47 Vedanta Excel in Mathematics Teachers' Manual - 9 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 Again, number of bricks (N) = Volume of wall (V) Volume of each brick (v) or, 4,700 = 4h – 1.2 0.004 or, 18.8 = 4h – 1.2 ? h = 5 m 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 ?x2 h = 288 …equation (i) Again, area of floor = Total cost of carpeting Rate of carpeting = 6720 105 or, x2 = 64 or, x = 8 ? l = b = 8 m Also, putting the value of x in equation (i), we get 82 h = 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]
Vedanta Excel in Mathematics Teachers' Manual - 9 48 Unit 6 Algebraic Expressions Allocated teaching periods 10 Competency - To factorize the algebraic expression of the forms a2 – b2 , a3 – b3 and a4 + a2 b2 + 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 + a2 b2 + 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 1. Knowledge (K) - To define factorization - 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 2. Understanding (U) - To factorize the algebraic expressions by using the formula a2 – b2 , a3 ± b3 - To factorize the formulae of the forms a4 + a2 b2 + b4 3. Application (A) - To factorize the expressions involving 5 or 6 terms by expressing in the formulae of (a± b)2 and a2 – b2 - 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)3 with blocks or ICT tools 5. Discuss upon the factorization of the form a4 + a2 b2 + b4 with examples 6. Engage the students to factorize the expressions given in the exercise 7. Focus on more practical problems related to factorization
49 Vedanta Excel in Mathematics Teachers' Manual - 9 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 = a2 c2 – a2 d2 – b2 c2 + b2 d2 + 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 p5 Solution: Here, p7 + 1 p5 = p7 + p × 1 p6 = p p6 + 1 p6 = p (p2 ) 3 + 1 p2 3 = p p2 + 1 p2 p4 – p2 × 1 p2 + 1 p4 = p p2 + 1 p2 p4 – 1 + 1 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
Vedanta Excel in Mathematics Teachers' Manual - 9 50 = (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: 1) 9a4 + 14a2 + 25 2) a4 b4 + a2 b2 + 1 3) x4 y4 + 1 + y4 x4 4) p6 – 1 5) x2 – 10x + 16 – 6y – y2 6) (9 – a2 ) (100 – b2 ) 7) 6(a + b)2 – (a + b) – 7 8) 2x2 y – 3 – 5y2 x2 Answers 1. (3a2 + 4a + 5) (3a2 – 4a + 5) 2. ( a2 b2 + a b + 1) ( a2 b2 – a b + 1) 3. (x2 y2 + 1 + y2 x2 ) (x2 y2 – 1 + y2 x2 ) 4. (p + 1) (p – 1) (p2 + p + 1) (p2 – p + 1) 5. (x + y – 2) (x – y – 8) 6. (ab + 10a + 3b + 30) (ab – 10a – 3b + 30) 7. (a + b + 1) (6a + 6b – 7) 8. (2x y – 5y x ) (x y + y x) 2. H.C.F., L.C.M. and simplification 1. Find the H.C.F. of x3 – y3 , x6 – y6 , x4 + x2 y2 + y4 Solution: Here, x3 – y3 , x6 – y6 , x4 + x2 y2 + 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 + x2 y2 + y4 = (x2 ) 2 + (y2 ) 2 + x2 y2 = (x2 + y2 ) 2 – 2x2 y2 + x2 y2 = (x2 + y2 ) 2 – x2 y2 = (x2 + y2 ) 2 – (xy)2 = (x2 + xy + y2 ) (x2 – xy + y2 ) ? H.C.F. = x2 + xy + y2
51 Vedanta Excel in Mathematics Teachers' Manual - 9 2. Simplify: 1 (a – b)2 – 1 a2 – b2 Solution: Here, 1 (a – b)2 – 1 a2 – b2 = 1 (a – b) (a – b) – 1 (a + b) (a – b) = (a + b) – (a – b) (a – b) (a – b) (a + b) = a + b – a + b (a – b)2 (a + b) = 2b (a + b) (a – b)2 3. Simplify: 2y + 5 y2 + 6y + 9 – 11 y2 – 9 – 16y 8y2 – 24y Solution: Here, 2y + 5 y2 + 6y + 9 – 11 y2 – 9 – 16y 8y2 – 24y = 2y + 5 (y + 3)2 – 11 (y + 3) (y – 3) – 16y 8y(y – 3) = (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 (y + 3)2 (y – 3) = –2y (y + 3)2 (y – 3) = 2y (3 – y) (y + 3)2 4. Simplify: a + 2 1 + a + a2 – a – 2 1 – a + a2 – 2a2 1 + a2 + a4 Solution: Here, a + 2 1 + a + a2 – a – 2 1 – a + a2 – 2a2 1 + a2 + a4 = (a + 2) (1 – a + a2 ) – (a – 2) (1 + a + a2 ) (1 + a + a2 ) (1 – a + a2 ) – 2a2 (1 + a2 ) 2 – 2a2 + a2 = a – a2 + a3 + 2 – 2a + 2a2 – a – a2 – a3 + 2 + 2a + 2a2 (1 + a + a2 ) (1 – a + a2 ) – 2a2 (1 + a + a2 ) (1 – a + a2 ) = 2a2 (1 + a + a2 ) (1 – a + a2 ) – 2a2 (1 + a + a2 ) (1 – a + a2 ) = 2a2 – 2a2 (1 + a + a2 ) (1 – a + a2 ) = 0 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 + x2 y2 + y4 d) x2 – 4, x3 + 8, x2 + 5x + 6 e) (a + b)2 – 4ab, a3 – b3 , a2 + ab – 2b2 2. Simplify: a) x – y xy + y – z yz + z – x zx b) a2 – ab + b2 a – b + a2 + ab + b2 a + b c) x xy – y2 + y xy – x2 d) a2 + b2 ab – b2 a(a + b) – a2 b(a + b) e) (a – b)2 – c2 a2 – (b + c)2 + (b – c)2 – a2 b2 – (c + a)2 + (c – a)2 – b2 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 + x2 y2 + 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 a2 – b2 c) x + y xy d) 1 e) 1
Vedanta Excel in Mathematics Teachers' Manual - 9 52 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 1. Knowledge (K) - To identify coefficient, base and power of expression - To recall the laws of indices - To express the product/quotient of expressions having same base in terms of single base 2. Understanding (U) - To evaluate the numerical problems by using laws of indices - To simplify/prove the simple given expressions - To solve the exponential equations 3. Application (A) - To simplify/prove the given rational expressions (involving roots as well) by applying the laws of indices - 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 indices. For, 64 4 = 26 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. Unit 7 Indices
