177 Geometry Questions3E LEARNING MATHS BY AMIYA : GEOMETRY To get more follow www.facebook.com/MathsByAmiya www.3elearning.in Join FB Groups Groups www.facebook.com/groups/MBAMathsByAmiya www.facebook.com/groups/MBAMathsByAmiya www.facebook.com/groups/CGLPO To Follow Amiya : https://www.facebook.com/kumar.amiya http://in.linkedin.com/in/kumaramiya Geometry Maths By Amiya, QUESTIONS & Solutions 1. 2. Maths By Amiya, 3E Learning, www.facebook.com/MathsByAmiya ©AMIYA KUMAR 3. 4. https://www.youtube.com/watch?v=NBZawFsTrvc 5. If in a ∆ ABC , AB= 7cm , BC=8cm , CA=12 cm, points D,E,F are on BC,CA, & AB respectively , such that AF=4 cm, CD=2cm and AE=5cm, Point O is on the intersection of line AD and EF, then what would be ratio of area of ∆ AFO to that of ∆AOE ? a. 49:36 b. 36:49 c. 20:21 d. 21:20 e. NoT ans: [e] 144:35 https://www.youtube.com/watch?v=NBZawFsTrvc 6. If in a ∆ ABC , points D,E,F are on BC,CA, & AB respectively , such that AE=3 cm, CD=2cm and AC=7cm & BD= 5cm, CF is angle bisector of ∠ACB. Point O is on the intersection of line DE and CF, then what would be ratio of area of ∆ EFO to that of ∆DOC ? Maths By Amiya, 3E Learning, www.facebook.com/MathsByAmiya ©AMIYA KUMAR a. 2:5 b. 5:2 c. 1:2 d. 2:1 e. NoT ans: [e] 13:4 https://www.youtube.com/watch?v=NBZawFsTrvc 7. If smallest side of an integral right angled triangle is 23 cm then what would be digital sum of perimeter of this triangle. a. 2 b. 3 c. 7 d. 8 e. NoT Sol: [b] 3, sides are 23, 264 , 265 8. If ∆ABC is a right angled triangle with hypotenuse AC=15 cm , points M and N trisect the side AC, then BM^2 + BN^2 =? a. 100 b. 125 c. 175 d. 225 e. Not Sol: [b] 125 , By apolloniusBM^2 + BN^2 =(5/9)*AC^2 => (5/9)*15^2 = 125, or use coordinate, with origin as B and take AB and BC as 12 and 9 cm, then use section formula and find coordinates of M & N and get answer. 9. In a triangle ABC, point F and D on side BC such that that BF:FD:DC = 1:2:3. Point E is on AB and AE:EB = 2:3. If G is mid point of ED, then what would be ratio of area of quad BEGF to that to that AEDC. Ans : 2:7 10. In a triangle ABC , D, E & F are points on BC, CA & BA (resp). If D is mid point of BC, CE=6 cm, EA = 4 cm ; AF : FB = 4:5 and area of quad BDEF is 47 cm^2 then what is the area of triangle DEC. Ans : 27 11. If AD = 18cm , BE= 24 cm and CF= 30 cm are medians of ∆ABC and G is centroid. then what would be area ∆ EFG Ans : 24 , https://www.facebook.com/MathsByAmiya/509385892446865 12. If in a ∆ABC, D is a point on side AB such that AD=4cm ,DB=5 cm and DC = 8 cm, then find the perimeter of ∆ABC if ∠ABC=∠DCA. a. 27 b. 28 c. 30 d. data inadequate based on previous year MBA question e. NoT Ans: 27, sides are 10,12,6 , use similarity , ∆ABC ~∆ACD Maths By Amiya, 3E Learning, www.facebook.com/MathsByAmiya ©AMIYA KUMAR NoT Sol: [b] 12/13 Relation = + 1 1 1 1 13 = + + = 2 3 4 12 12 = 13 + . 11/12 b. Data inadequate e. 12/13 c. where ℎ. If 2.3.13. 13/12 d. a.4 are altitudes of a triangle then what is the inradius of the triangle. NoT 19. . AB=BC. 18. If in a ∆ABC . ℎ &ℎ 14. If in a ∆ABC side BC makes an angle 132° at incentre then angle subtended by side BC at orthocentre is ???? a. 45 c. E is on AC and F is on AD such that ∆DEF is an equilateral triangle &∆ AFE is an isosceles triangle. NoT www. 2:1 c. c. 50 d. If in a ∆ABC . Point D is on BC such that AD is angle bisector and I is incentre then AI:ID = ? a. AB=BC= 6m and CD = 10cm then what is the distance between incentre and centroid of the triangle. 4:13 e. 86 e. 11:10 d. NoT 16. If is orthocentre of ∆ABC. line FG ||AB where G is on BC then ∠FGD ? a. 96 d. 17. AB=7cm . then which point would be orthocentre of ∆OAC. Ans :√13 15. 3E Learning. If in ∆ABC . 104 Maths By Amiya. BC=4cm and CA = 10CM . 100 b.com/MathsByAmiya ©AMIYA KUMAR . D is midpoint of BC. 30 b.facebook. Can't say e. 17:4 b. . so 137 common vertices .3 c. consider the ant is shape zero. Ans: Both are 10° 23. If ∠TAP=20° and ∠TBP = 30° . NoT Sol: [a]They have same common multiple of central angle. If in triangle distance between orthocentre and circum-centre is 9. If PT is tangent to both the circle & secant PA cuts bigger circle at A and C (PA>PC) and touches smaller circle at B.5* √401 e. d. There is a sugar cube of dimension 10cm*10cm*10cm. SoD(137) = 2 [By Sanket] !"# $%&'() *+.2 b.be/JEKl2VN-oNY 24.'. Ans: 6. If a circle touches another circle at point T internally.facebook. Then among options which one could be total distance travelled by the ant..NoT www.10 ∗ √401 c. central angle is 959 = 137*7 & 1781 =137*13 . 3E Learning.9 cm then what is the distance between centroid and orthocentre.20. ( -'. then find ∠BTA =&∠BTC. A regular polygon with 959 vertices and another regular polygon with 1781 sides has a common circumscribe circle if they have some common vertices then digital sum of number of common vertices is a. If an ant start moving with uniformly from one bottom corner to top corner of the same edge along all four faces of sugar cube uniformly in his entire journey. <QBS= 20 &<RAT=50 then <SPM = ? 22.5 d. √401 √0# 1 b.+'* http://youtu.com/MathsByAmiya ©AMIYA KUMAR . by euler line 21.6 . 9 e. a. Sol: [d] Maths By Amiya. sincee mo moves along four sides.com/ om/MathsByAmiya ©AMIYA KUMAR .facebook. Total distance = 10 ∗ √401 25. and for two rotation n its 80. which is created d by joining j the mid points of all 6 surfaces of a cu cubical room of dimension 10m*10m*10m 0m . if Jerry ry a rat ra eats the cheese cube from centre of all the su surfaces in such a way that the rest of the cheese chees cube has three similar air tunnels crossing sing ea each other at the centre of cube and dimension nsion of cross section of each tunnel is 1cm*1cm. www. 702 2 cm^2 cm^ e. 698 cm^2 d. 3E Learning. 