CLASS IX MID TERM EXAMINATION ( ) Subject: MATHS SOLUTIONS. Set B2. TIME :3hrs MAX.MARKS: 80


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1 CLASS IX MID TERM EXAMINATION (01718) Subject: MATHS SOLUTIONS Set B TIME :hrs MAX.MARKS: 80 General Instructions:Do not copy any question.make a rough figure wherever needed. Section A contains Q 1. to Q 6 carry one markeach,section B contains Q 7 to Q1 carry marks each. Section C contains Q 1 to Q carry marks each, Section D contains Q to Q0 carry 4 marks each. SectionA Q1. Find the area of a triangle whose Pythagorean triplet is ( 1, 5, 7 ). Solu: 10 sq units. Q.If C = P,then, ABC is congruent to which triangle in order? cm B C Q R 5 cm cm 5 cm A P Solu: RQP Q.Find the angles which differ by and complementary to each other. Solu: 61 0 and 9 Q4.Write a linear equation in two variables in standard form using,, 7, x, y only. Solu: x + 7y = ( or any ) Q5.Find the zero of the polynomial P(y) = y + 1 Solu: y = 1 Q6.Find the value of: Solu:
2 SectionB Q7. If side of a rhombus is 6cm and its diagonal is 8cm.Find its area. Solu: Ar of = s( s a)( s b)( s c) S = = = 10 Ar of = 10 (4)(4)() = 8 5 Ar of rhombus = X 8 5 = 16 5 cccc Q8. In the given figure, AD and BC are two equal perpendiculars to a line segment AB.Show that CD bisects AB. B C O D A Solu: In BOC and AOD B = A = 90, (Given) BOC = AOD (Vertically opposite angles) BC = AD (Given) BOC AOD by AAS Congruence rule. AO = BO by c.p.c.t Q9. In the given figure, AOB is a straight line.find BOC C (5x + 40) (x  60) A O B Solu: (5x + 40) + (x  60) = 180 ( linear pair) 8x  0 = 180 x = 5 BOC = = 15
3 Q10.In the given figure, ABC = ACB and = 4. Use Euclids axiom to show that 1 =. Solu: ABC = ACB (Given) = + But = 4 (Given) 1 = As equals are subtracted from equals, the remainders are equal. Q11. Find the coordinates of the point: (i) which lies on both xx axis and yy axis. (ii) whose ordinate is ( ) and abscissa is 5. Solu: (i) ( 0, 0 ) (ii) ( 5,  ) Q1. The polynomial xxp  k xxp + 7 xx 1 when divided by (xx 1),leaves the remainder.find the value of k. Solu: P(1) = k = k + 8 = k = 5 SectionC Q1. In the given figure E and F are respectively the midpoints of equal sides AB and AC of Δ ABC. Show that BF = CE. Solu: In Δ ABF and Δ ACE, AB = AC (Given) A = A (Common) AF = AE (Halves of equal sides) So, Δ ABF Δ ACE (SAS rule) Therefore, BF = CE (CPCT)
4 Q14. BA AC and DE DF such that BA = DE and BF = EC. Show that ABC DEF Solu: Given, To Prove Proof: In ABC and DEF A = D = 90, (Given) BF + FC = EC + FC BC = EF BA = DE (Given) ABC DEF by RHS Congruence rule. Q15. Prove that if two lines intersects each other,then the vertically opposite angles are equal. Solu: Figure, Given, Toprove and Proof. Q16.Do as directed: (i)find k if the line 4kx y 9k = 7 passes through origin. (ii) What is the equation of yaxis? (iii) Write an equation which is parallel to xaxis at a distance of units from xaxis. Solu: ( i) k = , (ii) x = 0 (iii) y = or y =  Q17.How many solutions of the equation 5 xx + yy = 18 can be found? Is (, 4 ) is one of the solution of the given equation? Solu: Infinitely many solutions. LHS = 5 + ( 4 ) = = 18 RHS = 18 its a solution. Q18.Plot the points A( 0, 4 ), B( 4, 4 ), C( 4,  4 ), D ( 0,  4 ).Name the figure formed by joining these points and find its area. Solu: Plotting of points, Figure formed is rectangle. Area = sq units. Q19.Find: xx + 1, if xxp xx Solu: (xx + 1 xx )P = xxp = 6 + = 64 xx + 1 xx = + 8 xx = 6. 1 xx +. 1 xx
