Solution:
Here
Here, the length of wall (l) = 20 m, the width (b) = 20 cm = 0.2 m
Now, volume of wall (V) = l b h= 20 m 0.2 m h = 4h m3
Also, the length of window (l1) = 2 m, the height (h1) = 1.5 m and the breadth (b1) = width
of wall= 20 cm = 0.2 m
?Volume space occupied by of 2 windows = 2(2 m 0.2 m 1.5 m)= 1.2 cm3
Volume of wall excluding the windows = (4h – 1.2) m3
Again, volume of each bricks (v) = 25 cm 16 cm 10 cm
= 0.25 m 0.16 m 0.1m = 0.004cm3
Volume of wall (V)
Again, number of bricks (N) = Volume of each brick (v)
or, 4,700 = 4h – 1.2
0.004
or, 18.8 = 4h – 1.2 ?h=5m
Hence, the height of the wall is 5 m.
3. A square room contains 288 m3 of air. The cost of carpeting the room at Rs 105 per
sq. metre is Rs 6,720. Find the cost of painting its walls at Rs 45 per sq. metre.
Solution:
Let the length of room (l) = breadth of the room (b) = x m
Now, volume of the room (V) = 288 m3
or, lbh = 288 or, x.x.h = 288 ?x2h = 288 …equation (i)
Again, area of floor = Total cost of carpeting = 6720
Rate of carpeting 105
or, x2 = 64 or, x = 8 ?l=b=8m
Also, putting the value of x in equation (i), we get
82h = 288 ? h = 4.5 m
Again, the area of walls (A) = 2h (l +b) = 24.5(8+8) = 144 m2
?Total cost of painting the walls (T) = Area (A) Rate (R) = 144 Rs 45 = Rs 6,480
Extra Questions
1. A square room contains 180 cu. metre of air. The cost of plastering its four walls at Rs 20
per sq. metre is Rs 2,400. Find the height of the room.[Ans:5m]
2. After destruction by massive earthquake, the wall of length 40 m, height 5 m and width
20 cm was reconstructed. It contains two windows each of 2 m 1.5 m and a gate of size
1.5 m 4 m.
(i) Find the number of bricks each of 25 cm 20 cm 4 cm required to construct the
wall.
(ii) Find the cost of the bricks at the rate of Rs 16,000 per 10000 bricks.
[Ans: (i) 23,500 (ii) Rs 3,76,000 ]
3. Mr. Gurung constructed a compound wall 40 m long and 20 cm wide by the bricks, each
measuring 20 cm 10 cm 5 cm. If he paid Rs 4,32,000 at the rate of Rs 18,000 per 1000
bricks, find the height of the wall. [Ans: 3m]
47 Vedanta Excel in Mathematics Teachers' Manual - 9
Unit Algebraic Expressions
6
Allocated teaching periods 10
Competency
- 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 and simplify the rations 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
- To simplify the rations expressions (up to three terms)
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 LCM of denominator of rational expressions
- To factorize the algebraic expressions by using the
formula a2 – b2, a3 ± b3
2. Understanding (U) - To factorize the formulae of the forms a4 + a2b2 + b4
- To factorize the expressions involving 5 or 6 terms by
expressing in the formulae of (a± b)2 and a2 – b2
3. Application (A) - To find the HCF and LCM of the given expressions
- To simplify the rational 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
A. Factorization of algebraic expressions
Teaching Activities
1. Recall the formulae through chart paper
2. 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.
3. 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
4. Present the derivation of the formulae of (a + b)3with blocks or ICT tools
5. Discuss upon the factorization of the form a4 + a2b2 + b4 with examples
6. Engage the students to factorize the expressions given in the exercise
7. Focus on more practical problems related to factorization
Vedanta Excel in Mathematics Teachers' Manual - 9 48
B. H.C.F., L.C.M. and simplification of the algebraic expressions
Teaching Activities
1. Recall about higher common factors (H.C.F.) and lowest common multiples
(L.C.M.)
2. Make the groups of students and encourage them to find the H.C.F. and L.C.M. of
the expressions with recalling factorization of expressions
3. Discuss about rational expressions and encourage the students to simplify the
rational expressions by giving examples
1. Factorisation.
Solution of selected problems from Vedanta Excel in Mathematics
1. Factorise: 16a4 – 4a2 – 4a – 1
Solution:
Here, 16a4 – 4a2 – 4a – 1
= 16a4 – (4a2 + 4a + 1)
= (4a2)2 – [(2a)2 + 2.2a.1 + (1)2]
= (4a2)2 – (2a + 1)2
= (4a2 + 2a + 1) (4a2 – 2a – 1)
2. Factorise: 16p2 – 72pq + 80q2 – 6qr – 9r2.
Solution:
Here, 16p2 – 72pq + 80q2 – 6qr – 9r2
= (4p)2 – 2.4p.9q2 – (9q)2 – (9q)2 + 80q2 – 6qr – 9r2
= (4p – 9q)2 – q2 – 6qr – 9r2
= (4p – 9q)2 – (q2 + 6qr + 9r2)
= (4p – 9q)2 – [q2 + 2.q.3r + (3r)2]
= (4p – 9q)2 – (q + 3r)2
= (4p – 9q + q + 3r) (4p – 9q – q – 3r) = (4p – 8q + 3r) (4p – 10q – 3r)
3. Resolve into factors: (a2 – b2) (c2 – d2) + 4abcd
Solution:
Here, (a2 – b2) (c2 – d2) + 4abcd
= a2c2 – a2d2 – b2c2 + b2d2 + 4abcd
= (ac)2 + 2abcd + (bd)2 – (ad)2 + 2abcd – (bc)2
= (ac + bd)2 – [(ad)2 – 2abcd + (bc)2]
= (ac + bd)2 – (ad – bc)2
= (ac + bd + ad – bc) (ac + bd – ad + bc) = (ac + ad – bc + bd) (ac – ad + bc + bd)
4. Resolve into factors: p7 + 1
Solution: p5
Here, p7 + 1 = p7 + p × 1 = p p6 + 1
p5 p6 p6
= p (p2)3 + 13 =p p2 + 1 p4 – p2 × 1 + 1
p2 p2 p2 p4
= p p2 + 1 p4 – 1 + 1
p2 p4
5. Factorise: a3 + b3 + c3 + 3abc
Solution:
Here, a3 + b3 + c3 – 3abc
= (a + b)3 – 3ab(a + b) + c3 – 3abc
= (a + b)3 + c3 – 3ab(a + b) – 3abc
49 Vedanta Excel in Mathematics Teachers' Manual - 9
= (a + b + c)3 – 3(a + b).c(a + b + c) – 3ab (a + b + c)
= (a + b + c) [(a + b + c)2 – 3(ac + bc) – 3ab]
= (a + b + c) (a2 + b2 + c2 + 2ab + 2bc + 2ca – 3ca – 3bc – 3ab)
= (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
6. The area of rectangular plot of land is (x2 + 13x + 40) sq. m.
(i) Find the length and breadth of the land.
(ii) If the length and breadth of the land are reduced by 2/2 m respectively, find
the new area of the land.
Solution:
Here, the area of rectangular plot of land = (x2 + 13x + 40) sq. m.
or, l × b = x2 +8x + 5x + 40
or, l × b = x(x + 8) + 5(x + 8)
or, l × b = (x + 8) (x + 5)
? length (l) = (x + 8) m and breadth (b) = (x + 5) m.
Again, New length (l') of the plot = (x + 8 – 2) m = (x + 6) m
New breadth (b') of the plot = (x + 5 – 2) m = (x + 3) m
? Area (A) = l' × b'
= (x + 6) (x + 3) m2
= (x2 + 9x + 18) m2
Extra questions: 2) a4 + a2 + 1 3) x4 + 1 + y4 4) p6 – 1
1) 9a4 + 14a2 + 25 b4 b2 y4 x4
5) x2 – 10x + 16 – 6y – y2 6) (9 – a2) (100 – b2) 7) 6(a + b)2 – (a + b) – 7
8) 2x2 – 3 – 5y2
y x2
Answers a2 a a2 a
b2 b b2 b
1. (3a2 + 4a + 5) (3a2 – 4a + 5) 2. ( + + 1) ( – + 1)
3. ( x2 + 1 + y2 ) ( x2 – 1 + y2 ) 4. (p + 1) (p – 1) (p2 + p + 1) (p2 – p + 1)
y2 x2 y2 x2
5. (x + y – 2) (x – y – 8) 6. (ab + 10a + 3b + 30) (ab – 10a – 3b + 30)
7. (a + b + 1) (6a + 6b – 7) 8. (2yx – 5xy) ( xy + y )
x
2. H.C.F., L.C.M. and simplification
1. Find the H.C.F. of x3 – y3, x6 – y6, x4 + x2y2 + y4
Solution:
Here, x3 – y3, x6 – y6, x4 + x2y2 + y4
The 1st expression = x3 – y3 = (x – y) (x2 + xy + y2)
The 2nd expression = x6 – y6 = (x3 + y3) (x3 – y3)
= (x + y) (x2 – xy + y2) (x – y) (x2 + xy + y2)
= (x + y) (x – y) (x2 + xy + y2) (x2 – xy + y2)
The 3rd expression = x4 + x2y2 + y4
= (x2)2 + (y2)2 + x2y2
= (x2 + y2)2 – 2x2y2 + x2y2
= (x2 + y2)2 – x2y2 = (x2 + y2)2 – (xy)2 = (x2 + xy + y2) (x2 – xy + y2)
? H.C.F. = x2 + xy + y2
Vedanta Excel in Mathematics Teachers' Manual - 9 50
2. Simplify: (a 1 – a2 1 b2
– b)2 –
Solution:
Here, (a 1 – a2 1 b2 = (a – 1 – b) – (a + 1 – b)
– b)2 – b) (a b) (a
= (a (a + b) – (a – b) b) = a +b–a+b = (a + 2b – b)2
– b) (a – b) (a + (a – b)2 (a + b) b) (a
3. Simplify: y2 2y + 5 9 – 11 – 16y
+ 6y + y2 – 9 8y2 – 24y
Solution:
Here, y2 2y + 5 9 – 11 – 16y
+ 6y + y2 – 9 8y2 – 24y
= 2y + 5 – (y + 11 – 3) – 16y 3)
(y + 3)2 3) (y 8y(y –
= (2y + 5) (y – 3) + 11(y + 3) – 2(y + 3)2
(y + 3)2 (y – 3)
= 2y2 – 6y + 5y – 15 + 11y + 33 – 2(y2 + 6y + 9)
(y + 3)2 (y – 3)
= 2y2 + 10y + 18 – 2y2 – 12y – 18 = –2y = 2y
(y + 3)2 (y – 3) (y + 3)2 (y – 3) (3 – y) (y + 3)2
4. Simplify: a+2 – a–2 – 2a2
1 + a + a2 1 – a + a2 1 + a2 + a4
Solution:
Here, 1 a+2 a2 – 1 a–2 a2 – 1 2a2 a4
+a+ –a+ + a2 +
= (a + 2) (1 – a + a2) – (a – 2) (1 + a + a2) – 2a2 + a2
(1 + a + a2) (1 – a + a2) (1 + a2)2 – 2a2
= a – a2 + a3 + 2 – 2a + 2a2 – a – a2 – a3 + 2 + 2a + 2a2 – 2a2
(1 + a + a2) (1 – a + a2) (1 + a + a2) (1 – a + a2)
= (1 + a + 2a2 – a + a2) – (1 + a + 2a2 – a + a2) = (1 + a 2a2 – 2a2 a + a2) = 0
a2) (1 a2) (1 + a2) (1 –
Extra questions:
1. Find the H.C.F. and L.C.M. of
a) a2 – b2, a3 – b3 and a4 – b4 b) a2 + 2ab + b2 – c2, b2 + 2bc + c2 – a2, c2 + 2ca + a2 – b2
c) x3 + y3, x4 + x2y2 + y4 d) x2 – 4, x3 + 8, x2 + 5x + 6
e) (a + b)2 – 4ab, a3 – b3, a2 + ab – 2b2
2. Simplify:
a) x–y + y–z + z–x b) a2 – ab + b2 + a2 + ab + b2
xy yz zx a–b a+b
x y a2 + b2 b2 a2
c) xy – y2 + xy – x2 d) ab – a(a + b) – b(a + b)
e) (a – b)2 – c2 + (b – c)2 – a2 + (c – a)2 – b2
a2 – (b + c)2 b2 – (c + a)2 c2 – (a + b)2
Answers
1. a) (a – b) (a4 – b4) (a2 + ab + b2) b) a + b + c, (a + b + c) (a + b – c) (b + c – a) (c + a – b)
c) (x2 – xy + y2), (x + y) (x4 + x2y2 + y4) d) (x + 2), (x – 2) (x + 3) (x3 + 8)
e) (a – b), (a – b)2 (a + 2b) (a2 + ab + b2) 2. a) 0 b) 2a3 c) x+ y d) 1 e) 1
a2 – b2 xy
51 Vedanta Excel in Mathematics Teachers' Manual - 9
Unit Indices
7
Allocated teaching periods 6
Competency
- To simplify the expressions involving indices and solve the exponential equations
Learning Outcomes
- To simplify the expressions by using the laws of indices related to negative and