600 cm^2 b. Maths By Amiya. then what is total surface area of the rest cheese eese cube. Height gain is height of cube 10cm 0cm but base is 10*4 =40cm in one rotation.2 -4 -44 = 702 26.Open the cube alon along faces . There is a solid cheese cube oof dimension 10cm*10cm*10cm. 1 So total distance = 43 ∗ 4101 + <1$> = 43 ∗ 1$ √16 ∗ 31 + 1 = 10√16 ∗ 31 + 1 = = For n=5 .40 ∗ 3@1 = 10 ∗ :1 + .43@1 for number of rotation =5 . Height gain is 10 but base movem movement is 40*(number of rotation) 67 839 = :101 + . What would be total volume lume ccreated by a 3-D figure..The height gain on each face ce wou would be 10/(4*n) where n is number of rotation tations in ants total journey.faces destroyed = 600 + 40 + 36 + 36 . ing. 5588 cm cm^2 c. total distance = 10√16 16 ∗ 51 + 1 = 10 ∗ √401 Alternate :. c a. NoT Sol : [d] New surface area = old surface face ar area + new faces created . 1:1 b. Then which range best describe N. 30. 500/3 m^2 c. AB and AD & DC respectively. 1:2 c. What would be maximum volume of a cube inside a semi-sphere of radius 10 cm 31. E over line AC. 28. NoT SOl: [e] Actually total mass would be much greater than given mass. 2:3 e.so can remove 729 cubes max. with base dimension 5√2 ∗ 5√2A1 and height 5m. midpoint of diagonal EF.67 m^2 e. 29.so D is the midpoint of GH.com/MathsByAmiya ©AMIYA KUMAR . 6.H are the projections of F. Then GD: DH = ? a. BD.com/MathsByAmiya/photos/745195878865864/ Maths By Amiya.Since I is the midpoint of EF . 150 to 300 e. What would be maximum volume of a cuboid inside a semi-sphere of radius 10 cm https://www.G. 7 to 16 c.a. if the total mass of the bucket with water is 1600 kg then what is the mass of plastic used in bucket. 2:1 d. 60 kg c.facebook.facebook. 166. 1000/3 m^3 d. 6 kg b. NoT Sol: [d] The figure is Bi-pyramidal . Less than 6 b. In a right triangle ABC. so volume = ∗ B C ∗ DEℎ = ∗ 5√2 ∗ 5√2 ∗ 5 = ! 1 1 ! =## ! A! 27. BC . we can cut from the cubical box such that the total surface area of remaining box would same as the original box. I. 8 kg e. DF. There is a solid cubical box of dimension 10cm*10cm*10cm. www. FGand EH are perpendicular to AC. 16 to 150 d. a. 3:2 Sol:[a] 1:1 FBED is a rectangle and diagonals BD and EF intersect at point I. D . If there is a frustum shaped plastic bucketwhose close bottom is a circle ofinner radius 70 cm and the top is a circle ofinner radius 140 cm and height is 100 cm. a. 250/3 m^3 b. If N is maximum number of small cubes of dimension 1cm*1cm*1cm. 3E Learning. More than 300 Sol: [e] Can remove a cube of 9*9*9 from a corner without affecting surface area.6 kg d. DE. NoT (???) Sol: [c] D if AB AB=15cm . www.com/ om/MathsByAmiya ©AMIYA KUMAR . m. Sides of triangle are 3 cm. tr a. 2cm. m. 48 Sol: [b] https://www. 2568 o triangles then add Sol: [c]JoinBD and get two right angled triangles BAD & BAC and find area of the values. b . In a quadrilateral ABCD 90°then its area is .circum radius of triangle angle ((required radius) . Then what at is ra radius of new circle a.. and their radii are a 1cm. BC=144cm. 2.55 cm c. so R = half of hypotenuse. 442 d. 36.facebook. 5 ccm & 4 cm ∗I∗9 3∗5∗4 H= = = 2. ing. 45 b. what would be area of equilateral teral triangle.th . 3.??? c. 1284 d.32.5 9A 4∗∆ 4∗6 R . 2cm & 3cm. & 33cm then. c are sides and ∆ is area ar of triangle Or triangle is a right angled gled triangle . a. NoT Sol: Circle would be circum circle of triangle formed by joining the centres of tangential t circle.33 cm e. CD=145cm and AD= 8 cm with ∠BC8 = 33. If the perimeter of an integral tegral sided triangle is 45 cm then how many different diffe triangles are possible c. Inside an equilateral triangle angle tthere is a point from which the length off perpendicular perp on all the three sides are 1cm.. 3E Learning. 3 cm d. 35.. If three tangential circles If we draw a circle such that tthe centre of all three given tangential circles ircles are on the perimeter of new circle. 34.. 12 ∗ √3 9A^2 d. 2 cm b.. 1224 b.facebook. If paint used by all colour are same and radius of smallest purpule color circle is 1 cm. 9 ∗ √3 cm^2 b 12cm^2 cm^2 c. NoT a. Data ata inadequate ina e. Data inadequate e.then what is the thickness of Maths By Amiya.com/M om/MathsByAmiya/photos/528057913912996/ es touc touches each other from outside. 56 e. NoT (???) a. In a triangle ABC ... D point AC=14 cm & AD= 5 cm m then what is the circumradius of triangle ABC.67*pi c. N NoT Sol[e] Area covered by horse = Area of sector (330°@ with radius 8cm + area at side BC C with radius 2 cm (extra length of rope) !!#° !"#° = ∗ Q ∗ 81 + 1#° !"#° ∗ Q ∗ 21 S common area 97AA3 =61*pi .com/ om/MathsByAmiya ©AMIYA KUMAR . (1+x)cm.1 O 1 By solving these O = √!M 0 . 64*pi . y cm & (y+x)cm . 17*pi d..58.. www. BC If AB= 40cm . b. 38. c. ????? a. then the maximum area covered cover by horse is . + O@1 S 11 + . √!L 0 NoT Sol :[d] Here we have 4 circles with rradius 1cm. + O@1 S 11 @ + . b.61*pi e. if thickness of bbalck colour is same with all circles L!∗√! 0 M!∗√! 0 a. where wher y is radius of yellow circle and x is thickne ickness then according to question Purple = Total Black = Total Yellow . If AB=AC C = 6 and ∠BAC= 30° .balck colour . then y = 2x (x is width of black) Alternate.1 ..1 + O@1 @ N ∗ 11 = N ∗ . There is a triangular park AB horse is threaded outsidee of th the part at vertex A with a rope of length 8 cm such that horse cannot entre inside the triangl triangle due to fence.1 + O@ 1 = .T = L!∗√! 0 yellow is y . A 37. has brick wall on its side.comm oint is on BC such that AD is perpendicular on BC.Take width of yell ABC. e..facebook..T + O@1 S T 1 = T 1 S . √!M 0 d.T 1 S .9 b. Maths By Amiya. ing. 3E Learning.T + O@1 S T 1 @@ = N ∗ . 70 cm d. 5 c. DE. There are how many different is 30 unit a. FG and EH are perpendicularr to AC. if ∠BC[ = ∠ 8B[ = ] then ] =? c. BD. AB and AD & DC respectively.b^2 = 30^2 [h & b must be even] Maths By Amiya. NoT N = 0 According to Sine Formula ula CB CB 2H = ⇒H= 2 ∗ 39 3[ H= 40 2∗ = 0 = 56 9A 39. AC=2x. ing. 66.com/ om/MathsByAmiya ©AMIYA KUMAR . ID = (FG+EH)/2 = (4+2)/2 2)/2 = 3 cm. www. then h^2 . 6 d. By sine law. 14 c. So. so it would be median of trapezium pezium. 7 e. Take DC=xx. 60° e. 13 b. and ID D || to parallel sides. 45° d.facebook.77 cm b. a.NoT e. 3E Learning. NoT Sol: [c] FBED is a rectangle andd diag diagonals BD and EF intersect at point I (say) ay) so I would be midpoint of BD & EF so BD=2*ID. NoT SOl: [e] 5 If 30 is not hypotenuse (take take 330 as perpendicular) . 