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NCERT Class 9 Maths Lab Manual – Verify the Algebraic Identity (a+b+c)² = a²+b²+c²+2ab+2bc+2ca
OBJECTIVE
To verify the algebraic identity (a+b+c)² = a²+b²+c²+2ab+2bc+2ca .
Materials Required
![Lab Lab](https://i.pinimg.com/originals/96/d2/e5/96d2e59021dc5075142d01d7e659ef43.png)
- Hardboard
- Coloured papers
- Adhesive
- White paper
- Scissors
- Geometry Box
Prerequisite Knowledge
- Square and its area.
- Rectangle and its area.
Theory
- For square and its area refer to Activity 3.
- For rectangle and its area refer to Activity 3.
Procedure
- Take a hardboard of suitable size and paste a white paper on it.
- From a coloured paper, cut out a square of side a units, (see Fig. 6.1)
- Further, cut out a square of sided units (b < a)from another coloured paper, (see Fig. 6.2)
- Also, cut out a square of sidec units (c < b)from different coloured paper.(see Fig. 6.3)
- Cut out two rectangles of dimensions b x a from different coloured paper, (see Fig. 6.4)
- Also, cut out two rectangles of dimensions c x b from different coloured paper, (see Fig. 6.5)
- Now further, cut out two rectangles of dimensions c x a from another coloured paper, (see Fig. 6.6)
- Paste the squares and rectangles on the hardboard as shown in Fig. 6.7.
Demonstration
From Fig. 6.7, it is clear that from the arrangement of sqaures and rectangle, square PQRS of side (a + b+c) units is obtained.
Area of square PQRS = (a + b +c)² [∴ area of square = (side)²] … (i)
Also, area of square PQRS = Sum of the areas of all the squares and rectangles, which are
used to make the square PQRS = a² + b² + c² + ab + ab + bc +bc +ca+ca = (a² + b² + c² + 2ab + 2bc + 2ca) …(ii)
From Eqs. (i) and (ii), we have
(a + b+c)² =(a² +b² +c² + 2ab + 2bc + 2ca) Here, area is in square units.
From Fig. 6.7, it is clear that from the arrangement of sqaures and rectangle, square PQRS of side (a + b+c) units is obtained.
Area of square PQRS = (a + b +c)² [∴ area of square = (side)²] … (i)
Also, area of square PQRS = Sum of the areas of all the squares and rectangles, which are
used to make the square PQRS = a² + b² + c² + ab + ab + bc +bc +ca+ca = (a² + b² + c² + 2ab + 2bc + 2ca) …(ii)
From Eqs. (i) and (ii), we have
(a + b+c)² =(a² +b² +c² + 2ab + 2bc + 2ca) Here, area is in square units.
Observation
On actual measurement, we get
a = ……. , b = ……. , c = ……. ,
So, a² = ……. , b² = ……. , c² = ……. ,
ab = ……. , bc = ……. , ca = ……. ,
2ab = ……. , 2bc = ……. , 2ca = ……. ,
a + b+c = ……. ,
and (a + b+c)² = ……. ,
Hence, (a + b+c)² = a² + b² + c² + 2ab + 2bc +2ca
On actual measurement, we get
a = ……. , b = ……. , c = ……. ,
So, a² = ……. , b² = ……. , c² = ……. ,
ab = ……. , bc = ……. , ca = ……. ,
2ab = ……. , 2bc = ……. , 2ca = ……. ,
a + b+c = ……. ,
and (a + b+c)² = ……. ,
Hence, (a + b+c)² = a² + b² + c² + 2ab + 2bc +2ca
Result
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Complete gamecube collection torrent. Identity (a + b+c)²= a² +b² +c² +2ab + 2bc + 2ca has been verified. Conan exiles star metal locations map.
Application
This identity may be used for
This identity may be used for
- calculating the square of a number which can be expressed as a sum of three convenient numbers.
- simplification and factorisation of algebraic expressions.
