HOTS Level 1 And Level2

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1 HOTS Level 1 And Level2 2 Mark Questions Q1 In the given figure PQ,PR and AB are tangents at points Q,R and S respectively of a circle. If PQ =8 cm.find the Perimeter of triangle P A B Q RRr R S Sol. AQ=AS BR=BS PQ=PR=8cm Perimeter of APB =AP+AB+PB = PQ-AQ+AS+BS+PR-BR =PQ+PR =8+8=16cm Q2. PT is a Tangent to circle with centre O. OT=56cm,TP=90 cm Find OP Sol. A Tangnt to the Circle is perpendicularto the radius at the point of contact So, OT TP Implies OTP is a rt angle Therefore Op 2 =OT 2 + TP = =11236 = OP=106cm Q3. L and M are two Tangents to a circle. How could the centre of Circle is Ascertained

L M Sol. Draw the perpendiculars on the point of contact of both Tangents.Let these perpendiculars meet at a point O then O will be the Centre of Circle 3 Mark Questions Q1.

2 L M Sol. Draw the perpendiculars on the point of contact of both Tangents.Let these perpendiculars meet at a point O then O will be the Centre of Circle 3 Mark Questions Q1.Theorem: The tangent at any point on a circle is perpendicular to the radius drawn to the point of contact. Given: -A tangent AB with point of contact P. To prove: OP AB Proof: Consider point C on AB other than P. C must lie outside the circle. ( A tangent can have only one point of contact with the circle) OC > OP ( C lies outside the circle) This is true for all positions of C on AB. Thus, OP is the shortest distance between point P and line segment AB. Hence, OP AB. Q2. Theorem :Tangents drawn to a circle from an external point are equal in length. Given:- Two tangents AB and AC from an external point A to points B and C on a circle.

(Common side) Hence, AB = AC (By C.P.C.T) C (RHS congr ence r le) Q3. In figure 5, the common tangents AB and CD to two circles with centres O and O intersect in E.

3 To prove: AB = AC Construction: Join OA, OB and OC. Proof:In triangles OAB and OAC, OBA = 90⁰ (Radius OB Tangent AB at B) OCA = 90⁰ (Radius OC Tangent AC at C) In triangles OBA and OCA, OBA = OCA = 90⁰ OB = OC (Radii of the same circle) OA = OA (Common side) Hence, AB = AC (By C.P.C.T) C (RHS congr ence r le) Q3. In figure 5, the common tangents AB and CD to two circles with centres O and O intersect in E. Prove that the points O, E and O are collinear. In figure 5, the common tangents AB and CD to two circles with centres O and O intersect in E. Prove that the points O, E and O are collinear. Sol. By the property of tangents drawn to a circle from an external point, we have 1= 2.(i) 3= 4.(ii) Also AED = CEB (Vertically opposite angles) (iii) Adding (i),(ii) and (iii), we get AED= CEB But ( AED) + ( CEB) = 360, (angles at a point) we must have AED= ½ (360 ) = 180 EO and EO are collinear O, E and O lie in the same line

4 Q4.From the given fig. A Circle touches all four sides of Quadrilateral ABCD. Prove that AB+CD=BC+DA A S P D R Q C Sol. From the fig. AS=AP, SD=DR, PB=BQ, CR=CQ(Tangents) AS+SD+BQ+CQ=AP+PB+CR+DR AD+BC=AB+CD Q5. Prove that the gm circumscribing a circle is a rhombu A M B Q D P C Sol. Given: ABCD gm touching the circle at M,N,P,Q To prove: ABCD is a rhombus Proof: AQ=AM DQ=DP BN=MB NC=PC Adding the above we get AD+BC=AB+CD AD=BC and AB=CD AD=AB=BC=CD It is a rhombus Hot questions for bright students

5 Area related to circles 2 marks Questions Q.1. A race track is in the form of a ring whose circumference is 352m, and the outer circumference is 396m. Find the width of the track? R o R r Sol. Let the outer and inner radii of the ring be R and r metres respectively. Then 2 R = 396 and 2 r = x 22/7 x R = 396 and 2 x 22/7 x r = 352 R= 396 x 7/22 x 1/2 and r= 352 x t/22 x 1/2 R = 63m and r = 56 m Hence, width of the track = (R r ) metres, =(63 56 ) metres Q.2. Sol. A wheel has diameter 84cm. Find how many complete revolution must it take to cover 792 meters. Let r be the radius of the wheel. Then, Diameter = 84 cm 2r = 84 r = 42cm Circumference of the wheel = 2 r cm = 2 x 22/7 x 42 cm = 264 cm = 2.64m So, the wheel comes 2.64 metres in one complete revolutions.

