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 If 2. a. 13/12 d.3. Data inadequate e. 11/12 b.13. NoT Sol: [b] 12/13 Relation = + 1 1 1 1 13 = + + = 2 3 4 12 12 = 13 + .4 are altitudes of a triangle then what is the inradius of the triangle. where ℎ. 12/13 c. If in a ∆ABC side BC makes an angle 132° at incentre then angle subtended by side BC at orthocentre is ???? a. 17:4 b. If in a ∆ABC . If is orthocentre of ∆ABC. NoT 16. D is midpoint of BC. 4:13 e. line FG ||AB where G is on BC then ∠FGD ? a. 2:1 c. c. NoT 19. AB=BC= 6m and CD = 10cm then what is the distance between incentre and centroid of the triangle.facebook. 18. Can't say e. 100 b. 3E Learning. AB=BC. NoT www. 50 d. If in a ∆ABC . 45 c. ℎ &ℎ 14. Ans :√13 15. 17. 86 e. then which point would be orthocentre of ∆OAC. E is on AC and F is on AD such that ∆DEF is an equilateral triangle &∆ AFE is an isosceles triangle.com/MathsByAmiya ©AMIYA KUMAR . 11:10 d. . 104 Maths By Amiya. 30 b. BC=4cm and CA = 10CM . 96 d. Point D is on BC such that AD is angle bisector and I is incentre then AI:ID = ? a. If in ∆ABC . AB=7cm . Ans: Both are 10° 23.. central angle is 959 = 137*7 & 1781 =137*13 . 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. by euler line 21.2 b. 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. Then among options which one could be total distance travelled by the ant. There is a sugar cube of dimension 10cm*10cm*10cm. If ∠TAP=20° and ∠TBP = 30° .10 ∗ √401 c. 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.'.5* √401 e. a. ( -'.facebook. d. SoD(137) = 2 [By Sanket] !"# $%&'() *+.3 c.be/JEKl2VN-oNY 24. √401 √0# 1 b. Ans: 6. consider the ant is shape zero.NoT www. <QBS= 20 &<RAT=50 then <SPM = ? 22. 3E Learning. If in triangle distance between orthocentre and circum-centre is 9..5 d.9 cm then what is the distance between centroid and orthocentre. so 137 common vertices .+'* http://youtu.6 . Sol: [d] Maths By Amiya.com/MathsByAmiya ©AMIYA KUMAR . then find ∠BTA =&∠BTC.20. If a circle touches another circle at point T internally. 9 e. NoT Sol: [a]They have same common multiple of central angle. Height gain is 10 but base movem movement is 40*(number of rotation) 67 839 = :101 + . and for two rotation n its 80.40 ∗ 3@1 = 10 ∗ :1 + . NoT Sol : [d] New surface area = old surface face ar area + new faces created . There is a solid cheese cube oof dimension 10cm*10cm*10cm. if Jerry ry a rat ra eats the cheese cube from centre of all the su surfaces in such a way that the rest of the cheese chees cube has three similar air tunnels crossing sing ea each other at the centre of cube and dimension nsion of cross section of each tunnel is 1cm*1cm. 698 cm^2 d. then what is total surface area of the rest cheese eese cube. which is created d by joining j the mid points of all 6 surfaces of a cu cubical room of dimension 10m*10m*10m 0m . 702 2 cm^2 cm^ e.. www. sincee mo moves along four sides. What would be total volume lume ccreated by a 3-D figure.com/ om/MathsByAmiya ©AMIYA KUMAR .facebook.43@1 for number of rotation =5 . 3E Learning. total distance = 10√16 16 ∗ 51 + 1 = 10 ∗ √401 Alternate :. Total distance = 10 ∗ √401 25. ing.2 -4 -44 = 702 26.Open the cube alon along faces . Maths By Amiya. Height gain is height of cube 10cm 0cm but base is 10*4 =40cm in one rotation. c a.The height gain on each face ce wou would be 10/(4*n) where n is number of rotation tations in ants total journey. 