fractional power
- To solve the exponential equations
Level-wise learning objectives
S.N. LEVELS OBJECTIVES
- To identify coefficient, base and power of expression
1. Knowledge (K) - To recall the laws of indices
- To express the product/quotient of expressions having same
base in terms of single base
- To evaluate the numerical problems by using laws of indices
2. Understanding (U) - To simplify/prove the simple given expressions
- To solve the exponential equations
- To simplify/prove the given rational expressions (involving
roots as well) by applying the laws of indices
3. Application (A) - To solve the exponential equations of the quadratic form
4. High Ability (HA) - To prove the rational expression under the given condition/s.
- To prepare the report about the use of indices
Required Teaching Materials/ Resources
Chart papers with laws of indices, scissors, ruler, glue-stick and computer/projector if
possible
Pre-knowledge: Laws of indices, basic operations
A. Indices
Teaching Activities
1. Give the practical examples of use of laws of indices.
For example
(i) The cost of 1 kg of apple is Rs 125. Find the cost of 5 kg of apples by using the
product law of indices. For, 5 × 125 = 51 × 53 = 51 + 3 = 54 = 625
(ii) Divide 64 copies are equally among 4 friends by using the quotient law of
64 26
indices. For, 4 = 22 = 26 – 2 = 24 = 16
2. Recall of indices by presenting in chart paper with proper examples
3. Table of law of indices discuss, give the way of solving the various problems and
involve the students in solving the problems from exercise
4. Under given condition, prove the expressions and give the same type problems to
the students and tell them to prove in the class.
Vedanta Excel in Mathematics Teachers' Manual - 9 52
5. Call the students randomly to solve the problems on the board in order to make
them confident to solve the problems
B. Exponential Equations
Teaching Activities
1. Ask the laws of indices
2. Discuss upon the exponential equations like 2x = 8, x =? etc.
3. With more examples, list he following ideas
(i) If ax = ap then x = p
(ii) If xn = kn then x = k
(iii) If ax = 1 then x = 0 = a0
4. Solved some equations and give same type of equations to solve in the class or at
home
5. Discuss upon the problems given in the exercise
Solution of selected problems from Vedanta Excel in Mathematics
1. Find the value of c) y 1 x 1 1 d) 3 u 4 1
x 1 y 1 100
a) (1 3 5) 1 (1 35) 1 b) (a b) 1 . (a 1 b 1) 1004
Solution:
a) (1 3 5) 1 (1 35) 1 b) (a b) 1 . (a 1 b 1)
= 1 1 1 = 1 11
35 a bu a b
(1 35) 1
= 35 1 1 1 = 1 b a
35 1 35 a bu ab
= 35 1 = 1
35 1 35 1 ab
= 35 1 =1
35 1
c) y 1 x 1 1 xy 1 = x2 y2 1 = xy
x 1 y 1 y x xy x2 y2
=
3 41 3 1 13 1 3 1 3 1 3 1
100 102
d) 1004 × = (102)4 u 4 = 102 u 102 u 1 = 102 u 1 = 102 u 10 2 = 10 2 = 10
4
102
aS2).olutioSn5im3:5n p 2nluiufy(65:2u5an2 )1)5 135n 2n u 625n 1 9x u 3x 1 3x
b) 32x 1 u 3x 2 3x
u (5 u 2) 1
= 5 n u (54)n 1 = 5 n 4n 4 3n 2 1 u 2 = 5 1 u 2 = 2
53n 2 u 5 1 u 2 1 5
b) 9x u 3x 1 3x
32x 1 u 3x 2 3x
32x 3x u 3 1 3x =3x3(3x(23x u2x3u u 3 2 1) 32x 1 1 32x 1 1
= 32x u u u 3x u 3 2 3x = 32x 1 2 1 = 32x 1 1 =1
3
3. Simplify: (p2q)13
(pq2)31 q13
p + p– + p31 a1 b1 c1
q x u x u xb c b a a b a c
a) × 1– b) c a c b
c) ax y x2 × ay z y2 × az x z2
ay2 az2 ax2
53 Vedanta Excel in Mathematics Teachers' Manual - 9
Solution: (pq2)31 + (p2q)31 × 1– q13
a) p + p– q p13
= p + p31q23 + p23q13 u p13 q31
p–q p13
= p13 (p32 q32 p31q31) u p31 q31
p–q p31
= (p31 q13)[(p31)2 p13 q13 (q31)2] = (p31)3 (q13)3 = p–q =1
p–q p – q p–q
b1 c1 a1
x u x u xb c b a a b a c
b) c a c b
bca
= x u x u x(b c)(b a)
(c a)(c b) (a b)(a c)
=x b (c c b) a
(b c)(b a) a)(c (a b)(a c)
b(c a) c(a b) a(b c) –bc + ab – ca + bc – ab + ca
(a b)(b c) (c a) (a b)(b c) (c a)
b
= x = x = x = x = 1(b c)(a b)
(c c c) a 0
a)(b (a b)(c a)
c) ax y x2 × ay z y2 × az x z2
ay2 az2 ax2
= x+y ax2 y2 u y z ay2 z2 u az x z2 x2
x2 y2 y2 z2 z2 x2
= a x y u a y z u a z x = = aax y y z z x 0 =1
4. (a) If xyz = 1, prove that 1 + 1 y–1 + 1 + 1 z–1 + 1 + 1 x–1 = 1.
x+ 1 y+ 1 z+ 1
(b) If a + b + c = 0, prove that 1 + xa + x–b + 1 + xb + x–c + 1 + xc + x–a = 1
Solution:
(a) Here, xyz = 1
L.H.S = 1 + 1 y–1 + 1 + 1 z–1 + 1 + 1 x–1
x+ y+ z+
= 1 1 1
x
1 + 1 1 + y 1 1 + z 1
y z x
1 1 1
= y xy 1 z yz 1 x zx 1
yzx
zuy z x
= z(y xy 1) z yz 1 x zx 1
= yz yz z z z 1 x x 1
xyz yz zx
= yz zx
yz 1 z z yz 1 x zx 1
= x(yz z) x = xyz zx x x = 1 zx x = 1 = RHS
x(yz z 1) x zx 1 xyz zx x zx 1 1 zx x
Vedanta Excel in Mathematics Teachers' Manual - 9 54
(b) Here, a + b + c = 0
LHS = 1 + 1 x–b + 1 + 1 x–c + 1 + 1 x–a = 1
xa + xb + xc +
= 1 1 1
xa
1 + 1 1 + xb 1 1 + xc 1
xb xc xa
xb u xc xc xa
= xc(xb xa b 1 xc xb c 1 xa xc a 1
= xb c xb c xc xc xa
xa b c xb c xc 1 xc a xa 1
= xb c xc xa
xb c x0 xc xb c xc 1 xc a xa 1
= xb c xc xa xc a xa 1
xb c xc 1 xc a xa 1
xc a xa 1
= xa(xb c xc) xa
xa(xb c xc 1)
xc a xa 1
= xa b c xc a xa
xa b c xc a
xa xc a xa 1
= x0 xc a xa
x0 xc a xa
xc a xa 1
= 1 xc a xa = = 1 = R.H.S
1 xc a xa xc a xa 1
5. a) If x = 1 + 2– 1 , prove that: 2x3 – 6x = 5.
3
23
12
b) If x – 2 = 3 3 + 3 3 , show that: x(x2 – 6x + 3) = 2.
Solution:
a) Here, 1 2– 1
x = 23 3
+
Cubing on both sides, we get,
3
1 2– 1
x3 = + 3
23
13 31
or, x3 = 2– 1 2– 1 1 + 2– 1 [ (a b)3 = a3 b3 3ab(a b)]
23 + 3 + 3 u 23 3 3
? u 23
?11
or, x3 = 2 221 1 3x3 u 1 u x [ + 2– 3 = x]
or, x3 = 2 23
or, x3 = 4 1 6x
2
or, 2x3 – 6x = 5 proved
b) Here,
12
x – 2 = 33 + 33
Cubing on both sides, we get
55 Vedanta Excel in Mathematics Teachers' Manual - 9
1 23
(x – 2)3 = 3 3 + 3 3
13 23 1 21 2
or, x3 – 23 – 3 u x u 2 (x – 2) = 3 3 + 3 3 + 3 u 3 3 u 3 3 3 3 + 3 3
or, x3 – 8 – 6x(x – 2) = 3 32 3 u 31 u (x – 2)
or, x3 – 8 – 6x2 12x = 12 9x – 18
or, x3 – 6x2 3x = 2
? x(x2 – 6x 3) = 2 proved.
6. Solve: 2x 3 u 3x 2 = 432
Solution:
Here, 2x 3 u 3x 2 = 432
or, 2x u 23 u 3x u 32 = 432
or, (2 u 3)x u 8 u 9 = 432
or, 6x = 6 ? x = 1
7. a) If xa = y, yb = z and zc = x, prove that abc = 1
11
b) If a x = b 3 and ab = 1, prove that x 3 = 0.
c) If ax = by and ay = bx, show that x = y
Solution:
a) Here, xa = y, yb = z and zc = x.
Now, xa = y or, (zc)a = y [? x = zc
or, zca = y or, (yb)ca = y [? z = yb]
or, yabc = y1 ? abc = 1 proved.
11 or, 1 x= 1x
b) Here, a x = b 3 ax b3
x
? a = b3
Now, ab = 1 x
or, bx 1 = b0 or, b 3 . b = 1
3 or, x 3 = 0 proved.
c) Here, ax = by ........ (i)
ay = bx or, bx = ay ......... (ii)
Multiplying equn (i) and (ii), we get
ax.bx = by.ay
or, (ab)x = (ab)y
? x = y proved.
8. a) If (a–1 b–1) (a b)–1 = ambn, prove that am–n = 1
b) If m–1n2 7 m3n–5 –5
Solution: m2n4 m–2n3
y = mxny, prove that mx–2y = 1.
a) Here, (a–1 b–1) (a b)–1 = ambn
1 1 1
or, a b a b = ambn
or, b a 1 = ambn
ab a b
or, a–1b–1 = ambn ? m = –1, n= –1
Again, am–n = a–1–(–) = a0 = 1 proved
Vedanta Excel in Mathematics Teachers' Manual - 9 56
b) m–1n2 7 m3n–5 –5
m2n4 m–2n3
y = mxny
or, (m–3n6)7 y (m5n–8)–5 = mxny
or, m–21n42 y m–25n40 = mxny
or, m4n2 = mxny
? x = 4 and y = 2
Again, mx–2y = m4–2u2 = m0 = 1 proved
9. If xm.xn = (xm)n, prove that xm(m–2) u xn(m–2) = 1
Solution:
Here, xm.xn = (xm)n
or, xm n = xmn
or, m n = mn ....... (i)
Now,
LHS = xm(n–2) u xn(m–2)
= xmn – 2m mn – 2n
= x2mn – 2(m n)
= x2mn – 2mn [From (i)]
= x0 = 1 = RHS Proved.