53 Vedanta Excel in Mathematics Teachers' Manual - 9 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 a) (1 35 ) 1 (1 35 ) 1 b) (a b)1 . (a1 b1 ) c) y1 x1 x1 y1 1 d) 100 3 4 u 1 100 4 Solution: a) (1 35 ) 1 (1 35 ) 1 b) (a b) 1 . (a1 b1 ) = 1 1 35 1 (1 35 ) 1 = 1 a b u 1 a 1 b = 35 1 35 1 1 1 35 = 1 a b u b a ab = 35 35 1 1 35 1 = 1 ab = 35 1 35 1 = 1 c) y1 x1 x1 y1 1 = x y y x 1 = x2 y2 xy 1 = xy x2 y2 d) 100 3 4 × 1 100 4 = (102 ) 3 4 u 1 102 1 4 = 10 3 2 u 1 102 u 1 4 = 10 3 2 u 1 10 1 2 = 10 3 2 u10 1 2 = 10 3 1 2 = 10 2. Simplify: a) 5n u 625n1 53n2 u (5 u 2)1 b) 9x u 3x1 3x 32x1 u 3x 2 3x Solution: a) 5n u 625n1 53n2 u (5 u 2)1 = 5n u (54 ) n1 53n2 u 51 u21 = 5n4n43n21 u 2 = 51 u 2 = 2 5 b) 9x u 3x1 3x 32x1 u 3x 2 3x = 32x u 3x u 31 3x 32x u 3 u 3x u 3 2 3x = 3x(32x u 3x(32x u 3 u 3 2 1) = 32x 1 1 32x 1 2 1 = 32x 1 1 32x 1 1 = 1 3. Simplify: a) p + (pq2 ) 1 3 + (p2 q)1 3 p – q × 1 – q 1 3 p 1 3 b) x a b c 1 b a u x b c a 1 c b u x c a b 1 a c c) xy ax2 ay2 × yz ay2 az2 × zx az2 ax2
Vedanta Excel in Mathematics Teachers' Manual - 9 54 Solution: a) p + (pq2 ) 1 3 + (p2 q) 1 3 p – q × 1 – q 1 3 p 1 3 = p + p 1 3q 2 3 + p 2 3q 1 3 p – q u p 1 3 q 1 3 p 1 3 = p 1 3 (p 2 3 q 2 3 p 1 3q 1 3) p – q u p 1 3 q 1 3 p 1 3 = (p 1 3 q 1 3)[(p 1 3) 2 p 1 3q 1 3 (q 1 3) 2 ] p – q = (p 1 3) 3 (q 1 3) 3 p – q = p – q p – q = 1 b) x b b c 1 b a u x c c a 1 c b u x a a b 1 a c = x b (b c)(b a) u x c (c a)(c b) u x a (a b)(a c) = x b (b c)(b a) c (c a)(c b) a (a b)(a c) = x b (b c)(a b) c (c a)(b c) a (a b)(c a) = x b(c a) c(a b) a(b c) (a b)(b c) (c a) = x –bc + ab – ca + bc – ab + ca (a b)(b c) (c a) = x0 = 1 c) xy ax2 ay2 × yz ay2 az2 × zx az2 ax2 = a x2 y x+y 2 u a y2 z yz 2 u a z2 x zx 2 = a x2 y2 x y u a y2 z2 y z u a z2 x2 z x = ax y y z z x = a0 = 1 4. (a) If xyz = 1, prove that 1 1 + x + y–1 + 1 1 + y + z–1 + 1 1 + z + x–1 = 1. (b) If a + b + c = 0, prove that 1 1 + xa + x–b + 1 1 + xb + x–c + 1 1 + xc + x–a = 1 Solution: (a) Here, xyz = 1 L.H.S = 1 1 + x + y–1 + 1 1 + y + z–1 + 1 1 + z + x–1 = 1 1 + x 1 y 1 1 + y 1 z 1 1 + z 1 x = 1 y xy 1 y 1 z yz 1 z 1 x zx 1 x = z u y z(y xy 1) z z yz 1 x x zx 1 = yz yz xyz z z z yz 1 x x zx 1 = yz yz 1 z z z yz 1 x x zx 1 = x(yz z) x(yz z 1) x x zx 1 = xyz zx xyz zx x x x zx 1 = 1 zx x 1 zx x = 1 = RHS
55 Vedanta Excel in Mathematics Teachers' Manual - 9 (b) Here, a + b + c = 0 LHS = 1 1 + x a + x –b + 1 1 + x b + x –c + 1 1 + x c + x –a = 1 = 1 1 + x a 1 x b 1 1 + x b 1 x c 1 1 + x c 1 x a = x b u xc x c (x b x ab 1 x c x c x bc 1 x a x a x ca 1 = x bc x bc x abc x c x c x bc x c 1 x a x ca x a 1 = x bc x bc x 0 x c x c x bc x c 1 x a x ca x a 1 = x bc xc x bc x c 1 x a x ca x a 1 = x a (x bc xc ) x a (x bc x c 1) x a x ca x a 1 = x abc xca x abc x ca x a x a x ca x a 1 = x 0 xca x 0 x ca x a x a x ca x a 1 = 1 x ca 1 x ca x a x a x ca x a 1 = x ca x a 1 x ca x a 1 = 1 = R.H.S 5. a) If x = 21 3 + 2– 1 3 , prove that: 2x3 – 6x = 5. b) If x – 2 = 31 3 + 32 3 , show that: x(x2 – 6x + 3) = 2. Solution: a) Here, x = 21 3 + 2– 1 3 Cubing on both sides, we get, x 3 = 2 1 3 + 2– 1 3 3 or, x 3 = 2 1 3 3 + 2 – 1 3 3 + 3 u 2 1 3 u 2– 1 3 2 1 3 + 2– 1 3 [ ? (a b)3 = a3 b3 3ab(a b)] or, x3 = 2 21 3 u 1 u x [ ? 2 1 3 + 2– 1 3 = x] or, x3 = 2 1 2 3x or, x3 = 4 1 6x 2 or, 2x3 – 6x = 5 proved b) Here, x – 2 = 31 3 + 32 3 Cubing on both sides, we get
Vedanta Excel in Mathematics Teachers' Manual - 9 56 (x – 2)3 = 3 1 3 + 32 3 3 or, x3 – 23 – 3 u x u 2 (x – 2) = 3 1 3 3 + 3 2 3 3 + 3 u 3 1 3 u 3 2 3 3 1 3 + 32 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: 2x3 u 3x2 = 432 Solution: Here, 2x3 u 3x2 = 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 b) If a1 x = b1 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. b) Here, a 1 x = b 1 3 or, a 1 x x = b 1 3 x ? a = b x 3 Now, ab = 1 or, b x 3 . b = 1 or, b x 3 1 = b0 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–1 n2 m2 n4 7 y m3 n–5 m–2n3 –5 = mx ny , prove that mx–2y = 1. Solution: a) Here, (a–1 b–1) (a b) –1 = ambn or, 1 a 1 b 1 a b = ambn or, b a ab 1 a b = ambn or, a–1b–1 = ambn ? m = –1, n= –1 Again, am–n = a–1–(–) = a0 = 1 proved
57 Vedanta Excel in Mathematics Teachers' Manual - 9 b) m–1n2 m2 n4 7 y m3 n–5 m–2n3 –5 = mx ny or, (m–3n6 ) 7 y (m5 n–8) –5 = mx ny or, m–21n42 y m–25n40 = mx ny or, m4 n2 = mx ny ? 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, xmn = 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 1. Simplify: a) (a + x) 3 –8× (a + x) 2 3 Ans: 1 (a + x)2 b) 20pq5 r 3 8× 50p4 qr 3 –2 Ans: 10pq2 r2 2. Prove that: a) 1 1 + xa – b + xc – b + 1 1 + xb – c + xa – c + 1 1 + xc – a + xb – a = 1 b) a2 – 1 b2 a × a – 1 b b – a b2 – 1 a2 b × b+ 1 a a – b = a b a + b 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
Vedanta Excel in Mathematics Teachers' Manual - 9 58 Unit 8 Simultaneous Linear Equations 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 - To define simultaneous linear equations - To tell the methods of solving pair of linear equations 2. Understanding (U) - To solve the pair of simultaneous equation (simple) by substitution method - To solve the pair of simple simultaneous equation by elimination method 3. Application (A) - 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 4. High Ability (HA) - 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