600 cm^2 b. 1 So total distance = 43 ∗ 4101 + <1$> = 43 ∗ 1$ √16 ∗ 31 + 1 = 10√16 ∗ 31 + 1 = = For n=5 . 5588 cm cm^2 c.faces destroyed = 600 + 40 + 36 + 36 . Then GD: DH = ? a. www. FGand EH are perpendicular to AC.G.Since I is the midpoint of EF . DF. BD.6 kg d. AB and AD & DC respectively. 150 to 300 e. 166. E over line AC. 28. 7 to 16 c. BC .67 m^2 e. More than 300 Sol: [e] Can remove a cube of 9*9*9 from a corner without affecting surface area. 500/3 m^2 c. If N is maximum number of small cubes of dimension 1cm*1cm*1cm. 8 kg e. 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.so can remove 729 cubes max. Less than 6 b. NoT SOl: [e] Actually total mass would be much greater than given mass.so D is the midpoint of GH. 60 kg c. D . 2:1 d.facebook.a. we can cut from the cubical box such that the total surface area of remaining box would same as the original box. 6 kg b.H are the projections of F. 1:1 b. 6. NoT Sol: [d] The figure is Bi-pyramidal . midpoint of diagonal EF. What would be maximum volume of a cube inside a semi-sphere of radius 10 cm 31.facebook. 30. 250/3 m^3 b. a.com/MathsByAmiya ©AMIYA KUMAR . 1000/3 m^3 d. 3:2 Sol:[a] 1:1 FBED is a rectangle and diagonals BD and EF intersect at point I. Then which range best describe N. 16 to 150 d. I. so volume = ∗ B C ∗ DEℎ = ∗ 5√2 ∗ 5√2 ∗ 5 = ! 1 1 ! =## ! A! 27. a. DE. with base dimension 5√2 ∗ 5√2A1 and height 5m.com/MathsByAmiya/photos/745195878865864/ Maths By Amiya. 1:2 c. 29. if the total mass of the bucket with water is 1600 kg then what is the mass of plastic used in bucket. What would be maximum volume of a cuboid inside a semi-sphere of radius 10 cm https://www. In a right triangle ABC. 3E Learning. There is a solid cubical box of dimension 10cm*10cm*10cm. 2:3 e. m. tr a. 56 e. If paint used by all colour are same and radius of smallest purpule color circle is 1 cm. NoT (???) Sol: [c] D if AB AB=15cm . ing. CD=145cm and AD= 8 cm with ∠BC8 = 33. 48 Sol: [b] https://www.55 cm c.. 1284 d. 2568 o triangles then add Sol: [c]JoinBD and get two right angled triangles BAD & BAC and find area of the values. c are sides and ∆ is area ar of triangle Or triangle is a right angled gled triangle . & 33cm then. 1224 b. 36. If the perimeter of an integral tegral sided triangle is 45 cm then how many different diffe triangles are possible c. and their radii are a 1cm. 2cm. In a quadrilateral ABCD 90°then its area is .circum radius of triangle angle ((required radius) .5 9A 4∗∆ 4∗6 R .33 cm e.. m. 2 cm b. NoT (???) a. Data inadequate e. 34..com/ om/MathsByAmiya ©AMIYA KUMAR . 35. 12 ∗ √3 9A^2 d. BC=144cm. 3. NoT Sol: Circle would be circum circle of triangle formed by joining the centres of tangential t circle. 45 b.th . 5 ccm & 4 cm ∗I∗9 3∗5∗4 H= = = 2. Inside an equilateral triangle angle tthere is a point from which the length off perpendicular perp on all the three sides are 1cm. what would be area of equilateral teral triangle. 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.. Then what at is ra radius of new circle a. 3E Learning.. 3 cm d. so R = half of hypotenuse.facebook. b .then what is the thickness of Maths By Amiya. www. Data ata inadequate ina e. 2. 442 d.32.??? c. 2cm & 3cm. Sides of triangle are 3 cm. NoT a. a.facebook. 