www.InyaTrust.com BELTHANGADY TALUK MATHS TEACHERS WORKSHOP MINIMUM STUDY LEVEL QUESTION PAPER Time: 2hour I) TOTAL: SUBJECT: MATHEMATICS 50 1x7=7 Choose the best alternative from following options 1) If Tn = 2n – 1, then S2 = (A) 1 (B) 3 (C) (D) 5 2) The number of diagonals to be drawn in a hexagon (A) 5 (B) 9 (C) 7 (D) 6 3) The subset of sample space is called (A) Trial (B) Event (C) Sample space (D) Sample point 4) The formula to find Standard Deviation by step deviation method (A) ∑ − ∑ (B) ∑ − ∑ xC (C) ∑ − ∑ (D) ∑ − ∑ 5) The formula to find the distance between two points in a plane d = (A) ( − ) + ( − ) (B) ( − ) − ( − ) ( ) ( − ) +( − ) (D) ( − ) − ( − ) , 6) If √3 tan = 1 ,then = (A) (B) 45 (C) 90 (D) 0 7) The degree of the polynomial x2 - 9x + 20 is (A) 1 (B) 3 (C) -1 (D) 2 II. Answer the following 1x5=5 8) What is composite number? A number greater than 1 and not a prime number is a composite number. 9) Draw Venn diagram to illustrate A U (B ∩ C). 10) State Thales Theorem. If a straight line is drawn parallel to a side of a triangle,then it divides the other two sides proportionally. 11) What are the angles in the Major segment? Acute angle 12) What is the formula to find the Volume of a solid hemisphere? V= r3 III. Answer the following 2x 10 = 20 13) Verify Euler’s formula for the given graph 14) 15) 16) 17) N = 7, R = 5, A = 10 N+R = 7+ 5 = 12 A+2 = 10 + 2 = 12 ∴ N+R = A+2 Typed by Yakub S,Ghs Nada,Belthangady. For MSTF Mangalore(Belthangady wing) www.InyaTrust.com Page 1 www.InyaTrust.com 14) Sketch out the field to the following notes from the field book (1 cm = 50m.) To E 350 100To D 300 75To C 250 150 50 To B 150 to F 100 To G 50 From A 1cm = 50m 50m =1cm, 150m = 3cm, 250m = 5cm, 300m = 6cm, 350m = 7cm, 100m = 2cm,75m=1.5cm Typed by Yakub S,Ghs Nada,Belthangady. For MSTF Mangalore(Belthangady wing) www.InyaTrust.com Page 2 www.InyaTrust.com 15) In a circle of radius 4.5cm draw two radii such that the angle between them is700 Construct tangents at the non-centre ends of the radii. 18) Rationalize the denominator and simplify = = = √ √ √ √ √ = √ √ √ √ √ √ √ √ √ √ √ √ = = √ x √ − √ 16) Find the product of √3 x √2 √3 x √2 √3 = 3 = 3 = √3 √2 = 2 = 2 = √2 √3 x2 Typed by Yakub S,Ghs Nada,Belthangady. For MSTF Mangalore(Belthangady wing) www.InyaTrust.com Page 3 www.InyaTrust.com √81x8 √648 17) If U = { 1, 2, 3, 4, 5, }, A = { 2, 4}, B = {3, 4, 5 } Prove that (AUB)1 = A1 ∩ B1 U = { 1, 2, 3, 4, 5, } A = { 2, 4} B = {3, 4, 5 } LHS =(AUB)1 =[{ 2, 4}U{3, 4, 5 }]1 18) How many 4 digit numbers can be formed using the digits 1, 2, 3, 7, 8, 9 without repeating any digit. 