59 Vedanta Excel in Mathematics Teachers' Manual - 9 1. Solve the following simultaneous equations graphically. a) x 2y = 14 and x – y = 5 b) x – 2 2 = y = 2x – 6 3 Solution: a) The given linear equation are x + 2y = 14 ........... (i) and x – y = 5 ........... (ii) Now from equation (i). x + 2y = 14 or, y = 14 – x 2 x 024 y 765 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 x 123 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 2 = y ......... (i) and y = 2x – 6 3 ......... (ii) Now, from equation (i), y = x – 2 2 x 024 y – 1 0 1 Plotting the point (0,–1), (2,0) and (4,1) and joining them to form a straight line. Also, from equation (ii), y = 2x – 6 3 x 036 y – 2 0 2 Plotting the point (1,–2), (3.0) and (6,2) and joining them to form a straight line. Since the graphs of equation y = x – 2 2 intersect at (6,2). Hence, x = 6 and y = 2 2. Solve the equation 2 x 6 y = 3 and 10 x – 9 y = 2 by elimination method. Solution: Here, The given equation are (8, 3) x + 2y = 14 x – y = 5 (6, 2)
Vedanta Excel in Mathematics Teachers' Manual - 9 60 2 x 6 y = 3 ......... (i) u 3 10 x – 9 y = 2 ......... (ii) u 2 Now, multiplying equation (i) by 3 and equation (ii) by 2 and adding them. 6 x 18 y = 9 20 x – 18 y =4 26 x = 13 or, 13x = 26 ? x = 2 Again, putting the value of x in equation (i), we get 2 2 6 y = 3 or, 6 y = 3 – 1 or, 2y = 6 ? y = 3 Hence, x = 2 and y = 3 3. Solve the given system of equation by substation method. x – 1 y 1 = 1 2 and x – 2 y 2 = 1 3 Solution: Here, The given equation are x – 1 y 1 = 1 2 or, 2x – 2 = y 1 ? y = 2x – 3 ...... (i) and x – 2 y 2 = 1 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 = 2 u 5 – 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
61 Vedanta Excel in Mathematics Teachers' Manual - 9 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 x – 4 y = 1 , 15 x 2 y = 6 [Ans: 3,2] 3. Solve each pair of simultaneous equations by substitution mehtod. i) x – 1 y 1 = 1 2 and x – 1 y 1 = 1 2 [Ans: 10,11] ii) x x – 1 y 1 = 6 and y – x – 1 y 1 = 2 [Ans: 4,3]
Vedanta Excel in Mathematics Teachers' Manual - 9 62 Unit 9 Quadratic Equation 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 1. Knowledge (K) - To define quadratic equation - 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 3. Application (A) - To solve the quadratic equation by factorization method - 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.
63 Vedanta Excel in Mathematics Teachers' Manual - 9 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 a + b a x = – c a (To make the coefficient of x2 unity, dividing both sides by a) or, x2 + b a x = – c a or, x2 + b a x + b 2a 2 = b 2a 2 – c a or. x2 + 2. b 2a x + b 2a 2 = b2 4a2 – c a or, x + b 2a 2 = b2 – 4ac 4a2 or, x + b 2a = ± b2 – 4ac 2a or, x = – b 2a ± b2 – 4ac 2a = – b ± b2 – 4ac 2a Thus, the required roots of x are – b + b2 – 4ac 2a and – b – b2 – 4ac 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 x – 5 x 5 = 2 1 2 b) x – 2 x 2 x 2 x – 2 = 2x 6 x – 3 c) x – 2 x – 3 3x – 11 x – 4 = 4x 13 x 1 Solution: a) x 5 x – 5 x – 5 x 5 = 2 1 2 or, (x 5)2 (x – 5)2 (x – 5) (x 5) = 5 2 or, x2 10x 25 x2 – 10x 25 x2 – 25 = 5 2 or, 2x2 50 x2 – 25 = 5 2 or, 5x2 – 125 = 4x2 100 (Adding the square of half of the coefficient of x to both sides)
Vedanta Excel in Mathematics Teachers' Manual - 9 64 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 x 2 x – 2 = 2x 6 x – 3 or, (x – 2)2 (x 2)2 (x 2) (x – 2) = 2x 6 x – 3 or, x2 – 4x 4 x2 4x 4 x2 – 4 = 2x 6 x – 3 or, 2x2 8 x2 – 4 = 2x 6 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 or, 3x – 4 = 0 ? x = 4 3 Hence, x = 0 or , 4 3 c) x – 2 x – 3 3x – 11 x – 4 = 4x 13 x 1 or, (x – 2) (x – 4) (x – 3) (3x – 11) (x – 3) (x – 4) = 4x 13 x 1 or, x2 – 4x – 2x 8 3x2 – 11x – 9x 33 x2 – 4x – 3x 12 = 4x 13 x 1 or, 4x2 – 26x 41 x2 – 7x 12 = 4x 13 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 or, 7x – 23 = 0 ? x = 23 7 Hence, x = 5 or 23 7 2. Solve these equation by factorization method. a) x x – 3 x – 3 x = 5 2 b) 2x – 3 x – 1 – 4 x – 1 2x – 3 = 3
65 Vedanta Excel in Mathematics Teachers' Manual - 9 Solution: a) Let x x – 3 = a, then x – 3 x = 1 a Now, a 1 a = 5 2 or, a2 1 a = 5 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 x – 3 = 2 or, x x – 3 = 4 or, 4x – 12 = x or, x = 12 3 = 4 When a = 1 2 , x x – 3 = 1 2 or, x x – 3 = 1 4 or, 4x = x – 3 or, 3x = 3 ? x = 1 Hence, x = 1 or 4 b) Let 2x – 3 x – 1 = a, then x – 1 2x – 3 = 1 a Now, a 4 u 1 a = 3 or, a2 – 4 a = 3 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 When a = 4, 2x – 3 x 1 = 4 or, 4x 4 = 2x 3 or, 2x = 1 ? x = 1 2 When a = 1, 2x – 3 x 1 = 1
Vedanta Excel in Mathematics Teachers' Manual - 9 66 or, 2x 3 = x 1 or, 3x = 4 ? x = 4 3 = 1 1 3 Hence, x = 1 2 , 1 1 3 3. Solve these equations. a) 1 a b x = 1 a 1 b 1 x b) ax2 bx c px2 qx r = ax b px q c) 1 (a b) (x b) 1 (c a) (x c) = 1 (x b) (x c) 1 (a b) (c a) Solution: a) 1 a b x = 1 a 1 b 1 x or, 1 a b x 1 x = 1 a 1 b or, x a b x x (a b x) = b a ab or, (a b) x (a b x) = (a b) 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 px2 qx r = ax b 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 ? x = br cq cp ar c) 1 (a b) (x b) 1 (c a) (x c) = 1 (x b) (x c) 1 (a b) (c a) 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
67 Vedanta Excel in Mathematics Teachers' Manual - 9 or, 3x2 3 5 3 x = 2 3 [Dividing both sides by 3] or, x2 5 3 x = 2 3 or, x2 5 3 x 5 6 2 = 2 3 5 6 2 or, x2 2 . x . 5 6 5 6 2 = 2 3 25 36 or, x 5 6 2 = 24 25 36 or, x 5 6 2 = – 7 6 2 Taking () ve sign Taking () ve sign x 5 6 = 7 6 x 5 6 = 7 6 ? x = 2 6 = 1 3 ? x = 12 6 = 5 6 Hence x = 1 3 or 2. b) 15x2 + 2ax = a2 Dividing each term by 15 15x2 15 2ax 15 = a2 15 or, x2 2ax 15 2a 2 u 15 2 = a2 15 2a 2 u 15 2 or, x2 2 . x . a 15 a 15 2 = a2 15 a2 225 or, x a 15 2 = 16a2 225 or, x a 15 2 = – 4a 15 2 Taking () ve sign Taking () ve sign x a 15 = 4a 15 x a 15 = 4a 15 ? x = 5a 15 = a 3 ? x = 3a 15 = a 5 Hence x = a 3 or a 5 . 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.
