Conversion of Algebraic Expressions into Tree Diagrams - 6th Maths Term 2

Conversion of Algebraic Expressions into Tree Diagrams

Chapter: 6th Maths : Term 2 Unit 5 : Information Processing

There is more fun with trees.

Conversion of Algebraic Expressions into Tree Diagrams

There is more fun with trees. Observe the following trees

The above tree is nothing but the familiar equation $a \times (b + c) = (a \times b) + (a \times c)$. Thus we can see the algebraic expressions as trees.

• • The tree on the left has less number of nodes and looks simple.
• • The tree on the right has more number of nodes
• • Can we conclude that the value of both the trees are different

Example 15:

Convert ‘5a’ into Tree diagram

Solution:

Example 16:

Convert '3a + b' into Tree diagram

Algebraic expression

3a+b

Tree diagram

Example 17:

'6 times a and 7 less’ Convert into a Tree diagram.

Algebraic expression

6a − 7

Tree diagram

Example 18:

Convert the tree diagram into an algebraic expression.

Tree diagram

Algebraic expression

8b ÷ 6

Example 19:

Convert the tree diagram into an algebraic expression.

Tree diagram

Algebraic expression

(7 + t) 5

Example 20:

Verify whether given trees are equal or not

Tree diagram

( a + b) + c = a + (b+c)

Yes, they are equal.

Try these

1. Check whether the Tree diagrams are equal or not

2. Check whether the following algebraic expressions are equal or not by using Tree diagrams

i) (x − y) + z and x − (y + z)

(x – y) + z ≠ x – (y + z). They are not equal.

ii) (p × q) × r and p × (q × r)

(p × q) × r = p × (q × r). They are equal.

iii) a − (b − c) and (a − b) − c

a – (b – c) ≠ (a – b) – c. They are not equal.

Do You Know

Consider the numerical expression 9 – 4. which means 4 is to be subtracted from 9. 9 – 4 can be represented as – 9 4 (so far we have come across with operation in between the operands)

Suppose the expression is 9 – 4 × 2. This can be represented as × – 9 4 2 gives the meaning of

Step 1: × 9 – 4 2

Step 2: (9 – 4) × 2

Take the expression + × − 9 4 2 5

Step 1: + × 9 − 4 2 5

Step 2: + (9 − 4 ) × 2 5

Step 3: [(9 − 4) × 2] + 5

This is reading an expression from “left to right”. Similarly, we can read expressions from “right to left” also

9 4 2 5 + × − can be read as “right to left” expression which gives the meaning of

• 9 4 2 5 + × => (9 − 4) 2 5 + ×
•          => (9 − 4) × 2 5 +
•          => [(9 − 4) × 2] + 5

Hence an expression can be read as “left to right” or “right to left” giving the same answer which is similar to name 4 as Naangu (நான்கு), Four, Nalagu (నాలుగు) and Char (चार), all of them representing the collection of four objects. Similarly the numerical expression

[ (9 – 4) × 8 ] ÷ [ (8 + 2) × 3] can be written as ÷ × – 9 4 8 × + 8 2 3 ( left to right) or 8 9 4 – × 3 8 2 + × ÷ ( right to left ).

Try these:

1) × – + 9 7 8 2

2) ÷ × + 2 3 8 5

6th Maths Term 2: Conversion of Tree Diagrams into Numerical Expressions

Conversion of Tree Diagrams into Numerical Expressions

Chapter: 6th Maths : Term 2 Unit 5 : Information Processing

When the nodes for addition and subtraction are interchanged the value remains the same which is represented using tree diagram as given below.

Conversion of Tree Diagrams into Numerical Expressions

For instance, consider the tree

We could first find as 10, then as 5.

When we multiply the results 10 and 5 we get 50. When the nodes for addition and subtraction are interchanged the value remains the same which is represented using tree diagram as given below.

Does it mean that the branches also can be interchanged? Yes, when the node is addition it is possible.

But it is not possible when the node represents subtraction.

Therefore from this tree diagram:

The expression can be converted into either (10−5) × (8 + 2) or (8 + 2) × (10 − 5) or (2 + 8) × (10 − 5) or (10 − 5) × (2 + 8) without changing the value.

Example 11:

Convert the tree diagram into numerical expression.