Extra Questions Ans: (a 1 x)2
+
2
Ans: 10pq2r2
1. Simplify: a) 3 (a + x)–8 × (a + x) 3
b) 3 20pq5r8 × 3 50p4qr–2
2. Prove that: a) 1 + 1 + xc – b + 1 + 1 + xa – c + 1 + 1 + xb – a = 1
xa – b xb – c xc – a
a2 – 1 a× a – 1 b–a a a+b
b2 b b
b) =
1 b× 1 a–b
b2 – a2 b + a
3. Solve: a) 2x + 2–x = 4 + 4–1 Ans: ± 2
b) 25x – 6 a 5x + 1 + 125 = 0 Ans: 1, 2
4. a) a + b + c = 0, prove that (1 + xa + x–b)–1 + (1 + xb + x–c)–1 + (1 + xc + x–a)–1 = 1
57 Vedanta Excel in Mathematics Teachers' Manual - 9
Unit Simultaneous Linear Equations
8
Allocated teaching periods 5
Competency
- To solve the real life problems based on linear simultaneous equations
Learning Outcomes
- To solve the linear simultaneous linear equations by substitution, elimination and graphical
methods
Level-wise learning objectives
S.N. LEVELS OBJECTIVES
1. Knowledge (K) - To define linear equation
2. Understanding (U) - To define simultaneous linear equations
- To tell the methods of solving pair of linear equations
3. Application (A) - To solve the pair of simultaneous equation (simple) by
4. High Ability (HA)
substitution method
- To solve the pair of simple simultaneous equation by
elimination method
- To solve the pair of simultaneous equation by substitution
method
- To solve the pair of simultaneous equation by elimination
method
- To make the equation for the given conditions and solve
them
- To identify the contextual problems as related to
simultaneous equations, make the equations and solve it
Required Teaching Materials/ Resources
Graph board, graph paper, charts with various pair of linear equations, Geo-gebra tool, list of grocery
items with rates of costs, audio-video materials etc.
Pre-knowledge: Linear equation, solving simultaneous equations by graphical method
Teaching Activities
1. Warm up the class with some mathematical logics, quiz questions, game etc.
2. Divide the students into four five groups and give some real life problems (if possible with
pictures) to solve 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 12000, 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 500 more than that of the shirt. If the total cost of the shirt and
pant is Rs 1600, 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
3. Discuss about their equations that the students made, solutions and methods of solving the
problems in the above activities
4. Discuss about the substitution method with proper examples
5. Explain about the elimination method with proper examples
6. Discuss about the graphical method with proper examples
7. 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
Vedanta Excel in Mathematics Teachers' Manual - 9 58
1. Solve the following simultaneous equations graphically.
a) x 2y = 14 and x – y = 5
x–2 2x – 6
b) 2 =y= 3
Solution: x + 2y = 14
a) The given linear equation are
x + 2y = 14 ........... (i)
and x – y = 5 ........... (ii) (8, 3)
Now from equation (i). x–y=5
x + 2y = 14
14 – x
or, y= 2
x024
y765
Plotting the point (0,7), (2,6) and (4,5) and joining them to form a straight line.
Also, from equation (ii), x – y = 5 ? y = x – 5
x123
y –4 –3 –2
Plotting the point (1,–4), (2.–3) and (3,–2) and joining them to form a straight line.
Since the graphs of equation x 2y = 14 and x – y = 5 intersect at (8,3).
So, x = 8 and y = 3
b) The given liner equation are
x– 2 =y ......... (i)
2 y ......... (ii)
= 2x – 6
and 3
x–2 (6, 2)
Now, from equation (i), y = 2
x024
y –1 0 1
Plotting the point (0,–1), (2,0) and (4,1) and
joining them to form a straight line.
2x – 6
Also, from equation (ii), y = 3
x036
y –2 0 2
Plotting the point (1,–2), (3.0) and (6,2) and joining them to form a straight line.
x–2
Since the graphs of equation y = 2 intersect at (6,2).
Hence, x = 6 and y = 2
2. Solve the equation 2 6 = 3 and 10 – 9 = 2 by elimination method.
x y x y
Solution:
Here,
The given equation are
59 Vedanta Excel in Mathematics Teachers' Manual - 9
2 6 = 3 ......... (i) u 3
x y ......... (ii) u 2
10 9
x – y = 2
Now, multiplying equation (i) by 3 and equation (ii) by 2 and adding them.
6 18 =9
x y =4
20 18
x – y
26 = 13
x
or, 13x = 26 ? x = 2
Again, putting the value of x in equation (i), we get
2 6
2 y = 3
or, 6 = 3 – 1
y
or, 2y = 6 ? y = 3
Hence, x = 2 and y = 3
3. Solve the given system of equation by substation method.
x–1 1 x–2 1
y 1 = 2 and y 2 = 3
Solution:
Here,
The given equation are
x–1 1
and xy – 21 = 12 or, 2x – 2 = y 1 ? y = 2x – 3 ...... (i)
y 2 = 3 or, 3x – 6 = y 2 ? 3x – y = 8 ...... (ii)
Now, substituting the value of y from equation (i) in equation (ii), we get
3x – (2x – 3) = 8 ? x = 5
Again, Substituting the value of x in equation (i), we get
y=2u5–3 =7
Hence, x = 5 and y = 7
4. The cost of 4 kg of chicken and 5 kg of mutton is Rs 7,200. If the cost of 4 kg of
chicken is the same as the cost of 1 kg of mutton, find the rate of cost of chicken and
mutton.
Solution:
Let the cost of chicken be Rs x per kg and that of the mutton be Rs y per kg.
Then,
According to the given first condition,
4x 5y = 7200 ........ (i)
According to the given second condition,
4x = y
Now, Substituting the value of y in equation (i) from equation (ii). We get
4x 5 u 4x = 7200
or, 24x = 7200 ? x = 300
Again Substituting the value of x in equation (ii). We get
Vedanta Excel in Mathematics Teachers' Manual - 9 60
4 u 300 = y ? y = 1200
Hence, the rate of cost of chicken is Rs 300 and that of mutton is Rs 1200.
5. Mother is three times as old as her daughter. Three years ago she was four times as
old as her daughter was. Find their present ages.
Solution:
Let the present age of the mother be x years and that of her daughter be y years
Then,
From the 1st condition;
x = 3y ..... (i)
From the 2nd condition;
x – 3 = 4 (y – 3) ..... (ii)
Now, Substituting the value of x from equation (i) in equation (ii). We get
3y – 3 = 4y – 12 ? y = 9
Again substituting the value of y in equation (i). We get,
x = 3 u 9 = 27
Hence, the present age of the mother is 27 years and that of daughter is 9 years.
6. The sum of the digits of a two digit number is 10. If 18 is subtracted from the number,
the places of the digits are reversed. Find the number.
Solution:
Let the digit at tens place be x and at ones place be y.
Then, the number is 10x y
When the digits are reversed, the new number is 10y x
From the first condition,
x y = 10 or, y = 10 – x ..... (i)
From the second condition,
10x y – 18 = 10y xor, x – y = 2 ....... (ii)
Now, Substituting the value of from equation (i) in equation (ii), We get
x – (10 – x) = 2 or, 2x = 12 ? x = 6
Again, Substituting the value of x in equation (i). We get
y = 10 – 6 = 4
Hence, the required number is 10x y = 10 u 6 4 = 64
Extra Questions
1. Solve each pair of simultaneous equation by graphical method.
i) x 2y = 4 and 3x – y = 5 [Ans: 2,1]
ii) 2x – y = 17 and 2x 3y 11 = 0 [Ans: 5, –7)
2. Solve each pair of simultaneous equations by elimination method.
i) x 3y = 7 and 3x – y = 11 [Ans: 4,1]
ii) 9 – 4 = 1 , 15 2 = 6 [Ans: 3,2]
x y x y
3. Solve each pair of simultaneous equations by substitution mehtod.
x–1 1 x–1 1
i) y 1 = 2 and y 1 = 2 [Ans: 10,11]
ii) x x–1 = 6 and y– x–1 =2 [Ans: 4,3]
y 1 y 1
61 Vedanta Excel in Mathematics Teachers' Manual - 9
Unit Quadratic Equation
9
Allocated teaching periods 5
Competency
- To solve the quadratic equations
Learning Outcomes
- To solve the quadratic equations by factorization, completing square methods and using
quadratic formula
Level-wise learning objectives
S.N. LEVELS OBJECTIVES
- To define quadratic equation
1. Knowledge (K)
- To write the roots of the equation ax2+bx + c = 0, a≠0
2. Understanding (U) - To solve the simple quadratic equations by factorization
method
- To solve the quadratic equation by factorization method
3. Application (A) - To solve the quadratic equation by completing square
method
- To solve the quadratic equation by using formula
4. High Ability (HA) - To identify the contextual problems as related to
quadratic equations and solve them
Required Teaching Materials/ Resources
Chart paper with various quadratic equations, ICT tools (if possible), audio-video materials
etc.
Pre-knowledge: Quadratic equation, factors of quadratic expressions etc
Teaching Activities
1. Write some equation on the board or show in chart paper and ask about the degree and
variable
2. Explain about quadratic equations with examples.
3. Recall the factorization and discuss about the solution of quadratic equation by
factorization method
For example: x2 – 5x + 6 = 0 or, x2 – 3x – 2x + 6 = 0 or, (x – 2) (x – 3) = 0
Either, x – 2 = 0 ? x = 2 or, x – 3 = 0 ?x=3
Hence, x = 2 or 3
4. Give the similar questions for class work and discuss on the solutions
5. To develop critical thinking in the students, give the values of variable and tell to make
the quadratic equations satisfying the values and discuss on the solutions again.
Vedanta Excel in Mathematics Teachers' Manual - 9 62
6. Recall the formulae a2 ± 2ab + b2 = (a ± b)2 then discuss on the solution of the quadratic
equation by completing square method
7. Discuss about the solution of the equation ax2 + bx + c = 0, a ≠ 0 by completing the
square method
ax2 + bx + c = 0
or, ax2 + bx = – c (Transposing c to R.H.S.)
or, ax2 + b x = – c (To make the coefficient of x2 unity, dividing both sides by a)
a a a
or, x2 + b x = – c
a a
or, x2 + b x + b 2= b 2 – c (Adding the square of half of the
a 2a 2a a coefficient of x to both sides)
or. x2 + 2. b x+ b2 = b2 – c
2a 2a 4a2 a
or, x + b 2= b2 – 4ac
2a 4a2
or, x + b = ± b2 – 4ac
2a 2a
or, x = – b ± b2 – 4ac = –b ± b2 – 4ac
2a 2a 2a
Thus, the required roots of x are – b + b2 – 4ac and – b – b2 – 4ac .
2a 2a
8. Solve some quadratic equation by using the above formula and ask some similar
equations ti be solved by formula.
Solution of selected problems from Vedanta Excel in Mathematics
1. Solve the following equations by factorization method.
a) x 5 x–5 = 221
x–5 x 5
b) x–2 x 2 = 2x 6
x 2 x–2 x–3
c) x– 2 3x – 11 = 4x 13
x– 3 x –4 x 1
Solution: = 221
a) x 5 x–5
x–5 x 5
5
or, (x 5)2 (x – 5)2 = 2
(x – 5) (x 5)
5
or, x2 10x 25 x2 – 10x 25 = 2
x2 – 25
5
or, 2x2 50 = 2
x2 – 25
or, 5x2 – 125 = 4x2 100
63 Vedanta Excel in Mathematics Teachers' Manual - 9
or, x2 = 225
or, x2 – 225 = 0
or, x2 – 152 = 0
or, (x 15) (x – 15) = 0
Either x 15 = 0 ? x = – 15
or, x – 15 = 0 ? x = 15
Hence, x = – 15
b) x–2 x 2 = 2x 6
x 2 x–2 x–3
or, (x – 2)2 (x 2)2 = 2x 6
(x 2) (x – 2) x–3
or, x2 – 4x 4 x2 4x 4 = 2x 6
x2 – 4 x–3
or, 2x2 8 = 2x 6
x2 – 4 x–3
or, (2x2 8) (x – 3) = (2x 6) (x2 – 4)
or, 2x3 – 6x2 8x – 24 = 2x3 – 8x 6x2 – 24
or, – 12x2 16x = 0
or, – 4x(3x – 4) = 0
Either – 4x = 0 ?x=0
4
or, 3x – 4 =0 4 ? x = 3
Hence, x=0 or , 3
c) x–2 3x – 11 = 4x 13
x–3 x–4 x 1
or, (x – 2) (x – 4) (x – 3) (3x – 11) = 4x 13
(x – 3) (x – 4) x 1
or, x2 – 4x – 2x 8 3x2 – 11x – 9x 33 = 4x 13
x2 – 4x – 3x 12 x 1
or, 4x2 – 26x 41 = 4x 13
x2 – 7x 12 x 1
or, 4x3 4x2 – 26x2 – 26x 41x 41 = 4x3 – 28x2 48x 13x2 – 91x 156
or, – 7x2 58x – 115 = 0
or, 7x2 – 58x 115 = 0
or, 7x2 – 35x – 23x 115 = 0
or, 7x(x – 5) – 23(x – 5) = 0
or, (x – 5) (7x – 23) = 0
Either x – 5 = 0 ?x=5
23
or, 7x – 23 = 0 ? x = 7
23
Hence, x = 5 or 7
2. Solve these equation by factorization method.
a) x x–3 = 5
x–3 x 2
b) 2x – 3 – 4 x–1 =3
x–1 2x – 3
Vedanta Excel in Mathematics Teachers' Manual - 9 64
Solution:
a) Let x = a, then x–3 = 1
x–3 x a
1 5
Now, a a = 2
or, a2 1 = 5
a 2
or, 2a2 2 = 5a
or, 2a2 – 5a 2 = 0
or, 2a2 – 4a – a 2 = 0
or, 2a(a – 2) – 1(a – 2) = 0
or, (a – 2) (2a – 1) = 0
Either a – 2 = 0 ? a = 2
or, 2a – 1 = 0 ? a = 1
2
When a = 2, x =2
or, x–3
x
x–3 =4
or, 4x – 12 = x
12
or, x = 3 = 4
When a = 21, x = 1
x–3 2
or, x = 1
x–3 4
or, 4x = x – 3
or, 3x = 3 ? x = 1
Hence, x = 1 or 4
b) Let 2x – 3 = a, then x–1 = 1
x–1 2x – 3 a
1
Now, a 4 u a = 3
or, a2 – 4 = 3
a
or, a2 – 3a 4 = 0
or, a2 – 4a a 4 = 0
or, a(a – 4) 1(a – 4) = 0
or, (a – 4) (a 1) = 0
Either a – 4 = 0 ?a=4
or, a 1 = 0 ?a= 1
2x – 3
When a = 4, x 1 = 4
or, 4x 4= 2x 3 1
or, 2x =1 ? 2
x =
When a = 1, 2x – 3 = 1
x 1
65 Vedanta Excel in Mathematics Teachers' Manual - 9
or, 2x 3 = x 1 4 = 131
or, ? 3
3x = 4 x =
Hence, x 1 131
= 2 ,
3. Solve these equations.
1 1 1
a) a 1 x = a b x b) ax2 bx c = ax b
b px2 qx r px q
c) (a 1 b) (c 1 c) = (x 1 c) (a 1 a)
b) (x a) (x b) (x b) (c
Solution: 1 1 1
1 a b x
a) a b x =
or, 1 1 = 1 1
a b x x a b
or, x a b x = b a
x (a b x) ab
or, (a b) = (a b)
x (a b x) ab
or, ax bx x2 = ab
or, x2 ax bx ab = 0
or, x(x a) (x b) = 0
Either x a = 0 ?x= a
or, x b = 0 ?x= b
Hence, x = a or b.
b) ax2 bx c = ax b
px2 qx r px q
or, apx3 aqx2 bpx2 bqx cpx cq = apx3 aqx2 arx bpx2 bqx br
or, cpx cq = arx br
or, x(cp ar) = br cq
br cq
? x = cp ar
c) (a 1 b) (c 1 c) = (x 1 c) (a 1 a)
b) (x a) (x b) (x b) (c
or, (c a) (x c) (a b) (x c) = (a b) (c a) (x b) (x c)
(a b) (c a) (x b) (x c) (a b) (c a) (x b) (x c)
or, (c a) (x c) (a b) (x c) = (a b) (c a) (x b) (x c)
or, (c a) (x c) (a b) (c a) (a b) (x c) (x b) (x c) = 0
or, (c a) (x c a b) (x c) (a b x b) = 0
or, (x a b c) (a b c x) = 0
Either x a b c = 0 ? x = a b c
or, a b c x = 0 ? x = a b c
Hence, x = a b c , a b c
4. Solve each equation by completing the square.
a) 3x2 5x 2 = 0 b) 15x2 2ax = a2
Solution:
a) 3x2 + 5x – 2 = 0
Vedanta Excel in Mathematics Teachers' Manual - 9 66
or, 3x2 5 x = 2 [Dividing both sides by 3]
3 3 3
or, x2 5 x = 2
3 3
2 52
or, x2 5 x 5 = 2 6
3 6 3
or, x2 2 . x . 5 5 2 = 2 25
6 6 3 36
or, x 5 2 = 24 25
6 36
or, x 5 2 = – 7 2
6 6
Taking ( ) ve sign Taking ( ) ve sign
x 5 = 7 x 5 = 7
6 6 6 6
? x = 2 = 1 ? x = 12 = 5
6 3 6 6
Hence x = 1 or 2.
3
b) 15x2 + 2ax = a2
Dividing each term by 15
15x2 2ax a2
15 15 = 15
or, x2 2ax 2a 2 = a2 2a 2
15 u 15 15 2 u 15
2
or, x2 2 . x . a a 2 = a2 a2
15 15 15 225
or, x a 2 = 16a2
15 225
or, x a 2 = – 4a 2
15 15
Taking ( ) ve sign Taking ( ) ve sign
x a = 4a x a = 4a
15 15 15 15
5a a 3a a
? x = 15 = 3 ? x = 15 = 5
Hence x = a or 5a.
3
5. Express 2x2 4x 5 in the form of a(x h)2 k, where a, h and k are whole numbers.
Hence find the root of x2 hx ak = 0
Solution:
Here, 2x2 4x 5 = 2(x2 2x) 5
= 2(x2 2 . x . 1 12 12) 5
= 2{(x 1)2 1} 5
= 2(x 1)2 2 5
= 2(x 1)2 3
Which is the form a(x h)2 k, where a = 2, h = 1 and k = 3.
67 Vedanta Excel in Mathematics Teachers' Manual - 9
Again,
x2 hx ak = 0
or, x2 1 . x 2 u 3 = 0
or, x2 x 6 = 0
or, x2 3x 2x 6 = 0
or, x(x 3) 2(x 3) = 0
or, (x 3) (x 2) = 0
Either
x 3=0 ?x= 3
or, x 2 = 0 ? x = 2
Hence, the required roots of x2 hx ak = 0 are 3 and 2 when h = 1, a = 2 and k = 3.
6. Solve: px2 qx r = 0
Solution:
Here, px2 qx r = 0
Dividing each term by p we get,
px2 qx r 0
p p p = p
or, x2 q x r = 0
p p
q 2 r q2
or, x2 q x 2p = p 2p
p
or, x q 2 = r a2
2p p 4p2
or, x q 2 = q2 4pr
2p 4p2
or, x q = ± q2 4pr
2p 4p2
or, x = q ± q2 – 4pr
2p 2p
? x = q ± q2 – 4pr
2p
7. Solve the equation 3x2 10 = 11x by using formula.
Solution:
Here, 3x2 10 = 11x
or, 3x2 11x 10 = 0
Comparing it with ax2 bx c = 0. we get a = 3, b = 11 and c = 10
We have b ± b2 – 4ac
x 2a
=
= 11 ± 112 – 4 u 3 u 10 = 11 ± 1 = 11 ± 1
2u3 6 6
Taking ( ) ve sign, Taking ( ) ve sign,
x= 11 1 x= 11 1
6 6
? x = 5 x= 2
3
5
Hence, x = 3 or 2.
Vedanta Excel in Mathematics Teachers' Manual - 9 68
c) (x 1) (x 2) = (x 3) (x 4)
(x 1) (x 2) (x 3) (x 4)
Solution: 1) (x 2) (x 3) (x 4)
(x 1) (x 2) (x 3) (x 4)
Here, (x =
or, (x 1) (x 2) (x 2) (x 1) = (x 3) (x 4) (x 4) (x 3)
(x 1) (x 2) (x 3) (x 4)
or, x2 3x 2 x2 3x 2 = x2 7x 12 x2 7x 12
x2 x 2 x2 x 12
or, 2x2 4 = 2x2 24
x2 x 2 x2 x 12
or, 2(x2 2) = 2(x2 12)
x2 x 2 x2 x 12
or, (x2 2) (x2 x 12) = (x2 12) (x2 x 2)
or, x4 x3 12x2 2x2 2x 24 = x4 x3 2x2 12x2 12x 24
or, 10x2 2x = 10x2 12x
or, 20x2 10x = 0
or, 10x(2x 1) = 0
Either 10x = 0 ?x=0
1
or, 2x 1 = 0 ? x = 2
1
Hence, x = 0 , 2
9. Solve: x2(a b) x(a b) 2b = 0
Solution:
Here, x2(a b) x(a b) 2b = 0
or, ax2 bx2 ax bx 2b = 0
or, ax2 ax bx2 bx 2bx 2b = 0
or, ax(x 1) bx(x 1) 2b(x 1) = 0
or, (x 1) (ax bx 2b) = 0
Either x 1 = 0 ?x=1
or, ax bx 2b = 0 2b
or, ? a b
x(a b) = 2b x =
Hence, 2b
x = 1 , a b
1. Solve: Extra Questions
a) x2 1 = 3 Ans : ±4 b) x = 4 Ans : ± 6
5 9 x
c) 6 5x x2 = 0 Ans : 2 , 3
2. Solve:
a) x 2 x = 112 Ans : 2 , 4 b) x x 1 = 261 Ans : 2 , 3
x 3 x
x 2 x 1
c) x 3 x 3 = 2x 3 Ans : 0 , 4
x 2 x 2 x 1
69 Vedanta Excel in Mathematics Teachers' Manual - 9
Unit Ratio and Proportion
10 Allocated teaching periods 6
Competency
- To solve the general problems on ratio and proportion, algebraic forms and behavioural
problems
Learning Outcomes
- To solve the problems related to ratio and proportion
Level-wise learning objectives OBJECTIVES
S.N. LEVELS To define ratio
- To identify antecedent and consequent of the ratio
1. Knowledge (K)
-
- To recall the types of ratios: compound, duplicate, sub-
duplicate, triplicate, sub-triplicate, inverse ratio
- To define proportion
- To recall the properties of proportion: invertendo,
alternaendo, acomponendo, dividendo and addendo
2. Understanding (U) - To find the simple ratio
- To solve the algebraic form of ratio and proportion
3. Application (A) - To prove the conditional relations on proportion
- To identify the contextual problems based on ratio and
proportion and solve them
4. High Ability (HA)
- To solve the higher level conditional problems on
proportion
Required Teaching Materials/ Resources
Chart paper with types of ratios and properties of proportions separately with examples
Pre-knowledge: ratio and proportion
Teaching Activities
1. Ask the weight of two students and ask to find their ratio, similarly collect the
prices of copies, pen or bags, umbrellas etc bought in the classroom by the
students and discuss on ratios.
2. Under discussion, list the following notes
(i) The comparison of quantities of same kind by division. The ratio represents
how many times a quantity is grater or smaller than another quantity of same
kind
a
(ii) If a and b are two quantities of same kind, the ratio of a and b is a:b or b
(iii) In the ratio a: b, a is called antecedent and b called consequent.
Vedanta Excel in Mathematics Teachers' Manual - 9 70
3. Discuss on the types of ratio with examples as follows
(i) Compound ratio:
A new ratio obtained by multiplying two or more ratios is called the compound
ratio of the given ratios.
ac
For two ratios a: b and c: d, the compound ratio = (a: b) × (c: d) = bd
Example: The compound ratio of 2:3 and 9:10 is
(ii) Duplicate and sub-duplicate ratio:
A new ratio obtained by multiplying a ratio by itself is called the duplicate ratio
of the given ratio.
Suppose a: b is a ratio then the duplicate ratio a: b× a: b =a2: b2
Example: The duplicate ratio of 4:5= 42: 52 = 16:25
A new ratio obtained by taking square root of a ratio is called the sub-duplicate
ratio of the given ratio.
Suppose a: b is a ratio then the sub-duplicate ratio is a : b
Example: The sub-duplicate ratio of 9: 49= 9 : 49 = 3:7
(iii) Triplicate and sub-triplicate ratio:
A new ratio obtained by multiplying a ratio three times by itself is called the
triplicate ratio of the given ratio.
Suppose a: b is a ratio then the triplicate ratio is a3: b3
Example: The triplicate ratio of 2:3= 23: 33 = 8:27
A new ratio obtained by taking cube root of a ratio is called the sub-triplicate
ratio of the given ratio.
Suppose a: b is a ratio then the sub-duplicate ratio 3 a : 3 b
Example: The sub-triplicate ratio of 125 : 64 = 3125 : 364 = 5 : 4
(iv) Inverse ratio:
A new ratio obtained by interchanging the antecedent and consequent is the
inverse ratio of the given ratio.
