SOLUTIONS SECTION A SECTION B


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1 SOLUTIONS SECTION A 1. C (1). A (1) 3. B (1) 4. B (1) 5. C (1) 6. B (1) 7. A (1) 8. D (1) SECTION B = = = = 10. To find: (1)³ + (7)³ + (5)³ Since = 0 Therefore (1)³ + (7)³ + (5)³ = 3(1)(7)(5) = We have q(x) = x x 3 + x 9 Put x = 3, we get q(3) = (3) (3) 3 + (3) 9 [1/] = = 0 Hence, x + 3 is a multiple of p(x). [1/] OR ( x 1) ( x) ( 1) 3( x) ( 1) ( x 1) + = + + +
2 3 3 3 ( ) = 8x x x + 1 = 8x x + 6x = 8x + 1x + 6x Using angle sum property, we have PPR = 180 o (40 o + 95 o ) = 45 o Therefore, STQ = 45 o (alternate opp. Angles) [1/] So, STQ = 180 o (75 o + 45 o ) = 60 o [1/] 13. We may observe that x + y + z + w = 360º (Complete angle) It is given that x + y = z + w x + y + x + y = 360º (x + y) = 360º x + y = 180º Since x and y form a linear pair, thus AOB is a line. 14. The coordinate are: A (, 3) [1/] B (4, 5) [1/] C (5, 6) [1/] D (3, 5) [1/] Section C = [1/] Let x = (i) [1/] 10x = [1/] 100x = (ii) [1/] Subtracting (ii) from (i) gives 99 x = 43 [1/]
3 x = [1/] 16. Rational number: A number that can be expressed in a form p q, where p and q are integers and q 0 is called rational number. Examples: 0.3 and 7. [1/] Irrational number: A number whose decimal representation is nonterminating and nonrepeating is called an irrational number. Example: and [1/] 17. 1x 7x + 1 The two numbers such that pq = 1 1 = 1 and p + q = 7. They are p = 4 and q = 3 Now, 1x 7x + 1 = 1x 4x 3x + 1 = 4x (3x 1) 1 (3x 1) = (3x 1) (4x 1) The zero of polynomial are given by 3x 1= 0 and 4x 1 = 0 i.e. x = 1 3 and x = We use the identity: ( + ) 3 = ( + ) Now (10) 3 = (100 + ) 3 a b a b ab a b. = (100) 3 + () 3 + 3(100) () (100 + ) = (10) = = Given that AC = BC AC + AC = BC + AC (equals are added on both sides) Here, (BC + AC) coincides with AB. We know that things which coincide with one another are equal to one another. BC + AC = AB () We know that things which are equal to the same thing are equal to one another. So, from equation (1) and equation (), we have AC + AC = AB AC = AB 1 AC = AB OR From the figure we may observe that
4 AC = AB + BC BD = BC + CD Given that AC = BD AB + BC = BC + CD (1) According to Euclid s axiom, we know that when equals are subtracted from equals, remainders are also equal. Subtracting BC from the equation (1), we have AB + BC BC = BC + CD BC AB = CD 0. As we know that sum of all interior angles of a triangle is 180º, so for XYZ Now, X + XYZ + XZY = 180º 6º + 54º + XZY = 180º XZY = 180º 116º XZY = 64º OZY = 64 = 3º (OZ is angle bisector of XZY) [1/] Similarly, OYZ = 54 = 7º [1/] Using angle sum property for OYZ, we have OYZ + YOZ + OZY = 180º 7º + YOZ + 3º = 180º YOZ = 180º 59º YOZ = 11º 1. Given that PQ SR and QR is a transversal line PQR = QRT (alternate interior angles) x + 30º = 60º x = 60º 30º x = 30º By using angle sum property for SPQ, we have SPQ + x + y = 180º 90º + 30º + y = 180º y = 180º 10º y = 60º Also, z = x = 30º (alternate opposite angles) OR Let us draw a line XY parallel to ST and passing through point R. PQR + QRX = 180º (cointerior angles on the same side of transversal QR) 110º + QRX = 180º QRX = 70º Now,