15 e. FGHE is a trapezium with ith I aas midpoint of its unparallel sides. DF.. 30° b.facebook. so BD = 6 cm. In a right triangle ABC.a.5 b. 15° ∆ABC~∆BDC. 80 cm Sol: [e] 56 cm In fig sin 9 = XY XZ = c. 77. https://www. then BC=√2 x. In a triangle ABC. AC BC .com/M om/MathsByAmiya/photos/550655228319931/ fferent integral sided right angled triangle are possible possi whose one side 41.66 cm e. 12 d. D is on AC such that BD is medianand∠BDC= 45°. 4. If FG G = 4 cm & EH = 2 cm then BD =? a. ] = 30° *+$_ ` = *+$0= √1` 40. c(. d e%&' () ). 1 b. 3E Learning. BC=24 cm & D is midpoint of BC then AI = ? a.com/MathsByAmiya ©AMIYA KUMAR .5 d.30 SO total required triangles = 4+1 = 5 By NitinGuppta Sirhttps://www. If P and Q are mid points of diagonals AC & BD then PQ = ? a. 5cm d. 4cm c. 36 cm^2 b. 3 cm b. 108 cm^2 c.24.5) so 30 as hypotenuse would have only one integral triplet which is 18. Ifexradii of a triangle are 12. If I is incetnre of∆ABC where AB=AC=15 cm . NoT Ans : [c] 45.. NoT Sol: 216 cm^2 1 1 1 1 1 1 =l = + + = O 12 18 36 6 3 ∆ = :Q79 7m 3 3 O = √6 ∗ 12 ∗ 18 ∗ 36 = 216 44.10152541092894878 &type=1 42. then its factor should be crude hypotenuse . 216 cm^2 e.Then find the area of triangle a. Not (???) Sol: [a] Another parallel side would be 8 unit so PQ = (difference of parallel side )/2 = 1 unit 43. 6 cm e. 40 cm e.facebook.com/photo. Cannot be determined e. 20 cm b.4. If area of a trapezium ABCD is 180 unit square whose height is 20 unit and one parallel side is 10 unit. 108√2 cm^2 d. 36 cm d.facebook. 18 & 36 cm. 2 c. 24 cm c. points E & are on sides AC & AB respectively such that AE:AC=2:5 & AF:FB=4:1 and if BO= 15 cm then OE =? . NoT Maths By Amiya.(* () By this case we would get 4 values = ab 1 fgh i jk = 4 [ ] is GIF If 30 is hypotenuse. www. and factor of 30 only crude hypotenuse is 5 (with 3. a. 2. If in ∆ABC . where O is point of intersection of CF & BE.php?fbid=10152541092894878&set=p. 2) . 8:13 c. NoT . 20 d.facebook.com/MathsByAmiya ©AMIYA KUMAR . 5:6 c. 3E Learning. In a ∆PQR . How many triangles are possible whose two altitudes are 6cm & 8cm and other altitude is also a natural number. 1:3 b.in a plane@ are possible in a regular Pentadecagon . 6:11 d. 6:5 b.6 . There are how many distinct equilateral triangles . 1:1. 4:15 e. 100 e. What is the ratio of BO:BE a. 11:8 e. 8/25 d. NoT Ans [e]1 S 1= S 0 n ! = op !1= 47. such that BD:DC=2:3 & AE:AC=1:4. d. 5:11 e. NoT www. 3:1 Ans: [b] Maths By Amiya. 200 d. NoT Ans: [c] 51. NoT Ans: [d] 50. 13 c. points S & T are on PQ . 6/25 e. a. then MN/QV=? a. 18 b. 5:16 c.Ans : [c] 46. 4/25 b.42 <x <24 so total 20 natural x be possible 48. It's given that all the distances between any two points are integer Directions 48-50 : If in a ∆ABC points D & E are on side BC & AC respectively.a 15 sided fig@ such that two vertices of equilateral triangle are also the vertices of Pentadecagon. such that QT=2*PS=2*ST=2 cm. NoT Sol: Altitudes are . 105 c. O is point of intersection of AD & BE then 49. 3:8 b. M & N are point of intersections of line "QV & RS" and "QV &UT" respectively. 21 e.375 d. 210 b. What is area ratio of ∆AOE to ∆BOD a. 8 & x (let) so side ratio is 8x : 6x : 48 By basic property 14O > 48 & 2O < 48 (sum of two sides is greater than third and difference is less than third side) ⇒3. 3/5 c. a. point U is on QR such that QU=3cm & UR = 2cm and point V is on PR such that RV=4cm & PV=3cm.???@ Sol: (c) 2*c(15.2*5 = 200 [ there would be If P. What is ratio of AO:AD a. Q and R are three points on a plane. 4:5 b. 1:5 c. 57. ing. BC=16 cm . O is point of intersection of lines AD & BF then BO:BF =? a. NoT Ans: [b] 81:235 . points D & E aare on side BC such that BD:DE:EC = 2:3:4 such that AQ:QC=3:4. NoT Ans [d] 56.:3:4 & Q is on side AC 52. Points O & P are points of intersections of linee "AD & BQ" and "AE & BQ" respectively. Maths By Amiya. If in a ∆ABC AB=6cm. 4/5 d.com/ om/MathsByAmiya ©AMIYA KUMAR . 28 b. AD is angle bisector of angle ang A .5 d. and oone more point at centre of gravity. Total number of point off inter intersections (PoI) of all diagonals of a regular octagon a.66 e. 33 d. there are 2 A regular polygon makes hexagon. If in a ∆ABC. which are 8-5=3 5=3 . just need to che check points on second octagon from outer side . 4 d. NoT AB=AC=10 cm . 3E Learning. Total number of point off inter intersections (PoI) of all diagonals of a regular hexagon Ans :. P is pointt of intersection int of AD & BE then PE = ? a. BC=7 cm. points D & E lies ies on sides BC & AC respectively. 0. 0. there are 4 octagon. CA= 8cm . and BE is median. 1:3 d. 316 c. In a ∆ABC . 0. 57 e. In an isosceles ∆ABC. 235 e. NoT Ans: [d] 54. 0. which are 6-5=1 . If OP:BQ :BQ = m:n then m+n = ? (m & n are co-primes) a. E lies on AD such that AE:AD A = 1:3 & F lies on CE such that CF:CE :CE = 1:4 . D lies on BC C suc such that BD:BC=1:2 . 64 c.facebook.√106 @/5 e. just need to check heck ppoints on second hexagon from outer side . 1:4 e. I and G are incentr ncentre& centroid then 55. if AB= what is the distance between ween I & G a. 2212 b. www.2*6 + 1*6 + 1 = 19 es sam same fig with PoI of diagonals.33 b.. 4 = n c. and one more point at centre of gravity.25 c.. sum=316 53. 2cm b. NoT Sol: 4*8 + 3*8 + 1 = 57 A regular polygon makes es sam same fig with PoI of diagonals. AD is median. x² + y² + z² = 2(R² + r² ) e. 59. 3E Learning. 5. x² + y² + z² = R² . 7 cm c. point A and C are ccentre of these two circles. Data ata ina inadequate e.r² ) c. NoT Sol: [c] Sides of triangle are. In the fig. NoT Maths By Amiya. 14 cm d. ing. ∠BAD AD = 50° &∠CAD= 80° then AD =? b. so AD=AE = 7 cm. and ∆ADE E is an isosceles ∆. x² + y² + z² = R² + r² d. 8 d.r² b. by Heron's formula. AB is chord of biggerr circle 61.facebook. 45° d. x² + y² + z² = 2(R² . D & E lies on the each circle and these two are point of intersections of AC and circle circles. NoT a. www.NoT Sol: ∠DBE 180 ∠BDE BED 180 1/2(180 ∠BE ∠A) 1/2(180 ∠C) 1/2(∠A ∠C) 45 60. There are two concentric which cuts smaller circle le at B and C. 1 b.If “r” & “R” are radii of smaller and bigger circles respectively ely and AB = x .30° c.com/ om/MathsByAmiya ©AMIYA KUMAR . In ∆ABC. point B lies on the bot both the circles such that ∠ABC=90° . Data inadequate equate e. by MPT AB||DE . (MPT) ic circ circles with centreO. point oint D is on BC such that. AC=14cm.12&13 12&13. then ∠DBE=? a. EB is a chord of smaller circle which is perpendicular on AB AB. If the area and perimeterr of a triangle have same numerical value which side of triangle is 13 unit it then what is the different between longest side and an smallest side of this triangle. 