Answer : (In a circle the angle between the radii through two points and angle between the tangents at these points are supplementary.
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1 Second Terminal Examination 2016 MATHEMATICS X STD 1. in the figure AD and AB are tangents to the circle with centre at O. If <BAD = 80 0, then find <BOD and <BCD. Mark 2 (In a circle the angle between the radii through two points and angle between the tangents at these points are supplementary.) By this theorem <BOD = = mark Angle subtended is half of the central angle so <BCD = 100/2 = mark 2. What is the slant height of a cone of base perimeter 12 cm and height 8 cm. Mark 2 2 r = 12 2r = 12 r = mark given h = 8 l, r, h are Pythagorean triples, so h = mark 3. The product of first two terms of an arithmetic sequence with common difference 6 is 135. Find the first term. Mark 2 let first term = n n(n+6) = mark n 2 +6n = 135 n 2 +6n-135 = 0 n = 9 or mark 4. ABCD is a parallelogram AB =8 cm, AD =6cm, and <D = find the area of parallelogram. Mark 3 Area of parallelogram = b x h ratio of sides of 1: 3:2 ----mark1 so, h = mark 1 Area = 8 x 3 3 = mark 1
2 5. A(1,2), B(6,4), C(8,9) are vertices of the parallelogram ABCD. Find the co-ordinates of D. Mark 3 O is the mid point of AC The coordinate of O p = (8+1)/2 =9/2 q= (9+2)/2 =11/ mark 1 O is the mid point of BD also So, 9/2 = (x+6)/2 9 = x+6 x = mark 1 and 11/2 = (y+4)/2 11= y+4 y = mark 1 ie co-ordinate of D(x,y) = D(3,7) (6) Perimeter of Triangle ABC =20 cm and AC = BC = 7cm, What is CP? Mark 3 Given 2(x+y+z) = mark1 ie x + y + z = 10 ie 7 + z = 10 ie z = mark1 now x + z = 7 ie x = 4 ie CP = mark 1 (7) Lateral surface area of a square pyramid with base area 196 sq.cm is 700 sq.cm. find its. a) Base edge. b) Slant height. c) Height. Mark 3
3 Lateral surface area of square pyramid = 2al = 700 sq.cm base area = 196 sq.cm a) so, base edge a = 14 cm ----mark1 b) Slant height l = 700/ 2x 14 = 700/28 = 25 cm ---mark 1 c) height h = l 2 -(a/2) 2. = (7) 2. =24 cm mark 1 (8) For a rectangle area is 40 sq.cm and perimeter 28 cm. Find the length and breadth of rectangle. Mark 3 L x b = mark 1 l + b = mark 1 l = 10, b = mark 1 (9) A circle with centre as origin passes through the point (6,0) a) find the radius of the circle b) Write the co-ordinate of A and B Mark 3 a) Radius of the circle 6 cm mark b) Angles of Triangle OCA 30 0, 60 0, 90 0 so AC = 3 cm and OC = 3 3 cm ie co-ordinate of A = A (3 3, 3) mark Triangle ODB have angles 45 0, 45 0, 90 0 so OD =-6/ 2 and BD = 6/ 2 ie co-ordinate of B = (- 6/ 2, 6/ 2) or B (-3/ 2, 3/ 2) mark (10) In triangle ABC, AC =BC =18 cm and < ACB = a) Find the perpendicular distance from C to AB b) What is the area of the triangle? c)what is the ratio of sides of the triangle with angles 30 0, 30 0, Mark 4 a) In triangle ADC angles are 30 0, 60 0, 90 0 ratio of sides are 1 : 3: 2
4 so CD = 9 cm mark b) Area of triangle = ½bh AD = 9 3 AB = 18 3 Area = ½ x 18 3 x mark = 81 x 3 = 243 sq.cm mark c) Ratio of sides of triangle = 18 : 18 : 18 3 = 1: 1: mark (11) Two circles touches the point B. AC and BM are common tangents to the circles a) Prove that M is the midpoint of AC b) prove that triangle ABC is right angled. Mark 4 a ) AM and MB are tangents and therefore equal. CM and MB are also tangents. Ie AM =CM, M is the midpoints mark b) In Quadrilateral APQC let <PAB = x, then <PBA = x, since triangle PAB is isosceless mark Let <BCQ =y then <QBC =y mark and <MAB = 90-x and <MCQ = 90-y since AC is a tangent to both circles. <ABC = 180-(90-x+90-y) = x+y ie considering linear pairs in the line PQ x + (x+y) + y = 180 ie 2 X (x+y) = 180 ie x+y = 90 <ABC = mark (12) Radius of a cylinder made by wax is 6 cm and its height is 8cm. A cone of maximum size is carved out from this cylinder.
