Unit Factorization, H.C.F. and L.C.M.
10
Allocated teaching periods: 11
Competency
- Factorizing the simple algebraic expressions
- Finding the HCF and LCM of given expressions
Learning Outcomes
- To factorize the algebraic expression of the forms a2 – b2, a3 – b3 and a4 + a2b2 + b4
- To find the HCF and LCM of given expressions
Level-wise learning objectives
S.N. LEVELS OBJECTIVES
- To define factorization
1. Knowledge (K) - To recall the formula of (a± b)2, a2 – b2, a3 ± b3
- To tell the HCF of monomials
- To find the LCM of monomials
- To factorize the algebraic expressions by taking common
2. Understanding (U) and using the formula a2 – b2, a3 ± b3
- To factorize the formulae of the forms a4 + a2b2 + b4
- To factorize the expressions involving 5 or 6 terms by
3. Application (A) expressing in the formulae of (a± b)2 and a2 – b2
- To find the HCF and LCM of the given expressions
4. High Ability (HA) - To connect daily life problems with factorizations and solve
them
Required Teaching Materials/ Resources
Chart papers with formulae, scissors, ruler, glue-stick, tiles for factorization, ICT tools (if
possible), audio-video materials etc
Pre-knowledge: Factorization of the form a2 – b2, HCF and LCM etc
Teaching Activities for factorization
1. Recall the formulae through chart paper
2. Make a discussion upon taking common factors from each term
3. Discuss upon the factorization and following steps of factorization
(i) Taking common
(ii) Use of formulae: (a± b)2, a2 – b2, a3 ± b3
(iii) Middle term splitting etc.
51 Vedanta Excel in Mathematics Teachers' Manual - 8
4. Give algebra tiles/manipulative materials for arranging to a rectangular shape and discuss
about the factorization. For example,
5. Divide the students into 5 groups. Provide chart paper, colourful marker to each group.
Provide the group works and engage them in the factorization of the algebraic expressions
practically then let them present in the classroom. The group works may be like Group-A:
(a+ b)2, Group-B: (a – b)2, Group-C: a2 – b2, Group-D: a2 + 3x + 2, Group-E: x2 + 5x + 6
6. Present the derivation of the formulae of (a + b)3 with blocks or ICT tools
7. Discuss upon the factorization of the form a4 + a2b2 + b4 with examples
8. Engage the students to factorize the expressions given in the exercise
9. Focus on more practical problems related to factorization
Solution of selected questions from Excel in Mathematics -Book 8
1. Resolve into factors: x2 – x(y + z) + yz
Solution:
Here, x2 – x(y + z) + yz = x2 – xy – xz + yz = x (x – y) – z(x – y) = (x – y) (x – z)
2. Resolve into factors: x2 – (y – 3)x – 3y
Solution:
Here, x2 – (y – 3)x – 3y = x2 – xy + 3x – 3y = x (x – y) + 3(x – y) = (x – y) (x + 3)
3. Resolve into factors: x(a2 –b2) + a(b2 – x2)
Solution:
Here, x(a2 –b2) + a(b2 – x2) = a2x –b2x + ab2 – ax2 = a2x – ax2 + ab2 –b2x
= ax(a – x) + b2(a – x) = (a – x)(ax + b2)
4. The area of a rectangular field is (x2+3xy+2xy+6y2) sq. unit. Find the perimeter of the
filed.
Solution:
Here, the area of the rectangular field = (x2+3xy+2xy+6y2) sq. units
or, l × b = {x (x + 3y) + 2y(x+3y)} sq. units
= (x + 3y) (x+ 2y) sq. units
Hence, length (l) = (x + 3y) units and breadth (b) = (x + 2y) units
Vedanta Excel in Mathematics Teachers' Manual - 8 52
Again, perimeter of the field = 2 (l + b)
= 2 [(x + 3y) + (x + 2y)] units
= 2(4x + 5y) units
= 8x + 10y units
5. Resolve into factors: 256x8 – y8
Solution:
Here,
256x8 – y8 = (16x4)2 – (y4)2 = (16x4 + y4) (16x4 – y4) = (16x4 + y4) [(4x2)2 – (y2)2]
= (16x4 + y4) (4x2 +y2) (4x2 –y2) = (16x4 + y4) (4x2 +y2) [(2x)2 – y2]
= (16x4 + y4) (4x2 +y2) (2x + y) (2x – y)
6. Resolve into factors: (a – 5)2 – (b – 4)2
Solution:
Here, (a – 5)2 – (b – 4)2 = [(a – 5) + (b – 4)] [(a – 5) – (b – 4)] = (a – 5 + b – 4) (a – 5 – b + 4)
= (a + b – 9) (a – b – 1)
7. Resolve into factors: 4a4 – 13a2b2 + 9b4
Solution:
Here, 4a4 – 13a2b2 + 9b4 = (2a2)2 + (3b2)2 – 13a2b2
= (2a2 + 3b2)2 – 2.2a2.3b2 –13a2b2
= (2a2 + 3b2)2 – 12a2b2 – 13a2b2
= (2a2 + 3b2)2 – 25a2b2
= (2a2 + 3b2)2 – (5ab)2
= (2a2 + 3b2 + 5ab) (2a2 + 3b2 – 5ab)
= (2a2 + 5ab + 3b2) (5a2 – 5ab + 3b2)
8. Resolve into factors: a3 – 1000b3
Solution:
Here, a3 –1000b3 = a3 – (10b)3 = (a – 10b)(a2 + a.10 + 102)= (a – 10b)(a2 + 10a + 100)
9. Factorize: (x – y)2 – 9(x – y) – 22
Solution:
Here, (x – y)2 – 9(x – y) – 22
Let, (x – y) = a.
Then, (x – y)2 – 9(x – y) – 22 = a2 – 9a – 22 = a2 – (11 – 2)a – 22
= a2 – 11a + 2a– 22 = a(a – 11) + 2(a – 11)
= (a – 11) (a + 2)
Replacing the value of a, we get
= (x – y – 11) (x –y + 2)
Hence, (x – y)2 – 9(x – y) – 22 = (x – y – 11) (x –y + 2)
53 Vedanta Excel in Mathematics Teachers' Manual - 8
10. The area of rectangular garden is x2 + 5x – 36 sq. units. If its length and breadths are
reduced by 2/2 units, find the new area of the ground.
Solution:
Here, Area of the ground = x2 + 5x – 36 = x2 + 9x – 4x – 36
or, l × b = x(x + 9) – 4(x + 9) = (x + 9)(x – 4)
Thus, length of the garden (l) = (x + 9) units and breadth (b) = (x – 4) units
When the length and breadth of the ground are reduced by 2/2 units,
Then, new length of the ground (L) = (x + 9 – 2) units = (x + 7) units
New breadth of the ground (B) = (x – 4 – 2) units = (x – 6) units
Now, the new area of the ground = A × B = (x + 7) (x – 6) units
= (x2 – 6x + 7x – 42) sq. units
= (x2 + x – 42) sq. units
11. Factorise: 25x2 + 2xy + y2
Solution: 25
Here, 25x2 + 2xy + y2 = (25x)2 + 2xy + y 2 = (5x)2 + 2.5x. y + y 2= 5x + y 2
25 5 5 5 5
1
12. Factorise: 9p2 – 1 + 36p2
Solution:
Here, 9p2 – 1 + 1 = (3p)2 – 1 + 1 2 (3p)2 – 2.3p.61p + 1 2 3p – 1 2
36p2 6p 6p 6p
= =
13. Factorise: a2 + b2 – c2 – d2 + 2ab – 2cd
Solution:
Here, a2 + b2 – c2 – d2 + 2ab – 2cd = a2 +2ab + b2 – c2 – 2cd– d2
= (a + b)2 – (c2 + 2cd + d2)
= (a + b)2 – (c + d)2
= [(a + b)+ (c + d)] [(a + b) - (c + d)]
= (a + b + c + d) (a + b – c – d)
14. Factorize: 3(a + b)2 – 10(a + b) + 8
Solution:
Here, 3(a + b)2 – 10(a + b) + 8
Let, (a + b) = x.
Then, 3(a + b)2 – 10(a + b) + 8 = 3x2 – 10x + 8 = 3x2 – (6 + 4)x + 8
= 3x2 – 6x – 4x + 8 = 3x(x – 2) – 4(x – 2)
= (x – 2) (3x – 4)
Replacing the value of x, we get
= (a + b – 2){3(a + b) – 4 }
Hence, 3(a + b)2 – 10(a + b) + 8 = (a + b – 2) (3a + 3b – 4)
Vedanta Excel in Mathematics Teachers' Manual - 8 54
Extra Questions
1. Factorise: -x2 + 5x – 6 [Ans: (2 – x) (x – 3)]
2. Resolve into factors: ab(x2 – y2) + (a2 – b2) xy [Ans: (bx + ay) (ax – by)]
3. The area of rectangular garden is (x2 + x – 56) sq. feet. Find the length and breadth of the
window. Also, find the perimeter of the garden. [Ans: (x + 8) ft., (x – 7) ft., (4x + 2) ft]
4. While factorizing x2 + 5x – 24, Soniaya got the answer (x + 8) (x – 5), Bhusan got the
answer (x – 6) (x + 4) and Dipson got the answer (x + 12) (x – 2). Factorize yourself and
find who got the correct answer. [Ans: Soniya]
5. The length and breadth of a rectangular piece of carpet are (2x + 1) ft. and (x + 2) ft.
If the area of the rectangular mat is 5 sq. ft. less than the area of the carpet, find the
dimensions of the mat. [Ans: (x+3) ft. × (2x – 1) ft.]
Teaching Activities for H.C.F and L.C.M
1. Discuss upon the higher common factors (H.C.F.) and lowest common multiples (L.C.M.)
of some arithmetic problems.
2. Ask the definitions of HCF and LCM of the numbers with examples.
3. Recall factorisation of the algebraic expressions.
4. Make a discussion about HCF and LCM of algebraic expressions.
5. Use Venn-diagram to show the HCF and LCM of the expressions.
6. Use ICT tool to visualizing the factorisation of expressions and their HCF and LCM.
7. Make the groups of students and encourage them to find the H.C.F. and L.C.M. of the
expressions.