5 RST + SRY = 180º (cointerior angles on the same side of transversal SR) 130º + SRY = 180º SRY = 50º XY is a straight line. RQ and RS stand on it. QRX + QRS + SRY = 180º 70º + QRS + 50º = 180º QRS = 180º 10º = 60º. Given: An isosceles triangle ABC, in which AB = AC. To Prove: B= C Construction: Draw the bisector AO of A. Proof: In ABO and ACO AB=AC (Given) AO=OA (Common) BAO= CAO (By Construction ) ABO ΑCO (By S.A.S Congruence criteria) Thus, B = C (By C.P.C.T.) 3. Given: A square ABCD in which AY = BX. To prove: BY = AX and BAY = ABX. Proof: In ABY and BAX, we have, AB = AB (common) AY = BX (given) and ABY = BAX (each angle of a square is a right angle) Therefore, by RHS criterion of congruency, we get,
6 ABY BAX So BY = AX and BAY = ABX (By CPCT) 4. Let s be the semi perimeter of the equilateral triangle of side length k We have, s = 1 ( ) 3 k k + k + k = 3k k 3k k 3k k (s a) = k, (s b) k, and (s c) k = = = = = So, by Heron's formula, we have Area = s(s a)(s b)(s c) square units 3k k k k 3k = sq units = square units. 4 3k Hence, the area of an equilateral triangle of side k is square 4 units. OR For each triangular piece ( ) cm Semi perimeter s = = 60 cm By Heron s formula Area of triangle = s ( s a)( s b)( s c) ( )( )( ) Area of each triangular piece = cm = 60( 10)( 10)( 40) cm = 00 6 cm Since, there are 5 triangular pieces made of two different colours cloth cm So, area of each cloth required ( ) = = cm Section D 5. Using Pythagoras Theorem: 5= +1
7 Taking positive square root we get 5 ( ) ( 1) = + 1. Mark a point A representing units on number line.. Now construct AB of unit length perpendicular to OA. Join OB 3. Now taking O as centre and OB as radius draw an arc, intersecting number line at point C. 4. Point C represents 5 on number line. OR [] 1. Mark a line segment OB = 9.3 on number line.. Take BC of 1 unit. 3. Find mid point D of OC and draw a semicircle on OC while taking D as its centre. 4. Draw a perpendicular to line OC passing through point B. Let it intersect semicircle at E. Length of perpendicular BE = Taking B as centre and BE as radius draw an arc intersecting number line at F. BF is 9.3 i.e point F represents 9.3 on number line 6. (a) = = = = ( ) m a m n Q = a n 3 = = 5 a ( 5 ) n Q a = n a (b) ( ) Q = = = 5 Q 15 = = 5 [1/] 1 m ( ) n m n 1 = 5 a = a where m = 3 & n = 3 3 Q [1/]
8 = 5 = 5 7. Let p(x) = x x + 3x + 0 By hit and trial method p( 1) = ( 1) ( 1) + 3( 1) + 0 = = = 0 As p( 1) is zero, so x + 1 is a factor of this polynomial p(x). Let us find the quotient while dividing x x + 3x + 0 by (x + 1) By long division x + 1x x + x + x + x x + x 3 1x + 3 1x + 1 x x 0x + 0 0x We know that Dividend = Divisor Quotient + Remainder x x + 3x + 0 = (x + 1) (x + 1x + 0) + 0 = (x + 1) (x + 10x + x + 0) = (x + 1) [x (x + 10) + (x + 10)] = (x + 1) (x + 10) (x + ) = (x + 1) (x + ) (x + 10) 8. (a) ( ) ( ) x + y + z 3xyz = x + y + z x + y + z xy yz zx 1 x y z x y z xy yz zx ( ) = = 1 ( x + y + z ) ( x y xy ) ( y z yz ) ( x z zx ) = ( x + y + z ) ( x y ) + ( y z ) + ( z x ) (b) ( x y) x y 3xy ( x y) + = + + +
9 ( ) ( ) x + y = x + y 3xy x + y ( ) ( ) = x + y x + y 3xy ( x y) ( x y xy 3xy) ( x y) ( x y xy) ( x y) ( x xy y ) = = + + = + + Given: In ABC, AD, BE and CF are the altitudes and they are all equal To prove: ABC is equilateral. Proof: In right triangles BCE and BFC, Hypotenuse BC = hypotenuse CB (common ) BFC = BEC = 90 o Side BE = side CF [given] Therefore, BCE CBF [by RHS theorem] Therefore, B = C [corresponding angles of congruent triangles] AB = AC (i) [sides opposite equal angles of a triangle are equal] Similarly, ABD CAF Therefore, B = A AC = BC (ii) From (i) and (ii) we get, BC = CA = AB Hence, ABC is an equilateral triangle. 30.