Suppose a: b is a ratio then the inverse ratio is b: a
Example: The inverse ratio of 4:7 is 7: 4
4. With proper guidelines, encourage the students to solve the problems related to
ratio given in the exercise
5. With proper examples, discuss upon the proportion, proportional and the
relation between the means and extremes
6. Discuss upon the properties of proportions
If a, b, c and d are in proportion then there are the following properties
(i) Invertendo
a c b d
If b = d , then a = c
Proof a c a c b d b d
b d b d a c a c
Here, = , then 1 ÷ = 1 ÷ or, 1 × = 1 × or, = Proved
Example: For 2:3 = 4:6, 3:2 = 2×3:2×2 = 6:4
(ii) Alternendo
a c a b
If b = d , then c = d
Proof a dc , bc ,
b
Here, = then multiplying both sides by we get,
a × b = c × b or, a = b Proved
b c d c c d
Example: For 2:5 = 4:10, 2:4 = 1:2 = 5×1:5×2 = 5:10
71 Vedanta Excel in Mathematics Teachers' Manual - 9
(iii) Componendo c
a c a + b + d
If b = d , then b = d
Proof a c
b d
Here, = , then adding 1 to both sides,
a +1 = c +1 or, a + b = c + d Proved
b d b d
Example: For 3:5 = 6:10, (3+5):5 = 8:5 and (6+10):10 = 16:10 = 8:5
Thus, 3:5 = 6:10 implies (3+5):5 = (6+10):10
(iv) Dividendo
a c a–b c–d
If b = d , then b = d
Proof a c
b d
Here, = , then subtracting 1 from both sides,
a –1 = c –1 or, a–b = c–d Proved
b d b d
Example: For 7:4= 21:12, (7 – 4):4 = 3:4 and (21 – 12):12 = 9:12 = 3:4
Thus, 7:4= 21:12 implies (7 – 4):4 = (21 – 12):12
(v) Componendo and Dividendo
If a = c , then a+b = c+d
b d a–b c–d
Proof
a c
Here, b = d , then by componendo, we have, Also, by dividendo , we have
a+b = c + d ........ (i) or, a–b = c–d ........ (ii)
b d b d
Dividing equation (i) by (ii), we get
a+b = c–d or, a+b = c+d proved
b d a–b c–d
a–b
b
Example: For 3:2= 9:6, (3 + 2): (3 – 2) = 5:1 and (9 + 6): (9 – 6) = 15:3 = 5:1
Thus, 3:2= 9:6 implies (3 + 2): (3 – 2) = (9 + 6): (9 – 6)
(vi) Addendo If a = c , then a = c = a + c
Proof b d b d b + d
Here, a = c , then by alternendo, we have, a = b
b d c d
and by componendo, we have = b+d
d
Again, by alternaendo, we have, a+c = c ? a = c = a+c proved.
b+d d b d b+d
Example: For 5:7= 10:14, (5 + 10): (7 + 14) = 15:21 = 5:7
Vedanta Excel in Mathematics Teachers' Manual - 9 72
Thus, 5:7= 10:14 = (5 + 10): (7 + 14)
7. Prove the conditional identities on proportion with discussion and give some
similar problems to prove in the class.
Solution of selected problems from Vedanta Excel in Mathematics
1. If a2 2ab b2 : a2 2ab b2 = 1 : , find a : b.
Solution: 1
a2 2ab b2 4
a2 2ab b2 =
or, (a b)2 = 1
(a b)2 4
or, a b 2 ±12 2
a b
=
Taking ( ) ve sign Taking ( ) ve sign
a b = 1 a b = 1
a b 2 a b 2
or, 2a 2b = a b or, 2a 2b = a b
or, a = 3b or, 3a = b
a 3 a 1
or, b = 1 or, b = 3
?a:b=3:1 ?a:b=1:3
2. If a : b = 3 : 2, find the value of
2a a2 ab
i) b 3 ii) ab b2
Solution:
a:b=3:2
Let a = 3x and b = 2x
i) b 2a = 2x u 2 u 3x =0
3 3
ii) a2 ab = (3x)2 3x u 2x
ab b2 3x . 2x (2x)2
= 9x2 6x2
6x2 4x2
= 3x2
10x2
3
= 10 = 3 : 10
3. If (x 4) : (3x 1) is the duplicate ratio of 3 : 4, find the value of x.
Solution:
Duplicate of 3 : 4 = 32 : 42 = 9 : 16
By question, 9
x 4 16
3x 1 =
or, 16x 64 = 27x 9 ?x=5
4. If (3a 7) : (4a 3) is the sub - triplicate ratio of 8 : 27, find the value of a.
Solution:
73 Vedanta Excel in Mathematics Teachers' Manual - 9
Here, sub - triplicate ratio of 8 : 27 = 2 : 3.
According to question,
Sub - triplicate ratio of 8 : 27 = (3a 7) : (4a 3)
or, 2 = 3a 7
3 4a 3
or, 9a 21 = 8a 6 ? a = 27
5. The ratio of two number is 2 : 3 and their lcm is 30, find the number.
Solution:
Let the required number be 2x and 3x.
According to the question;
LCM of 2x and 3x = 30
or, x u 2 u 3 = 30 ? x = 5
Hence, the first number = 2x = 2 u 5 = 10
and the second number = 3x = 3 u 5 = 15
6. In 50 l of milk, the ratio of pure milk and water is 2 : 3. How much pure milk should
be added to the mixture so that the pure milk and water will be 5 : 6 ratio ?
Solution:
Let the quantity of pure milk in 50 l of milk be 2x l and the quantity of water be 3x l.
Now, 2x 3x = 50 l ? x = 10 l
? Quantity of pure milk = 2x = 2 u 10 = 20 l
and Quantity of water = 3x = 3 u 10 = 30 l
Again,
Let the quantity of pure milk to be added in the mixture be y l.
Then, according to question:
20 y 5
30 = 6
or, y = 5
Hence, the quantity of milk to be added is 5 l.
7. If (a b), b and (a b) are in continuous proportion, show that a2 = 2.
Solution: b2
Here, (a b), b and (a b) are in continuous.
so, b2 = (a b) (a b)
or, b2 = a2 b2
or, 2b2 = a2
a2 a2
or, 2 = b2 ? b2 = 2 proved
8. What number should be added to each term 7, 10, 16 and 22. So that they will be in
proportion ?
Solution:
Let the number to be added be x.
7 x 16 x
Then, 10 x = 22 x
or, 154 7x 22x x2 = 160 10x 16x x2
or, 154 29x = 26x 160
or, 3x = 6 ? x = 2
Hence, the required number to be added is 2.
Vedanta Excel in Mathematics Teachers' Manual - 9 74
9. If a = ab, prove that:
b
a) a2 ab b2 = a
b2 bc c2 c
b) a3 b3 = a(a b)
b3 c3 c(b c)
c) a b c = (a b c)2
a b c a2 b2 c2
LSeotlutabio=n:bc = k then b = k i.e, b = ck
c
a
and b = k i.e, a = bk = ck.k = ck2
Now = a2 ab b2 = (ck2)2 ck2.ck (ck)2 = c2k4 c2k3 c2k2 = c2k2(k2 k 1) = k2
a) L.H.S b2 bc c2 (ck)2 ck.c c2 c2k2 c2k c2 c2(k2 k 1)
R.H.S = a = ck2 = k2
c c
Hence, L.H.S = R.H.S Proved
b) L.H.S = a3 b3 = (ck2)3 (ck)3 = c3k6 c3k3 = c3k3(k3 1) = k3
b3 c3 (ck)3 c3 c3k3 c3 c3(k3 1)
R.H.S = a(a b) = ck2(ck2 ck) = k2.ck(k 1) = k3
c(b c) c(ck c) c(k c)
Hence, L.H.S = R.H.S Proved
c) L.H.S = a b c = ck2 ck c = c(k2 k 1) = (k2 k 1)
a b c ck2 ck c c(k2 k 1) (k2 k 1)
R.H.S = (a b c)2
a2 b2 c2
= (ck2 ck c)2
(ck2)2 (ck)2 c2
= c2(k2 k 1)2
c2k4 c2k2 c2
= c2(k2 k 1)2
c2(k4 k2 1)
= (k2 k 1)2 = (k2 k 1)2 = (k2 (k2 k 1)2 1) = (k2 k 1)
(k2 1)2 2k2 k2 (k2 1)2 k2 k 1) (k2 k (k2 k 1)
10. If x, y, z are in continuous proportion. Prove that x2y2z2 1 1 1 = x3 y3 z3.
Solution: x3 y3 z3
Let x = y = k then y = k i.e y = zk, x =k i.e, x = yk = zk.k = zk2
y z 1z y
1 y3 1
Now, L.H.S = x2y2z2 x3 z3
= (3k2)2 (3k)2z2 1 1 1
(zk2)3 (zk)3 z3
= z2k4.z2k2.z2 1 1 1 = z6k6 1 k3 k6 = z3(1 k3 k6)
z3k6 z3k3 z3 z3k6
R.H.S = x3 y3 z3 = (3k2)3 (3k)3 z3 = z3k6 z3k3 z3 = z3(k6 k3 1) = z3(1 k3 k6)
Hence, L.H.S = R.H.S Proved
75 Vedanta Excel in Mathematics Teachers' Manual - 9
S11o.lutIifoabn:= b = c , prove that: a2 ab = b2 bc = c2 cd
c d b2 c2 d2
Let a = b = c = k then c = k i.e c = dk
b c d d
b
c = k i.e, b = ck = dk.k = dk2
and a = k i.e, a = bk = dk2.k = dk3
Now, b = (dk3)2 dk3.dk2
L.H.S = a2 ab
b2 (dk2)2
d2k6 d2k5
= d2k4 = d2k5(k 1) = k(k 1)
d2k4
(dk2)2 dk2.dk d2k4 d2k3
Middle Term (M.T.) = b2 bc = (dk)2 = d2k2 = d2k3(k 1) = k(k 1)
c2 d2k2
R.H.S = c2 cd = (dk)2 dk.d = d2k(k 1) = k(k 1)
d2 d2 d2
Hence, L.H.S. = M.T. = R.H.S. Proved
12. If a : b : : b : c : : c : d, prove that : (a c)2 (b c)2 (b d)2 = (a d)2
Solution: a = b = c
Here, b c d
? ac = b2, bd = c2 and ad = bc
LHS = (a c)2 (b c)2 (b d)2
= a2 2ac c2 b2 2bc c2 b2 2bd d2
= a2 2b2 c2 2b2 2ad c2 2c2 d2 = a2 2ad d2 = (a d)2
ax by by cz cz ax
13. If x : a : : y : b : : z : c, prove that: (a b)(x y) (b c)(y z) (c a)(z x) =3
SLeotl,utaxio=n: y = z = k ? x = ak, y = bk and z = ck
b c
ax by by cz cz ax
LHS = (a b)(x y) (b c)(y z) (c a)(z x)
= a.ak b.bk b.bk c.ck c.ck a.ak
(a b)(ak bk) (b c)(bk ck) (c a)(ck ak)
= k(a2 b2) k(b2 c2) k(c2 a2)
k(a b)(a b) k(b c)(b c) k(c a)(c a)
= 1 1 1 = 3 = R.H.S Proved that: x = y
14. If (a2 b2) (x2 y2) = (ax by)2, show a b
Solution:
Here, (a2 b2) (x2 y2) = (ax + by)2
or, a2x2 a2y2 b2x2 b2y2 = a2x2 2abxy b2y2
or, a2y2 2abxy b2x2 = 0
or, (ay bx)2 = 0
or, ay bx = 0
or, aHbyye=n=caxeb,xax
or,
= y Proved
b
15. If ay bx = cx az = bz cy , Prove that x = y = z
p q r a b c
Vedanta Excel in Mathematics Teachers' Manual - 9 76
Solutiaoyn : bx = cx az = bz cy
p q r
or, c(ay bx) = b(cx az) = a(bz cy)
cp bq ar
We know, = Sum of antecedents
Each ratio Sum of consequents
or, c(ay bx) = b(cx az) = a(bz cy) = c(ay bx) b(cx az) a(bz cy)
cp bq ap cp bq ap
or, c(ay bx) = b(cx az) = a(bz cy) =0
cp bq ap
or, c(ay bx) = 0, b(cx az) = 0 and a(bz cy) =0
cp bq ap
or, ay bx = 0, cx az = 0 and bz cy = 0
or, ay = bx, cx = az and bz cy
or, y = x , x = z and z = y
b a a c c b
or, x = y = z
a b c
Hence proved
Extra Question
1. If x = y = z , prove that: x3 y3 z3 = xyz
a b c a3 b3 c3 abc
2. If a = b , prove that: a b = a2(b c)
b c b c b2(a b)
3. If a, b, c are in continuous proportion, prove that (ab bc ca)3 = abc(a b c)3.
4. If a : b : : b : c : : c : d , prove that a3 b3 c3 = a3 b3 abc
b3 c3 d3 b3 c3 bcd
5. If a, b, c, d and e are in continuous proportion, prove that a : e = a4 : b4
77 Vedanta Excel in Mathematics Teachers' Manual - 9
Unit Geometry - Triangle
11
Allocated teaching periods 12
Competency
To verify experimentally and prove theoretically the properties of triangles
Learning Outcomes
- To differentiate between the theoretical proof and experimental verification on properties
of triangles
- To prove the theorems on triangles and verify the statements by inductive method
(theorems which cannot be proved theoretically)
- To prove the theorems based on mid-point theorem
Level-wise learning objectives
S.N. LEVELS OBJECTIVES
- To identify centroid and orthocentre of triangle
- To recall the sum of interior angles of triangle
- To tell the relation of exterior angle and the opposite
exterior angles of triangle
1. Knowledge (K) - To relate the sum of any two sides and the third side of
the triangle
- To sate the mid-point theorem
2. Understanding (U) - To find the unknown angles using the properties of
triangle
- To find the unknown length of sides of angles of triangles
by using mid-point theorem
3. Application (A) - To verify the properties of triangle experimentally
- To prove the theorems based on properties of triangle
4. High Ability (HA) - To prove the unseen theorems based on properties of
triangle
Required Teaching Materials/ Resources
Geo board, rubber-bands, mechano strip, sticks, colourful chart paper, models of triangles,
rectangular sheet of paper, ICT tools etc.