Tree diagram

Numerical Expression

15 − 6 + 9

Example 12:

Convert the Tree diagram into a numerical expression.

Tree diagram

Numerical Expression

8 × 9 + 10

Example 13:

Convert the Tree diagram into a numerical expression.

Tree diagram

Numerical Expression

10 × 6 + 6 ÷ 2

Example 14:

Convert the Tree diagram into a numerical expression.

Tree diagram

Numerical Expression

[7 × 9 + 2] + [6 – 3 + 3]

6th Maths Term 2 Chapter 5 Information Processing Study Material

Information Processing

Term 2 Chapter 5 | 6th Standard Mathematics

Learning Objectives

• To know how to represent numerical and algebraic expressions by tree diagrams.
• To know how to write numerical and algebraic expressions from tree diagrams.

Introduction

In today's digital era, it is almost impossible to imagine a day without computers. Right from small shops to big software companies, the use of computers is inevitable. If there are no computers, most of the works will be stopped. Computers are able to find solutions even for complicated numerical expression and algebraic expression in quick and easy way. The answer given by the computer will be very precise and need not to be recalculated. There will be a question, how the computer read these expression?.

Yes, Computers use Tree diagram to perform billions of operations in a uniform way and gives the answer. We will learn about the Tree diagram for both numeric and algebraic expressions in this chapter.

MATHEMATICS ALIVE – INFORMATION PROCESSING IN REAL LIFE

Consider the numerical expression [(9 − 4) × 8] ÷ [(8 + 2) × 3]. We can try to understand the expression in a better way through the tree diagram.

1) Let us consider e₁ = (9 − 4) × 8, e₂ = (8 + 2) × 3 we get:

2) e₁ = f₁ × f₂ Where, f₁ = 9 − 4 and f₂ = 8

3) f₁ = r₁ − r₂ where r₁ = 9 and r₂ = 4. f₁ is represented as:

Similarly, the trees can be developed from e₂.

4) Putting all together, we get the following tree diagram:

It is a picture which look like an upside-down tree! Every node has one or two branches. And the leaves are numbers. The branching nodes have operations on them. It is called tree diagram and the tree diagrams are general ways of representing arithmetical expressions. Here trees are drawn upside down.

The root is at the top, the leaves are at the bottom. Since all the arithmetical operations are binary (Involving two numbers) we have only 2 way branching in the tree.

Can you represent the addition of four numbers in the same way? Yes, there is a way for addition of 4 numbers.

Example 1

In the flower exhibition conducted at Ooty for 4 days the number of tickets sold on the first, second, third and fourth days are 1,10,010; 75,070; 25,720 and 30,636 respectively. Find the total number of tickets sold.

Solution:

• Number of tickets sold on the first day = 1,10,010
• Number of tickets sold on the second day = 75,070
• Number of tickets sold on the third day = 25,720
• Number of tickets sold on the fourth day = 30,636
• Total = 2,41,436

Total number of tickets sold = 2,41,436

Example 2

In one year, a paper company had sold 6,25,610 notebooks out of a stock of 7,50,800 notebooks. Find the number of notebooks left unsold.

Solution: Tree Diagram

• Number of Notebooks in stock = 7,50,800
• Number of Notebooks sold = 6,25,610
• Number of notebooks left unsold = 1,25,190

Example 3

Vani and Kala along with three other friends went to a butter milk shop. The cost of one butter milk is ₹ 6. If 9 more friends joined them, then how much money did they have to pay? Vani said they had to pay ₹ 84 whereas Kala said they had to pay ₹ 59. Who is correct?

Solution: Tree Diagram of (5 + 9) x 6

This confusion can be resolved by using the brackets in the correct places like (5+9) × 6. It is further clear from the tree diagram. Therefore Vani is correct.

Example 4

If a ration shop has distributed 1,00,000 kg of rice to 5000 families, then find the quantity of rice given to each family?

Solution: Tree Diagram

• Quantity of rice to be distributed to 5000 families = 1,00,000 kg
• Quantity of rice distributed to each family = 1,00,000 ÷ 5,000
• = 20 kg

Each family was given 20 kg of rice.