9 ∗ √3 cm^2 b 12cm^2 cm^2 c.com/M om/MathsByAmiya/photos/528057913912996/ es touc touches each other from outside. .T + O@1 S T 1 @@ = N ∗ . 3E Learning.1 O 1 By solving these O = √!M 0 . c.1 + O@1 @ N ∗ 11 = N ∗ . + O@1 S 11 + . Maths By Amiya. b. If AB=AC C = 6 and ∠BAC= 30° .T 1 S . 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 .. where wher y is radius of yellow circle and x is thickne ickness then according to question Purple = Total Black = Total Yellow .67*pi c. if thickness of bbalck colour is same with all circles L!∗√! 0 M!∗√! 0 a. then y = 2x (x is width of black) Alternate.61*pi e.balck colour . BC If AB= 40cm .com/ om/MathsByAmiya ©AMIYA KUMAR .facebook. D point AC=14 cm & AD= 5 cm m then what is the circumradius of triangle ABC. ing. b. www.58. 38. (1+x)cm.comm oint is on BC such that AD is perpendicular on BC.9 b.T + O@1 S T 1 = T 1 S . has brick wall on its side.1 .Take width of yell ABC. √!M 0 d. 64*pi . + O@1 S 11 @ + .. ????? a.1 + O@ 1 = . 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. then the maximum area covered cover by horse is . e. In a triangle ABC .. A 37..T = L!∗√! 0 yellow is y ... √!L 0 NoT Sol :[d] Here we have 4 circles with rradius 1cm. 17*pi d. y cm & (y+x)cm . facebook.NoT e. 3E Learning. ing. 12 d.66 cm e. then BC=√2 x. 15 e. 60° e. There are how many different is 30 unit a. DE. So..a. if ∠BC[ = ∠ 8B[ = ] then ] =? c. 15° ∆ABC~∆BDC. If FG G = 4 cm & EH = 2 cm then BD =? a. FGHE is a trapezium with ith I aas midpoint of its unparallel sides. AC=2x. then h^2 . 14 c. AC BC .com/ om/MathsByAmiya ©AMIYA KUMAR . NoT N = 0 According to Sine Formula ula CB CB 2H = ⇒H= 2 ∗ 39 3[ H= 40 2∗ = 0 = 56 9A 39. 6 d. www. 5 c.5 b.facebook. DF. AB and AD & DC respectively. FG and EH are perpendicularr to AC. By sine law. so BD = 6 cm. a. NoT SOl: [e] 5 If 30 is not hypotenuse (take take 330 as perpendicular) .77 cm b. In a triangle ABC. 45° d. In a right triangle ABC.com/M om/MathsByAmiya/photos/550655228319931/ fferent integral sided right angled triangle are possible possi whose one side 41. ] = 30° *+$_ ` = *+$0= √1` 40. 4. 66. D is on AC such that BD is medianand∠BDC= 45°. 70 cm d.b^2 = 30^2 [h & b must be even] Maths By Amiya. so it would be median of trapezium pezium. ID = (FG+EH)/2 = (4+2)/2 2)/2 = 3 cm. 13 b. Take DC=xx. BD. 30° b. 80 cm Sol: [e] 56 cm In fig sin 9 = XY XZ = c. https://www. 77. 7 e. and ID D || to parallel sides. 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. c(. d e%&' () ). com/MathsByAmiya ©AMIYA KUMAR .24. 20 cm b. 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 =? .(* () By this case we would get 4 values = ab 1 fgh i jk = 4 [ ] is GIF If 30 is hypotenuse. NoT Maths By Amiya. 1 b. If in ∆ABC . 4cm c. NoT Ans : [c] 45. Cannot be determined e. a. 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.com/photo. www.30 SO total required triangles = 4+1 = 5 By NitinGuppta Sirhttps://www. If I is incetnre of∆ABC where AB=AC=15 cm . 6 cm e.. Not (???) Sol: [a] Another parallel side would be 8 unit so PQ = (difference of parallel side )/2 = 1 unit 43. 108 cm^2 c. 2 c. 2.php?fbid=10152541092894878&set=p.Then find the area of triangle a.4. If P and Q are mid points of diagonals AC & BD then PQ = ? a.10152541092894878 &type=1 42. 