Axioms, Postulates and Theorems – Class VIII

Axioms:

There are certain elementary statements, which are self evident and which are accepted without any questions. These are called axioms.

Axiom 1: Things which are equal to the same thing are equal to one another.

For example:

Draw a line segment AB of length 10cm. Draw a second line CD having length equal to that of AB, using a compass. Measure the length of CD. We see that, CD = 10cm.

We can write it as, CD = AB and AB = 10cm implies CD = 10cm.

Axioms, Postulates and Theorems - class VIII


Axiom 2: If equals are added to equals, the wholes are equal.

Suppose we have two line segments AB and DE of equal length. Add BC to AB and add EF to DE. If BC = EF, then AC = DF.

Axioms, Postulates and Theorems - class VIII


Axiom 3: If equals are subtracted from equals, then the remainders are equal.

Suppose we have two line segments of equal length. Remove BC from AC and EF from DE respectively. If BC = EF, then AB = DE.

Axioms, Postulates and Theorems - class VIII


Axiom 4: Things which coincide with one another must be equal to one another.

This means that if two geometric figures can fit completely one in to another, they are essentially the same.


Axiom 5: The whole is greater than the part.

Take a container of water. Remove some water from it. Will the remaining volume of water the same as the original volume?


Geometric postulates – Axioms, Postulates and Theorems

Postulates in geometry are very similar to axioms, self-evident truths, and beliefs in logic, political philosophy and personal decision-making.

They are as follows:

  1. A straight line may be drawn from any given point to any other.
  2. A straight line may be extended to any finite length.
  3. A circle may be described with any given point as its center and any distance as its radius.
  4. All right angles are congruent.
  5. If a straight line intersects two other straight lines, and so makes the two interior angles on one side of it together less than two right angles, then the other straight lines will meet at a point if extended far enough on the side on which the angles are less than two right angles.

Axioms, Postulates and Theorems – EXERCISE 3.1.2

1. What are undefined objects in Euclid’s geometry?
Solution:

Points, line, plane are the undefined objects in Euclid’s geometry.


2. What are the difference between an axiom and a postulate?
Solution:

Axioms are elementary statements , which are self evident and which are accepted without questions

Postulates are statements which are particular to Geometry and accepted without question.


3. Give an example for the following axioms from your experience:
a) If equals are added to equals , the wholes are equal.
Solution:

4 + 2 = 5 + 1 ( equals)
Adding 3 both sides.
4 + 2+ 3 = 5 + 1 + 3 ( adding equals)
9 = 9 ( whole is equal)

b) The whole is greater than the part .
Solution:

Our hard is not part of our body. It is smaller than the body.


4. What is the need of introducing axioms?
Solution:

Greek mathematicians while developing geometry as a pure deductive science. They had to depend on certain primitive notions like points, straight lines, planes and space. But this was not enough to deduce everything . They had to set up certain statements, whose validity was accepted unquestionably thus there was a need of introducing axioms.


5. You have seen earlier that the set of all natural numbers is closed under additional (closure property). Is this an axiom or same thing you can prove?

Solution:

Yes


Lines and Angles – Axioms, Postulates and Theorems

Suppose we have a ray OA on a plane with end point O. With the same plane end point. Consider another ray OB in the same plane. We observe that OB is obtained from OA through suitable rotation around  the point O. We say OB subtends an angle with angle OA. The amount of rotation is the measure of this angle.jh

∠AOB

We use numerical measurement called degree to measure angles. We use the notation to denote a degrees. The rays OA and OB are called the sides of the angle and O is called the vertex if the angle. The angle subtended by the rays OA and OB is denoted by ∠AOB.

We know a geometrical instrument called protractor which is used to measure the angles.

We Know, different types of angles:

Straight line angle, right angle, acute angle, obtuse angle, reflex angle, complete angle, adjacent angle, complementary angles and supplementary angles.

Let us recall one by one,

Axioms, Postulates and Theorems - Class VIII

Two angles are said to be supplementary angles if their sum is 180˚. Similarly, two angles are said to be complementary if they add up to 90˚

Two angles are said to be adjacent angles, if both the angles have a common vertex and a common side.