Vedanta Excel in Mathematics Teachers' Manual - 9 68 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 p qx p r p = 0 p or, x2 q p x r p = 0 or, x2 q p x q 2p 2 = r p q 2p 2 or, x q 2p 2 = r p a2 4p2 or, x q 2p 2 = q2 4pr 4p2 or, x q 2p = ± q2 4pr 4p2 or, x = q 2p ± q2 – 4pr 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 x = b ± b2 – 4ac 2a = 11 ± 112 – 4 u 3 u 10 2 u 3 = 11 ± 1 6 = 11 ± 1 6 Taking () ve sign, Taking () ve sign, x = 11 1 6 x = 11 1 6 ? x = 5 3 x = 2 Hence, x = 5 3 or 2.
69 Vedanta Excel in Mathematics Teachers' Manual - 9 c) (x 1) (x 1) (x 2) (x 2) = (x 3) (x 3) (x 4) (x 4) Solution: Here, (x 1) (x 1) (x 2) (x 2) = (x 3) (x 3) (x 4) (x 4) or, (x 1) (x 2) (x 2) (x 1) (x 1) (x 2) = (x 3) (x 4) (x 4) (x 3) (x 3) (x 4) or, x2 3x 2 x2 3x 2 x2 x 2 = x2 7x 12 x2 7x 12 x2 x 12 or, 2x2 4 x2 x 2 = 2x2 24 x2 x 12 or, 2(x2 2) x2 x 2 = 2(x2 12) 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 or, 2x 1 = 0 ? x = 1 2 Hence, x = 0 , 1 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 or, x(a b) = 2b ? x = 2b a b Hence, x = 1 , 2b a b Extra Questions 1. Solve: a) x2 1 5 = 3 Ans : ± 4 b) x 9 = 4 x Ans : ± 6 c) 6 5x x2 = 0 Ans : 2 , 3 2. Solve: a) x 2 x x x 2 = 1 1 2 Ans : 2 , 4 3 b) x x 1 x 1 x = 2 1 6 Ans : 2 , 3 c) x 3 x 2 x 3 x 2 = 2x 3 x 1 Ans : 0 , 4
Vedanta Excel in Mathematics Teachers' Manual - 9 70 Unit 10 Ratio and Proportion 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 S.N. LEVELS OBJECTIVES 1. Knowledge (K) - To define ratio - To identify antecedent and consequent of the ratio - To recall the types of ratios: compound, duplicate, subduplicate, 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 4. High Ability (HA) - To identify the contextual problems based on ratio and proportion and solve them - 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 (ii) If a and b are two quantities of same kind, the ratio of a and b is a:b or a b (iii) In the ratio a: b, a is called antecedent and b called consequent.
71 Vedanta Excel in Mathematics Teachers' Manual - 9 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. For two ratios a: b and c: d, the compound ratio = (a: b) × (c: d) = ac 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 a 3 : b 3 Example: The sub-triplicate ratio of 125 : 64 = 125 3 : 64 3 = 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 If a b = c d , then b a = d c Proof Here, a b = c d , then 1 ÷ a b = 1 ÷ c d or, 1 × b a = 1 × d c or, b a = d c Proved Example: For 2:3 = 4:6, 3:2 = 2×3:2×2 = 6:4 (ii) Alternendo If a b = c d , then a c = b d Proof Here, a b = c d , then multiplying both sides by b c , we get, a b × b c = c d × b c or, a c = b d Proved Example: For 2:5 = 4:10, 2:4 = 1:2 = 5×1:5×2 = 5:10
Vedanta Excel in Mathematics Teachers' Manual - 9 72 (iii) Componendo If a b = c d , then a + b b = c + d d Proof Here, a b = c d , then adding 1 to both sides, a b + 1 = c d + 1 or, a + b b = c + d d Proved 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 If a b = c d , then a – b b = c – d d Proof Here, a b = c d , then subtracting 1 from both sides, a b – 1 = c d – 1 or, a – b b = c – d d Proved 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 b = c d , then a + b a – b = c + d c – d Proof Here, a b = c d , then by componendo, we have, Also, by dividendo , we have a + b b = c + d d ........ (i) or, a – b b = c – d d ........ (ii) Dividing equation (i) by (ii), we get a + b b a – b b c – d d = or, a + b a – b = c + d c – d proved 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 b = c d , then a b = c d = a + c b + d Proof Here, a b = c d , then by alternendo, we have, a c = b d and by componendo, we have = b + d d Again, by alternaendo, we have, a + c b + d = c d ? a b = c d = a + c b + d proved. Example: For 5:7= 10:14, (5 + 10): (7 + 14) = 15:21 = 5:7
73 Vedanta Excel in Mathematics Teachers' Manual - 9 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: a2 2ab b2 a2 2ab b2 = 1 4 or, (a b) 2 (a b) 2 = 1 4 or, a b a b 2 = ± 1 2 2 Taking () ve sign Taking () ve sign a b a b = 1 2 a b a b = 1 2 or, 2a 2b = a b or, 2a 2b = a b or, a = 3b or, 3a = b or, a b = 3 1 or, a b = 1 3 ? a : b = 3 : 1 ? a : b = 1 : 3 2. If a : b = 3 : 2, find the value of i) b 2a 3 ii) a2 ab ab b2 Solution: a : b = 3 : 2 Let a = 3x and b = 2x i) b 2a 3 = 2x u 2 u 3x 3 = 0 ii) a2 ab ab b2 = (3x) 2 3x u 2x 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, x 4 3x 1 = 9 16 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:
Vedanta Excel in Mathematics Teachers' Manual - 9 74 Here, sub - triplicate ratio of 8 : 27 = 2 : 3. According to question, Sub - triplicate ratio of 8 : 27 = (3a 7) : (4a 3) or, 2 3 = 3a 7 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 30 = 5 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 b2= 2. Solution: Here, (a b), b and (a b) are in continuous. so, b2 = (a b) (a b) or, b2 = a2 b2 or, 2b2 = a2 or, 2 = a2 b2 ? a2 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. Then, 7 x 10 x = 16 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.