Solution of selected questions from Excel in Mathematics -Book 8
1. Find the H.C.F. of a2 + 7a + 10, a2 + 6a + 8
Solution:
Here,
The first expression= a2 + 7a + 10 = a2 + (3 + 4) a + 10 = a2 + 3a + 4a + 10
= a (a + 3) + 4(a + 3) = (a + 3) (a + 4)
The second expression= a2 + 6a + 8 = a2 + (2 + 4) a + 8 = a2 + 2a + 4a + 8
= a (a + 2) + 4(a + 2) = (a + 2) (a + 4)
Hence, H.C.F. = common factors = a + 4
2. Find the H.C.F. of x4 – 1, x3 – 1
Solution:
Here,
The first expression= x4 – 1 = (x2)2 – 1 = (x2 + 1)(x2 – 1) = (x2 + 1)(x +1)(x – 1)
The second expression= x3 – 1 = (x – 1)(x2 + x + 1)
Hence, H.C.F. = common factors = x – 1
3. Find the H.C.F. of x4 – 125x, x3 – 3x2 – 10x
Solution:
Here,
55 Vedanta Excel in Mathematics Teachers' Manual - 8
The first expression= x4 – 125x = x(x3 – 125) = x (x3 – 53) = (x – 5) (x2 + x.5 + 52)
= x(x – 5) (x2 + 5x + 25)
The second expression= x3 – 3x2 – 10x = x (x2 – 3x – 10) = x [x2 – (5 – 2)x – 10]
= x (x2 – 5x + 2x – 10) = x [x (x – 5) + 2 (x – 5)]
= x (x – 5) (x + 2)
Hence, H.C.F. = common factors = x (x – 5)
4. Find the H.C.F. of x2 – 1, x2 + 2x – 3 and x2 – 3x + 2
Solution:
Here,
The first expression= x2 – 1 = (x + 1) (x – 1)
The second expression = x2 + 2x – 3 = x2 + (3 – 1) x – 3 = x2 + 3x – x – 3
= x (x + 3) – 1 (x +3) = (x + 3)(x – 1)
The third expression = x2 – 3x + 2 = x2 – (2 + 1)x – 3 = x2 – 2x – x + 2
= x (x – 2) – 1 (x – 2) = (x – 2)(x – 1)
Hence, H.C.F. = common factors = x – 1
5. Find the L.C.M. of 4x2 – 2x, 8x3 – 2x
Solution:
Here,
The first expression= 4x2 – 2x = 2x (2x – 1)
The second expression = 8x3 – 2x = 2x (4x2– 1) = 2x [(2x)2 – 12] = 2x (2x + 1)(2x – 1)
Hence, L.C.M. = common × remaining factors = 2x (2x – 1) (2x + 1) = 2x (4x2 – 1)
6. Find the L.C.M. of (a + b)2, a2 – b2 and 2a2 + ab – b2
Solution:
Here,
The first expression= (a + b)2 = (a + b) (a + b)
The second expression = a2 – b2 = (a + b) (a – b)
The third expression = 2a2 + ab – b2 = 2a2 + (2 – 1)ab – b2 = 2a2 + 2ab – ab – b2
= 2a(a + b) – b(a + b) = (a + b) (2a – b)
Hence, L.C.M. = common × remaining factors = (a + b) (a + b) (a – b) (2a – b)
= (a + b) (a2 – b2) (2a – b)
Extra Questions
1. Find the higher common factor of x3y – xy3, x3 – y3 and x2 – 2xy + y2 [Ans: x – y]
2. Find the H.C.F. of a2b2 – b4, ab2 – b3 and ab – b2 [Ans:b(a – b)]
3. Find the H.C.F. of a4 – 8a, 3a2 – 8a + 4 [Ans: a – 2]
4. Find the H.C.F. of 10p2 – 100p, 10p2 – 1000 and 10p2 – 200p + 1000 [Ans: 10(p – 10)
5. Find the lower common multiples of a2 – 3a + 2, a2 + a – 6 [Ans: (a – 1) (a – 2) (a + 3)]
6. Find the L.C.M. of m2 – 81, m2 – 18m + 81 [Ans: (m + 9)(m – 9)2]
Vedanta Excel in Mathematics Teachers' Manual - 8 56
Unit Rational Expression
11
Allocated teaching periods: 7
Competency
- Simplifying rational expressions
Learning Outcomes
- To multiply and divide rational expressions
- To simplify rational expressions
Level-wise learning objectives
S.N. LEVELS OBJECTIVES
1. Knowledge (K)
- To define rational expression
2. Understanding (U) - To tell the condition under which the rational expression is
3. Application (A) defined or undefined
4. High Ability (HA) - To reduce rational expression in the least term
- To factorize the numerator denominator of the rational
expression and reduce into the lowest term
- To multiply the rational expressions
- To divide the rational expressions
- To simplify the rational expressions
- To connect daily life problems with rational expressions and
simplify.
Required Teaching Materials/ Resources
Chart papers with formulae, audio-video materials etc
Pre-knowledge: Factorization of the expressions and L.C.M.
Teaching Activities for factorization
1. Recall factorization, H.C.F. and L.C.M.
2. Discuss about the rational expressions
3. Make the groups of students and give some rational expressions (numerator and denominator
are monomials) to reduce into lowest form.
4. Make a discussion about the expressions to be undefined.
5. Note that the rational expression having denominator zero is undefined.
6. Discuss about the multiplication and division of the rational expressions.
7. Simplify some problems regarding the lesson and encourage students to simplify themselves.
8. Recall the addition and subtraction of fractions and connect it to add and subtract the
rational expressions
9. Focus on collaborative learning
10. Give some daily life problems related to rational expressions and discuss over them.
57 Vedanta Excel in Mathematics Teachers' Manual - 8
Solution of selected questions from Excel in Mathematics -Book 8
x2 – 5x – 14
1. Reduce 2x3 – 13x2 – 7x in its lowest term
Solution:
Here, x2 – 5x – 14 = x2 – (7 – 2)x – 14 = x2 – 7x + 2x – 14 = x(x – 7) + 2(x – 7)
2x3 – 13x2 – 7x x(2x2 – 13x – 7) x{2x2 – (14 – 1)x – 7)} x{2x2 – 14x – 1x – 7)}
= (x – 7) (x + 2) 7)} = (x – 7) (x + 2) = x+2
x{2x(x – 7) + 1(x – x(x – 7) (2x + 1) x(2x + 1)
2. Simplify: a2 + 5a + 6 ÷ a2 – 9
a2 – 1 a2 – 2a –
Solution: 3
a2 + 5a + 6 ÷ a2 – 9
a2 – 1 a2 – 2a – 3
= a2 + 5a + 6 × a2 – 2a – 3 = a2 + (3 + 2)a + 6 × a2 – (3 – 1)a –3
a2 – 1 a2 – 9 a2 – 12 a2 – 92
= a2 + 3a + 2a + 6 × a2 – 3a – 1a – 3 = a(a + 3) + 2(a + 3) × a(a – 3) + 1(a – 3)
(a + 1) (a – 1) (a + 3) (a – 3) (a + 1) (a – 1) (a + 3) (a – 3)
= (a + 3) (a + 2) × (a – 3) (a + 1) = a+2
(a + 1) (a – 1) (a + 3) (a – 3) a–1
3. Simplify: a2 – a – 20 × a2 – a – 2 ÷ a+1
a2 – 1 a2 – 2a – 3 a2 + 5a
Solution:
a2 – a – 20 × a2 – a – 2 ÷ a+1
a2 – 1 a2 – 2a – 3 a2 + 5a
= a2 – (5 – 4)a – 20 × a2 – (2 – 1)a – 2 × a2 + 5a
(a + 5) (a – 5) a2 – (4 – 2)a – 8 a + 1
= a2 – 5a + 4a – 20 × a2 – 2a + a – 2 × a(a + 5)
(a + 5) (a – 5) a2 – (4 – 2)a – 8 a+1
= a(a – 5) + 4(a – 5) × a(a – 2) + 1(a – 2) × a(a + 5)
(a + 5) (a – 5) a(a – 4) + 2(a – 4) a+1
= (a – 5) (a + 4) × (a – 2) (a + 1) × a(a + 5)
(a + 5) (a – 5) (a – 4) (a + 2) a+1
= a(a +4) (a – 2)
(a – 4) (a + 2)
4. Simplify: x 2 1 + 2x – x2 + 3
+ x–1 x2 – 1
Solution:
x 2 1 + 2x – x2 + 3 = x 2 1 + 2x – (x x2 + 3 1)
+ x–1 x2 – 1 + x–1 + 1) (x –
= 2(x – 1) + 2x(x + 1) – (x2 + 3) = 2x – 2 + 2x2 + 2x – x2 – 3 = x2 + 4x – 5
(x + 1) (x – 1) (x + 1) (x – 1) (x + 1) (x – 1)
= x2 + (5 – 1)x – 5 = x2 + 5x – x – 5 = x(x + 5) – 1(x + 5) = (x + 5) (x –1) = x + 5
(x + 1) (x – 1) (x + 1) (x – 1) (x + 1) (x – 1) (x + 1) (x – 1) x + 1
Vedanta Excel in Mathematics Teachers' Manual - 8 58
5. Simplify: x2 + y2 + x2 y) – y2 y)
xy y(x + x(x +
Solution: x2 y2 (x + y) (x2 + y2) –
x2 + y2 y(x + x(x + xy(x + – x.x2 y.y2
xy + – = y)
y) y)
x3 + xy2 + x2y + y3 – x3 – y3 = xy2 + x2y = xy(x + y) = 1
xy(x + y) xy(x + y) xy(x + y)
6. Simplify: (a a c) + (b – b a) + (c – c – b)
– b) (a – c) (b – a) (c
Solution:
(a a – c) + (b – b a) + (c c =– (a – a a) – (b – b – b) – (c c – c)
– b) (a c) (b – – a) (c – b) b) (c – c) (a – a) (b
= –a(b – c) – b(c – a) – c(a – b) = –ab + ac – bc + ab – ac – bc = 0 = 0
(a – b) (b – c) (c – a) (a – b) (b – c) (c – a) (b –
(a – b) c) (c – a)
7. Simplify: x2 x–1 2 + x2 x–2 6 + x2 x–5 15
– 3x + – 5x + – 8x +
Solution: x–2 x–5
x–1 – 5x + – 8x +
x2 – 3x + 2 + x2 6 + x2 15
= x2 – x–1 + 2 + x2 – x – 2 + 6 + x2 – x–5 + 15
2x – x 3x – 2x 5x – 3x
= x(x – x – 1 – 2) + x–2 + x–5
2) – 1(x x(x – 3) – 2(x – 3) x(x – 5) – 3(x – 5)
= (x x–1 1) + (x x–2 2) + (x x–5 3)
– 2) (x – – 3) (x – – 5) (x –
= 1 + 1 + 1 = x – 3+ x–2 +x – 2 = 3x – 7
– – – (x – 2) (x – 3) – 2) (x –
(x 2) (x 3) (x 3) (x 3)
Simplify: y2 y +3 9 + 6 9 + 16y
+ 6y + y2 – 8y2 – 24y
Solution:
y2 y +3 9 + 6 + 16y = y+3 + 6 + 16y
+ 6y + y2 – 9 8y2 – 24y y2 + 2.y.3 + 9 y2 – 32 8y(y – 3)
= y+3 + (y + 6 – 3) + 2y = y 1 3 + (y + 6 – 3) + y 2 3
(y + 3)2 3) (y y–3 + 3) (y –
= y – 3 + 6 – 2(y + 3) = y – 3 + 6 – 2y – 6 = –y – 3
(y + 3) (y – 3) (y + 3) (y – 3) (y + 3) (y – 3)
= –(y + 3) = –1 = 1
(y + 3) (y – 3) y–3 3–y
59 Vedanta Excel in Mathematics Teachers' Manual - 8
2 Extra Questions
x–5
1. For what value of x, become undefined? [Ans: 5]
2. Simplify: a2 + ab ÷ ab [Ans: (a + b)2]
a–b a2 – b2 b
3. Simplify: b2 – 6b + 9 ÷ b2 – 5b + 62 [Ans: b – 1 ]
b2 – 2b – 3 b2 – 3b + b + 1
4. Simplify: (x – x – z) + (y – y – x) + (z – z – y) [Ans: 0]
y) (x z) (y x) (z
5. Simplify: 5 – x 4 y + 8x [Ans: x 9 y ]
x+y – x2 – y2 +
6. Simplify: x+1 – x–1 + yx 1 [Ans: 8x3 ]
x–1 x+1 x2 + x4 – 1
7. Ram can paint x2 – 1 + 2 ; x ≠ 1, 2 part of a wall in 1 day. Shiwanee can paint x2 1
3x – 5x + 6
; x ≠ 2, 3 part of a wall in 1 day. What part of the wall would be painted if they both paint
together? Find it. [Ans: (x – 2 – 3) ]
1) (x
Vedanta Excel in Mathematics Teachers' Manual - 8 60
Unit Equation, Inequality and Graph
12
Allocated teaching periods: 13
Competency
- Solving the inequalities and representing the solution on the graph
- Solving the pair of simultaneous equations
- Solving quadratic equation
Learning Outcomes
- To represent the inequality in the graph
- To solve the linear equation of single variable
- To solve the simultaneous equations
- To solve the quadratic equation of single variable
Level-wise learning objectives
S.N. LEVELS OBJECTIVES
- To define inequality
1. Knowledge (K) - To write the inequalities represented by the graph
- To find the point of intersection of two lines from the graph
- To define quadratic equation
- To solve the inequality and represent graphically
- To solve the linear equation of single variable
2. Understanding (U) - To fill up the table of values of dependent variable for linear
equation
- To solve the simple quadratic equation of a single variable.
- To make the inequality from the verbal problems and solve
then show graphically
- To solve the linear equations of single variable
3. Application (A) - To solve the simultaneous equations by substitution,
elimination and graphical methods
- To solved the quadratic equation.
4. High Ability (HA) - To mathematize real life problems on equations and solve.
Required Teaching Materials/ Resources
Graph board, graph chart/paper, scale, ICT tools like GeoGebra etc.
Pre-knowledge: Equation.
Teaching Activities for inequality
1. Discuss about true, false and open mathematical statements with examples and ask to
students to identify the types of statements.
61 Vedanta Excel in Mathematics Teachers' Manual - 8
2. Give more real life examples like, students securing more than 50 marks is examination,
students whose weight is less than 30 kg, time taking more than or equal to 20 minutes
to go to the school from home etc. and discuss about inequality.
3. Discuss about the trichotomy signs (<, >, ≤, ≥).
4. With discussion, state inequality as the open statements including trichotomy signs is
called inequality.
5. With examples, verify the following properties of inequality.
(i) If the same quantity/number is added to or subtracted from both sides of inequality,
the inequality relationship remains same.