5 a) Find the curved surface area of cylinder b) How many cylindrical candles of radius 1cm and height 8 cm can be made using the remaining wax? Mark 4 a) Curved surface area (CSA) of cone =πrl Triangle AOB is a Pythagorean triangle so AB = l = 10cm CSA = π X 6 X 10 = 60π sq.cm mark b) Volume of remaining part in Cylinder = 2/3 πr 2 h = 2/3 X π X 6 X 6 X 8 = 192π cu.cm mark Volume of one candle = π X 8 = 8π cu.cm mark Number of candles = 192π/8π = 24 candles mark (13) (A)Can you cut out a triangle having one angle 37 0 and side opposite to this angle as 9cm from a circular cardboard sheet of radius 14cm? [sin 37 0 = 0.60, cos 37 0 = 0.79, tan 37 0 = 0.75 ] Mark 4 We have a = d SinA mark 9 = dsin 37 ie d = 9/sin mark d = 8/ mark d = Diameter of cardboard is 28 cm. Diameter of cir-cum circle containing this triangle is so easily we can cut out a triangle mark (13) (B) in triangle ABC, AB =14cm, AC =15cm, Sin A=4/5, a) The perpendicular distance from C to AB b) The area of triangle c) The length of BC Mark 4 a) Given Sin A =4/ 5 Hypotenuse is 15
6 that is 4 changes to 12., The perpendicular distance from c to AB = 12 cm b) Area of triangle = ½bh = ½ X 14 X 12 = 7 X 12 = 84 sq.cm c) By Pingalas formula s(s-a)(s-b)(s-c) = 84 sides are 13, 14,15 So BC = 13 cm (14) The point A is at a distance of 13 cm from the centre of a circle of radius 5 cm. PA and AQ are tangents. BC is another tangent. a) Find the length of PA b) What is the perimeter of triangle ABC? Mark 4 a) AP = = 12 b) perimeter of triangle ABC = 12-x +x+x+12-x = =24 cm (15) Construct a triangle of in-radius 3cm and two angles 60 and70. Measure its sides Mark 4 Firstly draw a circle with radius 3cm. Draw a tangent to the circle Make angles 60 and 70 degrees in ends of tangents make sure that next arm is also a tangent to the circle. Complete the triangle or make central angles 120 and 110 in a circle of radius 3 cm and at the ends of the radii draw perpendicular lines. (16) (3,-1) is point on the circle with centre at (6,3) a) What is the radius of the circle b) is the circle cut the Y axis c)find the coordinate of the points of intersection of the circle with x axis. Mark 4
7 a) Radius is distance between these points. Distance = (x 1 -x 2 ) 2 + (y 2 -y 1 ) 2. = ( = 5 radius = 5 cm b) Since x coordinate is 6 cm and radius is 5 cm it will not cut y axis c) Hypotenuse is 5, altitude 3 so base is 4 cm ie the circle will cut on x axis at (2,0) and (10,0) (17) Find the ratio of base edge, slant height, and height of a square pyramid with all its edges equal. Mark 4 All edges equal. So let a = l = h l = e 2 -(a/2) 2 = a 2 -(a/2) 2 ` = 3a/2 h = ( 3a/2) 2 -(a/2) 2. = a/ 2 ratio of base : slant height : height = a : 3a/2 : a/ 2 = 2 : 3 : 2 = 4 : 3 : 2 (18) If in triangle ABC, AC =BC, <ACB = 80 0 and AB =16 cm, then compute the following. a) The perpendicular from C to AB b) The area of triangle ABC c) The length of the sides AC and BC Mark 4 [ Tan 50 =1.19, Sin 50 =0.77, Sin 80 =0.98 ]
8 a) Tan 50 = CD/AD = CD / 8 CD = 8 X 1.19 = 9.52 cm b) Area of triangle ABC = ½bh = 8 X 9.52 = sq. Cm c) Sin 50 = CD / AC = 9.52/ AC AC = 9.52 / 0.77 = cm = BC (19) (A) A ball starts to move along a straight line at a speed of 40 m/s and the speed decreases at the rate of 4 m/s every second. Write the algebra for the distance from the starting point to the ball after t seconds. At what time the ball is 150 meters away from the starting point? At what time the ball is maximum distance from the starting point? Mark 5 a) s = ut + ½at 2. = 40 t + ½(-4)t 2. = 40 t -2t 2. b) When s = = 40 t -2t 2. -2t2 +40t -150 = 0 2t2 40 t +150 =0 t 2-20t + 75 = 0 t = X t=( )/2 t = 15, 5 c) Initial velocity u =40m/s at first second speed =36m/s at 2 nd second speed =32m/s this is an arithmetic sequence 40, 36, 32,... Now question is at what time speed is 0. an = a+(n-1)d 0 = 40+(n-1) x -4 = 40-4n+4 4n = 44 n = 11
9 so at 10 th second the ball travels maximum. (19) )(B) While solving a second degree equation the coefficient of x was written as -5 instead of 5. The solution then found were 2 and 3. Find the solution of actual equation. Mark 5 the solution are all positive. So when changes the x coefficient, then Solution for actual equation is -2, -3 (20) A(-6,5), B(6,10), C(6-4) are the vertices of the triangle ABC a) Find the length of the sides AB, BC, and AC b) Fin the area of the triangle. Mark 5 Using Distance formulae ; AB = 13, AC =15, BC =14 Sides 13, 14,15 area = 84 sq.cm using heros formulae (21) (A) From the top a building one sees another top of taller building at an angle of elevation 70 degree and bottom at an angle of depression 40 degree. a) Draw a rough figure b) Find the height of the taller building. Mark 5 a) b) No other measures given (21)(B) In triangle ABC AD is the median from A. <BAD = 35, AB = 18 cm, AD =12 cm. Find a) The ratio of areas of triangle ADC and ABC b) The perpendicular distance from D to AB c) The length of BC Mark 5
10 described in above problems 22) Draw the figure using the given measurements PA =6 cm and AB = 4 cm a) Find the length of the tangents from P to the circle. b) Construct a tangent from P to the circle c) Construct a square of area 60 sq.cm Mark 5
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