75 Vedanta Excel in Mathematics Teachers' Manual - 9 9. If a b = a b, prove that: a) a2 ab b2 b2 bc c2 = a c b) a3 b3 b3 c3 = a(a b) c(b c) c) a b c a b c = (a b c)2 a2 b2 c2 Solution: Let a b = b c = k then b c = k i.e, b = ck and a b = k i.e, a = bk = ck.k = ck2 Now a) L.H.S = a2 ab b2 b2 bc c2 = (ck2 ) 2 ck2 .ck (ck) 2 (ck) 2 ck.c c2 = c2 k4 c2 k3 c2 k2 c2 k2 c2 k c2 = c2 k2 (k2 k 1) c2 (k2 k 1) = k2 R.H.S = a c = ck2 c = k2 Hence, L.H.S = R.H.S Proved b) L.H.S = a3 b3 b3 c3 = (ck2 ) 3 (ck) 3 (ck) 3 c3 = c3 k6 c3 k3 c3 k3 c3 = c3 k3 (k3 1) c3 (k3 1) = k3 R.H.S = a(a b) c(b c) = ck2 (ck2 ck) c(ck c) = k2 .ck(k 1) c(k c) = k3 Hence, L.H.S = R.H.S Proved c) L.H.S = a b c a b c = ck2 ck c ck2 ck c = c(k2 k 1) c(k2 k 1) = (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 c2 k4 c2 k2 c2 = c2 (k2 k 1)2 c2 (k4 k2 1) = (k2 k 1)2 (k2 1)2 2k2 k2 = (k2 k 1)2 (k2 1)2 k2 = (k2 k 1)2 (k2 k 1)(k2 k 1)= (k2 k 1) (k2 k 1) 10. If x, y, z are in continuous proportion. Prove that x2 y2 z2 1 x3 1 y3 1 z3 = x3 y3 z3 . Solution: Let x y = y z = k then y z = k i.e y = zk, x y = k i.e, x = yk = zk.k = zk2 Now, L.H.S = x2 y2 z2 1 x3 1 y3 1 z3 = (3k2 ) 2 (3k) 2 z2 1 (zk2 ) 3 1 (zk) 3 1 z3 = z2 k4 .z2 k2 .z2 1 z3 k6 1 z3 k3 1 z3 = z6 k6 1 k3 k6 z3 k6 = z3 (1 k3 k6 ) R.H.S = x3 y3 z3 = (3k2 ) 3 (3k) 3 z3 = z3 k6 z3 k3 z3 = z3 (k6 k3 1) = z3 (1 k3 k6 ) Hence, L.H.S = R.H.S Proved
Vedanta Excel in Mathematics Teachers' Manual - 9 76 11. If a b = b c = c d , prove that: a2 ab b2 = b2 bc c2 = c2 cd d2 Solution: Let a b = b c = c d = k then c d = k i.e c = dk b c = k i.e, b = ck = dk.k = dk2 and a b = k i.e, a = bk = dk2 .k = dk3 Now, L.H.S = a2 ab b2 = (dk3 ) 2 dk3 .dk2 (dk2 ) 2 = d2 k6 d2 k 5 d2 k4 = d2 k5 (k 1) d2 k4 = k(k 1) Middle Term (M.T.) = b2 bc c2 = (dk2 ) 2 dk2 .dk (dk) 2 = d2 k4 d2 k 3 d2 k2 = d2 k3 (k 1) d2 k2 = k(k 1) R.H.S = c2 cd d2 = (dk) 2 dk.d d2 = d2 k(k 1) d2 = k(k 1) 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: Here, a b = b c = 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 13. If x : a : : y : b : : z : c, prove that: ax by (a b)(x y) by cz (b c)(y z) cz ax (c a)(z x) = 3 Solution: Let, x a = y b = z c = k ? x = ak, y = bk and z = ck LHS = ax by (a b)(x y) by cz (b c)(y z) cz ax (c a)(z x) = a.ak b.bk (a b)(ak bk) b.bk c.ck (b c)(bk ck) c.ck a.ak (c a)(ck ak) = k(a2 b2 ) k(a b)(a b) k(b2 c2 ) k(b c)(b c) k(c2 a2 ) k(c a)(c a) = 1 1 1 = 3 = R.H.S Proved 14. If (a2 b2 ) (x2 y2 ) = (ax by)2 , show that: x a = y b Solution: Here, (a2 b2 ) (x2 y2 ) = (ax + by) 2 or, a2 x2 a2 y2 b2 x2 b2 y2 = a2 x2 2abxy b2 y2 or, a2 y2 2abxy b2 x2 = 0 or, (ay bx) 2 = 0 or, ay bx = 0 or, ay = bx or, y b = x a Hence, x a = y b Proved 15. If ay bx p = cx az q = bz cy r , Prove that x a = y b = z c
77 Vedanta Excel in Mathematics Teachers' Manual - 9 Solution: ay bx p = cx az q = bz cy r or, c(ay bx) cp = b(cx az) bq = a(bz cy) ar We know, Each ratio = Sum of antecedents Sum of consequents or, c(ay bx) cp = b(cx az) bq = a(bz cy) ap = c(ay bx) b(cx az) a(bz cy) cp bq ap or, c(ay bx) cp = b(cx az) bq = a(bz cy) ap = 0 or, c(ay bx) cp = 0, b(cx az) bq = 0 and a(bz cy) ap = 0 or, ay bx = 0, cx az = 0 and bz cy = 0 or, ay = bx, cx = az and bz cy or, y b = x a , x a = z c and z c = y b or, x a = y b = z c Hence proved Extra Question 1. If x a = y b = z c , prove that: x3 y3 z3 a3 b3 c3 = xyz abc 2. If a b = b c , prove that: a b b c = a2 (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 b3 c3 d3 = a3 b3 abc b3 c3 bcd 5. If a, b, c, d and e are in continuous proportion, prove that a : e = a4 : b4
Vedanta Excel in Mathematics Teachers' Manual - 9 78 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 1. Knowledge (K) - 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 - 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 Unit 11 Geometry - Triangle
79 Vedanta Excel in Mathematics Teachers' Manual - 9 - 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° Hence, P = 90°, Q = 70° and R = 20°. 3. Find the unknown sizes of angles in the given figure. Solution: i) In 'ABC, A B C = 180° 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° A B C D x x 65° y 35°
Vedanta Excel in Mathematics Teachers' Manual - 9 80 4. In the given figure, PQ = PR and QS is the bisector of PQR. If PQS = 30°, find PSQ. Solution: i) PQS = SQR = 30° [As QS the bisector of PQR] ii) PQR = PRQ = 60° [? PQ = PR] iii) In 'PQR; PQR PRS QPR = 180° 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 'ABC meet at O. Prove that BOC = 90° A 2 . Solution: Here, Given: In 'ABC, the bisectors of ABC and ACB meet at O. To prove: BOC = 90° A 2 Proof: Statements Reasons 1. ABC = 2OBC and ACB = 2OCB 1. Given 2. In 'ABC; ABC ACB BAC = 180° or, 2OBC 2OCB BAC = 180° or, 2(OBC OCB) = 180° BAC ? OBC OCB = 180° A 2 = 90° A 2 2. 3. In 'BOC; BOC OBC OCB = 180° or, BOC 90° A 2 = 180° ? BOC = 180° 90° A 2 = 90° A 2 3. From (ii) Proved 6. In the figure alongside, BP and CP are the angular bisectors of the exterior angles CBP and BCP of 'ABC. Prove that BPC = 90° A 2 . Solution: Here, Given: In 'ABC; BP and CP are the angular bisector of CBE adn BCD respectively. ? CBE = 2CBP and BCD = 2BCP. To prove: BPC = 90° A 2 Proof: Statements Reasons 1. CBE = 2CBP and BCD = 2BCP 1. Given 2. ABC = 180° CBE and ACB = 180° BCD 2. Supplementary angles P Q R S