Example 5

Convert into a Tree diagram: (9 × 5) + (10 × 12)

Solution:

Example 6

Convert into a Tree diagram: (10 × 9) − (8 × 2) + 3

Solution:

Example 7

Convert into a Tree diagram: [8 + (5 × 2)] – [(2 × 3) + 5]

Solution:

Example 8

Convert into a Tree diagram: [(9−4) × 8] + [(8+2) × 3]

Solution:

Example 9

Convert into a Tree diagram: {[(10 × 5) + 6] × [5 + (6 − 2)]} ÷ [8 × (4 + 2)]

Solution:

Example 10

Convert into a Tree diagram: 20 + [8 × 2 + {(6 × 3) – 10 ÷ 5} ]

Solution:

6th Maths Term 2 Unit 4 Geometry Summary & Key Properties

6th Maths : Term 2 Unit 4 : Geometry

Chapter Review and Core Concepts

Summary

•

A closed figure formed by three line segments is called a triangle.

[Image of different types of triangles]

•

A triangle has 3 sides, 3 angles and 3 vertices.

•

Based on the sides of triangles, we can classify triangles into 3 types as scalene triangle, isosceles triangle and equilateral triangle.

[Image of scalene, isosceles and equilateral triangles]

•

Based on the angles of triangles, we can classify triangles into 3 types as acute angled triangle, right angled triangle and obtuse angled triangle.

[Image of acute, right and obtuse angled triangles]

•

In a triangle, the sum of any two sides is greater than the third side. This is known as Triangle Inequality property.

•

Sum of three angles of a triangle is 180°.

•

Parallel and Perperndicular lines can easily be drawn using set squares.

•

The distance between a set of parallel lines always remains the same.

6th Maths Term 2 Geometry Exercise 4.3 Solutions & Challenge Problems

6th Maths : Term 2 Unit 4 : Geometry

Questions with Answers, Solution | Term 2 Chapter 4 | Exercise 4.3

Exercise 4.3

Miscellaneous Practice Problems

1. What are the angles of an isosceles right angled triangle?

45°, 45°, 90°

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.

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

4. If one angle of an isosceles triangle is 124° , then find the other angles.

One angle of an isosceles triangle = 124°
The sum of three angles of a triangle = 180°
The other two isosceles angles = 180° − 124° = 56°
One angle = 56 / 2 = 28°
The other two angles are 28°, 28°.

5. The diagram shows a square ABCD. If the line segment joins A and C, then mention the type of triangles so formed.

Both triangles are Isoceles Right angled triangle.

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?

Yes. These two perpendicular lines are parallel.

Challenge Problems

7. Is a triangle possible with the angles 90° , 90° and 0° Why?

90°, 90°, 0°
A triangle can not be formed with the given angles. Because, Atriangle can not have more than one right angle.

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.

'a' is true. Because an isosceles triangle need not have three equal sides.

9. If one angle of an isosceles triangle is 70°, then find the possibilities for the other two angles.

One angle of an isosceles triangle = 70°
If the other angle also = 70°
The third angle = 180° − (70° + 70°) = 180° − 140° = 40°
The other two angles are 70° and 40° (or)
One angle = 70°
The other two isosceles angles = (180° − 70°) = 110°
So, one angle = 110 / 2 = 55°
The other two isosceles angles are 55° and 55°
So, the two angles 70°, 40° (or) may be, 55°, 55°..

10. Which of the following can be the sides of an isosceles triangle?

a) 6cm, 3cm, 3cm

b) 5cm, 2cm, 2cm

c) 6cm, 6cm, 7cm

d) 4cm, 4cm, 8cm

a) 6 cm, 3 cm, 3 cm: Sum (3+3=6) is not greater than 6. Not possible.
b) 5 cm, 2 cm, 2 cm: Sum (2+2=4) is lesser than 5. Not possible.
c) 6 cm, 6 cm, 7 cm: Sum (6+6=12) is greater than 7. Possible.
d) 4 cm, 4 cm, 8 cm: Sum (4+4=8) is not greater than 8. Not possible.

