SOLUTIONS SECTION A [1] = 27(27 15)(27 25)(27 14) = 27(12)(2)(13) = cm. = s(s a)(s b)(s c)


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1 1. (A) = 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 = 3.36 o =108 o 3. (A) s = a + b + c = = 7 Apply the formula = area of a triangle = s(s a)(s b)(s c) = 7(7 15)(7 5)(7 14) = 7(1)()(13) = 18 6 cm 4. (C) Apply the formula (a b) = a + b ab. (3a 7b) = 9a + 49 b x 3 x 7ab 9a + 49b = (3a 7b) + x 3 x 7ab On substitution, we get 9a + 49b = (6) + x 3 x 7 x 5 = = (A) A nonzero constant polynomial has no zero.
2 6. (B) The highest power in the given polynomial is 7, so degree=7 7. (B) Perimeter of the rhombus = 36cm, hence each side is 9 cm. Area of rhombus = (area of the triangle) = 1 Base Height = Base Height 108 = 9.h h = 1cm 8. (B) x+5y=180 o.30 o +5y = 180 o Thus, y = 4 o SECTION B x = 180 o (Linear Pass) o = 180 o 1 = 180 o 61 o 1 = 119 o Now 1 & y are corresponding angle such that
3 1 = 119 o & y = 118 o i.e. 1 y Therefore, M is not parallel to n. 10. As the point lies on xaxis, its y coordinate is zero. So, it will be of the form (x,0) It is 9 units away from yaxis, so x= 9 Hence its coordinates will be (9,0) Similarly, any point lying on y axis is of the form (0,y) It is 9 units from x axis, so y = 9 Hence its coordinates will be (0,9) = ( 3 4) 8( 6 3) 15 5 ( ) ( 15 ) = =0 1. From the figure we may observe that AC = AB + BC BD = BC + CD Given that AC = BD AB + BC = BC + CD (i)
4 According to Euclid s axiom, we know that when equals are subtracted from equals, remainders are also equal. Subtracting BC from the equation (i), we have AB + BC BC = BC + CD BC AB = CD Or AOC= BOD (vertically opposite angles) BOD = 40 AOD + BOD = 180 o (Linear pair of angles) AOD =140 o BOC =140 o (vertically opposite angles) 13. x 1  a a = x (a + a +1) = x (a+1) = (x+a+1)(xa1) 14. Since (x1) is a factor of p(x) = kx + 1 k, p(1)=0 p(1) = k(1) + 1 k
5 k+1k = 0 k=1 SECTION C 15. Consider ABC and ABD AB is a common side AC = BD BC = AD (given that diagonals are equal) (Opposite sides of a parallelogram are equal) ABC BAD (By SSS congruency criterion) ABC= DAB...CPCT But, ABC+ DAB=180 o (adjacent angles of a parallelogram) ABC= DAB=90 o Next, if the diagonals intersect at O, Consider AOB and AOD AOB = AOD (each 90 o ) AO is the common side BO = OD (diagonals of a parallelogram bisect each other) AOB AOD (By SAS congruency criterion) AB = AD So we can say that ABCD is a square as its adjacent sides are equal and one angle is a right angle.
6 OR Consider triangles CAB and DBA A = B (each 90 o ) AB = AB (common) AD = BC (opposite sides of a parallelogram) Therefore, CAB DBA (SAS) AC = DB (c.p.c.t) Hence diagonals of a rectangle are equal. Consider triangle AOB and BOC AO=OC OB is common but we cannot locate 3 rd point to make triangles congruent. Hence we cannot prove AOB = BOC to make diagonals perpendicular. 16. x x x x 1 1 = x x x x 1 1 = x + x + x x 1 1 = x + x + x x OR
7 x 3 +13x +3x + 0 is a cubic polynomial, so it has three factors of which one is (x+). x + 11x x + x + 13x + 3x + 0 x 3 + x ( ) ( ) 11x 11x + 3x + x ( ) ( ) 10x x + 0 ( ) ( ) 0 [1½] Now, x +11x +10 = x +10x +x +10 = (x+1) (x+10) x 3 +13x +3x + 0 = (x+1)(x+)(x+10) [1½] 17. Here a = 10m, b = 80m, c = 50m a+ b+ c s= = = 15m Area of the park = s(s a)(s b)(s c) = 15(15 10)(15 80)(15 50) = m = 18. Let y = 3k & z = 7k m x = z (Alternate angles) x = 7k x + y = 180 o (Interior angles on same sets of transversal)
8 7k + 3k = 180 o k = 18 o x = 7k = 7 x 18 o = 16 o [1½] y = 3k = 3 x 18 o = 54 o z = x = 16 o 19. Construction: join the centre O to A, B, C, D Consider AOB and COD AO = CO and BO = DO (radii of the circle) AOB= COD (given) AOB COD (SAS congruency criterion) AB = CD (By CPCT) 0. Using Pythagoras Theorem: 5= +1 Taking positive square root we get 5 = ( ) + ( 1)