81 Vedanta Excel in Mathematics Teachers' Manual - 9 3. In 'ABC, ABC ACB BAC = 180° or, (180° CBE) (180° BCD) BAC = 180° or, CBE BCD BAC 180° = 0 ? BAC 180° = CBE BCD 3. Sum of angles of triangle and from (2). 4. A 180° = 2CBP 2BCP ?CBP + BCP = 90° + A 2 4. From (1) and (3) 5. In 'BCP, CBP BCP BPC = 180° or, 90° A 2 BPC = 180° ? BPC = 90° A 2 5. Sum of angles of triangle and form (4). Proved 7. In 'PQR, RX A PQ and QY A PR, RX and QY intersect at O. Prove that QOR = 180° P. Solution: Here, Given: In 'PQR, RX A PQ and QY A PR, RX and QY intersect at O. 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° or, XQR QRX 90° = 180° ? XQR = 90° QRX 2. Given 3. In 'QRY, QRY RYQ YQR = 180° or, QRY 90° YQR= 180° ? QRY = 90° YQR 3. 4. (90° QRX) (90° YQR) P = 180° or, P = QRX YQR 4. From (1), (2) and (3) 5. In 'QOR, QOR QRO OQR = 180° or, QOR P = 180° ? QOR = 180° P. 5. From (4) 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 P Q R X Y O
Vedanta Excel in Mathematics Teachers' Manual - 9 82 9. In the given figure, AM = BM = CM. Prove that 'ABC is a right 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° or, MBA MBC MBC MBA = 180° or, 2(MBA MBC) = 180° or, ABC = 90° 4. From (1), (2) and (3) ? MBA MBC = ABC Hence, 'ABC is a right angle triangle. 10. In the given figure ABCD, 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 CAD 3. Adding (1) and (2) 4. BCD = BAD ABC ADC 4. By whole part axiom Proved 11. In the adjoining star shaped figure, prove that A B C D E = 180 Solution: Given: ABCDE is a star shaped figure. To prove: A B C D E = 180 Proof: Statements Reasons 1. In 'ADQ, A D = DQC 1. Being ext. angle of ' equal to the sum of opposite interior angles 2. In 'BRE, B E = BRC 2. Same as (1) 3. In 'QCR, C DQC BRC = 180 3. Sum of angles of a triangle is 180° A B M C D B C A X A E P Q R S T B C D
83 Vedanta Excel in Mathematics Teachers' Manual - 9 4. C A D B E = 180 ? A B C D E = 180 4. From (1), (2) and (3) Proved 12. In the figure alongside, BE and CE are the angular bisector of ABC and ACD respectively. Prove that: BAC = 2BEC. Solution: 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 1. Given 2. In 'ABC, BAC ABC = ACD 2. Being ext. angle of ' equal to sum of opposite interior angles 3. BAC 2EBC = 2ECD or, ECD = 1 2 (BAC 2EBD) 3. From (1) and (2) 4. In 'BCE; EBC BEC = ECD 4. Same as (2) 5. EBC BEC = 1 2 (BAC 2EBC) or, 2EBC 2BEC = BAC 2EBC ? BAC = 2BEC 5. From (3) and (4) Proved 13. In the given figure, the bisector of ACU meets AU at O. Prove that: COT = 1 2 (CAT CUT) Solution: Given: In 'CAU, the bisector CO of ACU meets AT at O. To prove: COT = 1 2 (CAT CUT) Proof Statements Reasons 1. In 'CAO; COT = OCA CAT ? OCA = COT CAT 1. Being ext. angle ' equal to the sum of opposite interior angles 2. In 'CAU, CUT = UCA CAT 2. Same as (1) 3. UCA = 2OCA 3. Given 4. CUT = 2OCA CAT ? OCA = 1 2 (CUT CAT) 4. From (2) and (3) 5. COT CAT = 1 2 (CUT CAT) or, 2COT 2CAT = CUT CAT or, 2COT = CAT CUT ? COT = 1 2 (CAT CUT) 5. From (1) and (4) Proved A B C D E T C A O U
Vedanta Excel in Mathematics Teachers' Manual - 9 84 14. In the adjoining 'ABC, AY is the bisector of BAC and AX A BC. Prove that: XAY = 1 2 (B C) Solution: Given: In 'ABC, AY is the bisector of BAC and AX A BC. To prove: XAY = 1 2 (B C) Proof: Statements Reasons 1. BAC = 2BAY 1. Given 2. In 'ABC, BAC ABC ACB = 180° 2. Sum of angles of 'is 180° 3. 2BAY B C = 180° ? BAY = 1 2 (180° B C) 3. From (1) and (2) 4. In 'BAX, BAX ABX AXB = 180° or, BAX B 90° = 180° ? BAX = 180° B 90° = 90° B 4. Sum of angles of 'is 180° 5. BAY BAX =1 2 (180° B C) (90° B) or, XAY =1 2 (180° B C 180° 2B) ? XAY = 1 2 (B C) 5. Subtracting (4) from (3) Proved 15. In 'PQR, O is the interior point. Prove that OP OQ OR > 1 2 (PQ QR PR) Solution: Given: In 'PQR, O is the interior point. To prove: OP OQ OR > 1 2 (PQ QR PR) 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 or, 2OP 2OQ 2OR > PQ QR PR or, 2(OP OQ OR) > PQ QR PR ? OP OQ OR > 1 2 (PQ QR PR) 4. Adding (1), (2) and (3) Proved 16. Find the unknown sizes of angles in the following figures. a) D B x y A C E b) X 40° 30° 30° x U y Y Z V A B X Y C P Q R O
85 Vedanta Excel in Mathematics Teachers' Manual - 9 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° 17. In the given figure, AB = AC, BAC = 44° and ACD = 24°, show that BC = CD. Solution: 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 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 = 3 u 3 7 = 2 Hence, x = 2 and y = 3 A C 24° 44° B D P (x+8)cm (3y+1)cm (x+4)cm (y+3)cm Q R T
Vedanta Excel in Mathematics Teachers' Manual - 9 86 19. In the given figure, AB = AC, BD = EC and DAE = 30°. Prove 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. i) ii) iii) In 'ABD and 'AEC AB = AC (S) ABC = ACB (A) BD = EC (S) 1. i) ii) iii) Given Base angles of an isosceles triangle are equal 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 = 180° or, ADE ADE 30° = 180° ? ADE = 75° 5. Sum of angles of triangle is 180°. Proved 20. In the figure alongside, BO and CD are bisectors of ABC and ACB respectively. If BD = CD, prove that 'ABC is an isosceles triangle. Solution: 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) Proved 21. In the given figure PQRS, SP = RQ and RP = SQ. Prove that RT = ST. Solution: Given: In the figure PQRS; SP = RQ and RP = SQ To prove: RT = ST 30° A B C D R Q P S T