12 cm Sol: Draw DE (where E is midpoin idpoint of AC). 20° b. a.ich is 30 and the longest 58. BD= y & BE = z then which one is a correct relation a. Cant Say e. 7 c. 9:8 Maths By Amiya. What would be ratio of side oof a largest square inside a regular hexagon to that of regular hexagon a.47 c.facebook. 1 √"M√1 d.com/M om/MathsByAmiya/photos/584494198269367/ https://www.5^2 .6. !M√! d.com/M om/MathsByAmiya/photos/584494198269367/ gents from a outside point P on a given circle.com/ph om/photo. 26 cm b.r^2) = (pi^2)*(1.0 cm a. 3:4 c. If PQ & PR are two tangents AB on the same circle touche touches circle at C and points A & B lie on thee tangents tang PQ & PR respectively. ing.facebook. What is the approxcurvee surfa surface area of an uniform circular ring in cm^2 ^2 (wedding ring) if its inner diagonal is 2. What would be ratio of area oof pink coloured shape to that of blue coloured shape a.NoT Sol: Curve Surface Area = (pi^2) ^2)*(R^2 . What would be ratio of side of a largest regular hexagon inside a square √"L√1 1 a. 3 + √3 b.0cm and oouter diagonal is 3. 12.744000315659641&typ e=1 are to that of square 64. e. 3E Learning.93 b. 21 cm Sol:[b] A+PB+AB = PA+PB+AC+BC= PA+PB+AQ+BR= +BR= PQ + PR = 24cm Perimeter of ∆PAB = PA+PB 66.34 cm^2 cm 63.15. If Radius of the circle is 5 cm and CP is 8 cm then whatt is perimeter pe of ∆PAB a. www. Not Sol: [b] https://www.34 .facebook. 30 cm d. √"M√1 1 b.78 e. 4:3 b. 1 √"L√1 c. e.com/ om/MathsByAmiya ©AMIYA KUMAR .php?fbid=744000315659641&set=p. e. 1:1 c.facebook.Sol: [d] Hint:.Use Pythagoras & EC is diameter of smaller circle and AB=CD 62.744 =p. NoT N Sol: [b] https://www.34 d. 24 cm c. 3.1^2)= 12. And another tangent 65. 3 S √3 !L√! c. white area inside small circles.com/MathsByAmiya ©AMIYA KUMAR .facebook. www. 3√13 b.Sol: [c] Let side of the square : 4r . B &C are lines L. 2√14 c. 2 √13 d. Let L . M & N are three parallel lines . Points A. 3 √14 Sol: [c] Maths By Amiya. Then Total area of 4 small circles 4*(pi*r^2)=4(w + blue area) Where w . Now Area of big circle =>4*pi*r^2=4(w + pink area) => Pink area = Blue area. such that the perpendicular distance between LM & MN are 2cm &5 cm respectively. 67. M is in the middle on two. M & N respectively such that ABC form an equilateral triangle then what is the length of ∆ABC a. 3E Learning. facebook. Sol: b. 5. 3 e. Never Possible c. ∠ABC= . If in a triangle ∆ABC. NoT tan = cot ⇒ = 3 ∗ Q ± 45° Maths By Amiya. such that tan = cot and cosec S sec > 1 then how many integral value of k would be possible a.68. Not Possible for real K d. www. 3E Learning.com/MathsByAmiya ©AMIYA KUMAR . 20° b. If two circles of equal radius intersect at two different points C & D. isosceles triage property. such that∠EAB = 20° then ∠EBA = ? a. d. www.com/MathsByAmiya ©AMIYA KUMAR . C & D are in a plane such that AB=BC=CA=DA. 3E Learning. If A & B are centre of these two circle. 100° e. positive for = 3 ∗ Q ± 45° .facebook. There are four points A. ±1 69.NoT 70. 60° b. & line AC cuts other circle whose centre is B at E.cosec S sec > 1 . 80° Ans: [e]120° . 150° Sol: [d] http://youtu. where cosec S sec = √2 S S√2 = 2√2 > 1 = = 0. 120° c. B . If Line AD intersect Line BC at E such that ED:AE = 1:10 then ∠BDC = ? a. 70° c. 135° d. it shows angle is in second quadrilateral.be/Ouw21pvOyFA Maths By Amiya. we can safely assume ume triangle t ABC to be equilateral. 2 :9 b. we have y = . . Now the non-overlapping areas reas th that appear similarly placed about the three ree vertices ve will equal in area. .wikipedia. BY : YE and CZ : ZF is 3 : 4. 3E Learning. as shown s in figure. ¡L¡¡L=¡ . . http://en. 1 2O + T 1 = ⟹ 3O = . This will reducee our work considerably.1∗1L1L@. ¡ n Alternatively :.facebook.com/ om/MathsByAmiya ©AMIYA KUMAR .1∗1∗1M @h ¡ ¡L!`L! =¡ ! . www. 71. E and F are th Further. We need to find the ratio of . O= O + 2T + 2 3 = = Z ! 0 .XYZ@ X. 3 : 7 d. 4 : 9 Sol:[b] Since there is a unique answer wer (as the options suggest.XZ@ X. D. we have X.1∗1L1L@.X@ = .. . = YZ Y X. . `L L¡ = ⟹ 4O O + T = 3.XY@ Also. Assuming the areas as x. IF in ∆ABC.1∗1L1L@ = n By Routh's theorem.the points of trisection of respective sides. here re x=y=z = 2. 1 : 7 c. Find ind the ratio of the areas of triangle XYZ and triangle gle AB ABC a. each of the ratio io AX : XD.org/wiki/Routh's ki/Routh's_theorem Maths By Amiya. ! 0 Plugging x in terms of z. y and z ass shown show in the figure. ing. 0123 9A H4 73.com/MathsByAmiya/photos/528057913912996/ ph 0p ² = 6.5 ⇒ 552 < µ1 < 600 ⇒23.49 Maths By Amiya.6} By Formula :Total number of integral triangles = Nearest Integral value of ± https://www. There are how many different integral sided triangle possible whose perimeter is 18 unit a. 1.5<±0p² < 12.{4.{5.5<± ´h . 8 If P is even then 11.8}.{6. 1.{4. there are 12 different integral sided triangle are possible.49<.23456790 … A = 0.66 e.µ + 3@1<24. then what is the sum of digits of P ? Sol:[b] a.8}.com/MathsByAmiya ©AMIYA KUMAR . 2.facebook.6.8.7.5 ⇒ 552 < .6. NoT Sol:[e] This is case of mutually tangent circle .8}.µ + 3@1 < 600 ⇒23. 7 d.49<P<24.234 c. 3E Learning.facebook. www.c@ 7 By manual process: Different triangles are {2. 0. 6 c.72.7}.R3@ [Descartes' theorem] 1 1 1 1 1 = + + 2 ∗ ª« ¬ « ¬ H1 H2 H4 H1 H2 1 100 100 100 100 = + + 2 ∗ ª« ¬« ¬ 3 A H4 4 6.6. For given perimeter P unit of triangle .123 b.5 b. 7 d.25 4 6.5.75 = 7 74. 8 Sol: . 5 b.25 cm then what would be radius of smallest circle in cm.7.33 d.49 ⇒P = 24 If P is odd then 11.{5.7}.{3. looking for radius of smallest circle .´L!@h 0p ² < 12. 6 c. If three circles and a straight lines touches each other at six different points &radius of two bigger circle is 4cm and 6.25 1 = 81 3 A ⇒ H4 = 1.8}. a. R .4. www.4 5.. it should be multiple ple of 3.com/MathsByAmiya/photos/528057913912996/ 12996/ D is a sq square. 75.given@@ so sides w would be 1.8*.3.2 = 19.7*14 *14 = 952 cm^2 76.44 d. NoT .6 cm. 994 cm^2 d.8 .2 & 9. and common difference = 1.facebook. Sol: [c] a. Can Cannot be determined e. 19. ing. What would be area of right aangled triangle whose sides are in AP and difference d of longest side and smallest allest side is 3. So by solving w we will get R=35 cm So side of square = 70 cm Area of Shaded Region = Area of square .Area of circle Area of Rectangle = 70^2 . 10. Required area = .44 Maths By Amiya. 21. NoT Since sides are in AP of right angl angles triangle so . 3E Learning.com/Math