216 cm^2 e. 108√2 cm^2 d. 24 cm c. BC=24 cm & D is midpoint of BC then AI = ? a.facebook. If area of a trapezium ABCD is 180 unit square whose height is 20 unit and one parallel side is 10 unit. 3E Learning. 40 cm e. 36 cm^2 b.facebook. then its factor should be crude hypotenuse . 5cm d. 18 & 36 cm. 36 cm d. 3 cm b. Ifexradii of a triangle are 12. where O is point of intersection of CF & BE.5) so 30 as hypotenuse would have only one integral triplet which is 18.5 d. and factor of 30 only crude hypotenuse is 5 (with 3. 105 c. 100 e. NoT Ans [e]1 S 1= S 0 n ! = op !1= 47.???@ Sol: (c) 2*c(15. 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. 3E Learning. 5:11 e. 5:16 c. 6/25 e.com/MathsByAmiya ©AMIYA KUMAR . 1:3 b. How many triangles are possible whose two altitudes are 6cm & 8cm and other altitude is also a natural number. NoT www. NoT . a. 210 b. 8/25 d. 6:5 b. then MN/QV=? a. d. 21 e. What is area ratio of ∆AOE to ∆BOD a. NoT Ans: [d] 50.2) . a.375 d. What is ratio of AO:AD a. 200 d. 8:13 c. 3:1 Ans: [b] Maths By Amiya. 1:1. point U is on QR such that QU=3cm & UR = 2cm and point V is on PR such that RV=4cm & PV=3cm.Ans : [c] 46.42 <x <24 so total 20 natural x be possible 48. such that BD:DC=2:3 & AE:AC=1:4. 3:8 b.facebook.a 15 sided fig@ such that two vertices of equilateral triangle are also the vertices of Pentadecagon. 20 d. such that QT=2*PS=2*ST=2 cm. NoT Sol: Altitudes are . 5:6 c. O is point of intersection of AD & BE then 49. 11:8 e. There are how many distinct equilateral triangles . 4:15 e.6 . 3/5 c. Q and R are three points on a plane. 18 b. In a ∆PQR . What is the ratio of BO:BE a. 13 c. 4/25 b. points S & T are on PQ .2*5 = 200 [ there would be If P. M & N are point of intersections of line "QV & RS" and "QV &UT" respectively. 6:11 d. 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.in a plane@ are possible in a regular Pentadecagon . NoT Ans: [c] 51. 5 d. 1:3 d. 0. P is pointt of intersection int of AD & BE then PE = ? a. 2cm b. sum=316 53. there are 4 octagon. 0. I and G are incentr ncentre& centroid then 55. CA= 8cm . www.. 4 d. which are 6-5=1 . 4/5 d. 33 d. 0. In a ∆ABC . 4:5 b. Total number of point off inter intersections (PoI) of all diagonals of a regular hexagon Ans :. NoT Ans [d] 56. just need to che check points on second octagon from outer side . ing. BC=7 cm. and oone more point at centre of gravity. if AB= what is the distance between ween I & G a. Points O & P are points of intersections of linee "AD & BQ" and "AE & BQ" respectively. BC=16 cm . 316 c. which are 8-5=3 5=3 . and one more point at centre of gravity. 2212 b.:3:4 & Q is on side AC 52. points D & E lies ies on sides BC & AC respectively. NoT AB=AC=10 cm . 1:4 e. 28 b.√106 @/5 e. 4 = n c. 1:5 c. If OP:BQ :BQ = m:n then m+n = ? (m & n are co-primes) a. If in a ∆ABC AB=6cm. Maths By Amiya. 0. and BE is median. NoT Ans: [d] 54.33 b. If in a ∆ABC. E lies on AD such that AE:AD A = 1:3 & F lies on CE such that CF:CE :CE = 1:4 . AD is angle bisector of angle ang A . 57.facebook. D lies on BC C suc such that BD:BC=1:2 . Total number of point off inter intersections (PoI) of all diagonals of a regular octagon a.com/ om/MathsByAmiya ©AMIYA KUMAR . In an isosceles ∆ABC.66 e. NoT Sol: 4*8 + 3*8 + 1 = 57 A regular polygon makes es sam same fig with PoI of diagonals. O is point of intersection of lines AD & BF then BO:BF =? a.2*6 + 1*6 + 1 = 19 es sam same fig with PoI of diagonals. 