1,82,192, Second day 1,28,194, Third day 80,520 fourth day 92,004 and the fifth day 50,020. Find the total number of the book sold.
(ii) A water purification project cost 1,82,71,000. The machinery was bought for ₹ 69,12,000. What is the amount needed to complete the project?
(iii) The number of flowers needed to arrange in a flower pot is 62. Find the number of flowers needed to arrange in 55 pots?
(iv) If the total scholarship money sanctioned for 50 students are ₹ 62,000. Find the amount that each student can get?
Solution:  Question 2.
Convert into tree diagrams
(i) (10 × 5) + (2 × 16)
(ii) (5 × 3) – (8 × 6) + 9
(iii) [9 + (3 × 2)] – [(6 × 4) + 5]
(iv) [(4 – 1) × 16] + [(16 + 9) × 3]
(v) {[(10 × 6) + 5] × [ 4 + (3 – 2)]} ÷ [4 × (2 + 9)]
(vi) 4 + [8 × 6 + {(4 × 3) – (10 ÷ 4)}]
Solution:  Question 3.
Convert the following tree diagrams into numerical expressions. Solution:
(a) [4 × (6 + 2)] – [(4 – 2) ÷ 2]
(b) (12 × 6) + (6 ÷ 3)
(c) [4 × (10 – 2)] + [(4 + 9) × 3]

Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 5 Information Processing Ex 5.2

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Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 5 Information Processing Ex 5.2

Miscellaneous Practice Problems

Question 1.
Write the missing numbers. Solution:
(i) 15 × (9 ÷ 3) (ii) 65 ÷ (9 + 3) (iii) (8 + 5) – (9 + 2) Question 2.
Write the missing operations in the trees. Solution:
(i) 8 + (6 ÷ 2) (ii) 39 – (6 × 5) Question 3.
Check whether the Tree diagrams are equal or not. Solution:
c ÷ (a ÷ b) ≠ a(b ÷ c)

Challenge Problems

Question 4.
Convert the following questions into tree diagrams:
(i) The number of people who visited a library in the last 5 months was 1210, 2100, 2550, 3160 and 3310. Draw the tree diagram of the total number of people who had used the library for the 5 months.
(ii) Ram had a bank deposit of ₹ 7,55,250 and he had withdrawn ₹ 5,34,500 for educational purpose. Find the amount left in his account. Draw a tree diagram for this.
(iii) In a cycle factory, 1,600 bicycles were manufactured on a day. Draw tree diagram to find the number of bicycles produced in 20 days.
(iv) A company with 30 employees decided to distribute ₹ 90,000 as a special bonus equally among its employees. Draw tree diagram to show how much will each receive?
Solution:
(i) People who visited the library for the past 5 months = 1210 + 2100 + 2550 + 3160 + 3310 = 12,330
Total number of people visited = 12,330
Tree Diagram (ii) Total bank deposit of Ram = ₹ 7,55,250
Amount withdrawn for educational purpose = ₹ 5,34,500
Amount left in his account = ₹ 2,20,750
Amount left in Ram’s Account = ₹ 2,20,750
Tree Diagram (iii) Number of bicycles manufactured in a day = 1600
Number of bicycles manufactured in 20 days = 1600 × 20 = 32,000
Number of bicycles manufactured in 20 days = 32,000
Tree Diagram (iv) Total amount distributed to 30 employees = 90,000
Amount received by one employee = $$\frac { 90000 }{ 30 }$$ = 3000
Each employee receive Rs. 3,000
Tree Diagram Question 5.
Write the numerical expression which gives the answer 10 and also convert into a tree diagram.
Solution:
Numerical expression 15 – (3 + 2)
Tree Diagram Question 6.
Use brackets inappropriate place to the expression 3 × 8 – 5 which gives 19 and convert it into tree diagram for it.
Solution:
Numerical expression (3 × 8) – 5
Tree Diagram Question 7.
A football team gains 3 and 4 points for successive 2 days and loses 5 points on the third day. Find the total points scored by the team and also represent this in the tree diagram.
Solution:
Total points scored by the team = 3 + 4 – 5 = 2
Tree Diagram Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 5 Information Processing Ex 5.1

