Higher Order Thinking Skill questions TOPIC- Constructions (Class- X) 1. Draw a triangle ABC with sides BC = 6.3cm, AB = 5.2cm and ÐABC = 60. Then construct a triangle whose sides are times the corresponding sides of ABC. Solution: Steps of construction 1. Draw a line segment BC = 6.3cm 2. At B make < CBX = 60 3. With B as centre and radius equal to 5.2 cm, draw an arc intersecting BX at A 4. Join AC, Then ABC is the required triangle 5. Draw any ray by making an acute angle with BC on the opposite side to the vertex A 6. Locate the points B1, B2, B3, B4 on BY so thatb1b2= B2B3= B3B4, 7. Join B3 to C and draw al line through B4 parallel to B3 C intersecting the extended line segment BC at C. 8. Draw a line through C parallel to CA intersecting the extended line segment BA at A Thus A BC is the required triangle. 2. Let ABC be a right triangle in which AB = 6cm, BC = 8cm and ÐB = 90 BD is the perpendicular from B on AC. The circle through B,C and D is drawn construct the tangents from A to this circle. Solution: Steps of construction
1. Draw DABC with BC = 8cm, AB = 6cm and <B = 90 2. Draw perpendicular BD from B to AC 3. Let O be the mid point of BC. Draw a circle with centre O and radius OB = OC. This circle will pass through the point D 4. Join AO and bisect AO 5. Draw a circle with centre O' and O' Aas radius cuts the previous circle at B and P 6. Join AP, AP and AB are required tangents drawn from A to the circle passing through B,C and D. 3. Draw a circle of radius 4 cm. Take a point P outside the circle. Without using the centre of the circle, draw two tangents to the circle from point P. Solution: Steps of Construction: i) Draw a line segment of 4 cm. ii) Take a point P outside the circle and draw a secant PAB, intersecting the circle at A and B. iii) Produce AP to C such that AP=CP iv) Draw a semi-circle with CB as diameter. v) Draw PD perpendicular to CB, intersecting the semi-circle at D.
vi) vii) With P as centre and PD as radius draw arcs to intersect the given circle at and T. Join PT and PT. Then, PT and PT are the required tangents. 4. Construct a triangle ABC in which BC = 6cm, and the attitude through A is 4.5cm. Measure the length of median through A. Write the steps of construction. Solution:-
Steps of Construction: i) BC = 6cm is drawn and is made downwards with BC of any length. ii) is drawn iii) Perpendicular bisector RQ of BC is drawn which cut BC at M. and intersect BE at O. iv) Taking O as centre and OB as radius, a circle is drawn. v) ML = 4.5cm is cut from RQ. vi) A line XY, parallel to BC is drawn through L to intersect the circle at A and A'. AB, AC, A B and A C are joined. ABC and A BC are the required triangle Median AM = A'M = 5.5cm (app.) Solution:- 5. Construct a triangle ABC in which BC = 5cm, and median AD through A is of length 3.5cm. Also, determine the length of the altitude drawn from A on the side BC (Write the steps of construction also). Steps Of Construction: i) BC = 5cm is drawn and is constructed downwards. ii) BX is drawn perpendicular to BY.