Example: 5<8 implies that (a) 5+2<8+2 or, 7<10 (b) 5 – 2 < 8 – 2 or, 3 < 6
Thus, if x > y then x + a > y + a and x – a > y – a
(ii) If both the sides of an inequality is multiplied or divided by the same positive
number, the inequality relationship remains same.
Example: 6>4 implies that (a) 2×6 > 2×4 or, 12>8 (b) 6/2 >4/2 or,3> 2
Thus, if x > y then ax > ay and x/a > y/a where a ≠ 0.
(iii) If both the sides of an inequality is multiplied or divided by the same negative
number, the inequality relationship is reversed. Example: 9>6 implies that (a)
L.H.S. = (-3) × 9 = -27 and R.H.S. = (-3) × 6 = -18. Since,
-27 < - 18 and hence, (-3) × 9<(-3)×6 (b) 9/(-3) < 6/(-3) or,-3< -2
Thus, if x > y then for a negative integer ‘a’, ax < ay and x/a < y/a
6. Make a discussion with examples on the graphical representation of the inequalities in
the groups.
(i) Ask the inequalities from the graphs.
(ii) Give inequalities to draw in the graph.
7. Focus on collaborative learning.
Solution of selected questions from Excel in Mathematics -Book 8
1. Find the solution set of the inequality 2(x + 5) < 4 + 5x and represent graphically.
Solution:
2(x + 5) < 4 + 5x
or, 2x + 10 < 4 + 5x
or, 2x + 10 – 10 < 4 – 10 + 5x [Subtracting 10 from both sides]
or, 2x < -6 + 5x
or, 2x – 5x < -6 + 5x – 5x [Subtracting 5x from both sides]
or, -3x < -6
or, –3x > –6 [Dividing both sides by -3]
–3 –3
or, x > 2
∴ Required solution set = {3, 4, 5, …}
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
x > 2 (It does not include 2.)
Vedanta Excel in Mathematics Teachers' Manual - 8 62
2. Find the solution set of the inequality x + 2 – x ≥ x – 2 and represent graphically.
Solution: 6 3
x + 2 – x ≥ x – 2
6 3
or, x + 2 – 2x ≥ x – 2
6
or, 2 – x ≥ 6x – 12
or, 2 + 12 ≥ 6x + x
or, 14 ≥ 7x
or, 2 ≥ x ∴ x ≤ 2
∴Required solution set = {…-1, 0, 1, 2
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
x ≤ 2 (It includes 2.)
3. Find the solution set of the inequality – 1 ≤ x +2<1 and represent graphically.
3 3
Solution:
– 1 ≤ x + 2 < 1
3 3
1 x + 6
or, – 3 ≤ 3 < 1
or, – 1 × 3 ≤ x + 6 × 3 < 1 × 3 [Multiplying each term by 3]
3 3 [Subtracting 6 from each term]
or, –1 ≤ x + 6 < 3
or, –1 – 6 ≤ x + 6 – 6 < 3 – 6
or, –7 ≤ x <–3
∴Required solution set = {-7,-6,-5,-4}
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
–7 ≤ x < –3
(It includes – 7 but does not include –3.)
4. If x≤-3, find the value of y in y = 3x - 2
Solution:
Here, x ≤ -3 ∴x = {…,-6,-5,-4,-3} and y = 3x – 2
When x = -3 then y = 3(-3) – 2 = - 11
When x = -4 then y = 3(-4) – 2 = -14
When x = -5 then y = 3(-5) – 2 = -17
When x = -6 then y = 3(-6) – 2 = -20
Hence, y = { …, -20, -17, -14, -11}
Alternately, y = 3(-3) – 2 = -11. Thus, y ≤ - 11
63 Vedanta Excel in Mathematics Teachers' Manual - 8
5. If x>-1, find the value of y in y = 3 – 2x
Solution:
Here, x > -1 ∴x = {0, 1, 2, 3, …} and y = 3 – 2x
When x = 0 then y = 3 – 2(0) = 3
When x = 1 then y = 3 – 2(1) = 1
When x = 2 then y = 3 – 2(2) = 3=-1
When x = 3 then y = 3 – 2(3) = - 3
Hence, y = { …, -3, -1, 1, 3}
Alternately, y = 3 – 2(-1) = 5. Thus, y < 5
6. If x ≤ 5, find the value of y in 2x + 3y + 5 = 0
Solution:
Here, x ≤5 ∴x = {…, 2, 3, 4, 5} and 2x + 3y + 5
When x = 5 then 2(5) + 3y + 5 = 0 or, 3y = -15 ∴y = -5
When x = 2 then 2(2) + 3y + 5 = 0 or, 3y = -9 ∴y = -3
When x = -1 then 2(-1) + 3y + 5 = 0 or, 3y = -3 ∴y = -1
Hence, y = { …, -1, -3, -5}
Alternately,
2(5) + 3y + 5 = 0 or, 3y = - 15 or, y = -5. Thus, y ≥ -5
7. When 5 is subtracted from three times a number, the result is greater than 6. Find the
least whole number that satisfies the statement.
Solution:
Let, the required least whole number be x.
Then, according to the question;
3x – 5 > 6
or, 3x – 5 + 5 > 6 + 5
or, 3x > 11
or, 3x > 11
3 3
or, x > 3 2
3
Since, x is a whole number; the required least number is 4.
8. The cost of a pen is Rs 15 and the cost of a box is Rs 24. Shashwat has Rs 90 and he wants
to buy a box and a few numbers of pens. Find the maximum number of pens that he can
buy.
Solution:
Let, the required number of pen be x.
The cost of a box = Rs 24
The cost of x pens = Rs 15x
Then, according to the question;
Vedanta Excel in Mathematics Teachers' Manual - 8 64
15x + 24 ≤ 90
or, 15x + 24 – 24 ≤ 90 – 24
or, 15x ≤ 66
or, 15x ≤ 66
15 15
22
or, x ≤ 5
or, x ≤452
Since, x is a whole number; the required maximum number of pens is 4.
Extra Questions
1. Solve the inequality 5x – 1 ≤ 14 and show in a number line. [Ans: x≤3]
2. Solve the inequality 2 (x + 3) – 6 > 4 and show in a number line. [Ans: x> 2]
3. Solve the inequality x (x – 3) + 7 ≥ x2 + 4 and show in a number line. [Ans: x ≤ 1]
4. The cost of a copy is Rs 55 and that of book is Rs 300. If Rahul has Rs 1000 and he wants
to buy a book a few numbers of copies, find the maximum number of copies that he can
buy. [Ans: 12]
Teaching Activities for linear equation with one variable
1. Discuss upon equation and linear equation with proper examples.
2. Give real life examples related to linear equation. For example, the cost of 3 hens of equal
weight is Rs 6000, what is the cost of a hen?
3. Recall linear equations of one variable with examples.
4. With examples, make a discussion on the following points.
5. Explain with discussion the important terminologies used in making equations
(a) Terminologies based fundamental operations
(i) Addition: added, sum, exceed, altogether, all, more than, increased etc
(ii) Subtraction: subtracted, difference, left away, shorter than, less than, decreased by,
younger/cheaper than etc
(iii) Multiplication: product, times, multiplied by etc
(iv) Division: divided by, each, share, quotient etc.
(b) Terminologies based on ages: If the present age of an individual is x years
(i) Age before/ago: (x – t) years
(ii) Age after/ hence/ later : (x + t) years
(c) Terminologies based on consecutive numbers
(i) Consecutive numbers: 2, 4, 6, etc. i.e. x, x+1, x+2, ... etc
(ii) Consecutive odd/even numbers: 1, 3, 5 etc. i.e., x, x+2, x+4, ... etc
6. Under discussion, solve the problems given in exercise of the textbook and give a few
numbers of questions as classwork.
65 Vedanta Excel in Mathematics Teachers' Manual - 8
Solution of selected questions from Excel in Mathematics -Book 8
1. Solve: 2(x – 1) – 3(x – 2) = 1
3 4
Solution:
2(x – 1) – 3(x – 2) = 1
3 4
or, 8(x – 1) – 9(x – 2) = 1
12
or, 8x – 8 – 9x + 18 = 12
or, –x + 10 = 12
or, –x = 2
∴ x = –2
2. Solve: 1 + x 1 1 = x 2 1
x + –
Solution:
1 + x 1 1 = x 2 1
x + –
or, x+ 1+x = x 2 1
x(x + 1) –
or, 2x + 1 = x 2 1
x(x + 1) –
or, (2x + 1) (x – 1) = 2x(x + 1)
or, 2x2 – 2x + x – 1 = 2x2 + 2x
or, –x – 1 = 2x
or, –3x = 1
∴ x = – 31
3. The sum of three consecutive even numbers is 78. Find the numbers.
Solution:
Let the first even number be x.
The second consecutive even number = x + 2
The third consecutive even number = x + 4
Now, x + (x + 2) + (x + 4) = 78
or, 3x + 6 = 78 or, 3x = 72 or, x = 24
The first even number = x = 24
Vedanta Excel in Mathematics Teachers' Manual - 8 66
The second even number = x + 2 = 24 + 2 = 26
The third even number = x + 4 = 24 + 4 = 28
Hence, the required consecutive even numbers are 24, 26 and 28.
4. A sum of Rs 135 is divided in to two parts. If the greater part exceeds the smaller part by
Rs 25, find the parts of the sum.
Solution:
Let the smaller part be Rs x then the greater part = Rs (x + 25)
Now, x + (x + 25) = 135 or, 2x + 25 = 135 or, 2x = 110 or, x = 55
The smaller part = x = Rs 55 and the greater part = x + 25 = 55 + 25 = Rs 80
Hence, the parts of Rs 165 are Rs 55 and Rs 80.
5. A sum of Rs 140 is divided in to two parts. If the greater part is Rs 20 more than double of
the smaller part, find the parts of the sum.
Solution:
Let the smaller part be x then the greater part = 2x + 20
Now, x + (2x + 20) = 140 or, 3x = 120 or, x = 40
The smaller part = x = Rs 40 and the greater part = 2x + 20 = 2×40 + 20 = Rs 100
Hence, the parts of Rs 140 are Rs 40 and Rs 100.
6. A man spends 51supmarotfoRf shi8s4i0n0coinmae on his life insurance and 1 part on food. Iff he deposits
the remaining bank, find his income. 3
Solution:
Let the income of the man be Rs x. 5x, x
1 3
Then, expense for insurance = 5 of Rs x = Rs expense for food = Rs and deposit = Rs 8400
Now, x + x + 8400 = x or, 3x + 5x + 8400 = x or, 8x + 8400 = x
5 3 15 15
or, 8400 = x – 81x5 or, 8400 = 715x or, 126000 = 7x ∴ x = 18000
Hence, his income is Rs 18,000.
7. A rectangular field is three times longer than its breadth and its perimeter is 200 m. find
its length and breadth.
Solution:
Let, the breadth of the filed (b) = x m then the length (l) = 3x
Now, perimeter (P) = 2 (l + b) or, 200 = 2(3x + x) or, 200 = 8x ∴x = 25
The breadth (b) = x = 25 m and length (l) = 3x = 3×25 = 75 m
Hence, the length and breadth of the field are 75m and 25m respectively.
8. Ram is 9 years older than Hari. Three years ago, he was two times as old as Hari was. Find
his present age.
Solution:
67 Vedanta Excel in Mathematics Teachers' Manual - 8
Let, the present age of Hari be x years then the present age of Ram = (x + 9) years
Three years ago, the age of Hari = (x – 3) years
Three years ago, the age of Ram= (x + 9 – 3) years = (x + 6) years
Since, three years ago, Ram’s age was two times as old as Hari was.
So, x + 6 = 2 (x – 3) or, x + 6 = 2x – 6 ∴x = 12
The present age of Hari = x = 12 years
The present age of Ram = (x + 9) years = (12 + 9) years = 21 years
Hence, the present ages of Ram and Hari are 21 years and 12 years.
9. Two years hence, a boy will be two times as old as a girl. Before four years, he was three
times as old as the girl was. Find their present ages.
Solution:
Two years hence, let the age of the girl will be x years.
Then two years hence, the age of the boy will be 2x years.
The present ages of the girl = (x – 2) years and boy = (2x – 2) years
Four years ago, the age of the girl was (x – 2 – 4) years = (x – 6) years
Four years ago, the age of the boy was (2x – 2 – 4) years = (2x – 6) years
According to question, 2x – 6 = 3 (x – 6) or, 2x – 6 = 3x –18 ∴x = 12
The present age of the girl = (x – 2) years = (12 – 2) years = 10 years
The present age of the boy = (2x – 2) years = (2×12 – 2) years = 22 years
Hence, the present ages of the boy and girls are 22 years and 10 years respectively.