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Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 5 Information Processing Ex 5.1

Question 1.
Convert the following numerical expressions into Tree diagrams.
(i) 8 + (6 × 2)
(ii) 9 – (2 × 3)
(iii) (3 × 5) – (4 ÷ 2)
(iv) [(2 × 4) + 2] × (8 ÷ 2)
(v) [(6 + 4) × 7] ÷ [2 × (10 – 5)]
(vi) [(4 × 3) ÷ 2] + [8 × (5 – 3)]
Solution:
(i) 8 + (6 × 2) (ii) 9 – (2 × 3) (iii) (3 × 5) – (4 ÷ 2) (iv) [(2 × 4) + 2] × (8 ÷ 2) (v) [(6 + 4) × 7] ÷ [2 × (10 – 5)] (vi) [(4 × 3) ÷ 2] + [8 × (5 – 3)] The first expression goes to the right side branch. So that the value does not change.

Question 2.
Convert the following Tree diagrams into numerical expressions. Solution:
(i) The numerical Expression is 9 × 8
(ii) The numerical expression is (7 + 6) – 5
(iii) The numerical expression is (8 + 2) – (6 + 1)
(iv) The numerical expression is (5 × 6) – (10 ÷ 2)

Question 3.
Convert the following algebraic expressions into tree diagrams.
(i) 10 V
(ii) 3a – b
(iii) 5x + y
(iv) 20t × p
(v) 2(a + b)
(vi) (x × y) – (y × z)
(vii) 4x + 5y
(viii) (lm – n) ÷ (pq + r)
Solution:
(i) 10 V (ii) 3a – b (iii) 5x + y (iv) 20t × p (v) 2(a + b) (vi) (x × y) – (y × z) (vii) 4x + 5y (viii) (lm – n) ÷ (pq + r) Question 4.
Convert tree diagram into Algebraic expression. Solution:
(i) Algebraic Expression is p + q
(ii) Algebraic Expression is l – m
(iii) Algebraic Expression is (a × b) – c (or) (ab) – c
(iv) Algebraic Expression is (a + b) – (c + d)
(v) Algebraic Expression is (8 ÷ a) + [ (6 ÷ 4) + 3]

Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 4 Geometry Intext Questions

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Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 4 Geometry Intext Questions

Try These (Textbook Page No. 61)

Question 1.
Complete the following table. In any triangle. Solution: Try These (Textbook Page No. 62)

Question 1.
Complete the table. Solution: Try These (Textbook Page No. 64)

Question 1.
Can a triangle be formed with the given sides? If yes, state the type of triangle formed. Solution: Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 4 Geometry Additional Questions

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Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 4 Geometry Additional Questions

Question 1.
Name the type of the following triangles.
(a) ∆PQR with m∠Q = 90°
(b) ∆ABC with m∠B = 90° and AB = BC
Solution:
(a) One of the angles is 90°
It is a right-angled triangle
(b) Since two sides are equal.
It is an isosceles triangle. Also m∠B = 90°
It is an Isosceles right-angled triangle

Question 2.
Classify the triangles (scalene, isosceles, equilateral) given below.
(a) ∆ABC, AB = BC
(b) ∆PQR, PQ = QR = RP
(c) ∆ABC, ∠B = 90°
(d) ∆EFG, EF = 3 cm, FG = 4 cm and GE = 3 cm
Solution:
(a) Isosceles triangle
(b) Equilateral triangle
(c) Right angled triangle
(d) Isosceles triangle

Question 3.
In triangle ∆ABC, AB = BC = CA = 5 cm. Then what is the value of ∠A, ∠B and ∠C?
Solution:
Since AB = BC = CA
∠A = ∠B = ∠C
We know that ∠A + ∠B + ∠C = 180°
∴ ∠A = ∠B = ∠C = 60°