11. Study the given figure and identify the following triangles.

(a) Equilateral triangle: ΔABC
(b) Isosceles triangles: Δ ABC, Δ AEF
(c) Scalene triangles: Δ AEB, Δ AED, Δ ADF, Δ AFC, Δ ABD, Δ ADC, Δ ABF, Δ AEC
(d) Acute triangles: Δ ABC, Δ AEF, Δ ABF, Δ AEC
(e) Obtuse triangles: Δ AEB, Δ AFC
(f) Right triangles: Δ ADB, Δ ADC, Δ ADE, Δ ADF

12. Two sides of the triangle are given in the table. Find the third side of the triangle.

i) 7cm, 4 cm: The third side is, 4 cm or 5 cm or 6 cm.
ii) 8 cm, 8 cm: The third side is, 5 cm or 6 cm or 7 cm.
iii) 7.5 cm, 3.5 cm: The third side is 4.5 cm or 5 cm or 6 cm
iv. 10 cm, 14 cm: The third side is 5 cm or 7 cm or 8 cm

13. Complete the following table :

Quick Reference Answers

Exercise 4.3
1. 90°, 45° , 45°
2. c
3. c
4. 28°, 28°
5. Both are Isosceles Right angled triangles
6. Yes
7. No, A triangle cannot have more than one right angle
8. “a” is true, because an isosceles triangle need not have three equal sides
9. 70°,40° or 55°,55°
10. c
11. a) ∆ABC | b) ∆ABC , ∆AEF | c) ∆AEB, ∆AED, ∆ADF, ∆AFC, ∆ABD, ∆ADC, ∆ABF, ∆AEC | d) ∆ABC, ∆AEF, ∆ABF, ∆AEC | e) ∆AEB, ∆AFC | f) ∆ADB, ∆ADC, ∆ADE, ∆ADF
12. (i) between 3 and 11 (ii) between 0 and 16 (iii) between 4 and 11 (iv) between 4 and 24
13. i. Always acute angles | ii. Acute angle | iii. Obtuse angle

6th Maths Term 2 Geometry Exercise 4.2 Solutions & Constructions

6th Maths: Term 2 Unit 4: Geometry

Exercise 4.2 - Questions with Answers & Solutions

Exercise 4.2

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.

Step 1 :

Draw a line AB = 7 cm and take a point P on the line

Step 2 :

Place the set square on the line in such a way that the vertex which forms right angle coincides with P and one arm of the right angle coincides with the line AB.

Step 3 :

Draw a line PQ through P along the other arm of the right angle at the set square.

Step 4 :

The line PQ is perpendicular to the line AB at P.

That is PQ ⊥ AB.
∠APQ = ∠BPQ = 90°

2. Draw a line segment LM = 6.5 cm and take a point P not lying on it. Using a set square construct a line perpendicular to LM through P.

Step 1 :

Draw a line LM = 6.5 cm. Take a point P above the line LM.

Step 2 :

Place one of the arms of the right angle of a set square along the line LM and the other arm of its right angle touches the point P.

Step 3 :

Draw a line through the point P meeting LM at Q.

Step 4 :

The line PQ is perpendicular to the line LM at Q.

That is PQ ⊥ LM.

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.

Yes. They are parallel.

4. Draw a line segment measuring 7.8 cm. Mark a point B above it at a distance of 5 cm. Through B draw a line parallel to the given line segment.

Step 1 :

Draw a line segment xy = 7.8 cm and mark a point A on the line.

Step 2 :

Place the set square in such away that the vertex of the right angle coincides with A and one of the edges of right angle lies along XY mark the point B such that AB = 5 cm.

Step 3 :

Place the set square under the line XY and fix the scale touching the set square.

Step 4 :

Hold the scale firmly and slide the set square along the edge of the scale until the other edge touches the point B. Draw a line BC through B.

Step 5 :

The line BC is parallel to XY. That is, BC || XY.

5. Draw a line and mark a point R below it at a distance of 5.4 cm Through R draw a line parallel to the given line.

Step 1 :

Using a scale draw a line AB and mark a point C on the line.

Step 2 :

Place the set square upside down such that the vertex of the right angle coincides with C and one of the edges of right angle lies along AB. Mark a point R below AB such that CR = 5.4 cm

Step 3 :

Place the setsquare below AB and fix the scale touching the set square.

Step 4 :

Hold the scale firmly and slide the set square down until the other edge touches the point R. Draw a line RS through R.

Step 5 :

The line RS is parallel to AB. That is RS || AB.