Axioms, Postulates and Theorems - Class VIII

While measuring the lengths are segments of the angles we observe the following rules:

Rule 1: Every line segment has a positive length. (The length of the line segment AB is denoted by AB or |AB|.)

Rule 2: If a point C lies on a line segment |AB|, then the length of |AB| is equal to the sum of the lengths of |AC| and |CB|; that is AB = AC + CB.

Rule 3: Every angles has a certain magnitude. A straight angles measures 180°

Rule 4: If OA, OB and OC are such that OC lies between OA and OB then ∠AOB = ∠AOC + ∠COB

Rule 5: If the angle between two rays is zero then they coincide. Conversely,  if two rays coincide, the angle between them is either zero or an integral multiple of 360°

Note: While measuring the angles, if the angle is measured anti-clock-wise, it is positive. If it is measured clock-wise, then it is negative.


Axioms, Postulates and Theorems – EXERCISE 3.1.3

1. Draw diagram illustrating each of the following situation:
a) Three straight lines which do not pass through a fixed point.

Solution:

Axioms, Postulates and Theorems - Class VIII

b) A point and rays emanating from that point such that the angle.

Axioms, Postulates and Theorems - Class VIII

Solution: 

∠SPR acute angle

∠RPT acute angle

∠QPT acute angle

 

c) Two angles which are not adjacent angles, but still supplementary.

Solution:

4

d) Three points in the plane which are equidistant from each other.

Solution:

5


2. Recognise the type of angles the following figures:

(i)

6

Solution:

Adjacent angles BQX and XQA
BQX is obtuse and AQX acute

(ii)

Axioms, Postulates and Theorems - Class VIII

Solution:

∠YRP and ∠ROA are corresponding angles

∠ROA and ∠BOX are vertically opposite angles

∠BOR and ∠ROA, ∠YRQ and ∠YRP and like are adjacent angles

Axioms, Postulates and Theorems - Class VIII(iii)

reflex angle = ref∠AOB

∠AOB is acute angle

ref∠AOB + ∠AOB is complete angle


Axioms, Postulates and Theorems - Class VIII

3. Find value of x in each of the following diagrams:

Solution:

∠AOB = 180°

∠AOB = ∠AOC + ∠COB

180° = 2x + x

180° = 3x

x = 180°/3

x = 60°, ∠COB = 60°

2x = 120°

Therefore, ∠AOC = 120°


(ii)

3.2

Solution:

∠AOB = 180°

∠AOB = ∠AOC + ∠COD + ∠DOB = 180°

180° = 90° + 4x + x

180° – 90° = 5x

90° = 5x

x = 90°/5

x = 18°

Thus, 4x = 4×18° = 72°


(iii)

Axioms, Postulates and Theorems - Class VIII

Solution:

∠AOB = 180°

∠AOB = ∠AOC + ∠COB = 180°

180° = (x-y)+(x+y)

180° = 2x + 0

180° = 2x

x = 180°/2

Therefore, x = 90°


(iv)

Axioms, Postulates and Theorems - Class VIII

Solution:

∠AOC = ∠BOD = x + 30

∠AOE = ∠BOF = x

∠COF = ∠EOD = 3x

∠AOB = ∠AOC + ∠COF + ∠FOB

180° = x + 30° + x + 3x

180° – 30° = 5x

150° = 5x

x = 180°/5 = 36°

∠AOC = ∠BOD = 36° + 30° = 66°

∠AOE = ∠BOF = 36°

Therefore, ∠COF = ∠EOD = 3(36°) = 108°


(v)

Axioms, Postulates and Theorems - Class VIII

Solution:

∠AOB = 180°

∠AOC = ∠AOD  and ∠AOC +∠AOD =90°

∠AOC +∠AOC = 90°

2∠AOC = 90°

Thus, ∠AOC = 45°

∠AOB = ∠AOC + ∠COB

180° = 45° + x

x = 180° – 45°

Hence, x = 135°

 

(vi)

Axioms, Postulates and Theorems - Class VIII

Solution:

We have,

y + 65° = 180°

y = 180° – 65°

y = 115°

x + y  = 180°

x = 180° – 115°

Hence, x = 65°


4.Which pair of angles are supplementary in the following diagram ? Are there supplementary rays ?