iii) Q is drawn perpendicular bisector if BC intersecting BX at O and cutting BC at E. iv) Taking O as a centre and OB as radius, a circle is drawn. v) Taking E as centre and radius equal to 3.5cm, arc is drawn to cut the circle at A. vi) AC and AB are joined vii) AD is drawn perpendicular to BC from A to cut BC at D. viii) By measuring we find that AD = 3cm. 6. Draw a to a equilateral with side 5cm such that each of its sides is 6/7th of the corresponding side of Also draw the circumcircle of. Solution:- Steps Of Construction: i) A ray QX is drawn making any angle with QR and opposite to P. ii) Starting from Q, seven equal line segments QQ 1, Q 1 R 2, Q 2 Q 3, Q 3 Q 4, Q 4 Q 5, Q 5 Q 6, Q 6 Q 7 are cut of from QX. iii) RQ 7 is joined and a line CQ 6 is drawn parallel to RQ 4 to intersect QR at C. iv) Line CA is drawn parallel to PR. ABC is the required triangle. 7. Construct a triangle ABC in which BC = 6cm, and median AD = 5cm. Also construct another triangle BPQ similar to triangle BCA such that the side BP = 3/2BC. Solution:-
Steps Of Construction: i) A line segment BC of length 6cm is drawn. ii) At B, is drawn on downwards. iii) At B, is drawn iv) Perpendicular bisector of BC is drawn which intersect BY at O and BC at D. v) Taking O as a center and OB as a radius a circle passing through B and C is drawn. vi) Taking D as a centre and radius 5cm an arc is drawn to intersect the circle at A. vii) AB and AC are joined. The required triangle is ABC. viii) Taking C as centre and CD as radius an arc is drawn to intersect BC produced at P such that BP = 3/2BC. ix) Through P, PQ is drawn parallel to CA meeting BA produced at Q. x) BPQ is the required triangle similar to triangle BCA. 8. Construct a quadrilateral ABCD in which AB = 2.5cm, BC = 3.5cm, AC = 4.2cm, CD = 3.5cm and AD = 2.5cm. Construct another quadrilateral AB C D with diagonal AC = 6.3cm such that it is similar to quadrilateral ABCD. Solution:-
Steps Of Construction: i) A line segment Ac = 4.2cm is drawn. ii) With A as a centre and radius 2.5cm, two arcs, one above AC and one below AC are drawn. iii) With C as centre and radius 3.5cm, two arcs arc drawn intersecting previous arcs at B and D./li> iv) AB, AD, BC and CD are joined ABCD is the required quadrilateral. v) Taking A as a centre and radius 6.3cm an arc is drawn to intersect AC produced at C. vi) Through C, C B and C D are drawn parallel to CB and CD respectively. AB C D is the required quadrilateral similar to ABCD. ARITHMETIC PROGRESSIONS HOTS (2 MARKS) Q1. If the numbers n - 2, 4n - 1 and 5n + 2 are in A.P., then the value of n Sol. Since Here, n - 2, 4n - 1 and 5n + 2 are in A.P. => 4n - 1 - (n - 2) = 5n + 2 - (4n - 1) => 4n - 1 - n + 2 = 5n + 2-4n + 1 => 3n-n = 3-1 => 2n = 2 => n = 1 Q2. Which term of AP 7.3,6.9, 6.5. is first negative term. Sol. a = 7.3, d = 6.9-7.3 = - 0.4 a n < 0 => a + (n-1)d < 0 7.3 + ( n-1) (-0.4) < 0 ( n-1) (-0.4) < -7.3 n 1 > - 7.3/- 0.4 n >18.2 + 1 n > 19.2 n = 20
Q3. Write the next term of an AP 8, 18, 32,. Sol : a = d = = Therefore a 4 = = Q4. If t arithmetic mean between 3a,and 2a-7 is a+4, then find a. Sol : 5a -7 =2a + 8 5a 2a = 8 +7 3a = 15 a = 5 HOTS (3 MARKS) Q1. The sum of the 4 th and 8 th term of an A.P is 24 and the sum of the 6 th term and 10 th term is 44. Find the first three terms of the A.P. Sol. Let a be the first term and d is common difference According to statement a 4 +a 8 =24 and a 6 +a 10 =44 2a+10d=24 (1) 2a+14d=44 (ii) By subtraction (i) from (ii) we get, d =5, and put the value of d in (i) we get a=-13. There fore A.P is : -13,-8,-3.. Q2. Find the sum of