5 Q0.If xx = yy, then show that xxp Solu: Cubing both sides xxp xxp xxp = ( y ) = 8  yyp = 8  yyp 6 yy ( yy)  6xxxx + 6xxxx + yyp  8 = 0 xxp + 6xxxx + yyp  8 = 0 Q1.Represent 9. 6 on the number line. Solu: Construction on a graph paper. Q. Express in the form of pp,where p and q are integers and q 0. qq Solu. Let pp qq = pp qq = (1).() Subtracting (1) from () 99 pp qq = 159 pp qq = SectionD Q. Find the area of a trapezium whose parallel sides are 77 cm and 60 cm and non parallel sides are 5 cm and 6 cm. Solu: Draw BE AD BEC is an isosceles triangle and ABED is a parallelogram. A B D Ar. of = s( s a)( s b)( s c) For ABC, S = 68 = = 4 E C
6 Ar of ABC = 4 (8)(9)(17) = 17x (xx)(x)(17) = 17 x x x = 04 Ar of = 1 b x h 04 = 1 x 17 x h h = 4 Ar. Of trapezium = 1 x (a + b) x h = 1 x (60+77)(4) =1 x 17 = 1644 cm. Q4. AB is a line segment and P is its midpoint.d and E are points on the same side of AB such that BAD = ABE and EPA = DPB. Show that (i) (ii) DAP EBP AD = BE Solu. E D A P B Given,Toprove Proof: (i) In APD and BPE PAD = PBE (given) EPA + EPD = DPB + EPD AP = BP (given) APD BPE byasa congruence rule (ii) AD = BE ( by c.p.c.t)
7 Q5.In PQR, PR>PQ and PS bisects QPR where S lies on QR. Prove that PSR > PSQ. P Q S R Given : In PQR, PR > PQ i.e., 4> 1 and PS bisects QPR To Prove: PSR > PSQ Proof: PR> PQ (given) 4 > 1 (as greater angle has greater angle opp. To it ).(1) Also, 5 = 6.() (given) Adding (1) and (), we get > > [exterior angles of PSR and PSQ] PSR > PSQ Q6. Sides AB and AC of ABC are produced to points E and D respectively.if bisectors BO and CO of CBE and BCD respectively meet at point O,then prove that BOC = BAC Solu:
8 Given, To Prove Proof : CBO = 1 CBE ( as BO is a bisector of CBE given ) CBO + ABC= 1800 (as AE is a line) CBO = 1 ( ABC ) = ABC..(1) Similarly BCO = ACB..() Adding (1) and () CBO + BCO = ( ABC + ACB ) BOC = ( BAC ) ( ASP of ABC and BOC)  BOC= BAC BOC = BAC Q7. The taxi fare in a city is as follows.for the first kilometer,the fare is Rs 8 and for the subsequent distance,it is Rs 5 per km.taking the distance,covered as x km and the total fare Rs y,write a linear equation for this information and draw its graph. Write two values of taxi driver. Solu: y = 8 + 5( x 1 ) 5x y + = 0 y = 5x + x y  7 Graph on the graph paper. Any two values of taxi driver.
9 . Q8. Expand : ( 1 xx + yy ) and find the coefficient of yyp Solu: ( 1 xx + yy ) = ( 1 xx ) + ( 1 xx ) ( yy ) + (1 xx ) (yy ) + ( yy ) the coefficient of yyp = 1 xx + yy xx + yy xx + yy 7 1 = xx Q9. Factorise : xxp  xxp  17xxR + 0. Solu: Let P(xx) = xxp  xxp  17xxR + 0. P() = ()P  ()P 17()R + 0. = = 0 ( ) ia s factor of P(xx) By dividing by long division,we get P(xx) = (xx )( xxp + xx  15 ) = (xx )(xx + )(xx 5 ) Q0. Find the value of a and b such that : Solu = aa + b = aa + b = aa + b 6 = aa + b 6 a = 5 and b = 
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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