[email protected]/7@*35*35 . In the given fig. 900 cm^2 b.???@ Sol: [e] ∆FOP .4 * 7. ABCD A rectangle DEFG with ith EF=7c EF=7cm & FG=14cm. SO in ∆FO are in Pythagoras. whose sides ED & DG lies on side of square & one vertex F is on the circle. so sum of digits of SO only 24 Unit perimeter P is 6 https://www. circumscribing a circle . R-7 and R-14 Construct the same fig.5@ = 5. 7.2 e. 43.8 b.1/2@*5.facebook.4. Then find the area of sh shaded region [take pi=22/7] a.6 c. 936 cm^2 c.com/ om/MathsByAmiya ©AMIYA KUMAR .There is no such odd P ter we wil will get 12 different integral sided triangle.5 . facebook. 1980 d. CAbE .com/MathsByAmiya/photos/604561656262621/ 78. 3E Learning. x=20 Maths By Amiya. case SAS@ So FE=CE and ∠CFE= ∠ECF =30 In ∆½¼[ . 2700 e. www.facebook. There are how many different integral angled triangles are possible a.com/MathsByAmiya ©AMIYA KUMAR . NoT Sol: [d] https://www. Since triangle ABC is isosceles so ∠ABC=∠ACB=80 And∠EBF=40 and ∠FCE=30 ∆¼B½ ≅ ∆[B½ By SAS Triangle DfAE congruent to Tri.30 + x=50.77. 180 b.a@ . ∠CFE + x=50 . Ans: . 1800 c. 150° 80. ing. 30° b. 45° b. NoT 81.Direction for questionss 78 to 8 84 4:-There are four congruent circles les with centres ce A. ∠APD =? a. 75° d. 160° c.com/ om/MathsByAmiya ©AMIYA KUMAR .B. 60° c.facebook. ∠CPD =? a. C & D as shown in the fig then 79. 90° e. 75° d. 45° e. 3E Learning. NoT° Ans: [e] 150 Ans: [c] 75 Maths By Amiya. ∠BPC= ? a. 60° c. www. 130° b. 175° d. 105° e. 5° 84. 105° e. 7* sin45 Maths By Amiya. 27. 45° e. 30° b. 30° b. 15° c. 150° Ans: [b]60 Ans : [a]30 Ans: [d] 105 Ans: [e]150 edian & D lies on side BC such that ∠BAD = 80 & ∠DAC=20 ∠D and 86. 22. 75° d.com/ om/MathsByAmiya ©AMIYA KUMAR . 3E Learning.facebook. ∠PAQ =? a. ∠QPD =? a. 105° e. ∠SBP =? a. 60° c. 30° b.5° e. ing. NoT 83. c.Ans: [b] 60 82. 75° d. NoT www. If in a ∆ABC . 20° d. 7* sin90 d. 60° c. 7* sin60 d. AD is median AC= 14 cm then AD =? a. NoT 85. 75° d. 7* sin30 Sol: [d] 7 b. 30° b. 60° c. ∠QSC =? a. ∆PQR@: ar.87. CE:AE= 2:3 & AF:AB=3 =3 : 7 the then what would be ratio of area of quadrilateral quadrilat BDEF to area of triangle AFE a. points D D.31 d. www. ossible.sine formula@ Maths By Amiya.∆PQR@: ar. If two altitudes of a triangles aare 3cm and 4cm and magnitude of another anothe altitude is a natural number in cm m then ho how many different triangles are possible. ing. 30 e.then then ar.. 55: 29 d. 14:3 c.com/ om/MathsByAmiya ©AMIYA KUMAR . 29:55 b. 10 e.Q & R are o on the extended sides of BC. 9 d. 13:3 d. AC & AB such uch that BD:DC=1:2 B . 4:1 e.13 b.be/JEKl2VN-oNY 88. NoT Sol: [b] HCF (527. 40° d. 36° b. 8 c. BC=2*CP & 2*CA=3*QA 3*QA . NoT Ans: [d] =BC=CD=DE=EA then ∠BFD = ? 89. such that AB:BR =1:1 .17 c. 5 b. 27:50 c.NoT Ans : [b] 72 90. NoT Ans: [b] .sine formula@ 91. By ratio of side and aarea . A regular polygon with 527 vertices and another regular polygon with 221vertices 221 has a common circumscribe circle iif they have some common vertices thentotal ntotal number of common vertices are a. If in the given fig AB=BC=CD= a.facebook.E & F are on sides BC. If in the given fig P. 72° c. a. 221)=17http:/ http://youtu. If in a triangle ∆ABC . NoT Ans: [d] By ratio of side ide and area .∆ABC@= ? a. 11:3 b.80° e. 3E Learning. 50:27 e. CA & AB. whose length = 2*pi*. NoT Sol: [b] Area of entrancement = .E & F lies on side BC.14+7@ = 2*. NoT Ans : [d] By MPG AI:IG:GD = 3:1:2 93. If Centroid "G" lies on AD and points E & F are mid points of their respective sides. 2.*M. If the distance between meeting point and home of three antsare39 cm. 2.10@ = 120 m^2 [Shape is of an trapezoid] If we cut the entrancement and open wide it would be a prism . 2:1 c. If in a ∆ABC .if all of them cover distances in straight line and ants which covers distance 39 cm and 52 cm live in opposite corners@ a. where a.92.112 * 10^7 lit b.584 * 10^7 lit c. then what would be the maximum distance covered by any one of the three ants to reach the home of fourth ant from its home . To save a circular grass land of diameter 28 m from sheep there is an uniform entrenchment all around the grass land which is 14 m wide at top 10 m wide at bottom and 10 m deep. CA & AB . Can't be determined e. which is completely field with water. 52 cm &60 cm.14+10@*. If EF & AD intersect at point I. 1. then AI:GD =? a.1/2@*. a. a+b+c= 30 cm . 3:2 e. points D. 1:2 b. They live in the each hole of the board. 85 cm d. Then how much litter water is used to fill the entrenchment.64*10^7 lit d. If in a ∆ABC . 75 cm b. 2:3 d. . NoT Ans :[c] 95. A old rectangular carom board isa home of four ANTs.Carom has four holes one in each corner@ One day 3 of them decided to meet at one place then go to fourth home all together.b& c are sides of ∆ and s is semi perimeter then *h L.584 * 10^7 lit 94. 80 cm c.22/7@*21= 132m SO vol of entrenchment = 120*132 =15840 m^3 1m^3 = 1000 lit water So water capacity of entrenchment = 15840000 lit = 1. *M@h .@h L.*M@h L. =? c. 3E Learning.com/MathsByAmiya ©AMIYA KUMAR . 15 e. NoT www. 30 b. 1 d.h Lh L h a. 0 Ans: [c] Maths By Amiya.facebook. Ç Æ whereA. median AD= 2 cm then what would be length of median CF a. In a right angled ∆ABC . 24.5 cm Ans: [b] √601 = 24. a. 10 cm b. NoT Sol:DATA IN THE QUESTION IS WRONG In any triangle ÂÃ + ÄÃ + ÅÃ = < > ÈÃÂ + ÈÃÄ + ÈÃÅ .5 c. right angle is at B. 24 cm b. 27 cm 97. 11cm c. 26 cm d. If AC= 10 cm. If sides of a parallelogram is 12 cm and 13 cm and its area is 60 cm^2 then what is the approx length of its major diagonal. 