19) The number o Thousands Hundreds Tens ones 6 5 4 3 P1 P1 P1 P1 6 5 4 3 =6x5x4x3 = 30 20) Solve by using the formula x2 - 7x + 12 = 0 a = 1, b = -7 , c = 12 ( = )± ( ) ±√ = = ±√ = ± √ = ± √− 21) Write all Trigonometric ratios apply to a sin = cos = = cosec = sec = cot = Typed by Yakub S,Ghs Nada,Belthangady. For MSTF Mangalore(Belthangady wing) www.InyaTrust.com Page 4 www.InyaTrust.com 22) Prove that the line joining the points (4, 5) and (0,-2) is perpendicular to the line passing through (2,-3) and (-5, 1). − Slope = 2− 1 2 1 −2− 5 Slope of first line - m1 = 0−4 −7 m1 = −4 m1 = Slope of second line m2 = 1−(−3) −5−2 1+3 m2 = −7 m2 = m1xm2 = x m1xm2 =−1 Lines are perpendicular ∴ IV. Answer the following 23) Calculate the standard deviation of the following data X f 0-10 7 10-20 10 20-30 15 3 x 2= 6 30-40 8 40-50 10 CI X f d=x-25 fd d2 fd2 0-10 10-20 20-30 30-40 40-50 5 15 25 35 45 7 10 15 8 10 -20 -10 0 10 20 -140 -100 0 80 200 400 100 0 100 400 2800 1000 0 800 4000 50 = ∑ − = − = – . = √ . = 40 8600 ∑ . Typed by Yakub S,Ghs Nada,Belthangady. For MSTF Mangalore(Belthangady wing) www.InyaTrust.com Page 5 www.InyaTrust.com 24) Prove that the tangents drawn from an external point to a circle (a) Are equal (b) Subtend equal angles at the centre (c) Are equal inclined to the line joining the centre and the external point. Given:A is the centre.B is an external point.. BP and BQ are the tangents.AP, AQ and AB are joined To prove : (a). BP = BQ (b). ∠PAB = ∠QAB (c). ∠PBA = ∠QBA Proof: In ∆APB and ∆AQB, [ ∵ Radius of the same circle AP = AQ [ ∵ Radius drawn at the point of contact is ∠APB = ∠AQB =900 perpendicular to the tangent AB = ∴ ∴ ∆APB ≡ ∆AQB (a). AB [ ∵ RHS postulates BP = BQ (b). ∠PAB = ∠QAB [ ∵ CPCT (c). ∠PBA = ∠QBA Typed by Yakub S,Ghs Nada,Belthangady. For MSTF Mangalore(Belthangady wing) www.InyaTrust.com Page 6 www.InyaTrust.com V. Answer the following 4 x 3= 12 25) Draw direct common tangents to two circles of radii 5cm and 2.5cm having their Centers 10cm apart and measure their lengths. 26) Draw the graph of y = x2 + x – 6 x y -4 6 -3 0 -2 -4 -1 -6 0 -6 1 -4 2 0 Typed by Yakub S,Ghs Nada,Belthangady. For MSTF Mangalore(Belthangady wing) www.InyaTrust.com 3 6 Page 7 www.InyaTrust.com 27) Prove that “If the square on the longest side of a triangle is equal to the sum of the Squares on the other two sides then those two sides contain a right angle” In triangle,If a square of a side is equal to the sum of the squares of the other two sides,then it will be a reight angled triangle. Given: In the ∆ABC , AB2+ BC2 = AC2 To prove : ∠ABC = 900 Construction : At B draw AB⟘BC , extend BC to D such that DB = BC Join ‘A’ and ‘D’. Proof: ∆ABD ಯ ∠ABC = 900 [ ∵ Construction ∴ AD2 = AB2 + BC2 [∵Phythagoras theorem But In ∆ABC , AC2 = AB2 + BC2 [ ∵ Given ⇒ AD2 = AC2 ∴ AD = AC In ∆ABD and ∆ABC, AD = AC [ ∵ Proved BD = BC [ ∵ Construction AB = AB [ ∵ Common ∆ABD ≡ ∆ABC [ ∵ SSS Axiom ⇒ ∠ABD = ∠ABC But, ∠ABD +∠ABC =1800 [ ∵ BDC is straight line ⇒ ∠ABD = ∠ABC = 900 Typed by Yakub S,Ghs Nada,Belthangady. For MSTF Mangalore(Belthangady wing) www.InyaTrust.com Page 8
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