Extra Questions
1. There are 555 students in a school. If the number of boys is 55 more than that of girls, find
the number of boys and girls. [Ans: 305, 250]
2. If the number of three consecutive odd numbers is 111, find the numbers.
[Ans: 35, 37, 39]
3. The sum of two numbers is 50. If three times of the smaller number is equal to two times
of the greater number, find the numbers. [Ans: 20, 30]
4. The length of rectangular room is twice the breadth of the room. If the perimeter of the
room is 48 feet, find the length and breadth of the room. [Ans: 16 ft, 8 ft]
5. One-fourth of a pole is inside the mud, two-fifth part is inside the water and the remaining
length of 7 m is above the water surface. Find the length of the pole.
[Ans: 20 m]
6. A mother is five times as old as her daughter. Six years hence, she will be only three times
as old as her daughter. Find their present ages. [Ans: 30 years, 6 years]
Teaching Activities for linear equation with two variables
1. Recall linear equation with proper examples.
2. Discussion upon the linear equations with two variables
3. Visualize the linear equation in GeoGebra or other graphical apps or software.
4. Recall rectangular co-ordinates and plotting the points on the graph paper.
5. Tabulate the values of x and y for any linear equation and plot on the graph paper under
Vedanta Excel in Mathematics Teachers' Manual - 8 68
discussion.
6. Define simultaneous linear equations with examples.
7. Divide the students into four five groups and give some real life problems (if possible with
pictures) to make equations as fast as possible
Group A: The cost of a goat is five times the cost of a hen. If their total cost is Rs 9000, find
the cost of each.
Group B: The mother is thrice as old as her son. If their total age is 40 years, find their
present ages.
Group C: The cost of a pant is Rs 300 more than that of the shirt. If the total cost of the shirt
and pant is Rs 1900, find the cost of each.
Group D: The cost of umbrella is Rs 300 less than that of a bag. If the cost of two umbrellas
and a bag is Rs 1500, find the cost of each item
8. Discuss about their equations that the students made, solutions and following methods of
solving the problems in the above activities
(i) By graphical method.
(ii) Substitution method.
(iii) Elimination method.
9. Give the values of variables x and y and tell the students to make the equations satisfying
those values and discuss the solutions in the class
Solution of selected questions from Excel in Mathematics -Book 8
1. Solve the equations x + y = 5 and x – y = 1 graphically.
Solution:
Here, the given equations are x + y = 5 … eqn. (i) and x – y = 1 … eqn. (ii)
From eqn. (i), x + y = 5
or, y = 5 – x
x 123
y 432
The points (1, 4), (2, 3) and (3, 2) are
plotted on the graph and they are joined
to get a straight line.
From eqn. (ii), x – y = 1
or, x – 1 = y
or, y = x – 1
x 123
y 012
The points (1, 0), (2, 1) and (3, 2) are plotted on the graph and they are joined to get another
straight line.
In the graph, the lines (i) and (ii) intersect at (3, 2). ∴(3, 2) is the common solution of the
equations. Hence, x = 3 and y = 2.
69 Vedanta Excel in Mathematics Teachers' Manual - 8
2. Solve the equations 2x – y = 5 and x – y = 1 graphically.
Solution:
Here, the given equations are x + y = 5 … eqn. (i) and x – y = 1 … eqn. (ii)
From eqn. (i), 2x – y = 5
or, 2x – 5 = y
or, y = 2x – 5
x 123
y -3 -1 1
The points (1, -3), (2, -1) and (3, 1) are
plotted on the graph and they are joined
to get a straight line.
From eqn. (ii), x – y = 1
or, x – 1 = y
or, y = x – 1
x 123
y 012
The points (1, 0), (2, 1) and (3, 2) are plotted on the graph and they are joined to get another
straight line.
In the graph, the lines (i) and (ii) intersect at (4, 3). ∴(4, 3) is the common solution of the
equations. Hence, x = 4 and y = 3.
3. Solve the following simultaneous equations by elimination method: 2x – 3y = 1 and x +
2y = 18
Solution:
Here,
The given equations are 2x – 3y = 1 … (i)
x + 2y = 18 …(ii)
Multiplying equation (i) by 2 and (ii) by 3 and adding them, we get
4x – 6y = 2
3x + 6y = 54
7x = 56
∴x = 8
Substituting the value of x in equation (i), we get
2×8 – 3y = 1
or, 16 – 3y = 1
or, 15 = 3y
∴y = 5
Hence, x = 8 and y = 5
Vedanta Excel in Mathematics Teachers' Manual - 8 70
4. Solve the following simultaneous equations by elimination method: 3x + 4y = 16 and
x + 3y = 7
Solution:
The given equations are 3x + 4y = 16 … (i)
x + 3y = 7 …(ii)
Multiplying equation (ii) by 3 and subtracting (ii) from equation (i), we get
3x + 4y = 16
3x + 9y = 21
(-) (-) (-)
-5y = -5 ∴y = 1
Substituting the value of y in equation (i), we get
3x + 4×1 = 16
or, 3x + 4 = 16
or, 3x = 12
∴x = 4
Hence, x = 4 and y = 1
5. Solve the following simultaneous equations by substitution method: x + 2y = 19 and
x + y = 13
Solution:
Here,
The given equations are x + 2y = 9 … (i)
x + y = 13 …(ii)
From equation (i), x = 19 – 2y …(i)
Now, substituting the value of x from equation (ii) in equation (i), we get
19 – 2y + y = 13 or, - y = - 6 ∴y = 6
Again, substituting the value of y in equation (iii), we get
x = 19 - 2×6 = 7
Hence, x = 7 and y = 6
6. Solve the following simultaneous equations by substitution method: 2x – y = 2 and
x – y = -3
Solution:
Here,
The given equations are 2x – y = 2 … (i)
x–y=-3 …(ii)
From equation (i), 2x -2 = y …(i)
Now, substituting the value of y from equation (ii) in equation (i), we get
x – (2x – 2 = -3 or, x – 2x + 2 = - 3 or, -x = -5 x = 5
Again, substituting the value of x in equation (iii), we get
y = 2×5 – 2 = 8
Hence, x = 5 and y = 8.
71 Vedanta Excel in Mathematics Teachers' Manual - 8
7. The sum of two numbers is 39. If the greater number is 5 more than the smaller one, find
the numbers.
Solution:
Let the greater number be x and the smaller number be y.
Then,
From the first given condition, x + y = 39 or, x = 39 – y …(i)
From the second condition, x – y = 5 …(ii)
Now, substituting the value of x from equation (i) in equation (ii), we get
(39 – y) –y = 5 or, 39 – 2y = 5 or, -2y = -34 ∴y = 17
Again, substituting the value of y in equation (i), we get
x = 39 – 17 = 22
Hence, the required numbers are 22 and 17.
8. The difference of two numbers is 6. If three times the smaller number is equal to two times
the greater one, find the numbers.
Solution:
Let the greater number be x and the smaller number be y.
Then,
From the first given condition, x – y = 6 or, x = 6 + y …(i)
From the second condition, 2x = 3y …(ii)
Now, substituting the value of x from equation (i) in equation (ii), we get
2(6 + y) = 3y or, 12 + 2y = 3y ∴y = 12
Again, substituting the value of y in equation (i), we get
x = 6 + 12 = 18
Hence, the required numbers are 18 and 12.
9. The total number of students in a class is 36. If the number of girls is 6 more than the
number of boys, find the number of boys and girls.
Solution:
Let the number of boys be x and the number of girls be y.
Then,
From the first given condition, x + y = 36 …(i)
From the second condition, y = x + 6 …(ii)
Now, substituting the value of y from equation (ii) in equation (i), we get
x + (x + 6) = 36 or, 2x + 6 = 36 or, 2x = 30 ∴x = 15
Again, substituting the value of x in equation (ii), we get
y = 15 + 6 = 21
Hence, the number of boys and girls are 15 and 21 respectively.
10. The sum of age of father and his son is 40 years. If the father is 28 years older than his son,
find their ages.
Vedanta Excel in Mathematics Teachers' Manual - 8 72
Solution:
Let the present age of the father be x years and that of his son be y years.
Then,
From the first given condition, x + y = 40 …(i)
From the second condition, x = y + 28 …(ii)
Now, substituting the value of x from equation (ii) in equation (i), we get
(y + 28) + y = 40 or, 2y = 12 ∴y = 6
Again, substituting the value of y in equation (ii), we get
x = 6 + 28= 34
Hence, the present age of the father is 34 years and that of his son is 6 years.
11. The sum of age of father and his son is 40 years. If the father is 28 years older than his son,
find their ages.
Solution:
Let the present age of the father be x years and that of his son be y years.
Then,
From the first given condition, x + y = 40 …(i)
From the second condition, x = y + 28 …(ii)
Now, substituting the value of x from equation (ii) in equation (i), we get
(y + 28) + y = 40 or, 2y = 12 ∴y = 6
Again, substituting the value of y in equation (ii), we get
x = 6 + 28= 34
Hence, the present age of the father is 34 years and that of his son is 6 years.
12. The sum of age of father and his son is 48 years. Four years ago, the father was 9 times as
old as his son was. Find their present ages.
Solution:
Let the present age of the father be x years and that of his son be y years.
Then,
From the first given condition, x + y = 48 … (i)
From the second condition, (x – 4) = 9 (y – 4)
or, x – 4 = 9y – 36
or, x = 9y - 32 … (ii)
Now, substituting the value of x from equation (ii) in equation (i), we get
(9y- 32) + y = 48 or, 10y = 80 ∴y = 8
Again, substituting the value of y in equation (ii), we get
x = 9×8 – 32 = 40
Hence, the present age of the father is 40 years and that of his son is 8 years.
13. If twice the son’s age in years is added to his father’s age, the sum is 70. But if twice the
father’s age is added to his son’s age, the sum is 95. Find the present age of the father and
his son.
73 Vedanta Excel in Mathematics Teachers' Manual - 8
Solution:
Let the present age of the father be x years and that of his son be y years.
Then,
From the first given condition, x + 2y = 70 … (i)
From the second condition, 2x + y = 95
or, y = 95 – 2x … (ii)
Now, substituting the value of y from equation (ii) in equation (i), we get
x + 2 (95 – 2x) = 70
or, 190 – 3x = 70
or, 120 = 3x
∴x = 40
Again, substituting the value of x in equation (ii), we get
y = 95 – 2×40 = 15
Hence, the present age of the father is 40 years and that of his son is 15 years.
14. The length of a rectangular garden is two times the breadth. If the perimeter of the garden
is 72 m, find the length and breadth of the garden.
Solution:
Let the length of the garden be x m and the breadth be y m.
Then,
From the first given condition, x = 2y … (i)
From the second condition, perimeter (P) = 72
or, 2 (l + b) = 72
or, 2 (x+ y) = 72
or, x+ y)= 36 … (ii)
Now, substituting the value of x from equation (i) in equation (ii), we get
2y + y = 36
or, 3y = 36
∴y = 12
Again, substituting the value of y in equation (i), we get
x = 2×12 = 24
Hence, the length and breadth of the garden are 24 m and 12 m respectively.
Extra Questions
1. Solve graphically:
(i) 2x – y = 5 and x – y = 1 [Ans: x = 4, y = 3]
(ii) 2x – 1 = y and 3x – 2y = 0 [Ans: x = 2, y = 3]
(iii) 5x + 7y = 1 and x + 4y + 5 = 0 [Ans: x = 3, y = -2]
2. The cost of 1 kg of mutton same as four times the cost of 1 kg of chicken. If the cost of 2 kg
of mutton and 3 kg of chicken is Rs 3300, find the rate of cost of mutton and the chicken.
[Ans: Rs 1200, Rs 300]
3. The sum of two numbers is 20. If three times of the smaller number is equal to two times of
the greater number, find the numbers. [Ans: 12, 8]
Vedanta Excel in Mathematics Teachers' Manual - 8 74
4. The length of rectangular basketball court is twice its breadth. If the perimeter of the court
is 42 m, find the length and breadth of the room. [Ans: 14 m, 7 m]
5. The cost of four pencils and three pens is Rs 80. If two pencils and a pen cost Rs 30, find the
cost of each pencil and each pen. [Ans: Rs 5, Rs 20]
6. A mother is three times as old as her daughter. Six years ago, she was four times as old as
her daughter was. Find their present ages. [Ans: 54 years, 18 years]
Teaching Activities for Quadratic Equation
1. Recall degree of the polynomials.
2. Recall linear equation of single variable and double variables with proper examples.
3. Note that the linear equations are first degree equations and have a unique solution.
4. Write some equations of degree two and note them as second degree equations.
5. Define the quadratic equations as the second degree equations.
6. Ask to factorise the quadratic equations and discuss upon to find the solutions or roots.
7. Solve some related problems and give similar problems to be solved individually or group-
work.