64 c. points D & E aare on side BC such that BD:DE:EC = 2:3:4 such that AQ:QC=3:4. just need to check heck ppoints on second hexagon from outer side . 235 e. 57 e. there are 2 A regular polygon makes hexagon.25 c. NoT Ans: [b] 81:235 . 3E Learning.. ich is 30 and the longest 58. D & E lies on the each circle and these two are point of intersections of AC and circle circles. x² + y² + z² = 2(R² .facebook. There are two concentric which cuts smaller circle le at B and C.30° c. by Heron's formula. 7 cm c. ing. 1 b. (MPT) ic circ circles with centreO. 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. by MPT AB||DE . 12 cm Sol: Draw DE (where E is midpoin idpoint of AC).r² ) c. 59. In ∆ABC. point A and C are ccentre of these two circles. x² + y² + z² = 2(R² + r² ) e. 7 c. ∠BAD AD = 50° &∠CAD= 80° then AD =? b. AB is chord of biggerr circle 61. Data inadequate equate e. 45° d. x² + y² + z² = R² . a. 20° b.If “r” & “R” are radii of smaller and bigger circles respectively ely and AB = x . BD= y & BE = z then which one is a correct relation a. NoT Maths By Amiya. point oint D is on BC such that. 3E Learning. Cant Say e.NoT Sol: ∠DBE 180 ∠BDE BED 180 1/2(180 ∠BE ∠A) 1/2(180 ∠C) 1/2(∠A ∠C) 45 60. 14 cm d. x² + y² + z² = R² + r² d. Data ata ina inadequate e.r² b. so AD=AE = 7 cm. NoT Sol: [c] Sides of triangle are. then ∠DBE=? a. AD is median. In the fig. AC=14cm.12&13 12&13. EB is a chord of smaller circle which is perpendicular on AB AB. 5. point B lies on the bot both the circles such that ∠ABC=90° . 8 d. and ∆ADE E is an isosceles ∆.com/ om/MathsByAmiya ©AMIYA KUMAR . www. NoT a. 3. 9:8 Maths By Amiya.34 d.6.47 c. e.com/M om/MathsByAmiya/photos/584494198269367/ https://www. 4:3 b.15. !M√! d. 3 S √3 !L√! c. 1:1 c. What would be ratio of side oof a largest square inside a regular hexagon to that of regular hexagon a. And another tangent 65. 1 √"L√1 c.facebook. 24 cm c.com/ om/MathsByAmiya ©AMIYA KUMAR .NoT Sol: Curve Surface Area = (pi^2) ^2)*(R^2 . 26 cm b. 12.744000315659641&typ e=1 are to that of square 64.Use Pythagoras & EC is diameter of smaller circle and AB=CD 62. √"M√1 1 b. e. e. Not Sol: [b] https://www.1^2)= 12.0cm and oouter diagonal is 3.744 =p.0 cm a.34 .php?fbid=744000315659641&set=p.com/M om/MathsByAmiya/photos/584494198269367/ gents from a outside point P on a given circle. 3E Learning. If Radius of the circle is 5 cm and CP is 8 cm then whatt is perimeter pe of ∆PAB a. www. 1 √"M√1 d. What would be ratio of side of a largest regular hexagon inside a square √"L√1 1 a. ing. What would be ratio of area oof pink coloured shape to that of blue coloured shape a. NoT N Sol: [b] https://www.facebook.com/ph om/photo. 3:4 c.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.Sol: [d] Hint:.78 e. 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. 3 + √3 b.34 cm^2 cm 63. 30 cm d.facebook. 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.93 b.r^2) = (pi^2)*(1.5^2 . 67. M & N respectively such that ABC form an equilateral triangle then what is the length of ∆ABC a. 3E Learning. Points A. 2 √13 d. B &C are lines L. www. Then Total area of 4 small circles 4*(pi*r^2)=4(w + blue area) Where w . M & N are three parallel lines .Sol: [c] Let side of the square : 4r . such that the perpendicular distance between LM & MN are 2cm &5 cm respectively. Now Area of big circle =>4*pi*r^2=4(w + pink area) => Pink area = Blue area.white area inside small circles. 