Question 4.
In ∆PQR, ∠P = ∠Q = ∠R = 60°, then what can you say about the length of sides of ∆PQR? Also, write the name of the triangle?
Solution:
∠P = ∠Q = ∠R = 60°
So PQ = QR = RP
∆PQR is equilateral triangle

Question 5.
In ∆ABC, AB = BC and ∠A = 50°. Then find the value of ∠C?
Solution: It is an isosceles triangle.
∠A = ∠C
∠C = 50°

Question 6.
(a) Try to construct triangles using matchsticks.
(b) Can you make a triangle with?
(i) 3 matchsticks?
(ii) 4 matchsticks?
(iii) 5 matchsticks?
(iv) 6 matchsticks?
Name the type of triangle in each case. If you cannot make a triangle think of the reason for it.
Solution: (b) (i) With the help of 3 matchsticks, we can make an equilateral triangle. Since all three matchsticks are of equal length. (ii) With the help of 4 matchsticks, we cannot make any triangle because in this case, sum of two sides is equal to the third side and we know that the sum of the lengths of any two sides of a triangle is always greater than the length of the third side.
(iii) With the help of 5 matchsticks, we can make an isosceles triangle. Since we get two sides equal in this case. (iv) With the help of 6 matchsticks, we can make an equilateral triangle. Since we get three sides equal in length. Question 7.
A table is bought for ₹ 4500 and sold for ₹ 4800. Find the profit or loss.
Solution:
C.P = ₹ 4500
S.P. = ₹ 4800
Here S.P< C.P
Profit = S.P – C.P = ₹ 4800 – ₹ 4500 = ₹ 300

Question 8.
Draw any line segment $$\overline{\mathbf{P Q}}$$. Take any point R not on it. Through R, draw a perpendicular to $$\overline{\mathbf{P Q}}$$.
Solution:
Construction:
(i) Drawn a line segment $$\overline{\mathbf{P Q}}$$ using scale and taken a point R outside of $$\overline{\mathbf{P Q}}$$.
(ii) Placed a set-square on $$\overline{\mathbf{P Q}}$$ such that one arm of its right angle aligns along $$\overline{\mathbf{P Q}}$$.
(iii) Placed a scale along the other edge of the right angle of the set-square
(iv) Slide the set-square along the line till the point R touches the other arm of its right angle.
(v) Joined RS along the edge through R meeting $$\overline{\mathbf{P Q}}$$ at S.
Hence $$\overline{\mathbf{R S}}$$ ⊥ $$\overline{\mathbf{P Q}}$$. Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 4 Geometry Ex 4.3

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Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 4 Geometry Ex 4.3

Miscellaneous Practice Problems

Question 1.
What are the angles of an isosceles right-angled triangle?
Solution:
Since it is a right-angled triangle
One of the angles is 90°
Other two angles are equal because it is an isosceles triangle.
Other two angles must be 45° and 45°
Angles are 90°, 45°, 45°.

Question 2.
Which of the following correctly describes the given triangle. (a) It is a right isosceles triangle.
(b) It is an acute isosceles triangle.
(c) It is an obtuse isosceles triangle.
(d) it is an obtuse scalene triangle.
Solution:
(c) It is an obtuse isosceles triangle.

Question 3.
Which of the following is not possible?
(a) An obtuse isosceles triangle.
(b) An acute isosceles triangle.
(c) An obtuse equilateral triangle.
(d) An acute equilateral triangle.
Solution:
(c) An obtuse equilateral triangle.

Question 4.
If one angle of an isosceles triangle is 124°, then find the other angles.
Solution:
In an isosceles triangle, any two sides are equal. Also, two angles are equal.
Sum of three angles of a triangle = 180°
Given one angle = 124°
Sum of other two angles = 180° – 124° = 56°
Other angles are = $$\frac{56}{2}$$ = 28°
28° and 28°.