12 cm d. Can't Be e.96. . 2 É I 1 = « ¬ . A &A are respective medians so in right angled at B .A. com/MathsByAmiya/photos/650118671706919/ Maths By Amiya.1 + A1 + A1 @&ÈÄ = « ¬ ∗ Ä 3 Ã 1 1 1 1 1 ⇒ 10 = < > ∗ . MAIN REASON TO POST THIS QUESTION IS TO LET YOU KNOW ALL FOUR CONCEPTS 98.2 + 5 + A @ ! x= 11 Alternative :In In Right angled triangle at "B" ÈÃÂ + ÈÃÅ = Ê ∗ ÈÃÄ 21 + A1 = 5 ∗ 51 ⇒A = 11 But median of any triangle cant be more than the hypotenuse . √1 b.facebook. 30 & 120 so 3 = cos 30 + cos 30 + cos 120 S 1 = √3 S « ¬ 2 H 0 https://www. √3 + < > ! 1 c. 3E Learning. What would be ratio of inradius tocircum radius of a triangle whose angles are in the ratio of 1:1:4 a. so the data of the question is wrong. √3 S < > d.com/MathsByAmiya ©AMIYA KUMAR . www. ! 1 Sol: [c] Angles are 30.facebook. facebook. 56 Sol: [c] We know. OD= 3cm & BO=6 cm then among option which could be a possible length of altitude CE a. 30 cm b.10 + 12 = 2 ∗ + . 50 cm d. 4cm & 5cm are altitudes of a triangle then what would be in radius of that triangle a.01 comes in the range hence .com/MathsByAmiya/photos/794476340604484/ 100. Points M & N are n non parallel sides. 60/47 d. If 3cm. 3E Learning. AO*OD=BO*OE=CO*OF ⇒ 4*3 = 6*OF ⇒ OF = 2cm So. AD= 7cm & BF= 8cm Thus Range of CE would be 7∗8 7∗8 < [½ < ⇒ 3. 59 c. 54.√106 d. NoT And :d. 3. MN is||AB it divides trapezium in two equal halves then length of MN =? a. www.facebook. If two adjacent sides of a parallelogram is 10cm and 12 cm and one diagonal is 8 cm.733 < [½ < 56 7+8 8S7 By checking options only 55.com/MathsByAmiya/photos/528568450528609/ 102. such that AO= 4 cm. Then what is the length of another diagonal a. 47/30 Sol: [c] We know . = ⇒ r= "# 0n + + = + + = ! 0 = 0n "# 101. where a is half of another diagonal.4 Sol: By Apollonius . 40 cm c.facebook.55 d. 1/12 b. 8 b.99. 2 ∗ √106 c.com/MathsByAmiya ©AMIYA KUMAR . so diagonal = 2 ∗ √106 Maths By Amiya. If AD. Data adequate or NoT 1 1 1 1@ .7 b.c@ https://www. 30/47 c. BF & CE are the altitude of ∆ABC. NoT4 p#h L1#h 1 = 10√34 https://www. If ABCD is a trapezium such that AB||CD and AB= 80cm & CD= 20 cm. where x is less than 180 degree . c. If side of an equilateral triangle ABC is 3 cm and point D is on BC such that BD=1cm then AD=? c. ]2.com/MathsByAmiya ©AMIYA KUMAR . NoT .4] e.2] c. 3E Learning.103.3] d. then what would be the range of height of water surface inside the cone (from circular base & Assume all figures are under 100% spirit levelled) a. 4√2 cm b 8√2 cm Maths By Amiya. such that AD= 6 cm .[0. ]1. 2. NoT www.43 f # #M # 1 ∗ 10> + < # 1 #M # ∗ 10> ∗ 10¬ ∗ ℎ = ∗ Q ∗ < > ∗ 5 ! 1 ∗ 10> ∗ 10¬ ∗ ℎ 104. a. AC & DE are common tangents to two given circles whose centre are P & Q as shown in fig.7 cm b. If in a plan there are four points such that AB=BC=CA=DA and angle BDC = x degree . √8 cm d. ]3. If a cone of radius 10 cm and of height 10 cm is filled till 5cm from its close end (tip) then sealed with a circular sheet (of negligible width) then turned upside down. √7 9A Sol: [b] If B8 = B[ then 9 ∗ C81 = 7 ∗ CI 1 ! 105.facebook.5] Sol: # 1 Vol of water (tip is downside) = ! ∗ Q ∗ < > ∗ 5 1 Vol of water (circular base downside) = ∗ Q ∗ «101 + < ⇒ ∗ Q ∗ «101 + < #M ! ! # 1 ∗ 10> + < ℎ S 30ℎ + 300 ∗ ℎ S 125 = 0 ! #M ℎ = 5 ∗ 2 S √7 ≈ 0.???@ a. AE = 10 cm & DE= 8cm then what is the distance between P & Q. ]4. 4√5 cm d.1] b. then maximum possible value of x= ? Ans : 150 degree 106. If AB. facebook.com/MathsByAmiya/photos/812283092157142 92157142/ 107.-6. triangle = ] and Radius of circle of centre P = 6 cm [ ex radius = *MYÑ ] YÑ ∆ ∆ * Since ADE is a right angled tria triangle so we can work on this by co-ordina ordinate geometry .-6@ 6@ [IIIrdQuard] So QP = 4√5 cm [By Distance Formula] https://www. 2.4 . √6 e.-2@ 2@ [IVth quadrant] & co-ordinate of P = . NoT No = 2.com/M ok. If 6cm and 8cm are re length of diagonals of a rhombus then what hat would woul be length of radius of circle inscribed ibed in th this rhombus a. 2.4 c.Sol: [c] We have Radius of Circle ircle of ce centre Q = 2 cm [inradius of triangle.2. √5 d.5 Sol: [b] = 6∗8 2 ∗ √61 + 81 or can apply = b. Assume D is origin So coordinate of Q = . '. ing. www. 3E Learning.' Maths By Amiya. +&'.com/ om/MathsByAmiya ©AMIYA KUMAR .' *'&+MÐ'+&'.facebook. S 9@ ∗I∗9 = 4 ∗ . If in a triangle ABC . If in triangle ABC . D. S @ ∗ .com/MathsByAmiya ©AMIYA KUMAR . S 9@ For a=10x . Ö× www. 7:4 b. BE & CFal angle bisector and I in in-centre then AO:IO =? . NoT 111. D. CA=5cm& AB=3cm such that AD. NoT ∆ H ∗I∗9∗ ∗I∗9∗ ∗I∗9 & = ⇒ = = 4 ∗ ∆1 4∗∆ 4 ∗ ∗ . 4:7 c. 4:15 d.E & F are points on side BC. S @ ∗ . S
[email protected] angle A=30 or 150 degree. NoT 112. 14:5 Ans: [c] b. If side ratio of a triangle is 10:11:12 then what is the ratio of circum radius to inradius Sol: H= a.E & F are points on sides BC=4cm. 80:39 d. BD:DC=1:2 & BF:FA = 2:3 then BO:OE=? [if O is point of intersection of BE & DF a. and we dont know which two angles or sides are equal@ 109. S I@. If in triangle ABC . 4:19 c. 3E Learning. NoT Ans :[b] 110. 19:4 Ans : [c] b. D lies on the side BC and on line AD . [if O is point of intersection of EF&AD] a. b=11x& c=12x => = ÇØ Ô Õ Maths By Amiya. 40:19 c.facebook. CA & AB such that BE is median . 5:15 c. 13:7 b. 6:5 e.108. 3:1 d. O is point such that AO:OD=BD:DC=2:3 if we join points B & O and extend it to line AC then it cuts the line AC at E. Can't Say e. If in an isosceles triangle ABC . 15:19 e. 5:6 d. Then AE:EC=? a. BC=4cm and circum-radius is a square of a prime number who is even then angle C = ? Ans: 15or30or 75 or 120 degree. 1:3 e. Perimeter of this polygon po is 77 cm .3@@ d. If radii of all fourr circles ar a G.33 Length of BE a. Product of 2nd smallest est and 3rd smallest radius Sol: [d] ∆= :Õ ∗ ÕÂ ∗ ÕÄ ∗ ÕÅ wher wherer is inradius and rest are exradius of a triagnle. Sol:[c] For a given perimeter er . NoT d. 12 b. 20 b. Square of 2nd smallest st radius d. a.480 d.P series thenarea of ∆CB[ CB[ would be equal to a. 478. What is the maximum imum valu value of ratio of area of in-circle of a triangle to area of triangle.approx@ approx@ area of a regular convex polygon with 11 sided fig whose one side is 7 cm. 115. Length of BD = ? a. 25 Ans : [b] . 