the first 31 terms of an AP. Whose n th term is given by 3 + 2n/3. Sol. Given a n = 3 + 2n/3 Put n= 1, 2, 3 a =11/3, a 2 =13/3, a 3 = 15/3 and so on therefore a =11/3 and d = 13/3 11/3 = 2/3 S n = n/2[ 2a + (n-1)d] = 31/2 [ 2 X11/3 + 30X 2/3] = 31/2[ 22/3 + 60/3] = 31/2 X 82/3 =1271/3 (Ans) Q3. The angles of triangle are in A.P. The greatest angle is twice the least. Find all the angles of the triangle. Sol. Let the three angles of the triangle be x, y & z Where x is the smallest and z is the largest angle Acc. to given condition Z = 2x (1) 2y = x+ z 2y = x + 2x x + y + z = 180 0
Since x, y, z are in A.P. Y = 3 2 x Now put the value of z and y from eq. (i), (ii) in (iii) x 3 2 9 2 x 40 y x 180 80 3 2 0 0 z 2(40 x 2x 180 0 0 x 40 ) 0 60 Hence the angles of 0 0 0 0 are 40, 60 &80 0 (1) Q4. Find the common difference of an AP whose first term is 1 and the sum of the first four terms is one third to the sum of the next four terms. Sol.. Let a be the first term and d be the common-difference of an AP a = 1. 3 (t 1 + t 2 + t 3 + t 4 ) = (t 5 + t 6 + t 7 + t 8 ) Adding (t 1 + t 2 + t 3 + t 4 ) on b/sides of eq. (i) we get, (1) (t 1 + t 2 + t 3 + t 4 ) + 3 (t 1 + t 2 + t 3 + t 4 ) =(t 1 + t 2 + t 3 + t 4 + t 5 + t 6 + t 7 + t 8 ) 4 (t 1 + t 2 + t 3 + t 4 ) = (t 1 + t 2 + t 3 + t 4 + t 5 + t 6 + t 7 + t 8 ) 4S 4 = S 8 - (ii) (1) S 4 = 4 [2 1 (4 1) d] 4 6d 2 first term, a = 1 S 8 = 8 [2 1 (8 1) d] 8 28d 2
According to Question 4 (4 + 6d) = 8 +28d [from eq. in ii] 4d = 8 d = 2 Q5.If the sum of first n term of an A.P. is given =3-4n, then find its nth term. Sol. = 3 + 4n = 3-4(n-1) - = (3-4n) {3-4(n-1)} = (3-4n) (3( - 2n + 1) 4n + 4} = 3-4n {3-6n + 3 4n + 4} = 3-4n {3-10n + 7} = 3-4n - 3 + 10n 7 = 6n 7 Hence required term is 6n-7. Q6. The ratio of the sums of m and n terms of an AP is m 2 : n 2, so that the ratio of m th and n th terms is (2m-1): (2n-1) Sol.:- S m : S n = m 2 : n 2 => By cross multiplication Put d = 2a => 2 MARKS QUESTIONS HOTS QUESTIONS HEIGHT AND DISTANCES FOR BRIGHT STUDENTS Q1. A man is watching from the top of a tower a boat speeding away from the tower. The boat makes an angle of depression of 45 with the man's eye when at a distance of 100 metres from the tower. After 10 seconds, the angle of depression becomes 30. What is the approximate speed of the boat, assuming that it is running in still water? Solution.
Let AB be the tower. Let C and D be the positions of the boat ACB = 45, ADC = 30, BC = 100 m tan 45 = AB/BC =>1=AB/100 => AB = 100 ------(1) tan 30 = AB/BD =>1/ =100/BD ( Substituted the value of AB from equation 1) => BD = 100 CD = (BD - BC) = (100 100)=100( 1) given that the distance CD is covered in 10 seconds. Required speed = Distance Time=10(1.73 1)=7.3 m/s=7.3 3.6 km/hr=26.28 km/hr Q2. To a man standing outside his house, the angles of elevation of the top and bottom of a window are 60 and 45 respectively. If the height of the man is 180 cm and he is 5 m away from the wall, what is the length of the window? Solution.
Let AB be the man and CD be the window the height of the man, AB = 180 cm, the distance between the man and the wall, BE = 5 m, DAF = 45, CAF = 60 AF = BE = 5 m In right AFD, tan45 =DF/AF 1=DF/5 DF = 5------ (1) In right AFC, tan60 =CF/AF =CF/5 CF=5 ------ (2) Length of the window = CD = (CF - DF) [ Substitued the value of CF and DF from (1) and (2)] =5(1.73 1)=5 0.73=3.65 m Q3. The elevation of the summit of a mountain from its foot is 45. After ascending 2 km towards the mountain upon an incline of 30,the elevation changes to 60. What is the approximate height of the mountain? Solution.