Pre-knowledge: Types of triangles, basic properties of triangles
Teaching Activities
1. For warm-up, discuss upon the types of triangles based on sides or angles with models of
triangles on chart papers or geo-board or mechano strips or sticks
2. Show the centroid and orthocentre of the triangles by paper folding.
3. Discuss the following properties of triangles experimentally and theoretically
- The sum of angles of triangle is equal to two right angles
- The exterior angle of a triangle is equal to the sum of the two opposite interior angles
- The sum of any two sides of a triangle is greater than the third side
- In any triangle, the angle opposite to the longer side is greater than the angle opposite to
the shorter side
- In any triangle, the side opposite to the greater angle is longer than the side opposite to
the smaller angle
Vedanta Excel in Mathematics Teachers' Manual - 9 78
- Of all straight line segments drawn to the given line from a point outside it, the
perpendicular is the shorted one.
- Base angles of isosceles triangle are equal
- if two angles of triangle are equal, it is an isosceles triangle
- The bisector of vertical angle of an isosceles triangle is perpendicular bisector of the base.
- The straight line joining the vertex and the mid-0ppoint of the base of an isosceles
triangle is perpendicular to the base and bisects the vertical angle.
- The straight line joining the mid-points of any two sides of a triangle is parallel to the
third side and it is equal to half of the length of the third side.
4. The straight line segment drawn through the mid-point of one side of a triangle and
parallel to the another side bisects the third side
5. Divide the students into groups and give the theorems to prove them theoretically.
6. Focus on project work
Solution of selected problems from Vedanta Excel in Mathematics
A. Angles of Triangle.
1. In 'ABC, 2A = 3B = 6C, find A, B and C.
Solution: Here, 2A = 3B = 6C
? A = 3C and B = 2C
Now, A B C = 180° [? Being the sum of angle of ']
or, 3C 2C C = 180°
or, 6C = 180
? C = 30°
Also, A = 3C = 3 u 30° = 90° and B = 2C = 2 u 30° = 60°
Hence, A = 90° , B = 60° and C = 30
2. In 'PQR, P Q = 20° and Q R = 50°, find P, Q and R.
Solution: Here, In 'PQR, P Q = 20° ........... (i)
Q R = 50° ........... (ii)
But, P Q R = 180° ........... (iii)
Now, adding equations (i), (ii) and (iii) we get
(P Q) (Q R) (P Q R) = 20° 50° 180°
or, 2P Q = 250 ........... (iv)
Again adding equation (i) and (iv), we get
(P Q) (2P Q) = 20° 250°
or, 3P = 270° ? P = 90°
Substituting the value of P in equation (i), we get
90° Q = 20° ? Q = 70°
Substituting the value of Q in equation (ii), we get
70° R = 50° ? R = 20° A
xx
Hence, P = 90°, Q = 70° and R = 20°.
65° y
3. Find the unknown sizes of angles in the given figure. D
Solution:
i) In 'ABC, A B C = 180° B 35° C
or, (x x) 65° 35° = 180°
or, 2x = 80° or, ? x = 40°
ii) In 'ABD; BAD ABD ADB = 180°
or, x 65° y = 180°
or, 40° 65° y = 180 or, ? y = 75°
Hence, x = 40° and y = 75°
79 Vedanta Excel in Mathematics Teachers' Manual - 9
4. In the given figure, PQ = PR and QS is the bisector of PQR. If P
S
PQS = 30°, find PSQ. R
Solution:
i) PQS = SQR = 30° [As QS the bisector of PQR]
ii) PQR = PRQ = 60° [? PQ = PR]
iii) In 'PQR; PQR PRS QPR = 180° Q
or, 60° 60° QPR = 180°
? QPR = 60°
iv) In 'PQS; PQS PSQ QPS = 180°
or, 30 PSQ 60° = 180°
? PSQ = 90°
5. In the adjoining figure, the bisectors of ABC and ACB of
A
'ABC meet at O. Prove that BOC = 90° 2 .
Solution:
Here,
Given: In 'ABC, the bisectors of ABC and ACB meet at O.
A
To prove: BOC = 90° 2
Proof:
Statements Reasons
1. Given
1. ABC = 2OBC and ACB = 2OCB 2.
2. In 'ABC; ABC ACB BAC = 180° 3.
From (ii)
or, 2OBC 2OCB BAC = 180°
or, 2(OBC OCB) = 180° BAC
180° A A
? OBC OCB = 2 = 90° 2
3. In 'BOC; BOC OBC OCB = 180°
A
or, BOC 90° 2 = 180°
? BOC =
180° 90° A
= A 2
90° 2
Proved
6. In the figure alongside, BP and CP are the
angular bisectors of the exterior angles
A
CBP and BCP of 'ABC. Prove that BPC = 90° 2 .
Solution:
Here,
Given: In 'ABC; BP and CP are the angular bisector of CBE adn BCD respectively.
? CBE = 2CBP and BCD = 2BCP.
A
To prove: BPC = 90° 2
Proof:
Statements Reasons
1. CBE = 2CBP and BCD = 2BCP 1. Given
2. ABC = 180° CBE and ACB = 180° BCD 2. Supplementary angles
Vedanta Excel in Mathematics Teachers' Manual - 9 80
3. In 'ABC, ABC ACB BAC = 180° 3. Sum of angles of triangle
or, (180° CBE) (180° BCD) BAC = 180° and from (2).
or, CBE BCD BAC 180° = 0
? BAC 180° = CBE BCD
4. A 180° = 2CBP 2BCP 4. From (1) and (3)
A
?CBP + BCP = 90° + 2
5. In 'BCP, CBP BCP BPC = 180° 5. Sum of angles of triangle
=29A0° B2APC and form (4).
or, 90° = 180°
? BPC
Proved
7. In 'PQR, RX A PQ and QY A PR, RX and QY intersect at O. P
Prove that QOR = 180° P. XO Y
Solution: Here,
Given: In 'PQR, RX A PQ and QY A PR, RX and QY intersect at O. Q R
To prove: QOR = 180° P.
Proof
Statements Reasons
1. In 'PQR, PQR QRP QPR = 180° 1. Sum of angles of trianmgle is 180°
2. In 'QXR, XQR QRX QXR = 180° 2. Given
or, XQR QRX 90° = 180°
? XQR = 90° QRX
3. In 'QRY, QRY RYQ YQR = 180° 3.
or, QRY 90° YQR= 180°
? QRY = 90° YQR
4. (90° QRX) (90° YQR) P = 180° 4. From (1), (2) and (3)
or, P = QRX YQR
5. In 'QOR, QOR QRO OQR = 180° 5.
or, QOR P = 180° From (4)
? QOR = 180° P.
Proved
8. In the adjoining figure, ABC is an isosceles triangle. BO and CO are the bisector of
ABC and ACB respectively. Prove that BOC is an isosceles
triangle.
Solution:
Given: In 'ABC, AB = AC, BO and CO are the bisectors of ABC
and ACB respectively.
To prove: BOC is also the isosceles triangle.
Proof
Statements Reasons
1. ABC = ACB 1. AB = AC
2. ABC = 2OBC and ACB = 2OCB 2. Given
3. 2OBC = 2OCB ?OBC = OCB 3. From (1) and (2)
4. BOC is an isosceles triangle. 4. From (3), being base angles equal.
Proved
81 Vedanta Excel in Mathematics Teachers' Manual - 9
9. In the given figure, AM = BM = CM. Prove that 'ABC is a right A M
B C
angled triangle.
Solution:
Given: AM = BM = CM
To prove: 'ABC is a right angle triangle
Proof
Statements Reasons
1. MAB = MBA 1. AM = BM
2. MBC = MCB 2. BM = CM
3. In 'ABC, ABC BCA BAC = 180° 3. Sum of angles of triangle is 180°
4. (MBA MBC) MCB MAB = 180° 4. From (1), (2) and (3)
or, MBA MBC MBC MBA = 180°
or, 2(MBA MBC) = 180°
or, ABC = 90° ? MBA MBC = ABC
Hence, 'ABC is a right angle triangle.
D
10. In the given figure ABCD, A CX
B
Prove that: BCD = BAD ABC ADC.
Solution:
Given: ABCD is an arrowhead.
To prove: BCD = BAD ABC ADC
Construction: AC is produced to X.
Proof
Statements Reasons
1. In 'ABC; BCX = CAB ABC 1. Being ext. angle of ' equal to the
sum opposite interior angles
2. In 'ACD; DCX = ADC CAD 2. Same as (1)
3. BCX DCX = CAB ABC ADC 3. Adding (1) and (2)
CAD
4. BCD = BAD ABC ADC 4. By whole part axiom
Proved
11. In the adjoining star shaped figure, A E
prove that A B C D E = 180 T
Solution: PS
Given: ABCDE is a star shaped figure.
To prove: A B C D E = 180 B QR D
Proof:
C
Statements
1. In 'ADQ, A D = DQC Reasons
1. Being ext. angle of ' equal to the
2. In 'BRE, B E = BRC
3. In 'QCR, C DQC BRC = 180 sum of opposite interior angles
2. Same as (1)
3. Sum of angles of a triangle is 180°
Vedanta Excel in Mathematics Teachers' Manual - 9 82
4. C A D B E = 180 4. From (1), (2) and (3)
? A B C D E = 180
12. In the figure alongside, BE and CE are the angular bisector of Proved
AE
ABC and ACD respectively. Prove that: BAC = 2BEC.
Solution: B CD
Given: BE and CE are the angular bisectors of ABC and ACD respectively.
To prove: BAC = 2BEC
Proof
Statements Reasons
1. ABC = 2EBC and ACD = 2ECD
2. In 'ABC, BAC ABC = ACD 1. Given
3. BAC 2EBC = 2ECD 2. Being ext. angle of ' equal to sum
or, ECD = 21(BAC 2EBD) of opposite interior angles
3. From (1) and (2)
4. In 'BCE; EBC BEC = ECD 4. Same as (2)
5. EBC BEC = 12(BAC 2EBC) 5. From (3) and (4)
or, 2EBC 2BEC = BAC 2EBC
? BAC = 2BEC
13. In the given figure, the bisector of ACU meets AU at O. Proved
1
Prove that: COT = 2 (CAT CUT) T
Solution: U
O
Given: In 'CAU, the bisector CO of ACU meets AT at O. C
A
To prove: COT = 1 (CAT CUT)
2
Proof
Statements Reasons
1. Being ext. angle ' equal to the
1. In 'CAO; COT = OCA CAT
sum of opposite interior angles
? OCA = COT CAT 2. Same as (1)
3. Given
2. In 'CAU, CUT = UCA CAT 4. From (2) and (3)
3. UCA = 2OCA 5. From (1) and (4)
4. CUT = 2OCA CAT
? OCA = 12(CUT CAT)
5. COT CAT = 1 (CUT CAT)
2
or, 2COT 2CAT = CUT CAT
or, 2COT = CAT CUT
1
? COT = 2 (CAT CUT)
Proved
83 Vedanta Excel in Mathematics Teachers' Manual - 9
14. In the adjoining 'ABC, AY is the bisector of BAC and A
1 XY
AX A BC. Prove that: XAY = 2 (B C)
Solution:
Given: In 'ABC, AY is the bisector of BAC and AX A BC. B C
To prove: XAY = 1 (B C)
2
Proof:
Statements Reasons
1. BAC = 2BAY 1. Given
2. In 'ABC, BAC ABC ACB = 180° 2. Sum of angles of ' is 180°
3. ?2BBAAYY =B12 C = 180° 3. From (1) and (2)
(180° B
C)
4. In 'BAX, BAX ABX AXB = 180° 4. Sum of angles of ' is 180°
or, BAX B 90° = 180°
? BAX = 180° B 90° = 90° B
5. orB, AYXA Y=BA21X(1=8012°( 180B° BC 1C8)0 ° (90° B) 5. Subtracting (4) from (3)
2B)
? XAY = 1 (B C)
2
1 Proved
2
15. In 'PQR, O is the interior point. Prove that OP OQ OR > (PQ QR PR)
Solution:
P
Given: In 'PQR, O is the interior point.