Viva Voce
Question 1:
Is the identity (a+b+c)² = a²+b²+c²+2ab+2bc+2ca, holds for all real values of a, b and c?
Answer:
Yes
Question 1:
Is the identity (a+b+c)² = a²+b²+c²+2ab+2bc+2ca, holds for all real values of a, b and c?
Answer:
Yes
Rabobank debit card activation. Question 2:
Which identity should be used to expand (3x -√y + z)² ?
Answer:
(a+b+c)² = a²+b²+c² + 2ab + 2bc + 2ca .
Which identity should be used to expand (3x -√y + z)² ?
Answer:
(a+b+c)² = a²+b²+c² + 2ab + 2bc + 2ca .
Question 3:
What do you mean by a polynomial?
Answer:
An algebraic expression, in which the variables involved have only non-negative integral power, is called a polynomial.
What do you mean by a polynomial?
Answer:
An algebraic expression, in which the variables involved have only non-negative integral power, is called a polynomial.
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Class 9 Maths Ncert Pdf
Question 4:
What do you mean by zeroes of a polynomial?
Answer:
If p(x) is a polynomial, then a real number k is said to be zero of the polynomial p(x), If p(k) = 0.
What do you mean by zeroes of a polynomial?
Answer:
If p(x) is a polynomial, then a real number k is said to be zero of the polynomial p(x), If p(k) = 0.
Question 5:
What do you mean by a trinomial?
Answer:
A polynomial with three terms is called a trinomial.
What do you mean by a trinomial?
Answer:
A polynomial with three terms is called a trinomial.
Question 6:
Is the expansion of (x + y + z)², trinomial?
Answer:
No, because on expanding (x + y + z)², we get six terms.
Is the expansion of (x + y + z)², trinomial?
Answer:
No, because on expanding (x + y + z)², we get six terms.
Question 7:
Write the condition that the three variables in identity
(a+b+c)² = a²+b²+c²+2ab+2bc+2ca should be written in two variables form (x + y)² =x² + y² + 2xy.
Answer:
Anyone of the variable in the identity (a+b+c)² = a²+b²+c²+2ab+2bc+2ca should be zero.
Write the condition that the three variables in identity
(a+b+c)² = a²+b²+c²+2ab+2bc+2ca should be written in two variables form (x + y)² =x² + y² + 2xy.
Answer:
Anyone of the variable in the identity (a+b+c)² = a²+b²+c²+2ab+2bc+2ca should be zero.
Question 8:
If we take all the variables are equal in the identity
(a+b+c)² = a²+b²+c²+2ab+2bc+2ca, then it will give us a cube of one variable. What is the name of the solid figure in which it gives out a cube of one variable?
Answer:
Cubic solid figure
If we take all the variables are equal in the identity
(a+b+c)² = a²+b²+c²+2ab+2bc+2ca, then it will give us a cube of one variable. What is the name of the solid figure in which it gives out a cube of one variable?
Answer:
Cubic solid figure
Class 9 Maths Ncert Book
Suggested Activity
Verify the identity (a+b+c)² = a²+b²+c²+2ab+2bc+2ca by taking different values of a,b,c.
Verify the identity (a+b+c)² = a²+b²+c²+2ab+2bc+2ca by taking different values of a,b,c.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1.
Maths Ncert Solutions Class 8
Board | CBSE |
Textbook | NCERT |
Class | Class 9 |
Subject | Maths |
Chapter | Chapter 4 |
Chapter Name | Lines and Angles |
Exercise | Ex 4.1 |
Number of Questions Solved | 6 |
Category | NCERT Solutions |
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1
Question 1.
In figure, lines AB and CD intersect at 0. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.
Solution:
Here, ∠AOC and ∠BOD are vertically opposite angles.