9 1. We mark a point A representing units on number line.. Now we construct AB of unit length perpendicular to OA. Join OB 3. Now taking O as centre and OB as radius lets draw an arc, intersecting number line at point C. 4. Point C represents 5 on number line [1½] 1. BD = DC = BC/ (As AD is given to be a median) AD > BC/ (given) AD > BD from (i) In triangle ABD, B > BAD... (ii) (The side opposite to the bigger angle is longer) Similarly, in triangle ACD, C > CAD. (iii) From (ii) and (iii), we get, B+ C> A
10 . Let ABC be an equilateral triangle AB = BC = CA AB = BC B = A (Angles opposite to equal side are equal) (I) BC = CA A = B (Angles opposite to equal side are equal) (II) From (i) and (ii) A = B = C (III) A + B + C = 180 o 3 A = 180 o from (iii) A = 60 o A = B = C = 60 o Hence proved 3. (i) 3 3 x = x 3 x.1 3. x = x + x + x [1½]
11 (ii) ( ) 3a 7b c = 3a + 7b + c +.3a. 7b + 7b c + c 3a ( ) ( ) ( ) ( ) ( )( ) ( ) = 9a + 49b + c 4ab+ 14bc 6ac [1½] = = a + 3b ( ) ( ) = = = = a + 3b a =, b = [1½] [1½] Or x = = x + 3 Rationalizing the deno min ator, = x = = 3 [ 1 ] 3 ( ) ( ) ( ) 1 x + = = 4 x 1 x + = 4 = 16 x SECTION  D 5. Let p(x) = ax 3 + 4x + 3x 4,q(x) = x 3 4x+ a
12 The remainder when p(x) and q(x) are divided by x3 are p(3) and q(3) respectively p(3) = q (3) (given) a = a 7a = a [] 6a + 6 = 0 a = DCA = ECB (given) DCA + ECD = ECB + ECD ECA = DCB (I) In DBC & EAC DCB = ECA BC = AC DBC = EAC [from (I)] (given) (given) By ASA criteria DBC EAC DC = EC and BD = AE (CPCT) 7. (a+b+c) =a + b + c + (ab + bc+ ca) 9 = a + b + c = a + b + c + 80 a + b + c = 1
13 8. (a) Let A be the point having coordinates (3, ) and B be the point with coordinates (, 3). Now on plotting, we get the required plot as: The position of point A (3, ) and (, 3) is different on the Cartesian plane. So, (3, ) and (, 3) are the two distinct points. [] (b) Let A be the point having coordinates (1, 5) and B be the point with coordinate (1, 5). Now on plotting, we get the required plot as:
14 The position of point A(1, 5) and B (1, 5) is different on the Cartesian plane. So, (1, 5) and B (1, 5) are the two distinct points. [] 9. 16x 16x n+ 1 4x n n+ n+ x 4 n+ 1 n x x = x x 4 n+ n+ n+ 5 n+ = n+ 6 n+ 3 n+ 5 n+ =.. n+ 5 n+ n+ 5 n+ ( ) n+ 5 n+ ( ) 1 = = Or
15 x = = x = 3 ( x ) ( 3) = + 3 = = ( ) x 4x + 4 = 3 x 4x + 1 = 0 ( ) ( ) 3 x x 7x + 5 = x x 4x x 4x = x = (i) In BAD and CAD ADB = ADC AB = AC AD = AD (each 90º as AD is an altitude) (given) (common) BAD CAD (by RHS Congruence rule) [] BD = CD (by CPCT) Hence AD bisects BC (ii) Also by CPCT, BAD = CAD Hence, AD bisects A. [1½] Or In ABC
16 AB = AC ACB = ABC (given) (angles opposite to equal sides of a triangle are also equal) Now In ACD AC = AD ADC = ACD (angles opposite to equal sides of a triangle are also equal) Now, in BCD ABC + BCD + ADC = 180º (angle sum property of a triangle) ACB + ACB + ACD + ACD = 180º ( ACB + ACD) = 180º ( BCD) = 180º BCD = 90º 31. As PR > PQ PQR > PRQ (angle opposite to larger side is larger)... (1) PS is the bisector of QPR QPS = RPS... () Now, PSR is the exterior angle of PQS PSR = PQR + QPS... (3) Now, PSQ is the exterior angle of PRS PSQ = PRQ + RPS... (4) Now, adding equations (1) and (), we have PQR + QPS > PRQ + RPS PSR > PSQ [Using the values of equations (3) and (4)] 3.
17 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. 33. (i) a 7 b +ab 7 = ab(a 6 +b 6 ) = ab [(a ) 3 + (b ) 3 ] = ab (a + b )(a 4 a b + b 4 ) (ii) x 6 +8y 6 z 6 + 6x y z = (x ) 3 + (y ) 3 + (z ) 3 3 (x )(y )(z ) = (x + y  z ) (x 4 + 4y 4 +z 4  x y +y z +z x ) 34.
18 = (Rationalizing the denominator) = ( 1) ( 1) ( 1) ( + 1) ( 1) ( 1) = = 1 = ( 1) = = 0.414
19
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