1 O
To prove: OP OQ OR > 2 (PQ QR PR) Q R
Proof:
Statements Reasons
1. In 'POQ, OP OQ >PQ 1. The sum of any two sides of
a triangle is greater than the
third side
2. In 'QOR, OQ OR >QR 2. Same as (1)
3. In 'POR, OP OR > PR 3. Same as above
4. OP OQ OQ OR OP OR > PQ QR PR 4. Adding (1), (2) and (3)
or, 2OP 2OQ 2OR > PQ QR PR
?or,O2P(O PO QO QO RO>R)21>(PPQQ QQRR PPRR)
Proved
16. Find the unknown sizes of angles in the following figures.
E
a) D b) X
xy 40° xV
AB C Y 330°0° U y
Z
Vedanta Excel in Mathematics Teachers' Manual - 9 84
Solution:
a)
i) In 'ABD, DAB = ADB = x [ AB = BD]
ii) DBC = DAB ADB = x x = 2x [Being ext. angle of ' equal to sum of opposite
interior angles]
iii. In 'BCD, DBC = BCD = 2x [ BD = CD]
iv. In 'ACD, ACD CAD = ADE
or, x 2x = 90° ? x = 30°
v) In 'BCD, BCD CBD BDC = 180°
or, 2x 2x y = 180°
or, 4 u 30 y = 180°
? y = 60°
Hence, x = 30° , y = 60°
b)
i) In 'XYU, XUV = UXY UYX [Being ext. angle of ' equal to sum of
= 40° 30° opposite interior angles]
= 70°
ii) In 'XUY; XVU = XUV [ XU = XV]
? x = 70°
iii) In 'VYZ, VYZ YZV = YVX [Being ext. angle of ' equal to the sum of
or, 30° y = x opposite angles.
or, 30° y = 70° ? y = 40°
Hence, x = 70° and y = 40° A
17. In the given figure, AB = AC, BAC = 44° and ACD = 24°, show D 44°
that BC = CD. 24°
Solution: B C
i) In 'ABC; BAC ABC ACB = 180°
or, 44° ABC ABC = 180° [ ABC = ACB]
? ABC = 68°
ii) In 'ACD, CDB = DAC ACD = 44° 24° = 68°
Since, ABC = CDB = 68°.
Hence, BC = CD P(x+8)cm(3y+1)cm
Q(x+4)cm T (y+3)cmR
18. In the adjoining figure, find the value of x and y
Solution:
i) PQ = PR or, x 8 = 3y 1 ? x = 3y 7 ......... (i)
ii) QT = TR [ PQ = PR and PT A QR]
or, x 4 = y 3
or, x = y 1 ......... (ii)
Substituting the value of x in equn (ii), we get
3y 7 = y 1 or, 2y = 6 ? y = 3
Substituting the value of y in equn (i), we get
x=3u3 7 =2
Hence, x = 2 and y = 3
85 Vedanta Excel in Mathematics Teachers' Manual - 9
19. In the given figure, AB = AC, BD = EC and DAE = 30°. Prove 30°
that 'ADE is an isosceles triangle. Also calculate the size of
ADE.
Solution:
Given: AB = AC, BD = EC and DAE = 30°
To prove: 'ADE is an isosceles triangle
To find: ADE
Proof:
Statements Reasons
1. In 'ABD and 'AEC 1.
i) AB = AC (S) i) Given
ii) ABC = ACB (A) ii) Base angles of an isosceles triangle are equal
iii) BD = EC (S) iii) Given
2. 'ABD = 'AEC 2. By S.A.S. axiom
3. AD = AE 3. Corresponding sides of congruent triangle
are equal
4. ADE = AED 4. Base angles of isosceles triangle ADE
5. In 'ADE; ADE AED DAE 5. Sum of angles of triangle is 180°.
= 180°
or, ADE ADE 30° = 180°
? ADE = 75°
20. In the figure alongside, BO and CD are bisectors of ABC and Proved
ACB respectively. If BD = CD, prove that 'ABC is an isosceles
triangle. A
Solution: D
B C
Given: i) BD and CD are the bisectors of ABC and ACB respectively.
? ABC = 2DBC and ACB = 2DCB.
ii) BD = CD
To prove: 'ABC is an isosceles triangle
Proof:
Statements Reasons
1. DBC = DCB 1. BD = CD
2. ABC = 2DBC and ACB = 2DCB 2. Given
3. ABC = ACB 3. From (1) and (2)
4. 'ABC is an isosceles triangle 4. From (3)
21. In the given figure PQRS, SP = RQ and RP = SQ. Prove that R Proved
S
RT = ST. Q
Solution: T
Given: In the figure PQRS; SP = RQ and RP = SQ
To prove: RT = ST P
Vedanta Excel in Mathematics Teachers' Manual - 9 86
Proof:
Statements Reasons
1. In 'PRS and 'QRS
i. SP = RQ (S) 1.
ii. RP = SQ (S) i. Given
iii. RS = RS (S) ii. Given
2. 'PRS # 'QRS iii. Side common to both 'S
3. PRS = QSR 2. By S.S.S. axiom
3. Corresponding angles of congruent triangles
4. 'RTS is an isosceles triangle
5. RT = ST are equal
4. From (3), base angles are equal
5. From (4)
Proved
22. In the given figure, X is the mid - point of QR, XA A PQ, XB A PR
and XA = XB. Prove that 'PQR is an isosceles triangle.
Solution:
Given: X is the mid - point of QR i.e, QX = RX, XA A PQ, XB A PR
and XA = XB.
To prove: 'PQR is an isosceles triangle
Proof:
Statements Reasons
1. In ' AQX and 'BRX 1.
i. XAQ = XBR (R) i. XA A PQ and XB A PR
ii. QX = RX (H) ii. Given
iii. AX = BX (S) iii. Given
2. 'AQX # 'BRX 2. By R.H.S. axiom
3. AQX = BRX 3. Corresponding angles of congruent triangles
i.e, PQR = PRQ 4. From (3), base angles are equal
4. 'PQR is an isosceles triangle
Proved
23. In the given figure BN A AC, CM A AB and BN = CM. Prove that 'ABC is an iAsosceles
triangle.
Solution: MN
Given: BN A AC, CM A AB and BN = CM
To prove: 'ABC is an isosceles triangle.
Proof: BC
Statements Reasons
1. In 'MBC and 'NBC 1.
i. BMC = BNC (R) i. Both are right angles
ii. BC = BC (H) ii. Common side
iii. CM = BN (S) iii. Given
2. 'MBC # 'NBC 2. By R.H.S axiom
3. MBC = NCB 3. Corresponding angles of congruent triangle
i.e, ABC = ACB are equal
4. 'ABC is an isosceles triangle
4. From (3), being base angles equal
Proved
87 Vedanta Excel in Mathematics Teachers' Manual - 9
24. In the given triangle ABC, AB = AC, BP A AC and CQ A AB.
Prove that (i) BP = CQ (ii) OP = OQ
Solution:
Given: In 'ABC; AB = AC, BP A AC and CQ A AB.
To prove: (i) BP = CQ (ii) OP = OQ
Proof
Statements Reasons
1.
1. In 'QBC and 'PBC i. Both are right angles
ii. Base angles of isosceles 'ABC
i. BQC = BPC (A) iii. Common Side
2. By A.A.S axiom
ii. QBC = PCB (A) 3. Corresponding sides of congruent triangles
4. Corresponding angles of congruent triangles
iii. BC = BC (S) 5. From (4), base angles of 'OBC are equal
6. Subtracting (5) from (3)
2. 'QBC # 'PBC
3. CQ = BP i.e, BP = CQ
4. BCQ = PBC
5. OB = OC
6. BP OB = CQ OC
? OP = OQ
Proved
25. In the figure alongside, APB and AQC are equilateral
triangles. Prove that: PC = BQ.
Solution:
Given: APB and AQC are equilateral triangles
To prove: PC = BQ
Proof
Statements Reasons
1. PAB = QAC = 60° 1. Angles of equilateral triangles are equal
2. PAB BAC = QAC BAC 2. Adding BAC on both sides of (1)
3. PAC = QAB 3. By whole part axiom
4. In 'PAC and 'QAB 4.
i. AP = AB (S) i. Sides of equilateral triangle APB
ii. PAC = QAB (A) ii. From (3)
iii. AC = AQ (S) iii. Sides of equilateral triangle AQC
5. 'PAC # 'QAB 5. By S.A.S. axiom
6. PC = BQ 6. Corresponding sides of congruent triangles
26. In the figure alongside PABQ and CYX are squares. Proved
Prove that: PC = BX
X
Solution:
PA Y
Given: PABQ and ACYX are squares
QB
To prove: PC = BX
Proof
Statements Reasons
1. PAB = CAX = 90° 1. Angles of square are equal
Vedanta Excel in Mathematics Teachers' Manual - 9 88
2. PAB BAC = CAX BAC 2. Adding BAC on both sides of (1)
3. PAC = BAX 3. Whole part axiom
4. In 'PAC and 'BAX 4.
i. PA = AB (S) i. Sides of square PABQ
ii. PAC = BAX (A) ii. From (3)
iii. AC = AX (S) iii. Being the sides of square ACYX
5. 'PAC # 'BAX 5. By S.A.S. axiom
6. PC = BX 6. Corresponding sides of congruent triangle are
equal
27. In the adjoining figure ABC is an equilateral triangle and BCDE is a E Proved
square. Prove that: AE = AD B
D
Solution: A
Given: ABC is an equilateral triangle and BCDE is a square C
To prove: AE = AD
Proof
Statements Reasons
1. EBC = BCD = 90° 1. Angles of square are equal
2. ABC = BCA = 60° 2. Angles of equilateral triangle
3. EBC ABC = BCD BCA 3. Subtracting (1) from (2)
? EBA = ACD 4.
4. In 'EBA and 'ACD i. Sides of the square BCDE
i. BE = CD (S) ii. From (3)
ii. EBA = ACD (A) iii. Sides of equilateral 'ABC
iii. AB = AC (S) 5. By S.A.S. axiom
5. 'EBA # ACD 6. Corresponding sides of
6. AE = AD
congruent triangle are equal
28. In the figure alongside, PQRS is a square in which PA and SB S Proved
intersect at O. If PA = SB, prove that PA and SP are perpendicular P
to each other at O R
Solution: A
Given: PQRS is a square. PA = SB and intersect at O. O Q
B
To prove: PA A SB at O. i.e, SOA = 90°
Proof
Statements Reasons
1. In 'SPB and 'PAQ 1.
i. SPB = PQA i. Both are right angles
ii. SB = PA ii. Given
iii. SP = PQ iii. Sides of square PQRS
2. 'SPB # 'PAQ 2. By R.H.S. axiom
3. PSB = APQ 3. Corresponding angles of congruent triangles
89 Vedanta Excel in Mathematics Teachers' Manual - 9
4. SOA = SPO PSO 4. Being ext. angle of triangle SOP equal to the
sum of opposite interior angles
5. SOA = SPO OPB
6. SOA = SPB = 90° 5. From (3) and (4)
6. From (5), by whole part axiom
Proved
29. In the figure alongside, AB = AC and AD bisects CAE. Prove
that AD // BC.
Solution:
Given: AB = AC and AD bisects CAE
To prove: AD // BC
Proof:
Statements Reasons
1. ABC = ACB 1. AB = AC
2. CAE = 2CAD 2. CAD = DAE
3. CAE = ABC BCA 3. Being ext. angle of ' equal to the sum of
4. 2CAD = BCA BCA opposite interior angles
? CAD = BCA 4. From (1), (2) and (3)
5. AD // BC 5. From (4) alternate angles being equal
Proved
30. In the triangle given alongside, PQ = PR. The bisector of PQR
meets PR at S. Prove that PSQ = 3PQS.