∴ ∠AOC = ∠BOD
⇒ ∠AOC = 40° [∵ ∠BOD = 40°(Given)] …(i)
We have, ∠AOC + ∠ BOE = 70° (Given)
40°+ ∠BOE = 70° [From Eq. (i)]
⇒ ∠BOE = 30°
Also, ∠AOC + ∠COE + ∠BOE = 180° (Linear pair axiom)
⇒ 40° + ∠COE + 30° = 180°
⇒ ∠COE = 110°
Now, ∠COE + reflex ∠COE = 360° (Angles at a point)
110°+reflex ∠COE = 360°
⇒ Reflex ∠COE = 250°
In figure, lines AB and CD intersect at 0. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.
Solution:
Here, ∠AOC and ∠BOD are vertically opposite angles.
∴ ∠AOC = ∠BOD
⇒ ∠AOC = 40° [∵ ∠BOD = 40°(Given)] …(i)
We have, ∠AOC + ∠ BOE = 70° (Given)
40°+ ∠BOE = 70° [From Eq. (i)]
⇒ ∠BOE = 30°
Also, ∠AOC + ∠COE + ∠BOE = 180° (Linear pair axiom)
⇒ 40° + ∠COE + 30° = 180°
⇒ ∠COE = 110°
Now, ∠COE + reflex ∠COE = 360° (Angles at a point)
110°+reflex ∠COE = 360°
⇒ Reflex ∠COE = 250°
Question 2.
In figure, lines XY and MN intersect at 0. If ∠POY = 90° , and a : b = 2 : 3. find c.
Solution:
We have, ∠POY = 90°
⇒ ∠POY + ∠POX = 180° (Linear pair axiom)
⇒ ∠POX = 90°
⇒ a+b = 90°
Also, a : b = 2 : 3 (Given)
⇒ Let a = 2k,b = 3k
Now, from Eq. (j), we get
2k + 3k = 90°
⇒ 5k = 90°
⇒ k = 18°
∴ a = 2 x 18°=36°
and b=3 x 18°=54°
Now, ∠MOX + ∠XON = 1800 (Linear pair axiom)
b+ c = 180°
⇒ 540 + c= 180°
⇒ c = 126°
In figure, lines XY and MN intersect at 0. If ∠POY = 90° , and a : b = 2 : 3. find c.
Solution:
We have, ∠POY = 90°
⇒ ∠POY + ∠POX = 180° (Linear pair axiom)
⇒ ∠POX = 90°
⇒ a+b = 90°
Also, a : b = 2 : 3 (Given)
⇒ Let a = 2k,b = 3k
Now, from Eq. (j), we get
2k + 3k = 90°
⇒ 5k = 90°
⇒ k = 18°
∴ a = 2 x 18°=36°
and b=3 x 18°=54°
Now, ∠MOX + ∠XON = 1800 (Linear pair axiom)
b+ c = 180°
⇒ 540 + c= 180°
⇒ c = 126°
Question 3.
In figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.
Solution:
∵ ∠PQS+ ∠PQR = 180° (Linear pair axiom) ,.(i)
and ∠PRT + ∠PRQ = 180° (Linear pair axiom). .(ii)
From Eqs. (i) and (ii), we get
∠PQS + ∠PQR =∠PRT + ∠PRQ
∠PQS + ∠PRQ =∠PRT + ∠PRQ
[Given, ∠PQR = ∠PRQ]
⇒ ∠PQS = ∠PRT
In figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.
Solution:
∵ ∠PQS+ ∠PQR = 180° (Linear pair axiom) ,.(i)
and ∠PRT + ∠PRQ = 180° (Linear pair axiom). .(ii)
From Eqs. (i) and (ii), we get
∠PQS + ∠PQR =∠PRT + ∠PRQ
∠PQS + ∠PRQ =∠PRT + ∠PRQ
[Given, ∠PQR = ∠PRQ]
⇒ ∠PQS = ∠PRT
Question 4.
In figure, if x + y = w + z, then prove that AOB is a line.
Solution:
∵ x+ y+w+ z = 360° (Angle at a point)
x + y = w + z (Given)…(i)
∴ x+ y+ x+ y = 360° [From Eq. (i)]
2(x + y) = 360°
⇒ x + y = 180° (Linear pair axiom)
Hence, AOB is a straight line.