8. Call the students to solve the quadratic equations on the board which boosts up the
confident level.
9. Give 2/2 values of x and ask the students to get the quadratic equations.
Solution of selected questions from Excel in Mathematics -Book 8
1. Solve: 100 – 81x2 = 0
Solution:
100 – 81x2 = 0
or, (10)2 – (9x)2 = 0
or, (10 + 9x) (10 – 9x)= 0
Either, 10 + 9x = 0 … (i)
Or, 10 – 9x = 0 … (ii) –10
9
Solving equation (i), 10 + 9x = 0 or, 10 = -9x ∴ x =
Solving equation (ii), 10 – 9x = 0 or, 10 = 9x ∴ x = 10
10 9
Hence, x = ± 9
2. Solve: x2 + 10x = 75 … (i)
Solution:
x2 + 10x = 75
or, x2 + 10x – 75 = 0
or, x2 + (15 – 5)x – 75 = 0
or, x2 + 15x – 5x – 75 = 0
or, x (x + 15) – 5 (x + 15) = 0
or, (x + 15) (x – 5) = 0
Either, x + 15 = 0
75 Vedanta Excel in Mathematics Teachers' Manual - 8
Or, x – 5 = 0 … (ii)
Solving equation (i), x + 15 = 0 ∴x = -15
Solving equation (ii), x – 5 = 0 ∴x = 5
Hence, x = -15 or 5
3. Solve: 3x2 – 10x + 8 = 0
Solution:
3x2 – 10x + 8 = 0
or, 3x2 – (6 + 4) x + 8 = 0
or, 3x2 – 6x – 4x + 8 = 0
or, 3x (x – 2) – 4 (x – 2 )= 0
or, (x – 2) (3x – 4) = 0
Either, x – 2 = 0 … (i)
Or, 3x – 4 = 0 … (ii)
Solving equation (i), x – 2 = 0 ∴x = 2 4
3
Solving equation (ii), 3x – 4 = 0 or, 3x = 4 ∴ x =
Hence, x = 2 or 4
3
4. The sum of two numbers is 12 and their product is 32. Find the numbers.
Solution:
Let, one of the numbers be x. Then, the other number = (12 – x)
Now, their product = 32
or, x (12 – x) = 32
or, 12x - x2 = 32
or, 0 = x2 – 12x + 32
or, x2 – 12x + 32 = 0
or, x2 – (8 + 4) x + 32 = 0
or, x2 – 8x – 4x + 32 = 0
or, x (x – 8) – 4 (x – 8)= 0
or, (x – 8) (x – 4) = 0
Either, x – 8 = 0 ∴x = 8
Or, x – 4 = 0 ∴x = 4
When x = 8 then 12 – x = 12 – 8 = 4
When x = 4 then 12 – x = 12 – 4 = 8
Hence, the required numbers are 8 and 4 or 4 and 8.
5. In a two digit number, the product of the digit is 18 and their sum is 9. Find the number.
Solution:
In a two digit number; let, x be the digit at tens place and y be the digit at ones place. Then the
number = 10x + y
Now,
From the first condition: xy = 18 … (i)
Vedanta Excel in Mathematics Teachers' Manual - 8 76
From the second condition: x + y = 9
or, x = 9 – y … (ii)
Now, substituting the value of x from equation (ii) in equation (i), we get.
(9 – y) y = 18
or, 9y – y2 = 18
or, 0 = y2 – 9y + 18
or, y2 – 9y + 18 = 0
or, y2 – (6 + 3) y + 18 = 0
or, y2 – 6y – 3y + 18 = 0
or, y (y – 6) – 3 (y – 6)= 0
or, (y – 6) (y – 3) = 0
Either, y – 6 = 0 ∴y = 6
Or, y – 3 = 0 ∴y = 3
When y = 6, from equation (ii), x = 9 – 6 ∴x = 3. Thus, 10x + y = 10×3 + 6 = 36
When y = 3, from equation (ii), x = 9 – 3 ∴x = 6. Thus, 10x + y = 10×6 + 3 = 63
Hence, the required number is 36 or 63.
6. The difference of the ages of two sisters is 4 years and the product of their ages is 45. Find
the age of two sisters.
Solution:
Let the age of the elder sister be x years and that of the younger sister be y years.
Now,
From the first condition: x – y = 4
∴x = y + 4 … (i)
From the second condition: xy = 45 … (ii)
Now, substituting the value of x from equation (i) in equation (ii), we get.
(y + 4) y = 45
or, y2 + 4y = 45
or, y2 + 4y – 45 = 0
or, y2 + (9 – 5) y – 45 = 0
or, y2 + 9y – 5y – 45 = 0
or, y (y + 9) – 5 (y + 9)= 0
or, (y + 9) (y – 5) = 0
Either, y + 9 = 0 ∴y = - 9 which is impossible because the age cannot be negative.
Or, y – 5 = 0 ∴y = 5
Substituting the value of y in equation (i), x = 5 + 4 = 9
Hence, the ages of two sisters are 9 years and 5 years.
7. A room is 4 m longer than its breadth. If the area of the floor of the room is 96 sq. m, find
the length and breadth of the room.
Solution:
Let the length of the room be x m and its breadth be y m.
77 Vedanta Excel in Mathematics Teachers' Manual - 8
Now,
From the first condition: x – y = 4
∴x = y + 4 … (i)
From the second condition: area = 96
∴xy = 96 … (ii)
Now, substituting the value of x from equation (i) in equation (ii), we get.
(y + 4) y = 96
or, y2 + 4y = 96
or, y2 + 4y – 96 = 0
or, y2 + (12 – 8) y – 96 = 0
or, y2 + 12y – 8y – 96 = 0
or, y (y + 12) – 8 (y + 12)= 0
or, (y + 12) (y – 8) = 0
Either, y + 12 = 0
∴y = - 12 which is impossible because the breadth cannot be negative.
Or, y – 8 = 0 ∴y = 8
Substituting the value of y in equation (i), x = 8 + 4 = 12
Hence, the length of the room is 12 m and the breadth is 8 m.
Extra Questions
1. What are the roots of the quadratic equation x2 – 4 = 0? [Ans: ± 2]
2. Find the roots of the quadratic equation x2 – 5 = 20? [Ans: ± 5]
3. What are the roots of the quadratic equation x2 – 5x + 6 = 0? [Ans: 2, 3]
4. Solve: x2 – 6x – 72 = 0 [Ans: 12, -6]
5. Solve: 2y2 – y – 6 = 0 [Ans: 2, -3/2]
6. The sum and product of two numbers are 10 and respectively, find the numbers.
[Ans: 6, 4 or 4, 6]
Vedanta Excel in Mathematics Teachers' Manual - 8 78
Unit Transformation
13
Allocated teaching periods: 6
Competency
- To reflect, rotate and displace the geometric figure using coordinates
Learning Outcomes
- To reflect the geometric figure using coordinates
- To rotate the geometric figure using coordinates
- To displace the geometric figure using coordinates
Level-wise learning objectives
S.N. LEVELS OBJECTIVES
1. Knowledge (K) - To define transformation
- To tell the name of fundamental transformation
- To define reflection.
- To recall the formulae of image of the point P(x, y) under
the reflection about x and y-axes.
- To define rotation.
- To recall the formulae of image of the point P(x, y) under
the rotation through +900, -900 or 1800 about origin.
- Define displacement
- To reflect the given geometric figures about a given line
using set-squares and ruler.
- To rotate the geometric figures through the given angle and
2. Understanding (U) centre.
- To displace the geometric figures with the given vector.
- To find the image of the vertices of the given figures under
the given reflection and present in the graph.
- To draw the object and image figures on the same graph
3. Application (A) paper after rotation with the given centre, angle and
direction.
- To find the image of the vertices of the figures under the
given displacement and show in the graph.
4. High Ability (HA) - Prepare and present the project works on reflection, rotation
and displacement.
Required Teaching Materials/ Resources
Graph board, graph chart/paper, scale, set-square, mirror, ICT tools like GeoGebra etc.
Pre-knowledge: Co-ordinates of the points.
79 Vedanta Excel in Mathematics Teachers' Manual - 8
Teaching Activities for Transformation
1. Give some real life examples representing the various types of transformations. For
example; mirror reflects the objects on it, clean water level reflects the objects, flipping
images on the computer, rotating the ceiling fan, swing, pendulum, rotating the wheels of
cycle/car while moving, hitting a nail with hammer, throwing a stone, moving a train in
circular track, revolving the Earth, lifting the book in a direction, movement of aircraft as it
moves across the sky, pushing the table on the straight path, sewing the cloth with sewing
machine, resizing the photos, projecting the scenes/pictures on the screen in the cinema
hall etc.
2. Define the transformation as a rule which changes the position or size or both of an object.
A. Reflection
1. Define reflection with examples.
2. Reflect some pictures using GeoGebra tool or reflect
the geometric figures about the given line.
3. Make a discussion on the properties of reflection.
4. Draw a triangle on a graph board/paper or in device via GeoGebra and reflect it about x-axis
and discuss about relation of the vertices and their corresponding images and conclude the
result.
5. Draw a triangle on a graph board/paper or in device via GeoGebra and reflect it about y-axis
and discuss about relation of the vertices and their corresponding images and conclude the
following results.
(i) Reflection about x-axis, P (x, y) → P’ (x, -y)
(ii) Reflection about y-axis, P (x, y) → P’ (-x, y)
6. Ask the image of the points with coordinates under reflection about x-axis and y-axis.
7. Focus on the general exercise in group or individually.
8. Find the image of the vertices of geometric figure under reflection about x-axis or y-axis
and draw the object and the image on the same graph paper.
9. Give some project work on describing about the reflection and its properties then call to
present in the class.
Vedanta Excel in Mathematics Teachers' Manual - 8 80
Solution of selected questions from Excel in Mathematics -Book 8
1. Plot the points A (-2, 4), B (3, 3) and C (-5, 7) in graph paper and draw triangle joining the
points in order. Draw the image of each triangle under the reflection about x-axis. Write
the coordinates of the vertices of image.
Solution:
Here,
The vertices of the ∆ ABC are A (-2, 4), B (3, 3)
and C (-5, 7).
When the triangle ABC is reflected about x-axis,
A (-2, 4) A’ (-2, -4)
B (3, 3) B’ (3, -3)
C (-5, 7 C’ (-5, -7)
Hence, A’ (-2, -4), B’ (3, -3) and C’ (-5, -7) are the
vertices of the ∆ A’B’C’.
2. P (-7, -8), Q (4, 6) and R (-6, 2) are the vertices of ∆PQR. Find the coordinates of its imager
under the reflection about y-axis.
Solution:
Here,
The vertices of the ∆ PQR are P (-7, -8), Q (4, 6) and R (-6, 2).
When the ∆ PQR is reflected about y-axis, we have
P (x, y) P’ (-x, y)
P (-7, -8) P’ (7, -8)
Q (4, 6) Q’ (-4, 6)
R (-6, 2) R’ (6, 2)
Hence, P’ (7, -8), Q’ (-4, 6) and R’ (6, 2) are the vertices of the ∆ P’Q’R’.
3. W (4, 2), X (-3, 6) and Y (-1, -4) are the vertices of ∆WXY. Find the coordinates of the
vertices of ∆W’X’Y’ under the reflection x-axis. Also, find the coordinates of ∆W’’X’’Y’’
when ∆W’X’Y’ is reflected about y-axis
Solution:
Here,
The vertices of the ∆ WXY are W (4, 2), X (-3, 6) and Y (-1, -4).
When the ∆ WXY is reflected about x-axis, we have
P (x, y) P’ (x, -y)
W (4, 2) W’ (4, -2)
X (-3, 6) X’ (-3, -6)
Y (-1, -4) Y’ (-1, 4)
81 Vedanta Excel in Mathematics Teachers' Manual - 8
Hence, W’ (4, -2), X’ (-4, 6) and Y’ (-1, 4) are the vertices of the ∆ W’X’Y’.