Question 5.
The diagram shows a square ABCD. If the line segment joints A and C, then mention the type of triangles so formed. Solution:
For a square all sides are equal and each angle is 90°.
∆ABC and ∆ADC are isosceles right-angled triangles.

Question 6.
Draw a line segment AB of length 6 cm. At each end of this line segment AB, draw a line perpendicular to the line AB. Are these lines parallel?
Solution:
Here CA and DB are perpendicular to AB.
Yes CA and DB are parallel. Construction:
(i) Drawn a line segment AB of length 6 cm.
(ii) Place the set square on the line in such a way that the vertex of its right angle coincides with B first and A next and one arm of the right angle coincides with the line AB.
(iii) Drawn lines DB and CA through B and A, the other arm of the right angle of the set square.
(iv) The line CA and DB are perpendicular to AB at A and B.

Challenge Problems

Question 7.
Is a triangle possible with the angles 90°, 90° and 0°, Why?
Solution:
No, a right-angled triangle cannot have more than one right angle.

Question 8.
Which of the following statements is true. Why?
(a) Every equilateral triangle is an isosceles triangle.
(b) Every isosceles triangle is an equilateral triangle.
Solution:
(a) It is true
In an equilateral triangle, all three sides are equal.
It can be an isosceles triangle also, which has two sides equal.
(b) But every isosceles triangle need not be an equilateral triangle.

Question 9.
If one angle of an isosceles triangle is 70°, then find the possibilities for the other two angles.
Solution:
(i) Given one angle = 70°
Also, it is an isosceles triangle.
Another one angle also can be 70°.
Sum of these two angles = 70° + 70° = 140°
We know that the sum of three angles in a triangle = 180°.
Third angle = 180° – 140° = 40°
One possibility is 70°, 70° and 40°
(ii) Also if one angle is 70°
Sum of other two angles = 180° – 70° = 110°
Both are equal. They are $$\frac{110}{2}$$ = 55°.
Another possibility is 70°, 55° and 55°.

Question 10.
Which of the following can be the sides of an isosceles triangle?
(a) 6 cm, 3 cm, 3 cm
(b) 5 cm, 2 cm, 2 cm
(c) 6 cm, 6 cm, 7 cm
(d) 4 cm, 4 cm, 8 cm
Solution:
In a triangle sum of any two sides greater than the third side
(a), (b) and (d) cannot form a triangle.
(c) can be the sides of an isosceles triangle.

Question 11.
Study the given figure and identify the following triangles. (a) equilateral triangle
(b) isosceles triangles
(c) scalene triangles
(d) acute triangles
(e) obtuse triangles
(f) right triangles
Solution:
(a) BC = 1 + 1 + 1 + 1 = 4 cm
AB = AC = 4 cm
∆ABC is an equilateral triangle.
(b) ∆ABC and ∆AEF are isosceles triangles.
Since AB = AC = 4 cm Also AE = AF.
(c) In a scalene triangle, no two sides are equal.
∆AEB, ∆AED, ∆ADF, ∆AFC, ∆ABD, ∆ADC, ∆ABF and ∆AEC are scalene triangles.
(d) In an acute-angled triangle all the three angles are less than 90°.
∆ABC, ∆AEF, ∆ABF and ∆AEC are acute-angled triangles.
(e) In an obtuse-angled triangle any one of the angles is greater than 90°.
∆AEB and ∆AFC are obtuse angled triangles.
(f) In a right triangle, one of the angles is 90°.