In figure, if x + y = w + z, then prove that AOB is a line.
Solution:
∵ x+ y+w+ z = 360° (Angle at a point)
x + y = w + z (Given)…(i)
∴ x+ y+ x+ y = 360° [From Eq. (i)]
2(x + y) = 360°
⇒ x + y = 180° (Linear pair axiom)
Hence, AOB is a straight line.
Class 9 Maths Ncert Lab Manual Class 11 Maths
Question 5.
In figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that
Solution:
We have,
∠POR = ∠ROQ = 90° (∵ Given that, OR is perpendicular to PQ)
∴ ∠POS + ∠ROS = 90°
⇒ ∠ROS = 90° – ∠POS
On adding ∠ROS both sides, we get
2 ∠ROS = 90° – ∠POS + ∠ROS
⇒ 2 ∠ROS = (90° + ∠ROS) – ∠POS
⇒ 2∠ROS = ∠QOS – ∠POS (∵ ∠QOS = ∠ROQ + ∠ROS = 90° + ∠ROS)
⇒ ∠ROS = (frac { 1 }{ 2 }) (∠QOS – ∠POS)
Hence proved.
In figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that
Solution:
We have,
∠POR = ∠ROQ = 90° (∵ Given that, OR is perpendicular to PQ)
∴ ∠POS + ∠ROS = 90°
⇒ ∠ROS = 90° – ∠POS
On adding ∠ROS both sides, we get
2 ∠ROS = 90° – ∠POS + ∠ROS
⇒ 2 ∠ROS = (90° + ∠ROS) – ∠POS
⇒ 2∠ROS = ∠QOS – ∠POS (∵ ∠QOS = ∠ROQ + ∠ROS = 90° + ∠ROS)
⇒ ∠ROS = (frac { 1 }{ 2 }) (∠QOS – ∠POS)
Hence proved.
Question 6.
It is given that ∠XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ZYP, find ∠XYQ and reflex ∠QYP.
Solution:
Here, YQ bisects ∠ZYP.
Hence, ∠ZYQ = ∠QYP = (frac { 1 }{ 2 }) ∠ZYP …….(i)
Given, ∠XYZ = 64° ….(ii)
∵ ∠XYZ + ∠ZYQ + ZQYP = 180° (Linear pair axiom)
⇒ 64° + ∠ZYQ + ∠ZYQ = 180° [From Eqs. (i) and (ii)]
⇒ 2 ∠ZYQ = 180° – 64°
⇒ ∠ZYQ = (frac { 1 }{ 2 }) x 116°
⇒ ∠ZYQ = 58°
∴ ∠XYQ = ∠XYZ + ∠ZYQ = 64° + 58° = 122°
Now, ∠QYP + reflex ∠QYP = 360°
58° + reflex ∠QYP = 360°
⇒ reflex ∠ QYP = 302°
It is given that ∠XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ZYP, find ∠XYQ and reflex ∠QYP.
Solution:
Here, YQ bisects ∠ZYP.
Hence, ∠ZYQ = ∠QYP = (frac { 1 }{ 2 }) ∠ZYP …….(i)
Given, ∠XYZ = 64° ….(ii)
∵ ∠XYZ + ∠ZYQ + ZQYP = 180° (Linear pair axiom)
⇒ 64° + ∠ZYQ + ∠ZYQ = 180° [From Eqs. (i) and (ii)]
⇒ 2 ∠ZYQ = 180° – 64°
⇒ ∠ZYQ = (frac { 1 }{ 2 }) x 116°
⇒ ∠ZYQ = 58°
∴ ∠XYQ = ∠XYZ + ∠ZYQ = 64° + 58° = 122°
Now, ∠QYP + reflex ∠QYP = 360°
58° + reflex ∠QYP = 360°
⇒ reflex ∠ QYP = 302°
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