Total number of revolutions in covering 792 metres = 792/2.64 = 300 Hence, the wheel takes 300 revolution in covering 792 metres. Q.3. A chord AB of a circle of radius 15cm makes and angle of 60 0 at the center of the circle.

6 Total number of revolutions in covering 792 metres = 792/2.64 = 300 Hence, the wheel takes 300 revolution in covering 792 metres. Q.3. A chord AB of a circle of radius 15cm makes and angle of 60 0 at the center of the circle. Find the area of the minor segment. (take =3.14) 3=1.73) Sol. We know that area of a minor segment of angle in a circle of radius r is given by: A = { } r 2 A = { } (15) 2 A = { } x 225 cm2 O A = ( ) 225 cm 2 15cm = 225 x cm 2 = cm 2 A B 3 marks Questions Q1. In the given figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region. (Answer = 98 cm 2 ) Solution: BC 2 =AB 2 +AC 2 =

7 BC= 14 2 cm. Area of shaded portion = (Area of semi circle BC as diameter ) [Area of quadrant with AB as radius Area of triangle ABC] = = =98 cm 2 Q2. The area of an equilateral triangle is cm 2. Taking each angular point as center a circle is described with radius equal to half the length of the side of triangle as shown in figure. Find the area of triangle not included in circle.( ) Solution: AREA = cm 2 3 /4 a 2 = a 2 /4= / 3 A=3{60/360x3.14x(a 2 /2)} cm 2 A= cm 2 Required Area =Area of triangle-area of 3 sectors each of angle 60 0 = cm 2 Q3. Find the area of the shaded region in the figure given here, if BC = BD = 8 cm, AC = AD = 15 cm, and O is the centre of the circle.

2 2 Solution: AB= ( AC BC ) Radius= 17/2 cm. Area of Shaded Region = r 2 2xArea of triangle = 3.14 x17/2x17/2-2x1/2x8x15 = 106.87 cm 2 Q4. Find the area of the region shown shaded in the given figure.

8 2 2 Solution: AB= ( AC BC ) Radius= 17/2 cm. Area of Shaded Region = r 2 2xArea of triangle = 3.14 x17/2x17/2-2x1/2x8x15 = cm 2 Q4. Find the area of the region shown shaded in the given figure. Solution: Area of 1 semicircle = 2 1 r 2 = 2 1 x 7 22 x2 2 = 44/7 cm 2 Area of 4 semi circles= 4x44/7 =176/7 cm 2 Area of Small Square =4x4 =16 cm 2 Area of Shaded Region = 14x14 [16+176/7] = 196-[288/7] = = cm 2 Q5. In the given figure, find the area of the shaded design, where ABCD is a square of side 1- cm and semi-circle are drawn with each side of the square as diameter (use )

Solution A I B Let us mark the four unshaded regions as I,II, III, and IV IV II D III C Area of I + Area of III = Area of ABCD Area of two semicircles of each of radius 5 cm.

9 Solution A I B Let us mark the four unshaded regions as I,II, III, and IV IV II D III C Area of I + Area of III = Area of ABCD Area of two semicircles of each of radius 5 cm. = (10x10-2x 2 1 x x5 2 ) cm 2 = ( x25) cm 2 =( ) cm 2 = 21.5 cm 2 Similarly, Area of II + Area of IV = 21.5 cm 2 So, Area of shaded design = Area of ABCD Area of (I+II+III+IV) = (100-2x21.5) cm 2 =(100-43) cm 2 = 57 cm 2. HOTS QUESTIONS LEVEL A vessel is in the form of hollow hemisphere mounted by a hollow cylinder.the diameter of hemisphere is 14 cm and total height of vessel is 13 cm. Find the inner surface area of vessel. Sol. Here the radius of hemisphere = radius of cylinder = r cm =7 cm And height of cylinder, h= 13-7=6 cm of Now, inner S.A of vessel = curved surface area of cylindrical part + curved surface area cylindrical part = (2 2 ) = 2