Again, when the ∆ W’X’Y’ is reflected about y-axis, we have
P (x, y) P’ (-x, y)
W’ (4, -2) W’’ (-4, -2)
X’ (-3, -6) X’’ (3, -6)
Y’ (-1, 4) Y’’ (1, 4)
Hence, W’’ (-4, -2), X’’ (4, 6) and Y (1, 4) are the vertices of the ∆ W’’X’’Y’’.
B. Rotation
1. Discuss about the rotation with examples.
2. Rotate some pictures using
GeoGebra tool or rotate the
geometric figures about the given
line.
3. Make a discussion on the properties of reflection.
4. Draw a triangle on a graph board/paper or in device via GeoGebra and rotate it through
+900 or, -900 or +1800 about origin and discuss about relation of the vertices and their
corresponding images and conclude the following result results.
(i) Under rotation through + 900, P (x, y) → P’ (-y, x)
(ii) Under rotation through - 900, P (x, y) → P’ (y, -x)
(iii) Rotation through 1800, P (x, y) → P’ (-x, -y)
5. Ask the image of the points with coordinates under rotation through +900 or, -900 or +1800
about the centre at origin.
6. Find the image of the vertices of geometric figure under the rotation +900 or, -900 or +1800
about origin and draw the object and the image on the same graph paper.
Solution of selected questions from Excel in Mathematics -Book 8
1. P (-4, 2), Q (3, 7) and R (-1, -6) are the vertices of ∆PQR. Find the coordinates of its image
under the rotation through 900 in
(i) anti-clock wise (ii) clockwise direction about origin.
Also, draw the graphs of these transformations.
Solution:
Here,
The vertices of the ∆ PQR are P (-4, 2), Q (3, 7) and
R (-1, -6).
(i) When the ∆PQR is rotated though 900 in anti-
Vedanta Excel in Mathematics Teachers' Manual - 8 82
clockwise direction about origin,
P (x, y) P’ (-y, x)
P (-4, 2) P’ (-2, -4)
Q (3, 7) Q’ (-7, 3)
R (-1, -6) R’ (6, -1)
Thus, P’ (-2, -4), Q’ (-7, 3) and R’ (6, -1) are the
vertices of the ∆ P’Q’R’
(ii) When the ∆PQR is rotated though 900 in
clockwise direction about origin,
P (x, y) P’ (y, -x)
P (-4, 2) P’’ (2, 4)
Q (3, 7) Q’’ (-7, 3)
R (-1, -6) R’’ (-6, 1)
Thus, P’ (2, 4), Q’ (7, 3) and R’ (-7, 3) are the
vertices of the ∆ P’’Q’’R’’.
2. K (1, 4), L (-3, -5) and M (5, -2) are the vertices of ∆KLM. Find the coordinates of ∆K’L’M’
under the rotation through – 900 about origin. Also, find the coordinates of ∆K’’L’’M’’
when ∆K’L’M’ is rotated through +1800 about origin.
Solution:
Here,
The vertices of the ∆ KLM are K (1, 4), L (-3,-5) and M (5, -2).
When the ∆ KLM is rotated through – 900 about origin, we have
P (x, y) P’ (y, -x)
K (1, 4) K’ (4, -1)
L (-3, -5) L’ (-5, 3)
M (5, -2) N’ (-2, -5)
Hence, K’ (4, -1), L’ (-5, 3) and M’ (-2, -5) are the vertices of the ∆ K’L’M’.
Again,
When the ∆ K’L’M’ is rotated through +1800 about origin, we have
P (x, y) P’ (-x, -y)
K’ (4, -1) K’’ (-4, 1)
L’ (-5, 3) L’’ (5, -3)
M’ (-2, -5) N’’ (2, 5)
Hence, K’’ (-4, -1), L’’ (5, -3) and M’’ (2, 5) are the vertices of the ∆ P’’Q’’R’’.
83 Vedanta Excel in Mathematics Teachers' Manual - 8
3. D (-4, -1), E (0, -6) and F (5, 0) are the vertices of ∆DEF. Find the coordinates of ∆D’E’F’
under the rotation through +900 about origin. Also, find the coordinates of ∆D’’E’’F’’
under the reflection about x-axis.
Solution:
Here,
The vertices of the ∆ DEF are D (-4, -1), E (0, -6) and F (5, 0).
When the ∆ DEF is rotated through +900 about origin, we have
P (x, y) P’ (-y, x)
D (-4,-1) D’ (1, -4)
E (0, -6) E’ (6, 0)
F (5, 0) F’ (0, 5)
Hence, D’ (1, -4), E’ (6, 0) and F’ (0, 5) are the vertices of the ∆D’E’F’.
Again,
When the ∆ D’E’F’ is reflected about x-axis, we have
P (x, y) P’ (x, -y)
D’ (1, -4) D’’ (1, 4)
E’ (6, 0) E’’ (6, 0)
F’ (0, 5) F’’ (0, -5)
Hence, D’’ (-1, 4), E’’ (6, 0) and F’’ (0, 5) are the vertices of the ∆ D’’E’’F’’.
4. W (4, 2), X (-3, 6) and Y (-1, -4) are the vertices of ∆WXY. Find the coordinates of the
vertices of ∆W’X’Y’ under the reflection x-axis. Also, find the coordinates of ∆W’’X’’Y’’
when ∆W’X’Y’ is reflected about y-axis
Solution:
Here,
The vertices of the ∆ WXY are W (4, 2), X (-3, 6) and Y (-1, -4).
When the ∆ WXY is reflected about x-axis, we have
P (x, y) P’ (x, -y)
W (4, 2) W’ (4, -2)
X (-3, 6) X’ (-3, -6)
Y (-1, -4) Y’ (-1, 4)
Hence, W’ (4, -2), X’ (-4, 6) and Y’ (-1, 4) are the vertices of the ∆ W’X’Y’.
Again, when the ∆ W’X’Y’ is reflected about y-axis, we have
P (x, y) P’ (-x, y)
W’ (4, -2) W’’ (-4, -2)
X’ (-3, -6) X’’ (3, -6)
Y’ (-1, 4) Y’’ (1, 4)
Hence, W’’ (-4, -2), X’’ (4, 6) and Y (1, 4) are the vertices of the ∆ W’’X’’Y’’.
Vedanta Excel in Mathematics Teachers' Manual - 8 84
C. Displacement
1. Give some examples related to displacement and define it under discussion.
2. Translate some pictures using GeoGebra tool or reflect the geometric figures about the
given line.
S
A'
PR
A
B' C' Q S’
P’ R’
B
C
Q’
3. Make a discussion on the properties of displacement.
4. Discuss about the component of translation/displacement vector.
If a be the number of units of right or left and b be the number of units of up or down
displacement, then
(i) a units right → x-component is +a (ii) a units left → x-component is -a
(iii) b units up → y-component is +b (iv) b units down→ y-component is -b
Thus,
(i) the displacement/translation vector for a units right and b units up = a
b
(ii) the displacement/translation vector for a units left and b units up = –a
b
(iii) the displacement/translation vector for a units right and b units down = a
–b
(iv) the displacement/translation vector for a units left and b units down = –a
–b
5. Draw a geometric figure on a graph board/paper or in device via GeoGebra and displace
it by the vector and discuss about relation of the vertices and their corresponding
images and conclude the result.
(i) P (x, y) →P’(x + a, y + b) [Right and up displacement]
(ii) P (x, y) →P’(x - a, y + b) [Left and up displacement]
(iii) P (x, y) →P’(x + a, y - b) [Right and down displacement]
(iv) P (x, y) →P’(x - a, y - b) [Left and down displacement]
(v)
6. Find the image of the vertices of geometric figure under the given displacement and
draw the object and the image on the same graph paper.
Solution of selected questions from Excel in Mathematics -Book 8
1. Plot the points A (2, 5), B (-2, 0), C (-8, 4) and D (-5, 7) in the graph paper and join them.
Displace the figure so formed by 4 units left and 6 units up. Find the coordinates of
the image of vertices of image so formed. Also, check your answer by using left and
up rule.
85 Vedanta Excel in Mathematics Teachers' Manual - 8
Solution:
Here,
The vertices of the quadrilateral ABCD are A (2, 5), B
(-2, 0), C (-8, 4) and D (-5, 7).
When the quadrilateral ABCD is displaced by 4 units
left and 6 units up then the image of vertices are A’ (-2,
11), B’ (-6, 6), C’ (-12, 10) and D’ (-9, 13).
NDioswp,labcyemapepnltyvinegcttohre=rul–eabo=f le–f6t4up displacement,
We have, P (x, y) P’ (x + a, y + b)
A (2, 5) A’ (2 – 4, 5 + 6) = A’ (-2, 11)
B (-2, 0) B’ (-2 – 4, 0 + 6) = B’ (-6, 6)
C (-8, 4) C’ (-8 – 4, 4 + 6) = C’ (-12, 10)
D (-5, 7) D’ (-5 – 4, 7 + 6) = D’ (-9, 13)
2. Find the unit of displacement of the point (4, 2) to get image of (7, 7).
Solution:
Here,
The displacement vector = image – object = (7, 7) – (4, 2) = (3, 5)
Then the displacement is 3 unit right and 5 unit up.
Extra Questions
1. The vertices of triangle ABC are A (2, 5), B (1, 2) and C (6, 3), sketch it in a graph paper and
reflect it about x-axis. Also, write down the co-ordinates of image.
[Ans: A′ (2, -5), B′ (1, -2), C′ (6, -3)]
2. Draw a triangle PQR with vertices P (3, 2), Q (4, 5) and R (6, -3) on a graph paper. Reflect it
on y-axis and show the image on the same graph paper. [Ans: P′ (-3, 2), Q′ (-4, 5), R′ (-6, -3)]
3. A (-5, 3), B (-3, 0), C (-1, 1) and D (-4, 5) are the vertices of a quadrilateral ABCD. Reflect
it about the line y = 0 and find the coordinates of image. Also, present the object and the
image on the same graph paper. [Ans: A′ (-5, -3), B′ (-3, 0), C′ (-1, -3), D′ (-4, -5)]
4. Plot ∆ABC on a graph paper whose vertices are A (5, 0), B (2, 8) and C (-1, 4). Reflect it on
the line x = 0 and write the vertices of image ∆A′B′C′. [Ans: A′ (-5, 0), B′ (-2, 8), C′ (1, 4)]
5. Determine the vertices of ∆F′L′Y′ formed when ∆FLY with vertices F (-3, 6), L (-2, 4) and Y
(0, 1) is rotated through +900 about origin. Also, draw both the triangles on the same graph
paper. [Ans: F′ (-6, -3), L′ (-4, -2), Y′ (-1, 0)]
6. Rotate a ∆ABC with vertices A (-2, 3), B (3, 4) and C (2, 2) through -900 about (0, 0) in anti-
clockwise direction and show the object and image on the same graph paper.
[Ans: A′ (3, 2), B′ (4, -3), C′ (2, -2)]
7. Draw a ∆ABC with vertices A (2, 4), B (1, 3) and D (4, 2) on a graph paper. Rotate it through
+1800 about origin and represent both the triangles on the same graph paper.
[Ans: A′ (-2, -4), B′ (-1, -3), C′ (-4, -2)]
8. Plot ∆PEN with vertices P (3, 2), E (4, 4) and N (2, 5) in a graph paper. Translate it by 3
units right and 2 units upward and represent the triangle and its image on the same graph
paper. [Ans: P′ (6, 4), E′ (7, 6), N′ (5, 7)]
9. Plot ∆XYZ with vertices X (-1, 0), Y (3, 2) and Z (0, 4) in a graph paper. Translate it by 4
units left and 3 units downward and represent the triangle and its image on the same graph
paper. [Ans: X′ (-5, -3), Y′ (-1, -1), Z′ (-4, 1)]
Vedanta Excel in Mathematics Teachers' Manual - 8 86
Unit Geometry: Angles
14
Allocated teaching periods: 6
Competency
- Solving the problems based on the pair of angles formed by two intersecting lines and
made by a transversal with parallel lines
Learning Outcomes
- To verify experimentally the relationship of
(i) Linear pair
(ii) Vertically opposite angles.
- To explore the relationship between the following pair of angles made by a transversal
with the parallel lines
(i) Alternate angles
(ii) Corresponding angles
(iii) Co-interior angles
Level-wise learning objectives
S.N. LEVELS OBJECTIVES
1. Knowledge (K) - To define adjacent angles, linear pair, vertically opposite
angles, complementary and supplementary angles
- To write the sum of adjacent angles formed by two
intersecting lines
- To tell the relation between the vertically opposite angles
- To identify the alternate, corresponding and co-interior
angles made by the transversal with parallel lines.