18 Ans :[b] .facebook. a. 114. 3E Learning.are consecutive terms of 113. BC= 40 & CA = 50 . 26. c. 1 b. then 116. for equilateral triang triangle. 23.3* sqrt. What would be . pi / . 18 √5 e. Area o of a circle is maximum . ing. iagnle.916 b. NoT www. Product of smallest radius adius and 3rd largest radius c.169 c. D and E are two points p on side AC such that Perimeter of ∆ABE = Perimeter of ∆BEC & Area of ∆ABD ABD = Area Are of ∆BDC. Largest radius among all b. Best is coordinate rdinate ge geometry Maths By Amiya.com/ om/MathsByAmiya ©AMIYA KUMAR . 472.916 d. hence [c] Direction : If in a triangle angle AB = 30. 2 e. consider it as a circlee of circum circumference = 77cm so the area of this his circle is i approx472. e. pi c.Median 117. 458. so area of polygon should uld be less than 472 . NoT Ans : [c] . 12√5 c.66 d. c@ e.right turn| = 4 120. 3 . we get l=24 Now.n-2@ . One day he started from a point and reach at the same point and facing same direction as initially he was after taking 20 right turns that how many left turns he has taken. NoT 119. 3 .facebook. and we know sum of all internal angles = 180. 30 b. 45 c. 9√3 d. NoT Ans : [b] 122. Take Left turn as 270 degree and right turn as 90 as internal angle of a closed polygon.n-2@ .root 2 c. 3E Learning. 4 c. NoT Ans : [b] .5.118. 6√3 b.in degre@ a. 12√3 c.b@ or .5@ then what is its area a. In a ∆ µÙHPQ=QR . 15 e. .24 d. n is vertex Here 270*l + 90*r = 180 .com/MathsByAmiya ©AMIYA KUMAR . 20 b.root 2 e. 2 .root 3 b. by question r=20. 16 c.l+r-2@ . Best is coordinate geometry d.l+r-2@ . If the equation of one side of an equilateral triangle is 3x+4y=5 and its one vertex is . NoT Ans : [e] Maths By Amiya. 60 d. Direct formula = |left turn . Length of DE = ? a. 6 e. solve by options or coordinate or sine formula 121. www. 75 e.root 3 d. if he never repeated or crossed a path which is already marked and never take more than one turn on a point. by question r=20. we get l=16 So either 24 or 16. NoT Sol: [d] Take Left turn as 90 degree and right turn as 270 as internal angle of a closed polygon. 3 b. 5 Ans : [c] 5. n is vertex Here 90*l + 270*r = 180 . If a robot only moves in straight line and only takes either left or right turn and his movement path is marked by a LASER. a. 2 .angle PQR = 90°and S and T are points on PR such that PS^2 + TR^2 = ST^2 then angle SQT = ? . What is the ratio of side of a hexagon to a square of maximum area inside the same hexagon a. and we know sum of all internal angles = 180. What would be diameter iameter o of circle whose two chords AB and d CD make angle of 90° at point E such that . https://www.com cebook. What would be radius adius of ccircle of maximum area inside a sector of a circle c whose radius is 12 cm and central an angle of sector is 60° a. 2pi -11 ra rad c. √122 9A c.com/ om/MathsByAmiya ©AMIYA KUMAR . 4 cm c.facebook.then angle 124.facebook.https://www. NoT No Ans: [b] 126.6 degree Ans : [c] 125. √12 e. CE=2 E=2 cm. 1 rad b. www. If perimeter of a triangle two sides are odd integer er rest is even. NoT Ans: [b]4 Maths By Amiya. C CD= 9 cm & EB= 3 cm.(1/2) rad d. a.com/MathsByAmiya/photos/840413596010 13596010758/ 127.12 b. pi . ing. a. 3E Learning. 3 d.com/M ok.60 d. √129 9A d. If length of minor arc OAB = ? (where O is thee cent centre of circle and A & B are on the centre) a. 6 c. 2√3 d. NoT Ans : [d] rc crea created of a chord AB is equal to radius of the circle ci .com/MathsByAmiya/photos/584494198269367 98269367/ angle iis 24 cm then how many different triangles gles possible po where only 123. 6cm b.facebook. 12 cm b. www. There are how many convex polygon possible such that integral angles are in AP and integer and the smallest angle is 30 a.128.com/MathsByAmiya/photos/509985899053531/ 129. here a=30For convex polygon we get only 2 set of values http://www.facebook. 3 d. 4 e.n/2@. 17 c. 8 Sol: [e]NoT b. NoT 180*.wolframalpha. 3E Learning.com/input/?i=180*%28n2%29%3D%28n%2F2%29%282a%2B%28n1%29d%29%2C+a%3D30%2C+n%3E0%2C+d%3E0 Maths By Amiya.n-2@=. Ans: 10 cm^2 https://www.com/MathsByAmiya ©AMIYA KUMAR .2a+.n-1@
[email protected]. com/ om/MathsByAmiya ©AMIYA KUMAR .com/Math /MathsByAmiya/photos/532772120108242/ 133.facebook. Not Maths By Amiya.BF=7 cm. 1∗√! √!L1 c.facebook. 11440 cm^2 c. 8 cm b. NoT Ans: [e] cannot be determined 134. www.com ok. ing. 4 cm b. 3. NoT N inradius = 6 .com/MathsByAmiya/photos/859153280803456 03456 Ans: [b] 135.6)) and A=(0. DE= 9 cmThen EF =? a. Data D inadequate e. 6 cm d. BC= 30 cm & CA= 34 cm and I is incetre etre of o the triangle then IA =? a.6 Ans: [c] https://www. AB is also a side of a regular hexagon and AC is side of a regular pentagon then what is the he mea measurement (in degree) of angle BAC ? a.16) so IA= √61 + 101 = √136 = 2√ √34 132. so centre (6. angle ABC=30°. If the semi-perimeter er of ri right angled triangle is 154 cm and smallest lest median m is 72 cm then what would be area of the tria triangle. 114 c. 3E Learning. 3√5 c.1600 cm^2 b. 60 b.BCA. 108 e.facebook. CE= 5 130. 3 cm d. What would be side of ins inscribed square PQRS having maximum area inside i a triangle ABC such PQ lies on side ide BC BC.com/Math /MathsByAmiya/photos/853093624742755/ 131. If in a triangle ABC. √!M1 e. Sol: [c] √!M1 1∗√! b. √!L0 1∗√! d. 120 d. angle ACB=60°and side AC= A 6 cm 1∗√! a. a. AB=16 cm cm.facebook. https://www. NoT https://www. If in a ∆ABC.. N NoT c. If angle ABC=anglee BCA cm . 33. 2√17 Sol: [e] e.4 cm e. 1540 cm^2 d. 3E Learning.facebook.facebook. www.com/MathsByAmiya/photos/861355437249907/ Maths By Amiya.Sol: [c] Area = (s – c)*s = 1540 cm^2 https://www.com/MathsByAmiya ©AMIYA KUMAR . Only one set of data b. 137. A= 60° . NoT Ans : [a] only II is possible 138. a = 4cm. a=b= 12 cm. B. A= 60° III. By all we can construct unique triangle e. given below is (are) sufficient to constructt an unique un triangle ABC. bb= 3 cm a. C = 90° IV. A .b & c are sides to corresponding nding angles] I. Two set of data c. How many set of data giv [P= Perimeter. P = 16 cm . c = 9 cm. A= 60° . B = 45° . A= 60 60° II.136. How many statements nts are correct I. B = 30° . sin 1° . Three sets of data d. & C are angles & a. sin 1 II. ° . III. ° . IV ° . sec 5° . V. sec 5 VI ° . Ans: 4 Maths By Amiya.facebook. 