10 = 2 x x 7 x (6 +7) =2x22x13 =572cm 2 2. The largest possible sphere is curved out of a wooden solid cube of side 7 cm. Find the volume of wood left. Sol. Diameter of sphere carved out = side of cube = 7 cm R= 3.5 cm Volume of cube = a 3 =7 3 =343 cm 3 Volume of sphere carved out = 4/3 = cm 3 Volume of wood left = = cm 3 3. A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m of uniform thickness. Find the thickness of wire. Sol. The volume of the rod = x(1/2) 2 x 8 cm 3 = 2 cm 3 The length of new wire of the same volume= 18 m=1800 cm If r is the radius of cross-section of wire its volume= xr 2 x1800 xr 2 x1800=2 R=1/30 The thickness of wire of cross section is 1/15 cm. =0.67 mm 3 LEVEL A right triangle, whose side are 15cm and 20cm (other than hypotenuse) is made to revolve about it hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of pie as found appropriate). Sol. Let ABC be the right angled triangle such that AB=15cm and AC=20 cm. Using Pythagoras theorem, we have BC 2 =AB 2 +AC 2 BC 2 = BC 2 = = 625 BC=25cm Let OB=x and OA=y

11 Applying Pythagoras theorem in triangles OAB and OAC, we have AB 2 =OB 2 +OA 2 and AC 2 =OA 2 +OC =x 2 +y 2 and 20 2 =y 2 + (25-x 2 ) x 2 +y 2 = 225 and (25-x) 2 +y 2 = 400 {(25-x) 2 +y 2 } { x 2 +y 2 } = (25-x) 2 -x 2 = 175 (25-x-x) (25-x+x)= 175 (25-2x) X 25= x=7 2x=18 X=9 Putting x=9 in x 2 +y 2 =225, we get 81+y 2 =225 y 2 =144 y=12 Thus, we have OA=12cm and OB=9cm Volume of double cone = vol. of cone CAA + vol. of cone BAA = (OA 2 ) OC + (OA 2 ) OB = = 144 (16+9) = cm 3 =3768 cm 3 Surface area of double cone = curved surface area of cone CAA +curved surface area of cone BAA = OA AC + OA AC = =420 cm 3 = = cm 2

12 2. A cistern, internally measuring 150cm X 120cm X 110cm, has cm 3 of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being 22.5cm X 7.5cm X 6.5cm? Sol. We have Volume of cistern =150 X 120 X 110 = cm 3 Volume of water in cistern = cm 3 Volume of one brick = 22.5 x 7.5 x 6.5 = cm 3 Volume of water absorb by one brick = X cm 3 Let n be the total no, of brick which can be put in the cistern without water overflowing then, Volume of water absorbed by n bricks = n x X cm 3 Volume of water left in cistern = ( X ) cm 3 Since the cistern is filled up to the brim Therefore volume of water left in cistern + volume of brick = volume of cistern = X n x = n x X = x (n - ) = x = x = n = = is congruent = The height of a right circular cone is trisected by two planes drawn parallel to the base. Show that the volumes of the three portions from the top are in ratio 1:7:19. Sol. Let VAB a right circular cone of height 3h and base radius r. This cone is cut by planes parallel to its base at points O - and L, such that VL = LO - =h Since triangles VOA and VO - A - are similar V

13 = r1 = = C L D Also, = A O B A r 0 O B = r2 = Let v1 be the volume of cone VCD Then, v1 = r 2 2 h = 2 h = r 2 h Let v2 be t e vol me of fr t m DC. en V2 = (r 2 1+ r 2 2+ r1r2)h V2 = ( + + )h V2= 2 h Let V3 be t e vol me of fr t m. en V3 = (r 2 + r 2 1+ r1r)h V3 = (r )h V3= 2 h Required ratio = V1:V2:V3= r 2 h: 2 h: 2 h=1:7: The radius of the base of a right circular cone is r. It is cut by a plane parallel to the base at the height h from the base. The distance of the boundary of the upper surface from the centre of the frustum is. Show that the volume of the frustum is 2 h.