Question 12.
Two sides of the triangle are given in the table. Find the third side of the triangle? Solution: Question 13.
Complete the following table. Solution:
(i) Always acute angles
(ii) Acute angle
(iii) Obtuse angle

Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 4 Geometry Ex 4.2

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Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 4 Geometry Ex 4.2

Question 1.
Draw a line segment AB = 7 cm and Mark a point P on it. Draw a line perpendicular to the given line segment at P.
Solution:
Construction:
(i) Drawn a segment $$\overline{\mathrm{AB}}$$ such that $$\overline{\mathrm{AB}}$$ = 7 cm and took a point P anywhere on the line.
(ii) Placing the set square on the line in such a way that the vertex of its right angle coincides with P and one arm of the right angle coincides with the line AB.
(iii) Drawn a line PQ through P along the other arm of the right angle of the set square,
(iv) The line PQ is perpendicular to the line AB at P. ie PQ ⊥ AB and ∠APQ = ∠BPQ = 90° Question 2.
Draw a line segment LM = 6.5 cm and take a point P not lying on it. Using set square construct a line perpendicular to LM through P.
Solution:
Construction:
(i) Drawn a line segment $$\overline{\mathbf{L M}}$$ such that $$\overline{\mathbf{L M}}$$ = 6.5 cm and marked a point P anywhere above $$\overline{\mathbf{L M}}$$
(ii) Placing one of the arms of the right angle of the set square along the line segment LM.
(iii) Sliding the set square along the line segment in such a way that the other arm of its right angle touches the point P. Draw a line along this side, passing through point P meeting $$\overline{\mathbf{L M}}$$ at Q.
(iv) The line PQ is perpendicular to the line segment LM. ie LM ⊥ PQ. Question 3.
Find the distance between the given lines using a set square at two different points on each of the pairs of lines and check whether they are parallel. Solution:
Making the points P, Q, R, S and A, B, C, D on the given lines
PQ = RS = 0.9 cm
AB = CD = 1 cm
Distance between the two lines are equal.
They are parallel lines. Question 4.
Draw a line segment measuring 7,8cm. Mark a point B above it at a distance of 5 cm. Through B draw a line parallel to the given line segment.
Solution:
Construction:
(i) Using a scale drawn a line segment $$\overline{\mathrm{PQ}}$$ = 7.8 cm. Marked a point A on the line.
(ii) Placing the set square in such a way that the vertex of the right angle coincides with A and one of the edges of right angle lies along the line segment PQ. Mark a point B. Such that AB = 5 cm above the line PQ.
(iii) Placed the scale and the set square in such a way the set square is below PQ and one edge that form right angle with PQ. Placed the scale along the other edge of the right angle.
(iv) Holding the scale firmly and sliding the set square along the edge of the scale until the edge touches the point B. Drawn the line BC through B.
(v) Now the line BC is parallel to PQ i.e, BC || PQ. Question 5.
Draw a line. Mark a point R below it at a distance of 5.4 cm. Through R draw a line parallel to the given line.
Solution:
(i) Using a scale drawn a line AB and marked a point Q on the line.
(ii) Placing the set square in such a way that the vertex of the right angle coincides with Q and one of the edges of the right angle lies along AB. Marked the Point R such that QR = 5.4 cm.
(iii) Placing the set square above AB in such a way that one of the edges that form a right angle with AB. Placed the scale along the other edge of the right angle.
(iv) Holding the scale firmly and slide the set square along the edge of the scale until the edge touches the point R. Drawn the line RS through R. The line RS is parallel to AB i.e RS || AB. Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 4 Geometry Ex 4.1

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Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 4 Geometry Ex 4.1

Question 1.
Fill in the blanks.
(a) Every triangle has at least _____ acute angles.
(b) A triangle in which none of the sides equal is called a _____.
(c) In an isosceles triangle ______ angles are equal.
(d) The sum of three angles of a triangle is ______.
(e) A right-angled triangle with two equal sides is called ______.
Solution:
(a) Two
(b) Scalene Triangle
(c) Two
(d) 180°
(e) Isosceles right-angled triangle

Question 2.
Match the following:

 (i) No sides are equal Isosceles triangle (ii) One right angle Scalene triangle (iii) One obtuse angle Right-angled triangle (iv) Two sides of equal length Equilateral triangle (v) All sides are equal Obtuse angled triangle

Solution:

 (i) No sides are equal Scalene triangle (ii) One right angle Right-angled triangle (iii) One obtuse angle Obtuse angled triangle (iv) Two sides of equal length Isosceles triangle (v) All sides are equal Equilateral triangle

Question 3.
In ∆ABC, name the (a) Three sides: ____, _____, _____
(b) Three Angles: _____, _____, _____
(c) Three Vertices: _____, _____, _____
Solution:
(a) $$\overline{\mathrm{AB}}, \overline{\mathrm{BC}}, \overline{\mathrm{CA}}$$
(b) ∠ABC, ∠BCA, ∠CAB or ∠A, ∠B, ∠C
(c) A, B, C

Question 4.
Classify the given triangles based on its sides as scalene, isosceles or equilateral. Solution:
(i) Equilateral Triangle
(ii) Scalene Triangle
(iii) Isosceles Triangle
(iv) Scalene Triangle

Question 5.
Classify the given triangles based on its angles as acute-angled, right-angled or obtuse-angled. Solution:
(i) Acute angled triangle
(ii) Right angled triangle
(iii) Obtuse angled triangle
(iv) Acute angled triangle

Question 6.
Classify the following triangles based on its sides and angles. Solution:
(i) Isosceles Acute angled triangle
(ii) Scalene Right angled triangle
(iii) Isosceles Obtuse angled triangle
(iv) Isosceles Right angled triangle
(v) Equilateral Acute angled triangle
(vi) Scalene Obtuse angled triangle

Question 7.
Can a triangle be formed with the following sides? If yes, name the type of triangle.
(i) 8 cm, 6 cm, 4 cm
(ii) 10 cm, 8 cm, 5 cm
(iii) 6.2 cm, 1.3 cm, 3.5 cm
(iv) 6 cm, 6 cm, 4 cm
(v) 3.5 cm, 3.5 cm, 3.5 cm
(vi) 9 cm, 4 cm, 5 cm
Solution:
(i) 8 cm, 6 cm, 4 cm
Sum of smaller sides = 6 cm + 4 cm = 10 cm > 8 cm, third side
∴ Triangle can be formed with this sides
None of the sides is equal.
∴ Yes, it is a scalene triangle

(ii) 10 cm, 8 cm, 5 cm.
Sum of two smaller sides = 8 cm + 5 cm = 13 cm > 10 cm, third side
∴ Triangle can be formed
None of the sides is equal.
∴ Yes, it is a scalene triangle

(iii) 6.2 cm, 1.3 cm, 3.5 cm.
Here sum of two smaller sides = 1.3 cm + 3.5 cm = 4.8 cm < 6.2 cm, third side.
∴ No, The triangle cannot be formed

(iv) 6 cm, 6 cm, 4 cm
Sum of two smaller sides = 4 cm + 6 cm = 10 cm > 6 cm, third side.
∴ Triangle can be formed. Also two sides are equal.
∴ Yes, it is an isosceles triangle.

(v) 3.5 cm, 3.5cm, 3.5cm.
Here sum of two sides = 3.5 cm + 3.5 cm = 7 cm > 3.5 cm (Third side)
∴ Triangle can be formed.
Also, all three sides are equal.
∴ Yes, it is an equilateral triangle.

(vi) 9 cm, 4 cm, 5 cm.
Sum of two smaller sides = 4 cm + 5 cm = 9 cm (the third side).
∴ No, The triangle cannot be formed.

Question 8.
Can a triangle be formed with the following angles? if yes, name the type of triangle.
(i) 60°, 60°, 60°
(ii) 90°, 55°, 35°
(iii) 60°, 40°, 42°
(iv) 60°, 90°, 90°
(v) 70°, 60°, 50°
(vi) 100°, 50°, 30°
Solution:
(i) 60°, 60°, 60°
Sum of three angles = 60° + 60° + 60° = 180°
Yes, a triangle can be formed.
∴ It is Acute angled triangle. [∵ all the angles < 90°]

(ii) 90°, 55°, 35°.
Sum of three angles = 90° + 55° + 55° = 180°
Yes, a triangle can be formed.
∴ It is a right-angled triangle, [∵ one angle is 90°]

(iii) 60°, 40°, 42°.
Sum of three angles = 60° + 40° + 42° = 142° ≠ 180°
No, The triangle cannot be formed.