87 Vedanta Excel in Mathematics Teachers' Manual - 9 Proof: Statements Reasons 1. i. ii. iii. In 'PRS and 'QRS SP = RQ (S) RP = SQ (S) RS = RS (S) 1. i. ii. iii. Given Given Side common to both 'S 2. 'PRS # 'QRS 2. By S.S.S. axiom 3. PRS = QSR 3. Corresponding angles of congruent triangles are equal 4. 'RTS is an isosceles triangle 4. From (3), base angles are equal 5. RT = ST 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. i. ii. iii. In ' AQX and 'BRX XAQ = XBR (R) QX = RX (H) AX = BX (S) 1. i. ii. iii. XA A PQ and XB A PR Given Given 2. 'AQX # 'BRX 2. By R.H.S. axiom 3. AQX = BRX i.e, PQR = PRQ 3. Corresponding angles of congruent triangles 4. 'PQR is an isosceles triangle 4. From (3), base angles are equal Proved 23. In the given figure BN A AC, CM A AB and BN = CM. Prove that 'ABC is an isosceles triangle. Solution: Given: BN A AC, CM A AB and BN = CM To prove: 'ABC is an isosceles triangle. Proof: Statements Reasons 1. i. ii. iii. In 'MBC and 'NBC BMC = BNC (R) BC = BC (H) CM = BN (S) 1. i. ii. iii. Both are right angles Common side Given 2. 'MBC # 'NBC 2. By R.H.S axiom 3. MBC = NCB i.e, ABC = ACB 3. Corresponding angles of congruent triangle are equal 4. 'ABC is an isosceles triangle 4. From (3), being base angles equal Proved A B C M N
Vedanta Excel in Mathematics Teachers' Manual - 9 88 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. i. ii. iii. In 'QBC and 'PBC BQC = BPC (A) QBC = PCB (A) BC = BC (S) 1. i. ii. iii. Both are right angles Base angles of isosceles 'ABC Common Side 2. 'QBC # 'PBC 2. By A.A.S axiom 3. CQ = BP i.e, BP = CQ 3. Corresponding sides of congruent triangles 4. BCQ = PBC 4. Corresponding angles of congruent triangles 5. OB = OC 5. From (4), base angles of 'OBC are equal 6. BP OB = CQ OC ? OP = OQ 6. Subtracting (5) from (3) 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. i. ii. iii. In 'PAC and 'QAB AP = AB (S) PAC = QAB (A) AC = AQ (S) 4. i. ii. iii. Sides of equilateral triangle APB From (3) Sides of equilateral triangle AQC 5. 'PAC # 'QAB 5. By S.A.S. axiom 6. PC = BQ 6. Corresponding sides of congruent triangles Proved 26. In the figure alongside PABQ and CYX are squares. Prove that: PC = BX Solution: Given: PABQ and ACYX are squares To prove: PC = BX Proof Statements Reasons 1. PAB = CAX = 90° 1. Angles of square are equal X Y A B P Q
89 Vedanta Excel in Mathematics Teachers' Manual - 9 2. PAB BAC = CAX BAC 2. Adding BAC on both sides of (1) 3. PAC = BAX 3. Whole part axiom 4. i. ii. iii. In 'PAC and 'BAX PA = AB (S) PAC = BAX (A) AC = AX (S) 4. i. ii. iii. Sides of square PABQ From (3) 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 Proved 27. In the adjoining figure ABC is an equilateral triangle and BCDE is a square. Prove that: AE = AD Solution: Given: ABC is an equilateral triangle and BCDE is a square 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 ? EBA = ACD 3. Subtracting (1) from (2) 4. i. ii. iii. In 'EBA and 'ACD BE = CD (S) EBA = ACD (A) AB = AC (S) 4. i. ii. iii. Sides of the square BCDE From (3) Sides of equilateral 'ABC 5. 'EBA # ACD 5. By S.A.S. axiom 6. AE = AD 6. Corresponding sides of congruent triangle are equal Proved 28. In the figure alongside, PQRS is a square in which PA and SB intersect at O. If PA = SB, prove that PA and SP are perpendicular to each other at O Solution: Given: PQRS is a square. PA = SB and intersect at O. To prove: PA A SB at O. i.e, SOA = 90° Proof Statements Reasons 1. i. ii. iii. In 'SPB and 'PAQ SPB = PQA SB = PA SP = PQ 1. i. ii. iii. Both are right angles Given Sides of square PQRS 2. 'SPB # 'PAQ 2. By R.H.S. axiom 3. PSB = APQ 3. Corresponding angles of congruent triangles E D A B C S P Q R O A B
Vedanta Excel in Mathematics Teachers' Manual - 9 90 4. SOA = SPO PSO 4. Being ext. angle of triangle SOP equal to the sum of opposite interior angles 5. SOA = SPO OPB 5. From (3) and (4) 6. SOA = SPB = 90° 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 opposite interior angles 4. 2CAD = BCA BCA ? 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 ? PSQ = 3PQS v. From (iii) and (iv) 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
91 Vedanta Excel in Mathematics Teachers' Manual - 9 2. BCE = ACE 2. Given 3. In 'AEC, BEC = EAC ACE 3. Being ext. angle of ' equal to the sum of opposite interior angles 4. BEC = BAC ABC 4. From (1), (2) and (3) 5. In 'ABC, ACD = BAC ABC 5. Same as (3) 6. BEC = ACD 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 Proved 33. In 'ABC, ABC = 90° and M is the mid - point of AC. Prove that: AM = BM = CM. Solution: Given: In 'ABC, ABC = 90° and AM = CM To prove: AM = BM = CM Construction: BM is produced to D such that BM = MD and CD is joined. Proof Statements Reasons 1. i. ii. iii. In 'ABM and 'CDM AM = CM (S) AMB = CMD (A) BM = MD (S) 1. i. ii. iii. Given Vertically opposite angles are equal By construction 2. 'ABM # 'CDM 2. By S.A.S. axiom 3. AB = CD and ABM = CDM 3. Being corresponding sides and angles of congruent triangles 4. AB // DC 4. Being alternate angles ABM and CDM equal from (3) 5. ABC BCD = 180° or, 90° BCD = 180° 5. AB // DC and co-interior angles ? BCD = 90° A B C M A B C M D