14 Sol We have, O r 2 and =. In h 2 + =h A B h Therefore Volume of frustum = {r 2 +( ) 2 +r x }h A r O B Volume of frustum = {r 2 +( )+ }h= 2 h. 5. Water in a canal, 6m wide and 1.5m deep, is flowing with a speed of 10km/h. How much Area will it irrigate in 30 min if 8 cm of standing water is needed? Solution: We have, width of the canal =6m Depth of the canal =1.5m Now, length of water column per hour =10km Therefore length of water column in half hour = 5km= 5000m Therefore volume of water flow in 30min= 1.5x6x5000 = 45000m 3 Here, standing water needed in 8cm = 0.08m Therefore area irrigated in 30 min = = = m 2 (1 hectare= m 2 ) =56.25 hectare HOTS ON PROBABILITY (2 marks questions) 1. Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on (i) the same day? (ii) Consecutive days? Solution: The total number of days is 5 and hence both of them can reach the shop in 5 ways. Hence, total number of outcomes = 5 x 5 = 25 They can reach on the same day in 5 ways, i.e. (Tue, Tue), (Wed,Wed), (Thur,Thur), (Fri,Fri) and (Sat,Sat)

15 (i)p(reaching on same day) (ii) They can reach on consecutive days in following 8 ways: (tue,wed), (wed, tue), (wed,thur), (thur wed), (thurfri), (frithu), (fri,sat), (sat,fri) P(Reaching on consecutive days) 2. A die is numbered in such a way that its faces show the numbers 1, 2, 2, 3, 3, 6. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score on the two throws: What is the probability that total score is (i) even? (ii) 6? Solution: Following table shows the sample space: Total number of outcomes = 36 (i)number of even score = 18 Hence, P (even score) (ii)number of times 6 comes = 4

16 Hence, P (score of 6) 3. A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game. Solution: Possible outcomes of 3 tosses of coin 1st = HHH 2nd = HHT or HTH or THH 3rd = TTH or THT or HTT 4th = TTT Total number of events = 8 Number of favorable events = 6 (refer to 2nd and 3rd case) (3 marks questions) 1.Find the probability of having 53 Sundays in 2 1, (i) a leap year (ii) a non leap year (Ans: 7 7 ) Ans: An ordinary year has 365 days i.e. 52 weeks and 1 day This day can be any one of the 7 days of the week. 1 P(that this day is Sunday) = 7 1 Hence, P(an ordinary year has 53 Sunday) = 7 A leap year 366 days i.e. 52 weeks and 2 days This day can be any one of the 7 days of the week 2 P (that this day is Sunday) = 7 Hence, P(a leap year has 53 Sunday) = A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag. (Ans:10) Ans: Let the number of blue balls is the bag be x Then total number of balls is the bag = 5 + x Number of all possible outcomes = 5 + x Number of outcomes favourable to the event of drawing a blue ball = x ( there are x blue balls)

17 x Probability of drawing a blue ball 5 x Similarly, probability of drawing a red ball = 5 5 x According to the answer x 5 5 x = 2 ( 5 x ) x = A number x is selected from the numbers 1,2,3 and then a second number y is randomly selected from the numbers 1,4,9. What is the probability that the product xy of the two numbers will be less than 9? Ans:Numberx can be selected in three ways and corresponding to each such way there are three ways of selecting number y. Therefore, two numbers can be selected in 9 ways as listed below: (1,1), (1,4), (2,1), (2,4), (3,1) Favorable number of elementary events = 5 Hence, required probability = 5/9 4. Suppose you drop a die at random on the rectangular region shown below. What is the probability that it will land inside the circle with diameter 1m? Solution: Area of rectangle = 6 sq m Area of circle Area of rectangle gives the total number of events and area of circle gives the number of favourable events.

18 5. A die is thrown twice. What is the probability that (i) 5 will not come up either time? Solution: Possible outcomes of 2 throws of a die can be shown by following table: 1, 1 1, 2 1, 3 1, 4 1, 5 1, 6 2, 1 2, 2 2, 3 2, 4 2, 5 2, 6 3, 1 3, 2 3, 3 3, 4 3, 5 3, 6 4, 1 4, 2 4, 3 4, 4 4, 5 4, 6 5, 1 5, 2 5, 3 5, 4 5, 5 5, 6 6, 1 6, 2 6, 3 6, 4 6, 5 6, 6 Total number of events = 36 Number of times when 5 does not come up either of times = 25 (ii) 5 will come up at least once? Solution: Number of times 5 comes at least once = 11

CBSE CLASS X MATH -SOLUTION Therefore, 0.6, 0.25 and 0.3 are greater than or equal to 0 and less than or equal to 1.

CBSE CLASS X MATH -SOLUTION Therefore, 0.6, 0.25 and 0.3 are greater than or equal to 0 and less than or equal to 1. CBSE CLASS X MATH -SOLUTION 011 Q1 The probability of an event is always greater than or equal to zero and less than or equal to one. Here, 3 5 = 0.6 5% = 5 100 = 0.5 Therefore, 0.6, 0.5 and 0.3 are greater