Solution:

4.1

COD and BOA are supplementary angles , they are not supplementary rays.


5. Suppose two adjacent angles are supplementary. Show that if one of them is an obtuse angle, then the other angle must be acute.

Solution:

6

It is given that, ∠BQX and ∠XQA are complementary angles, ∠BQX is obtuse angle and ∠XQA is acute angle.

We know, ∠AQB = 180° = ∠BQX + ∠XQA and ∠BQX is obtuse angle , therefore, ∠BQX > 90°

Thus,

∠AOB = obtuse angle + ∠XQA

∠AOB = (<90) + ∠XQA

Therefore, ∠XQA = ∠AOB – (<90°)

Hence, ∠XQA = acute angle


Axioms, Postulates and Theorems – Exercise 3.1.4

  1. Find all the angles in following figure.

Solution:

Axioms, Postulates and Theorems - Class VIII

Solution:

It is given that, ∠PMD = 135°

We have, ∠PMD + ∠PMC = ∠CMD

135° + ∠PMC = 180°

∠PMC = 180° – 135° = 45°

∠PMD = 135° = ∠CML [vertically opposite angles]

∠PMC = 45° = ∠DML [vertically opposite angles]

∠CML =  135° = ∠BLM [corresponding angles]

∠DML = 45° = ∠ALM [corresponding angle]

∠BLM = 135° = ∠ALQ [vertically opposite angles]

∠ALM = 45° = ∠ALM [vertically opposite angles]

a


2. Find the value of x in the diagram below:

31.1.4 2

Solution:

It is given that, ∠MKI = 90°

∠MKH = ∠MKI + ∠IKH = 180°

∠IKH = ∠MKH – ∠MKI = 180° – 90°

∠IKH = 90°

It is also given that, ∠ADE = 130°

we know, ∠ADE = 130° = ∠DHI  [corresponding angles]

Therefore, ∠DHK = ∠DHI + ∠IHK = 180°

180°  = 130°+ ∠IHK

∠IHK = 50°

∠IKH + ∠HIK + ∠IHK = 180°

180° – 50° – 90° =∠IHK

Hence, ∠IHK = 40°


3.Show that if a straight line is perpendicular to one of the two or more parallel lines , then it is also perpendicular t the reminding lines.

Solution:

gfgh

Data: AB||CD and EH⊥ AB

To prove: EH ⊥ CD

∠AFE = 90° = ∠EFB [data]

∠AFE = 90° = ∠CGF [corresponding angles , AB||CD]

∠FGD = 90° = ∠EFB [corresponding angles, AB||CD]

Therefore, EF ⊥ CD


4.Let AB and CD be two parallel lines and PQ be transversal. Show that the angle bisectors of a pair of two interior angles on the same side of the transversal are perpendicular to each other.

Solution:

lksjkd

Data : AB||CD
PQ is a transversal , cuts AB and CD at M and N respectively MX and NY are angular bisectors of
BMN and MND respectively.
To prove that : MON = 90˚
Proof – In the figure
BMN + MND = 180˚ * Prepositions+
Let BMN = 2a˚ and MND = 2b˚
⇒ 20˚ + 2b˚ = 180˚
2 ( a + b ) = 180˚
a+b = 180/2 = 90°

a+b = 90˚
in the Δ MNO
LMN + ONM + MON = 180˚

[Sum of the interior angles of a triangle]
a˚ + b˚ + MON = 180˚
90˚ + MON = 180˚
MON = 180˚ – 90˚
MON = 90˚
Therefore,  angle bisector of a pair of two internal angles on the same side of the transversed are perpendicular to each other.


 

 

 

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