Solution:
Given: In PQR, PQ = PR, PQS = SQR
To prove: PSQ = 3PQS
Proof:
Statements Reasons
i. PQR = PRQ i. ? PQ = PR
ii. PQR = 2PQS ii. ? PQS = SQR
iii. PRQ = 2PQS iii. From (i) and (ii)
iv. In 'QSR; PSQ = SQR SRQ iv.
v. PSQ = PQS 2PQS v. From (iii) and (iv)
? PSQ = 3PQS
Proved
31. In the given figure, BE = EC and CE is the bisector of ACB.
Prove that BEC = ACD.
Solution:
Given: BE = EC and ACE = BCE
To prove: BEC = ACD
Proof:
Statements Reasons
1. EBC = BCE 1. BE = EC
Vedanta Excel in Mathematics Teachers' Manual - 9 90
2. BCE = ACE 2. Given
3. In 'AEC, BEC = EAC ACE
3. Being ext. angle of ' equal to
4. BEC = BAC ABC the sum of opposite interior
5. In 'ABC, ACD = BAC ABC angles
6. BEC = ACD
4. From (1), (2) and (3)
5. Same as (3)
6. From (4) and (5)
Proved
32. In the adjoining figure, AD is the bisector
of BAC and AD // EC. Prove that AC = AE.
Solution:
Given: BAD = CAD and AD // EC
To prove: AC = AE
Proof
Statements Reasons
1. BAD = CAD 1. Given
2. BAD = BEC 2. AD // EC and corresponding angles
3. CAD = ACE 3. AD // EC and alternate angles
4. BEC = ACE 4. From (i), (ii) and (iii)
5. AC = AE 5. From (iv), being base angles equal
A Proved
33. In 'ABC, ABC = 90° and M is the mid - point of AC.
Prove that: AM = BM = CM. M
Solution:
Given: In 'ABC, ABC = 90° and AM = CM B C
To prove: AM = BM = CM A D
Construction: BM is produced to D such that BM = MD M
and CD is joined.
Proof BC
Statements Reasons
1. In 'ABM and 'CDM
i. AM = CM (S) 1.
ii. AMB = CMD (A) i. Given
iii. BM = MD (S) ii. Vertically opposite angles are equal
2. 'ABM # 'CDM iii. By construction
3. AB = CD and ABM = CDM
2. By S.A.S. axiom
4. AB // DC
3. Being corresponding sides and angles of
5. ABC BCD = 180° congruent triangles
or, 90° BCD = 180°
4. Being alternate angles ABM and CDM
equal from (3)
5. AB // DC and co-interior angles
? BCD = 90°
91 Vedanta Excel in Mathematics Teachers' Manual - 9
6. In 'ABC and 'BCD, 6.
i. AB = CD (S) i. From (3)
ii. ABC = BCD (A) ii. Both are right angles
iii. BC = BC (S) iii. Common side
7. 'ABC # 'BCD 7. By S.A.S axiom
8. ACB = DBC 8. Corresponding angles of congruent triangles
9. BM = CM are equal
10. AM = BM = CM 9. From (8), being base angles of 'MBC equal
10. Given AM = CM and from (9)
A
Alternative process: (Effective after teaching mid - point theorem M
Construction: MN // CB is drawn through M N
Proof BC
Statements Reasons
1. AN = BN 1. In 'ABC, AM = CM and MN // CB
2. ANM = ABC 2. MN // CB, corresponding angles
3. In 'AMN and 'BMN 3.
i. AN = BN (S) i. From (1)
ii. ANM = BNM (A) ii. From (2), both are right angles
iii. MN = MN iii. Common side
4. 'AMN # 'BMN 4. By S.A.S axiom
5. AM = BM 5. Corresponding sides of congruent triangles
6. AM = BM = CM are equal
6. From (5) and given AM = CM
Alternative process: A Proved
D
Construction: BM is produced to D such that BM = MD and M
quadrilateral ABCD is completed.
Proof BC
Statements Reasons
1. AM = CM and BM = MD 1. Given and by construction
2. ABCD is a parallelogram 2. From (1), being diagonals bisect each other
3. ABCD is a rectangle 3. From (2) adn being ABC = 90°
4. AC = BD 4. Diagonals of rectangle are equal
5. AM = CM = BM = MD 5. From (1) and (4)
? AM = BM = CM
34. In the perpendiculars drawn from any two vertices to their A Proved
opposite sides of a triangle are equal. Prove that the triangle N
is an isosceles triangle. M
B C
Solution:
Given: In 'ABC; BM A AC , CN A AB and BM = CN
Vedanta Excel in Mathematics Teachers' Manual - 9 92
To prove: 'ABC is an isosceles triangle
Proof
Statements Reasons
1. In 'NBC and 'MBC, 1.
i. BNC = BMC (R) i. Both are right angles
ii. BC = BC (H) ii. Common sides
iii. CN = BM (S) iii. Given
2. 'NBC # 'MBC 2. By R.H.S. axiom
3. NBC = MCB 3. Corresponding angles of congruent triangle
i.e, ABC = ACB are equal
4. 'ABC is an isosceles triangle
4. From (3), Both angles are equal
35. If the perpendiculars drawn from the mid - point of any side of a P Proved
triangle to its other two sides are equal. Prove that the triangle
is an isosceles triangle A
Q
Solution:
MC
Given: In 'ABC, M is the mid point of side BC, MP A AB, MQ A AC B
and MP = MQ
To prove: 'ABC is an isosceles triangle
Proof
Statements Reasons
1. In 'PBM and QMC 1.
i. BPM = CQM (R) i. Both are right angles
ii. BM = CM (H) ii. M being mid point of side BC
iii. MP = MQ (S) iii. Given
2. 'PBM # 'QMC 2. By R.H.S. axiom
3. PBM = QCM 3. Corresponding angles of congruent
i.e, ABC = ACB triangle
4. 'ABC is an isosceles triangle
4. From (4), base angles are equal
36. Prove that the line joining the point of intersection of two Proved
angular bisectors of the base angles of an isosceles triangle to
the vertex bisects the vertical angle. A
Solution: O
C
Given: In isosceles 'ABC, AB = AC, OB and OC are bisectors of B
base angles ABC and ACB respectively.
To prove: OA bisect BAC i.e, OAB = OAC
Proof
Statements Reasons
1. ABC = ACB 1. Base angles of isosceles triangle are
equal
2. ABC = 2OBC and ACB = 2OCB 2. Given
3. 2OBC = 2OCB ? OBC = OCB 3. From (1) and (2)
93 Vedanta Excel in Mathematics Teachers' Manual - 9
4. OB = OC 4. Base angles of 'OBC are equal
5. In 'AOB and 'AOC
i. AB = AC (S) 5.
ii. OA = OA (S) i. Given
iii. OB = OC (S) ii. Common side
6. 'AOB # 'AOC iii. From (4)
7. OAB = OAC
6. By S.S.S. axiom
7. Corresponding sides of congruent
triangles
37. In the angular bisector of an angle of triangle bisects the opposite Proved
side, prove that the triangle is an isosceles triangle.
A
Solution:
Given: In 'ABC, BAM = CAM and BM = CM BM C
To prove: 'ABC is an isosceles triangle i.e, AB = AC
Construction: AM is produced to D such that AM = MD and CD is joined
Proof D
Statements Reasons
1. In 'ABM and 'CDM 1.
i. AM = MD (S) i. By construction
ii. AMB = CMD (A) ii. Vertically opposite angles
iii. BM = CM (S) iii. Given
2. 'ABM # 'CDM 2. By S.A.S. axiom
3. AB = CD and BAM = CDM 3. Corresponding part of congruent triangles
4. BAM = CAM 4. Given
5. CDM = CAM 5. From (3) and (4)
6. AC = CD 6. From (5), base angles of 'ACD are equal
7. AB = AC 7. From (3) and (6)
Proved
38. In the given rectangle PQRS, M is the mid - point of RS. Prove that PQM is an
isosceles triangle. S MR
Solution:
Given: In rectangle PQRS, M is the mid - point of RS.
To prove: PQM is an isosceles triangle PQ
Proof
Statements Reasons
1. In 'PMS and 'QRM 1.
i. SP = RQ i. Opposite sides of rectangle are equal
ii. PSM = QRM (A) ii. Both are right angles
iii. SM = RM (S) iii. Given
2. 'PMS # QRM 2. By S.A.S. axiom
3. PM = QM 3. Corresponding sides of congruent triangle
4. 'PQM is an isosceles triangle 4. From (3)
Proved
Vedanta Excel in Mathematics Teachers' Manual - 9 94
39. In the given figure, prove that QS is the perpendicular bisector Q P S
of PR. O
Solution: R
Given: PQ = QR and PS = RS
To prove: QS is perpendicular bisector of PQ
Proof
Statements Reasons
1. In 'PQS and 'QRS 1.
i. PQ = QR (S) i. Given
ii. PS = RS (S) ii. Given
iii. QS = QS (S) iii. Common side
2. 'PQS # QRS 2. By S.S.S. axiom
3. PQS = RQS 3. Corresponding angles of congruent triangle
4. In 'POQ and 'ROQ 4.
i. PQ = QR (S) i. Given
ii. PQO = RQO (A) ii. From (3)
iii. QO = QO iii. Common Side
5. 'POQ # 'ROQ 5. By S.A.S axiom
6. OP = OR and POQ = ROQ 6. Corresponding parts of congruent triangles
7. POQ = ROQ = 90° 7. Being adjacent angles on linear pair
8. OP = OR and QS A PR 8. From (6) and (7)
Hence, QS is perpendicular bisector of PR. DZC
40. In the figure alongside, ABCD is a square. X, Y and Z are the points A Y
on the sides AB, BC and CD respectively. Such that AX = BY = CZ. XB
Prove that XYZ is an isosceles traingle.
Solution:
Given: ABCD is a square. X, Y and Z are the point on the sides AB, BC and CD respectively.
To prove: XYZ is an isosceles triangle.
Proof
Statements Reasons
1. AB = BC 1. Adjacent sides of the square ABCD
2. AX = BY 2. Given
3. BX = CY 3. Subtracting (2) from (1)
4. In 'BXY and 'CYZ 4.
i. BX = CY (S) i. From (3)
ii. XBY = YCZ (A) ii. Both are right angles
iii. BY = CZ (S) iii. Given
5. 'BXY # 'CYZ 5. By S.A.S axiom
6. XY = YZ 6. Corresponding sides of congruent triangles
7. XYZ is an isosceles triangle 7. From (6)
Proved
95 Vedanta Excel in Mathematics Teachers' Manual - 9
41. In the given figure, PQ // SR. ST and RT bisect PSR and SRQ P T Q
R
respectively. Prove that : PQ = PS QR.
Solution:
Given: PQ // SR. ST and RT bisect PSR and SRQ respectively. i.e,
PST = TSR and SRT = QRT S
To prove: PQ = PS QR
Proof
Statements Reasons
1. PST = TSR = PTS 1. Given and alternate angles
within PQ // SR
2. PT = PS 2. From (1), PST = PTS
3. QRT = SRT = QTR 3. Given and alternate angles
within PQ // SR
4. TQ = QR 4. From (3), QRT = QTR
5. PT PQ = PS QR 5. Adding (2) and (4)
6. PQ = PS QR 6. From (5), whole part axiom
42. In the given figure, BA A AC, RQ A PQ, AB = QR and BP = CR. A Proved
Prove that AC = PQ.
Solution: B C R
Given: BA A AC, RQ A PQ, AB = QR and BP = CR P
To prove: AC = PQ
Proof Q
Statements Reasons
1. BP = CR 1. Given
2. BC = PR 2. Adding PC on both side of (1)
3. In 'ABC and 'PQR 3.
i. BAC = PQR (R) i. Both are right angles
ii. BC = PR (H) ii. From (2)
iii. AB = QR (S) iii. Given
4. 'ABC # PQR 4. By R.H.S. axiom
5. AC = PQ 5. Corresponding sides of congruent triangle are
equal
43. In the figure alongside, OAD = ODA and OBC = OCB. A Proved
Prove that AB = DC.
D
Solution:
O
Given: OAD = ODA and DBC = OCB C
To prove: AB = DC B
Proof
Statements Reasons
1. OB = OC 1. ?OBC = OCB
2. OD = OA 2. ? ODA = OAD
3. BD = AC 3. Adding (1) and (2)
Vedanta Excel in Mathematics Teachers' Manual - 9 96
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