10 Since BD=BC (Given) This implies that BDC is an isosceles triangle. 1= (Angle opposite to equal sides of triangle are equal) 3= =10 (As exterior angle is equal to the sum of two interior opposite angle). In BDC, =180(Sum angle property of triangle) 1= (As we have proved above) 1= =30 In ADC ACD= ACB + 1= =80. ( ACD is angle of the greatest measure in triangle ADC) Hence, AD > CD. (Since in triangle larger side is across from the larger angle) OR Given: Two triangles ABC and DEF such that B = E, C = F and BC = EF. To prove: ABC DEF Proof: Case I: If AB = DE then in ABC and DEF, AB = DE [by supposition] BC = EF [given] and B = E [given] Thus, ABC DEF [SAS criterion] Case II: If AB < DE Take a point G on ED such that EG = AB. Join GF. In ABC and GEF, we have AB = GE [by supposition] BC = EF [given] [1/] B = E [given] Thus, ABC GEF [SAS criterion] So, ACB = GFE [corresponding parts of congruent triangles are equal] But ACB = DFE [given] This gives, GFE = DFE, This is only possible when FG coincides with FD or G coincides with D. Now, AB must be equal to DE and hence, ABC DEF (by SAS) Case III: If AB > ED With a similar argument (as in case II), we may conclude that
11 ABC DEF (by SAS) Thus, ABC DEF. [1/] 31. Construction: Draw perpendicular PO to the base QR. Given: PA=PB To prove: PQR is an isosceles triangle Proof: PAB= PBA(Angles opposite to the equal sides of triangle are equal) Also, PAB+ PAQ=180 (Linear pair) PBA+ PBR=180 (Linear pair) PAB+ PAQ= PBA+ PBR PAQ= PBR In PAQ and PBR PA=PB (Given) QA=BR (Given) PAQ= PBR (Proved above) PAQ PBR (By S.A.S congruence rule) PQ=PR (By C.P.C.T) PQR is an isosceles triangle. 3. Given a o + b o = 116 o and a o  b o = 4 o a = 140 o a = 70 o Therefore, 70 o  4 o = b b = 46 o [1/] Also, by angle sum property of triangles, we have a o + b o + c o = 180 o [1/] 70 o + 46 o + c o = 180 o c o = 180 o o = 64 o Therefore, angles of the triangle are 70 o, 46 o, 64 o. 33. Let p(x) = x 3 + px + 5x+ q Since (x + 1) and (x  3) are factors of p(x), by factor theorem p(1) and p(3) are equal to zero.
12 Now p(1) = (1) 3 + p(1) + 5(1) + q = 1 + p q = p + q 6 Q p(1) = 0, p + q 6 = 0 or, p + q = 6 ( 1 ) And p(3) = (3) 3 + p(3) + 5(3) + q = p q = 9p + q + 4 Q p(3) = 0, 9p + q + 4= 0 or, 9p + q =  4 () subtracting () from (1), we get 8 p + 0 = 48 or, p = 48 8 =  6 Putting the value of p in equation ( 1), we get :  6+ q = 6 q = = 1 p =  6 and q = [ ]
SOLUTIONS SECTION A [1] = 27(27 15)(27 25)(27 14) = 27(12)(2)(13) = cm. = s(s a)(s b)(s c)
1. (A) 1 1 1 11 1 + 6 6 5 30 5 5 5 5 6 = 6 6 SOLUTIONS SECTION A. (B) Let the angles be x and 3x respectively x+3x = 180 o (sum of angles on same side of transversal is 180 o ) x=36 0 So, larger angle=3x
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