- To define the alternate, corresponding and co-interior angles.
- To state the relationships of alternate, corresponding and co-
interior angles
- To solve the simple problems based on linear pair, vertically
opposite angles etc.
2. Understanding (U) - To find the unknown angles from the figure using the
relationships of alternate, corresponding and co-interior
angles
- To prove the simple theorems by the virtue of adjacent
angles, vertically opposite angles
- To establish the more relation of the angles using based on
3. Application (A) alternate, corresponding and co-interior angles.
4. High Ability (HA) - To prepare and present the project works on the relationships
of vertically opposite angles, alternate, corresponding and
co-interior angles.
87 Vedanta Excel in Mathematics Teachers' Manual - 8
Required Teaching Materials/ Resources
Ruler, protractor, strips, sticks, chart paper, scissors, threads etc.
Pre-knowledge: Types of angles, adjacent angles, complementary and supplementary angles
Teaching Activities for Different pairs of angles formed by two intersecting lines
1. Make a discussion on the Geometry and its study.
2. Ask the students about the angles, types of angles, complementary and supplementary
angles etc.
3. Make the group discussion on adjacent angles.
4. Verify experimentally the relationship between the adjacent angles formed on the
straight angles.
5. With discussion, explore the relationship between the vertically opposite angles.
6. List out the following points.
(i) Pair of angles having the common vertex and a common arm are called adjacent
angels.
(ii) Adjacent angles on the straight line are supplementary.
(iii) The vertically opposite angles are always equal.
(iv) The sum of the parts of the straight angles is 1800.
(v) The sum of the parts of the complete angles is 3600.
Solution of selected questions from Excel in Mathematics -Book 8
1 If xo and x° are supplementary angles, find them.
Solution: 4
Here, x°
4
x° + = 180°
or, 4x° + x° = 180°
4
or, 5x° = 720°
or, x = 144° and x° = 144° = 36°
4 4
Hence, the required angles are 360 and 1440. q° x°
2. Find the unknown sizes of angles. 3q° x+25°
Solution: p° x+35°
Here,
(i) x0 + (x + 250) + (x + 350) = 1800 [The sum of parts of a straight angle is 1800]
or, 3x = 1200 ∴x = 400
(ii) x + 250 = 400+ 250 = 650 and x + 350 = 400+ 350 = 750
(iii) p0 = x0 = 400 [The vertically opposite angles are equal]
(iv) q0 + 3q0 + p0 = 1800 [The sum of parts of a straight angle is 1800]
or, 4q0 + 400 = 1800
or, 4q0 = 1400 q0 = 350
(v) 3p0 = 3×350 = 1050
Vedanta Excel in Mathematics Teachers' Manual - 8 88
3.
Solution: 1
2
Here, ∠p = ∠q q p
Now, ∠p +∠q = 1800 [The sum of adjacent angles on a straight line
is 1800]
or, 1 ∠q +∠q = 1800 [As, ∠p = 1 ∠q]
2 2
or, ∠q + 2∠q = 1800
2
or, 3∠q = 3600 ∴∠q = 1200 and ∠p = 1 ∠q = 1 × 1200 = 600
2 2
Hence, ∠p = 600 and ∠q = 1200.
4. c
Solution:
… (i) xa b
Here, ∠a + ∠b + ∠c = 1800 [Given] … (ii)
Now, ∠a + ∠x = 1800 [Being linear pair]
From (i) and (ii), we get
∠a + ∠b + ∠c = ∠a + ∠x
Hence, ∠x = ∠b + ∠c
Teaching Activities for Different pairs of angles made by the transversal with parallel lines
1. For warm up, ask about the parallel lines.
2. Draw a transversal to a pair of parallel lines and explain about the interior and exterior
angles.
3. Construct the parallel lines and a transversal line upon them on a chart paper. Cut out
the angles in the Z-shape and rotate it to fit the pair of angles. Then call it as the alternate
angles.
4. Define alternate angles as the pair of non-adjacent interior angles lying in the opposite
sides of the transversal.
5. Similarly, define alternate exterior angles as the pair of non-adjacent exterior angles lying
in the opposite sides of the transversal.
6. Explore the relationship between the alternate angles with the students under discussion.
7. Construct the parallel lines and a transversal line upon them on a chart paper. Cut out
the angles in the F-shape and translate it to fit the pair of angles. Then call it as the
corresponding angles.
8. Under discussion, conclude that the corresponding angles as the pair of non-adjacent
angles, one interior and another exterior, lying in the same side of the transversal.
9. Explore the relationship between the corresponding angles with the students under
discussion.
10. Construct the parallel lines and a transversal line upon them on a chart paper. Cut out the
89 Vedanta Excel in Mathematics Teachers' Manual - 8
angles in the C-shape and then call it as the co-interior angles.
11. Under discussion, define the co-interior angles as the pair of interior angles lying in the
same side of the transversal.
12. Also, define the co-interior exterior angles as the pair of exterior angles lying in the same
side of the transversal.
13. Explore the relationship between the co-interior angles with the students under discussion.
14. Note down the following points.
(i) The alternate angles are always equal.
(ii) The alternate exterior angles are always equal.
(iii) The corresponding angles are always equal.
(iv) The sum of co-interior angles is always 1800.
(v) The sum of co-exterior angles is always 1800.
15. Also, list out the following conditions of being two straight lines parallel to each other.
(i) When the alternate angles are equal, the lines are parallel.
(ii) When the alternate exterior angles are equal, the lines are parallel.
(iii) When the corresponding angles are equal, the lines are parallel.
(iv) When the sum of co-interior angles is 1800, the lines are parallel.
(v) When the sum of co-exterior angles is 1800, the lines are parallel.
Solution of selected questions from Excel in Mathematics -Book 8
1. From the figures given below, find the value of x. (3x+10)°
Solution: (4x–10)°
Here,
(3x + 10)0= (4x – 10)0 [The alternate angles between parallel lines are equal]
or, 3x – 4x = -100 – 100
or, – x = – 200
∴ x = 200
2. From the figures given below, find the value of x. (2x–30)°
Solution: x°
Here,
(2x – 30)0= x0 [The alternate exterior angles in parallel lines are equal]
or, 2x – x = 300
x = 300
(x+40)°
3. From the figures given below, find the value of x. 3x°
Solution:
Here,
(x + 40)0+ 3x0 = 1800 [The sum of co-exterior angles in parallel lines]
or, 4x = 1400
∴x = 350
Vedanta Excel in Mathematics Teachers' Manual - 8 90
4. Find the unknown sizes of angles. y 60° x
Solution: z 80°
Here,
(i) 800+600 +x = 1800 [The sum of the parts of straight angle is 1800]
or, 1400 + x = 1800
∴x = 400
(ii) (600+x) + y = 1800 [The sum of the co-interior angles within parallel lines]
or, (600+400) + y = 1800
or, 1000 + x = 1800
∴x = 800
(iii) z = 600 [The alternate angles between parallel lines are equal]
58°
5. Find the unknown sizes of angles.
Solution: 152°
Here, a
b
(i) b + 1520 = 1800 [The sum of the co-interior angles within parallel lines]
∴b = 280
(ii) (a+b) = 580 [The alternate angles within the parallel lines are always equal]
or, a+280 = 580
∴a = 300
6. Find the unknown sizes of angles. A 36° B
Solution: a F
Through H, EF parallel to AB and CD are drawn. E xb
C 28°
(i) a = 360 [The alternate angles within the parallel lines]
(ii) b = 280 [The alternate angles within the parallel lines]
(iii) x = a + b [By whole part axiom]
∴x = 360 + 280 = 640
7. Find the unknown sizes of angles. A J I
Solution: E 44°B
Through J and K, EF and GH parallel to AB and CD are drawn. F
(i) a = 440 [The alternate angles within the parallel lines]
(ii) a + b = 880 [By whole part axiom]
or, 440 + b = 880 b = 440
(iii) b + c = 1800 [The sum of co-interior angles within parallel lines]
or, 440 + c = 1800
∴c = 1360
(iv) d = 1550 [The corresponding angles within the parallel lines are equal]
(v) x = c + d [By whole part axiom]
91 Vedanta Excel in Mathematics Teachers' Manual - 8
= 1360 + 1550 = 2910 PQ R
∴x = 2910 x
R
8. In the adjoining figure, PQ//MR ∠NMR= 150° and M T
∠QNM = 40°, calculate the value of x. 40° 150°
Solution: N
Through N, NT parallel to MR and PQ is drawn. PQ
(i) a + 1500 = 1800 [Being co-interior angles] x
∴a = 300
(ii) x = a + 400 [Being the alternate angles] M
40° 150°
= 300 + 400 N
∴x = 700
wx ab
9. 8. In the figure alongside, show that zy dc
(i) ∠q = ∠d (ii) ∠f = ∠z
(iii) ∠x = ∠h (iv) ∠a = ∠r pq ef
(v) ∠p = ∠c (vi) ∠b = ∠s sr hg
Solution:
(i) ∠q = ∠f [The corresponding angles within parallel lines are equal] … (a)
∠f = ∠d [The alternate angles within parallel lines are equal] … (b)
From (a) and (b), we get
∠q = ∠d
(ii) ∠f = ∠q [The corresponding angles within parallel lines are equal] … (a)
∠q = ∠z [The alternate angles within parallel lines are equal] … (b)
From (a) and (b), we get
∠f = ∠z
(iii) ∠x = ∠d [The alternate angles within parallel lines are equal] … (a)
∠d = ∠h [The corresponding angles within parallel lines are equal] … (b)
From (a) and (b), we get
∠x = ∠h
(iv) ∠a = ∠y [The alternate angles within parallel lines are equal] … (a)
∠y = ∠r [The corresponding angles within parallel lines are equal] … (b)
From (a) and (b), we get
∠a = ∠r
(v) ∠p = ∠y [The alternate angles within parallel lines are equal] … (a)
∠y = ∠c [The corresponding angles within parallel lines are equal] … (b)
From (a) and (b), we get
∠p = ∠c
Vedanta Excel in Mathematics Teachers' Manual - 8 92
(vi) ∠b = ∠h [The alternate exterior angles within parallel lines are equal] … (a)
∠h = ∠s [The corresponding angles within parallel lines are equal] … (b)
From (a) and (b), we get
∠b = ∠s
10. In the adjoining figure show that ∠x + ∠y + ∠z = 180°. x
Solution:
(i) ∠x = ∠a [The alternate angles within parallel lines are equal] y za
(ii) ∠y = ∠b [The alternate angles within parallel lines are equal]
(iii) ∠a + ∠z + ∠b = 1800 [The sum of the parts of the straight lines is 1800] b
(iv) ∠x + ∠y + ∠z = 1800 [From (i), (ii) and (iii)]
11. In the figure alongside, show that yc b
∠a + ∠b + ∠c + ∠d = 4 right angles. d ax
Solution:
(i) ∠x = ∠b [The alternate angles within parallel lines are equal]
(ii) ∠y = ∠d [The alternate angles within parallel lines are equal]
(iii) ∠x + ∠a = 1800 [Being linear pair]
(iv) ∠y + ∠c = 1800 [Being linear pair]
(v) (∠x + ∠a) + (∠y + ∠c)= 1800+1800 [Adding (iii) and (iv)]
(vi) (∠b + ∠a) + (∠d + ∠c)= 3600 [From (i), (ii) and (iv)]
∴ ∠a + ∠b + ∠c + ∠d = 3600 A 50° B
12. In the adjoining figure, AB//CD. Show that, AC//BD.
Solution: 70° 60° D
(i) ∠ACB = 700 [Given] C
(ii) ∠ABD + ∠BDC = 1800 [AB // CD and the sum of co-interior angles]
or, (500 + ∠CBD) +600 = 1800
or, ∠CBD = 700
(iii) AC // BD [From (i) and (ii), the alternate angles are equal]
13. In the given figure, show that, PQ//TR. PT
100°
Solution: [Given] Q 45° 35° S
(i) ∠QPR = 1000 R
(ii) ∠QRP + ∠PRT +∠TRS = 1800 [The sum of parts of straight angle is 1800]
or, 450 + ∠PRT + 350 = 1800
or, ∠PRT = 1000
(iii) PQ // TR [From (i) and (ii), the alternate angles are equal]
93 Vedanta Excel in Mathematics Teachers' Manual - 8
Extra Questions
1. From the given figure; write the corresponding E HF B
angle of ∠AGH. AG D
[Ans: ∠CHF] C
2. Write down alternate angle of ∠MUT from the given R Q
figure. PT
[Ans: ∠QTS] M UN
S
3. Are the lines AB and CD parallel in the given figure?
Write with reason. n
2m
[Ans: Yes, because the alternate angles are equal]
3p m
4. Find the value of x from the figure given alongside. 7x – 40°
[Ans: 200] x + 80°
5. Find the value of y from the given figure. 2y – 15°
[Ans: 300] y + 15°
6. From the given figure, find the value of p and q. q p + 30°
[Ans: 300, 600]
2p
7. Find the value of m, n and p from the figure given alongside.
E
[Ans: 600, 1200, 1200]
AG 100° B
CH D
100°
F
Vedanta Excel in Mathematics Teachers' Manual - 8 94
Unit Geometry: Triangles
15
Allocated teaching periods: 9
Competency
- Verifying experimentally the properties of triangles
- Solving the problems based on congruency of triangles
- Solving the problems related to similarity triangles
Learning Outcomes
- To explore experimentally the sum of interior angles of triangle
- To verify that the base angles of triangle are equal.