Let A be the foot and C be the summit of a mountain. Given that CAB = 45 CB is the height of the mountain. Let CB = x Let D be the point after ascending 2 km towards the mountain such that AD = 2 km and DAY = 30 CDE = 60 In right ABC, tan45 =CB/AB =>1= x AB[ CB = x (the height of the mountain)]=>ab = x------ (1) From the right AYD, sin30 =DY/AD =>1/2=DY/2 ( AD = 2) => DY=1------ (2) cos30 =AY/AD => /2=AY/2 => AY= ------ (3) In right CED, tan60 =CE/DE =>tan60 =(CB - EB)/YB [CE = (CB - EB) and DE = YB)] =>tan60 =(CB - DY)/(AB - AY)[ EB = DY and YB = (AB - AY)] =>tan60 =(x - 1)/(x - ) ( CB = x, DY = 1, AB=x and AY = ) => =(x - 1)/(x - ) => x=2.73=2.7 i.e., the height of the mountain = 2.7 km
3 marks questions Q1. Two persons are on either sides of a tower of height 50 m. The persons observes the top of the tower at an angle of elevation of 30 and 60. If a car crosses these two persons in 10 seconds, what is the speed of the car? Solution. Let BD be the tower and A and C be the positions of the persons. Given that BD = 50 m, BAD = 30, BCD = 60 in right ABD, tan 30 = BD/BA 1/ =50BA BA=50 tan 60 = BD/BC =50 /BC BC=50 / Distance between the two persons = AC = BA + BC =50 +50/ =200/ 3 m the distance travelled by the car in 10 seconds = 200/ 3 m Speed of the car = Distance Time=20/ 3 m/s=20/ 3 3.6 km/hr=24 km/hr Q2. From the foot and the top of a building of height 230 m, a person observes the top of a tower with angles of elevation of b and a respectively. What is the distance between the top of these buildings if tan a = 5/12 and tan b = 4/5 Solution.
Let ED be the building and AC be the tower. Given that ED = 230 m, ADC = b, AEB = a given that tan a = 5/12 and tan b = 4/5 Let AC = h Required Distance = Distance between the top of these buildings = AE in right ABE, tan(a)=ab/be =>5/12=(h-230)/BE =>BE =12(h-230)/5------ ( 1) in right ACD, tan(b)=ac/cd =>4/5=h/CD =>CD = 5h/4------ ( 2) now BE = CD =>12(h-230)/5=5h/4 from (1) & (2) =>48h (4 12 230)=25h =>23h=(4 12 230) =>h=(4 12 230)/23=480 m ------ ( 3) in right ABE AB = (AC - BC)
= (480-230) = 250 m tan(a) = 5/12. Firstly find sin(a) now by using pyth.thm sin(a) = 5/13=>AB/AE=5/13=>AE = AB 13/5=250 13/5=650 m i.e., Distance between the top of the buildings = 650 m Q3. A boy standing on a horizontal plane finds a bird flying at a distance of 100m from him at an elevation of 30 0. A girl standing on the roof of 20 meter high building finds the angle of elevation of the same bird to be 45 0. Both the boy and the girl are on opposite sides of the bird. Find the distance of the bird from the girl. Solution. In right ACB Sin 30 0 AC = AB 1 AC = 2 100 2 AC = 100 AC = 50m AF = (50 20) = 30m In right AFE Sin 45 0 AF = AE 1 30 = 2 AE AE = 30 2 = 30 x 1.414 = 42.42m Q4. Two ships are sailing in the sea on either side of a lighthouse; the angles of depression of two ships as observed from the top of the lighthouse are 60 0 and 45 0 respectively. If the distance between the ships is meters, find the height of the lighthouse. Solution.
In right ABC Tan 60 o = BC h 3 = BC h h = 3 BC In right ABD Tan 45 o = BD h h = BD BC + BD = 200 BC + 3 BC = 200 BC = h = 200(1 3 BC 3(1 1 3 3 3) 1 3 3 3) 200 = 3 3 = 200m height of light house = 200m Q5. The angle of elevation of the top of a tower standing on a horizontal plane from a point D is. After walking a distance d towards the foot of the tower the angle of elevation is found to be. Find the height of the tower. Solution. Let BC = x
AB tan = CB tan = x h x = h tan x = h cot ---------(1) tan = tan = d + x = AB DC CB d h x h = hcot tan x = h cot - d ----------(2) from 1 and 2 h cot = h cot - d h (cot - cot ) = d h = cot d cot GROUP 15 PAWAN KUMAR (KV NO. 1 AKHNOOR) SUMAN KUMAR (KV SARAIKHAS) B B RATHORE (KV HIRANAGAR) SANJAY THAKUR (KV KISHTWAR)