(iv) 60°, 90°, 90°.
Sum of three angles = 60° + 90° + 90° = 240° ≠ 180°
∴ No, The triangle cannot be formed. [∵ one angle is > 90°]

(v) 70°, 60°, 50°.
Sum of three angles = 70° + 60° + 50° = 180°
Yes, A triangle can be formed.
∴ It is an acute-angled triangle.

(vi) 100°, 50°, 30°.
Sum of three angles = 100° + 50° + 30° = 180°
Yes, A triangle can be formed.
∴ It is an obtuse-angled triangle.

Question 9.
Two angles of the triangles are given. Find the third angle
(i) 80°, 60°
(ii) 75°, 35°
(iii) 52°, 68°
(iv) 50°, 90°
(v) 120°, 30°
(vi) 55°, 85°
Solution:
(i) 80°, 60°
Sum of the two angles = 80° + 60° = 140°
We know that sum of three angles of a triangle = 180°
Third angle = 180° – 140° = 40°

(ii) 75°, 35°
In the triangle sum of two angles = 75° + 35° = 110°
In a triangle sum of three angles = 180°
Third angle = 180° – 110° = 70°

(iii) 52°, 68°
In a triangle sum of 3 angles = 180°
Here sum of two given angles = 52° + 68° = 120°
Third angle = 180° – 120° = 60°

(iv) 50°, 90°
In a triangle sum of three angles = 180°
Sum of two given angles = 50° + 90° = 140°
Third angle = 180° – 140° = 40°

(v) 120°, 30°
In a triangle sum of three angles = 180°
Sum of two given angles = 120° + 30° = 150°
Third angle = 180° – 150° = 30°

(vi) 55°, 85°
In a triangle sum of three angles = 180°
Sum of two given angles = 55° + 85° = 140°
Third angle = 180° – 140° = 40°

Question 10.
I am a closed figure with each of my three angles is 60°. Who am I?
Solution:
Equilateral Triangle.

Question 11.
Using the given information, write the type of triangle in the table given below. Solution: Objective Type Questions

Question 12.
The given triangle is _____. (a) a right angled triangle
(b) an equilateral triangle
(c) a scalene triangle
(d) an obtuse angled triangle
Solution:
(b) an equilateral triangle

Question 13.
If all angles of a triangle are less than a right angle, then it is called _____.
(a) an obtuse-angled triangle
(b) a right-angled triangle
(c) an isosceles right-angled triangle
(d) an acute-angled triangle
Solution:
(d) an acute-angled triangle

Question 14.
If two sides of a triangle are 5 cm and 9 cm then the third side is _____.
(a) 5 cm
(b) 3 cm
(c) 4 cm
(d) 14 cm
Solution:
(a) 5 cm

Question 15.
The angles of a right-angled triangle are
(a) acute, acute, obtuse
(c) acute, right, right
(c) right, obtuse, acute
(d) acute, acute, right
Solution:
(d) acute, acute, right

Question 16.
An equilateral triangle is
(a) an obtuse-angled triangle
(b) a right-angled triangle
(c) an acute-angled triangle
(d) scalene triangle
Solution:
(c) an acute-angled triangle

Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 3 Bill, Profit and Loss Intext Questions

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 2 Chapter 3 Bill, Profit and Loss Intext Questions

Try These (Textbook Page No.49)

Question 1.
Arrange in ascending order
(i) C.P, M.P, Discount
(ii) M.P., S.P., Discount
Solution:
(i) Discount < C.P < M.P.
(ii) Discount < S.P < M.P

Question 2.
Which is greater S.P or M.P?
Solution:
M.P is greater than S.P