- To explore experimentally the base angles of the isosceles right angled triangle
- To verify experimentally that the line drawn from the vertex of an isosceles triangle to
the mid-point of the base is perpendicular to the base.
- To verify experimentally that the interior angles of the equilateral triangle are equal
and each of 600
- To solve the simple problems on congruent and similar triangles
Level-wise learning objectives
S.N. LEVELS OBJECTIVES
1. Knowledge (K) - To recall the sum of interior angles of triangle
- To tell the relation of base angles of isosceles triangle.
- To write the axiom of congruency of triangles
- To tell the relation between the corresponding angles of
similar triangle
- To solve the problems based on properties of triangle
- To find the unknown angles / sides from the congruent
2. Understanding (U) triangle.
- To solve the simple problems on similarity of triangle
- To explore experimentally the sum of interior angles of
triangle
- To verify experimentally that the base angles of isosceles
3. Application (A) triangle are equal.
- To explore experimentally the relationship between the base
angles of the isosceles right angled triangle
- To verify that the line drawn from the vertex of an isosceles
triangle to the mid-point of the base is perpendicular to the
base.
- To verify that the interior angles of the equilateral triangle
are equal and each of 600
- To solve the simple problems on congruent and similar
triangles
4. High Ability (HA) - To prove some theorems on the properties of triangles.
95 Vedanta Excel in Mathematics Teachers' Manual - 8
Required Teaching Materials/ Resources
Ruler, protractor, strips, sticks, chart paper, scissors, threads, ICT tools like GeoGebra etc.
Pre-knowledge: Different types of pair of angles, types of triangles and the sum of its interior
angles
Teaching Activities for Experimental Verification of Properties of Triangles
1. Show some models of triangles made up of strips, sticks etc. and discuss on the types of
triangles on the basis of sides and angles.
2. Recall the measurement of any angles by using protractor
3. Use triangular paper sheet or visualize in GeoGebra to show that the sum of angles of
triangle is 1800
4. Verify experimentally the sum of interior angles of triangle under discussion.
5. With discussion, explore experimentally the relationship between the base angles of
isosceles triangles.
6. Under discussion, present the experimental verification of base angels of isosceles right
angled triangle.
7. Explore experimentally the relationship between the base angles of equilateral triangle.
8. Make the group of student and give the same work to verify that the line drawn from the
vertex of an isosceles triangle to the mid-point of the base is perpendicular to the base.
9. Experimentally verify that the relationship between the exterior angle and the sum of
opposite interior angles.
10. Give the questions from General Section as classwork.
11. Make the group of students and the questions as given in the exercise.
Solution of selected questions from Excel in Mathematics -Book 8
1. Find the unknown sizes of angles in the given figure. P R
Solution:
x 2x
87°
Here, O S
(i) ∠OSR = x [PQ//RS and alternate angles]
y
Q
(ii) In ∆ROS, ∠OSR + ∠ORS = ∠POR [The exterior angel angle of triangle equal to the
sum of two opposite interior angles]
or, x + 2x = 870
or, 3x = 870
∴ x = 290 and 2x = 2× 290 = 580
(iii) y = 2x [PQ//RS and the alternate angles]
∴ y = 580 P
q
2. Find the unknown sizes of angles in the given figure.
Solution: A
Here,
(i) In ∆AQR, ∠AQR + ∠QAR + ∠ARQ = 1800 r pp R
48° 52°
QB
Vedanta Excel in Mathematics Teachers' Manual - 8 96
or, 480 + 2p + 520 = 1800 [PQ//AR and the alternate angles]
or, 2p + 1000 = 1800 [PQ//AR and the corresponding angles]
or, 2p = 800
∴ p = 400
(ii) r = p = 400 and
(iii) q = p = 400
3. Find the unknown sizes of angles in the given figure. P O Q
ab B
Solution:
z
Here, xy
(i) a = x [PQ//AR and the alternate angles] A
(ii) x = y
[ Base angles of isosceles triangle are equal]
(iv) In ∆AOB, x + y + z = 1800
or, a + a + z = 1800 [From (i) and (ii)]
or, 2a = 1800 – z
∴ a = 180° – z = 900 – z
2 2
Hence, proved
4. Find the unknown sizes of angles in the given figure. A G B
Solution: H D
P
Here, P C
B
(i) ∠AGH = 2∠PGH [PG is the bisector of ∠AGH] C
(ii) ∠CHG = 2∠PHG [PH is the bisector of ∠AGH]
(iii) ∠AGH + ∠CHG = 1800 [AB//CD and co-interior angles]
(iv) 2∠PGH + 2∠PHG = 1800 [From (i), (ii) and (iii)]
or, 2 (∠PGH + ∠PHG) = 1800
∴ ∠PGH + ∠PHG = 900
(v) In ∆PHG, ∠PGH + ∠PHG + ∠GPH = 1800
or, 900 + ∠GPH = 1800 [ From (iii)]
∴∠GPH = 900
Hence, ∠GPH = 1 right angle proved
D
5. Find the unknown sizes of angles in the given figure. A
Solution:
Here,
(i) ∠BAD = 2∠BAP [AP is the bisector of ∠BAD]
(ii) ∠ABC = 2∠ABP [BP is the bisector of ∠ABC]
(iii) ∠BAD + ∠ABC = 1800 [AD//BC and co-interior angles]
(iv) 2∠BAP + 2∠ABP = 1800
[From (i), (ii) and (iii)]
or, 2 (∠BAP + ∠ABP) = 1800
∴ ∠BAP + ∠ABP = 900
97 Vedanta Excel in Mathematics Teachers' Manual - 8
(v) In ∆ABP, ∠BAP + ∠ABP + ∠APB = 1800
or, 900 + ∠APB= 1800 [ From (iii)]
∴ ∠APB = 900
Hence, ∠APB = 1 right angle proved
6. Find the unknown sizes of angles in the given figure. A E
D
Solution:
Here,
(i) ∠ABC = ∠BAC = ∠ACB [AB = BC = AC] B C
(ii) ∠ABC = ∠ECD [AB//EC and corresponding angles]
(iii) ∠BAC = ∠ACE [AB//EC and alternate angles]
(iv) ∠ACE = ∠ECD [From (i), (ii) and (iii)]
Hence, proved
Practice Questions
1. Verify experimentally that the sum of interior angles of a triangle is always equal to two
right angles (1800). (Two figures of different measurement are necessary)
2. Verify experimentally that the base angles of an isosceles triangle are equal. (Two figures
of different measurement are necessary)
3. Explore experimentally the relationship of the angles of equilateral triangle. (Two figures
of different measurement are necessary)
4. Experimentally verify that the line drawn from the vertex of an isosceles triangle to the
mid-point of the base is perpendicular to the base.
5. Verify experimentally that the base angles of the isosceles right angle triangle are equal
and each of 450
Teaching Activities for Congruent Triangles
1. Discuss with the following activities
(i) Tell the students to do “Namste” and ask whether the palms fit to each other or not.
(ii) Take out two adjacent pages of a book and ask whether they fit to each other or not.
(iii) Take a rectangular sheet of paper and fold it along the diagonal. Then cut out the folded
edge by using scissors, place two triangular pieces and ask whether they fit to each other
or not. C D D' C D'
CA
DD' B
A BB' A B B' B'
D
After performing different such activities, define congruent figures. Ask the students to
observe, think and tell more examples of congruent figures which are available around
their surroundings.
Vedanta Excel in Mathematics Teachers' Manual - 8 98
2. Discuss upon the congruent triangles, notation and the corresponding parts. If possible,
make clearer about congruent triangles by copying, reflecting or translating or rotating a
triangle via GeoGebra tool.
A A'
B C C' B'
∆ABC ≅ ∆A′B′C′
Meaning of ≅ is ∼ and =. That is “similar in shape and equal in size”.
3. Show the experiments on conditions of congruency of triangles by construction
4. Discuss upon the following axioms of congruency of triangles.
(i) S.S.S. axiom
(ii) S.A.S. axiom
(iii) A.S.A. axiom
(iv) R.H.S. axiom
(v) A.A.S. axiom
5. Under discussion with experiments of congruent triangles, note out the following points.
(i) The sides opposite to equal angles are called Corresponding Sides.
(ii) The angles opposite to equal sides are called Corresponding Angles.
(iii) The corresponding sides are equal.
(iv) The corresponding angles are equal.
6. Make the groups of students and support to solve the problems from exercise.
Solution of selected questions from Excel in Mathematics -Book 8
1. Mention the necessary axioms to show the following pair of triangles 5 cm A
congruent. Also, find the value of x. x
Solution:
Here, BD C
In ∆ABD and ∆ADC; [Common side]
(i) AD = AD (S) [Both are right angles]
(ii) ∠ADB = ∠ADC (A) [Given]
(iii) BD = CD (S) [By S.A.S. axiom]
∴∆ABD ≅ ∆ADC [Corresponding sides of congruent triangles are equal]
Also, AC = AB
∴x = 5 cm
A C
2. Mention the necessary axioms to show the following pair of triangles
congruent. Also, find the value of x.
Solution: BD
Here,
In ∆AOB and ∆COD;
(i) ∠BAO = ∠CDO (A) [Given]
99 Vedanta Excel in Mathematics Teachers' Manual - 8
(ii) AO = DO (S) [Given]
(iii) ∠AOB = ∠COD (A) [Vertically opposite angles are equal]
∴∆AOB ≅ ∆COD [By A.S.A. axiom]
Also, AB = CD [Corresponding sides of congruent triangles are equal]
∴x = 6.5 cm
3. Mention the necessary axioms to show the following pair of A xD
triangles congruent. Also, find the value of x.
Solution: 50° C
Here, B
In ∆ABC and ∆ACD;
(i) AB = CD (S) [Given]
(ii) BC = AD (S) [Given]
(iii) AC= AC (S) [Common side]
∴∆ABC ≅ ∆ACD [By S.S.S. axiom]
Also, ∠ABC = ∠ADC [Corresponding angles of congruent triangles are equal]
∴x = 500
AP
4. Find the value of x in the following pairs of (x+315.3°)cm 85° 3.2cm
congruent triangles. Also, write all the sizes of x cm
35° 60°
corresponding angles and side. R
B 5.5cm C Q
Solution:
Here, ∆ABC ≅ ∆PQR;
As the corresponding angles of congruent triangles are equal
(i) ∠A = ∠P = 850
(ii) ∠C = ∠R = 600
As the corresponding sides of congruent triangles are equal
(i) BC = QR = 5.5 cm
(ii) AC = PR ∴x = 3.2 cm
(iii) AB = PQ = (x + 1.3) cm = (3.2 + 1.3) = 4.5 cm CE
5. Find the value of x in the following pairs of B (x+2)cm30° 30° 6.5cm
congruent triangles. Also, write all the sizes 86°
of corresponding angles and side. 3.1cm (2x–2)cm 64° F
A 3.1cmD
Solution:
Here, ∆ABC ≅ ∆DEF;
As the corresponding angles of congruent triangles are equal
(i) ∠A = ∠D = 640
(ii) ∠B = ∠F = 860
As the corresponding sides of congruent triangles are equal
(iii) BC = EF = 6.5 cm
or, x + 2 = 6.5 ∴x = 4.5
(iv) AC = DE = (2x – 2) cm = (2×4.5 – 2) cm = 7 cm
Vedanta Excel in Mathematics Teachers' Manual - 8 100
The words you are searching are inside this book. To get more targeted content, please make full-text search by clicking here.
Vedanta Excel in Mathematics Book 8 TG Final (2079)
Related Publications
Discover the best professional documents and content resources in AnyFlip Document Base.
Search