Vedanta Excel in Mathematics Teachers' Manual - 9 92 6. i. ii. iii. In 'ABC and 'BCD, AB = CD (S) ABC = BCD (A) BC = BC (S) 6. i. ii. iii. From (3) Both are right angles Common side 7. 'ABC # 'BCD 7. By S.A.S axiom 8. ACB = DBC 8. Corresponding angles of congruent triangles are equal 9. BM = CM 9. From (8), being base angles of 'MBC equal 10. AM = BM = CM 10. Given AM = CM and from (9) Alternative process: (Effective after teaching mid - point theorem Construction: MN // CB is drawn through M Proof Statements Reasons 1. AN = BN 1. In 'ABC, AM = CM and MN // CB 2. ANM = ABC 2. MN // CB, corresponding angles 3. i. ii. iii. In 'AMN and 'BMN AN = BN (S) ANM = BNM (A) MN = MN 3. i. ii. iii. From (1) From (2), both are right angles Common side 4. 'AMN # 'BMN 4. By S.A.S axiom 5. AM = BM 5. Corresponding sides of congruent triangles are equal 6. AM = BM = CM 6. From (5) and given AM = CM Proved Alternative process: Construction: BM is produced to D such that BM = MD and quadrilateral ABCD is completed. Proof 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 ? AM = BM = CM 5. From (1) and (4) Proved 34. In the perpendiculars drawn from any two vertices to their opposite sides of a triangle are equal. Prove that the triangle is an isosceles triangle. Solution: Given: In 'ABC; BM A AC , CN A AB and BM = CN A B C M N A B C M D A B C N M
93 Vedanta Excel in Mathematics Teachers' Manual - 9 To prove: 'ABC is an isosceles triangle Proof Statements Reasons 1. i. ii. iii. In 'NBC and 'MBC, BNC = BMC (R) BC = BC (H) CN = BM (S) 1. i. ii. iii. Both are right angles Common sides Given 2. 'NBC # 'MBC 2. By R.H.S. axiom 3. NBC = MCB i.e, ABC = ACB 3. Corresponding angles of congruent triangle are equal 4. 'ABC is an isosceles triangle 4. From (3), Both angles are equal Proved 35. If the perpendiculars drawn from the mid - point of any side of a triangle to its other two sides are equal. Prove that the triangle is an isosceles triangle Solution: Given: In 'ABC, M is the mid point of side BC, MP A AB, MQ A AC and MP = MQ To prove: 'ABC is an isosceles triangle Proof Statements Reasons 1. i. ii. iii. In 'PBM and QMC BPM = CQM (R) BM = CM (H) MP = MQ (S) 1. i. ii. iii. Both are right angles M being mid point of side BC Given 2. 'PBM # 'QMC 2. By R.H.S. axiom 3. PBM = QCM i.e, ABC = ACB 3. Corresponding angles of congruent triangle 4. 'ABC is an isosceles triangle 4. From (4), base angles are equal Proved 36. Prove that the line joining the point of intersection of two angular bisectors of the base angles of an isosceles triangle to the vertex bisects the vertical angle. Solution: Given: In isosceles 'ABC, AB = AC, OB and OC are bisectors of 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) A B C M P Q A B C O
Vedanta Excel in Mathematics Teachers' Manual - 9 94 4. OB = OC 4. Base angles of 'OBC are equal 5. i. ii. iii. In 'AOB and 'AOC AB = AC (S) OA = OA (S) OB = OC (S) 5. i. ii. iii. Given Common side From (4) 6. 'AOB # 'AOC 6. By S.S.S. axiom 7. OAB = OAC 7. Corresponding sides of congruent triangles Proved 37. In the angular bisector of an angle of triangle bisects the opposite side, prove that the triangle is an isosceles triangle. Solution: Given: In 'ABC, BAM = CAM and BM = CM 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 Statements Reasons 1. i. ii. iii. In 'ABM and 'CDM AM = MD (S) AMB = CMD (A) BM = CM (S) 1. i. ii. iii. By construction Vertically opposite angles 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. Solution: Given: In rectangle PQRS, M is the mid - point of RS. To prove: PQM is an isosceles triangle Proof Statements Reasons 1. i. ii. iii. In 'PMS and 'QRM SP = RQ PSM = QRM (A) SM = RM (S) 1. i. ii. iii. Opposite sides of rectangle are equal Both are right angles 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 M B C D A P Q S R M
95 Vedanta Excel in Mathematics Teachers' Manual - 9 39. In the given figure, prove that QS is the perpendicular bisector of PR. Solution: Given: PQ = QR and PS = RS To prove: QS is perpendicular bisector of PQ Proof Statements Reasons 1. i. ii. iii. In 'PQS and 'QRS PQ = QR (S) PS = RS (S) QS = QS (S) 1. i. ii. iii. Given Given Common side 2. 'PQS # QRS 2. By S.S.S. axiom 3. PQS = RQS 3. Corresponding angles of congruent triangle 4. i. ii. iii. In 'POQ and 'ROQ PQ = QR (S) PQO = RQO (A) QO = QO 4. i. ii. iii. Given From (3) 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. 40. In the figure alongside, ABCD is a square. X, Y and Z are the points on the sides AB, BC and CD respectively. Such that AX = BY = CZ. 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. i. ii. iii. In 'BXY and 'CYZ BX = CY (S) XBY = YCZ (A) BY = CZ (S) 4. i. ii. iii. From (3) Both are right angles 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 P R Q S O A XB D Z C Y
Vedanta Excel in Mathematics Teachers' Manual - 9 96 41. In the given figure, PQ // SR. ST and RT bisect PSR and SRQ 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 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 Proved 42. In the given figure, BA A AC, RQ A PQ, AB = QR and BP = CR. Prove that AC = PQ. Solution: Given: BA A AC, RQ A PQ, AB = QR and BP = CR To prove: AC = PQ Proof Statements Reasons 1. BP = CR 1. Given 2. BC = PR 2. Adding PC on both side of (1) 3. i. ii. iii. In 'ABC and 'PQR BAC = PQR (R) BC = PR (H) AB = QR (S) 3. i. ii. iii. Both are right angles From (2) Given 4. 'ABC # PQR 4. By R.H.S. axiom 5. AC = PQ 5. Corresponding sides of congruent triangle are equal Proved 43. In the figure alongside, OAD = ODA and OBC = OCB. Prove that AB = DC. Solution: Given: OAD = ODA and DBC = OCB To prove: AB = DC Proof Statements Reasons 1. OB = OC 1. ?OBC = OCB 2. OD = OA 2. ? ODA = OAD 3. BD = AC 3. Adding (1) and (2) A B C R P Q A D B C O P T Q S R