3E Learning. www. ing.com/ .com/MathsByAmiya ©AMIYA KUMAR . question is of triangle ngle w whose perimeter is 80. ing.facebook.QA:AB :AB = RB:BC =PC:PA = 1:3 .be/PvMh PvMhMly9cFY un and third side is a 140. In given ∆ ). PR:RJ Ans: 1:7 Maths By Amiya. ? [in cm^2] Ans: 39 cm^2 :RJ = 11:3 . www. NoT Ans: [c] . 22 e. 0123 *4+ . In the given fig . if S is set having three element S are positive integers inn whic which two are odd and one is even. If in a right angled ∆ )*+ AD= 4 cm & CD = 9 cm m then area of ∆ )*+ . 25 b. AQ:QP = 3:4 then QO:OJ = ? 144. 90° 143. then ! ∆ #$% . 80 e.? ! ∆ &'( Ans: 36:13 )*+. NoT Ans: [d] 141. a. then ! ∆ #$% . 33 c.-. 24 c.com/ .lements. How many differentt set S are possible. 133 b. 3E Learning. point D lies on AC such that angle A*+ .com/MathsByAmiya ©AMIYA KUMAR . 100 d. all elements of set 139.? ! ∆ &'( 142. a. it is known own that th sum of any two elements is greater than third element and sum of all elements is 80. 23 d. How many triangless are ppossible whose two sides are 50 unit and 47 unit prime number. https://youtu. In the given fig .QA:AB :AB = RB:BC =PC:PA = 1:3 . -.-.145.-.-. AQ:QP = 3:5 & AS:SJ = 2:3 then 5∆'7#9 .facebook.-.? 5 #&89 5∆#&8 Ans: 25:52 147. In given ∆ ).-. AQ:QP = 3:5 & AS:SJ = 2:3 5 &'7(9 5∎&'7( . In given ∆ ).com/ . ? :QR:RJ = 2:2:3 . AO:OR = 4:3 then AQ:QP 151. PR:RJ Ans: 8:15 :OJ = 3:2 . PQ:QR:R :QR:RJ = 2:2:3 . AQ:QP = 3:4 then AO:AR = ? Ans: 1:1 In given ∆ ). AO:OR = 4:3 then AQ:QP = ? In given ∆ ). AS:SJ = 4:3 then PM:MN = ? Ans: 10 : 7 153. ing. PQ:QR:R Maths By Amiya. AQ:QP = 3:5 then 5 &'7(9 5∎&'7( . PR:RJ :RJ = 55:3 . In given ∆ ).? 5∆(789 5∆ Ans: 1:1 146.-.-. 3E Learning.? 5∆#7:9 Ans: 5:4 150. In given ∆ ).? 5 #&89 5∆#&8 then 149. AQ:QP = 3:5 then 5∆#7'9 5∆ .-. 148. PR:RJ :RJ = 55:3 . PR:RJ :RJ = 44:3 . In given ∆ ). AS:SJ = 4:3 then PS:MN In given ∆ ). PR:RJ :RJ = 11:3 .com/MathsByAmiya ©AMIYA KUMAR = . www. In given ∆ ). :RJ = 33:2 . QO:OJ =? Ans: 1:15 152. PR:RJ :RJ = 4:3 . com/MathsByAmiya ©AMIYA KUMAR .-. 3 e.com/ . 7 e. NoT Ans: [b] 158. ∆ )*+ is a right angled gled tr triangle at B.<(9 . CBD e. 6√2 c. If are three altitudess of a ttriangle is √3 cm then what is the perimeter a. AS:SJ = 4:3 then In given ∆ ).5 cm b. OH 159.-. NoT N Ans: [d] 161.5 e. NoT Ans: [c] eter of o triangle 157.< '. NoT Ans: [b] 160. 4. :QR:RJ = 2:2:3 . 2. 0. If 3. √5 b. 6 b. QR:RJ = 2:2:3 . How many differentt triang triangle are possible whose two altitudes aree 7 cm & 6cm and third altitude is an even number. 2 & 2 are three altitu altitudes of a triangle then what is the perimeter eter of the triangle a. 6 d. O is the mind point of line segment DE then what is the distance length gth of BO (in cm) a. and AD. 12 b. cm Point D & E lie on AB & BC such that AD:D AD:DB = BC:BE = 2:1 . 162. in which AB= 9 cm & BC = 8 cm. 1 cm c.. 9/(4√7 7) c. 1 d.facebook. such that AO= 3 cm. 3E Learning. 4/3 b. 2 & 2 are three altitu altitudes of a triangle then what is in-radius of the triangle a. ing. If O is orthocentre off trian = 2 cm. triangle ABC . CBD e. AS:SJ = 4:3 then In given ∆ ). √7 c. 1. Ans: 19 If 3. PQ:QR:R 5∆&<(9 5∆ .? 5∆#<:9 5∆ 155. 66√3 d. (9√2) / 4 e.5 cm d. & BE are altitudes.5 cm d. NoT Ans: [a] If 3. 3/4 c. PQ:QR:R 5∎'. 6 cm c. NoT Maths By Amiya.154. 3 cm b. BO = 6 cm thenn OE = ? a.5 d. 2 & 2 are three altitu altitudes of a triangle then what is the area off the triangle t a. 1. www.? 5∎(<: (<:89 Ans: 8:17 156. If in ∆)*+ . 40. 150 b. 85 c..5% e. 45% Ans: [c] Maths By Amiya. 5:4 b. 40 cm & 24 cm then hen largest la angle of 164. 8 cm^2 c. 20 % b. BE & CF F are angle a bisector & I is the in centre of the triangle gle th then what is the ratio of area of ∆)CB to area of o ∆*4B? a. AD. NoT Ans: [c] 165. then what at is th the approx probability to hit coloured (shaded) haded) area. AB= 100 cm . B BC= 4 cm & CA = 3 cm.com/ . 3E Learning. If ABCD is a squaree targe target and all arcs are quarter circle whose radius adius is equal to side of square ABCD. NoT Ans: [e] NoT it should be obt obtuse gth of altitudes are 30 cm . BE & CF are altitudes such that. If in a ∆ )*+ . BE = 9 cm & CF = 12 cm then angle A = ? a. BC= 6 cm & CA = 8 cm & I is the in centre of the triangle then what is the area of ∆)*B ? a.facebook. 30 e. ing. NoT Ans: [c] 166. 67 b.com/MathsByAmiya ©AMIYA KUMAR . NoT Ans: [c] 167. If in ∆)*+ .Ans : 6√2 163. AD = 6 cm. 31. 6 cm^2 b. AD. www. 90 d. 115 c.5 % d. 50% c. 13: 30 c. a. AB= 2 cm . If in a @A0123 length triangle is a. 15:28 d. 90 d. 12 cm^2 e. 5:6 e. 10 cm^2 d. 120 e. EF||BC ||BC & GH|| AC. ED = ? 170. ing. Ratio of Area of F!GHIJ EKLMND to Area of ∆ #$% ly two types of internal angles.Directions : If in given image circle is in-circle of ∆)*+. What is the length off PQ 174.com/ . 3E Learning. one is 60 degree ree another ano is 300 degree. 300 degree ree an angles then there are how many 60 degrees es angles ang are in this polygon. Ratio of Area of ∆ %DE to Area of ∆ #$% 171. EF = ? 169. Ans: 33 If ABCD is a square off 10 cm then 173. If ABC is a quarter circle and PQRS is square of side "a". If AB= 6 cm . 172.com/MathsByAmiya ©AMIYA KUMAR . What is the measurement ement of ∡PDQ 177. EF & GH are tangents to the circle and DI || AB . and lines DI. www. Find the he radius rad of quarter circle in terms of "a" if CS =PB CHECK FACEBOOK K PAG PAGE OR GROUP FOR SOLUTIONS OF QUESTIONS QUE WHICH ARE NOT GIVEN HER Maths By Amiya. BC = 5 cm & AC = 7 cm m then find 168. If a polygon has only If there are 30. What is the length off PR 175.facebook. What is the length off PB 176. 3rd Floor.facebook.com/MathsByAmiya www.amiya http://in. Anand Complex.facebook. Above Dominos.facebook.com/kumar.com/groups/MBAMathsByAmiya To Follow Amiya : https://www.linkedin.com/in/kumaramiya To Join Classroom Course For CAT-16/17 XAT-17/18 SSC/BANK PO 3E Learning. HB HB Road Ranchi.in Join Fb Group www.To get more follow www.3elearning. 9534 002244 .