2√14 c. M is in the middle on two. Let L .facebook. 3√13 b.com/MathsByAmiya ©AMIYA KUMAR . 3 √14 Sol: [c] Maths By Amiya. 68. If in a triangle ∆ABC. NoT tan = cot ⇒ = 3 ∗ Q ± 45° Maths By Amiya. 5. Sol: b. 3 e. Never Possible c.com/MathsByAmiya ©AMIYA KUMAR . www. such that tan = cot and cosec S sec > 1 then how many integral value of k would be possible a.facebook. 3E Learning. Not Possible for real K d. ∠ABC= . facebook.NoT 70. 20° b. 135° d. d. 60° b. 120° c. C & D are in a plane such that AB=BC=CA=DA. www. where cosec S sec = √2 S S√2 = 2√2 > 1 = = 0. 3E Learning. If two circles of equal radius intersect at two different points C & D. If Line AD intersect Line BC at E such that ED:AE = 1:10 then ∠BDC = ? a. ±1 69. 150° Sol: [d] http://youtu. 80° Ans: [e]120° . & line AC cuts other circle whose centre is B at E. it shows angle is in second quadrilateral. B . 70° c. There are four points A.com/MathsByAmiya ©AMIYA KUMAR .be/Ouw21pvOyFA Maths By Amiya. isosceles triage property.cosec S sec > 1 . 100° e. such that∠EAB = 20° then ∠EBA = ? a. If A & B are centre of these two circle. positive for = 3 ∗ Q ± 45° . = YZ Y X. here re x=y=z = 2. 1 2O + T 1 = ⟹ 3O = . we have X. 1 : 7 c.1∗1L1L@ = n By Routh's theorem. each of the ratio io AX : XD.com/ om/MathsByAmiya ©AMIYA KUMAR .1∗1∗1M @h ¡ ¡L!`L! =¡ ! . ¡ n Alternatively :.1∗1L1L@. 71. BY : YE and CZ : ZF is 3 : 4. 3E Learning. http://en.XYZ@ X.X@ = .org/wiki/Routh's ki/Routh's_theorem Maths By Amiya. .the points of trisection of respective sides. ¡L¡¡L=¡ . . we have y = .1∗1L1L@. . www. ing. E and F are th Further. .. Assuming the areas as x. Find ind the ratio of the areas of triangle XYZ and triangle gle AB ABC a. D. This will reducee our work considerably. 3 : 7 d. Now the non-overlapping areas reas th that appear similarly placed about the three ree vertices ve will equal in area. we can safely assume ume triangle t ABC to be equilateral. We need to find the ratio of . as shown s in figure. IF in ∆ABC. ! 0 Plugging x in terms of z. `L L¡ = ⟹ 4O O + T = 3.XY@ Also. y and z ass shown show in the figure. O= O + 2T + 2 3 = = Z ! 0 .wikipedia.facebook. 4 : 9 Sol:[b] Since there is a unique answer wer (as the options suggest. . 2 :9 b.XZ@ X. then what is the sum of digits of P ? Sol:[b] a. 6 c.8}.7}. For given perimeter P unit of triangle .7. 6 c.72.49<. 7 d. 8 If P is even then 11.6.5<± ´h . a.6.facebook.49 Maths By Amiya. www.75 = 7 74.8}.{3. NoT Sol:[e] This is case of mutually tangent circle .49 ⇒P = 24 If P is odd then 11.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.{4.´L!@h 0p ² < 12. 0.facebook. looking for radius of smallest circle .5 ⇒ 552 < . If three circles and a straight lines touches each other at six different points &radius of two bigger circle is 4cm and 6. there are 12 different integral sided triangle are possible.{5. 1.µ + 3@1 < 600 ⇒23.com/MathsByAmiya ©AMIYA KUMAR .8.6} By Formula :Total number of integral triangles = Nearest Integral value of ± https://www. 3E Learning.25 1 = 81 3 A ⇒ H4 = 1.com/MathsByAmiya/photos/528057913912996/ ph 0p ² = 6.123 b.23456790 … A = 0.5.8}. 1.µ + 3@1<24. There are how many different integral sided triangle possible whose perimeter is 18 unit a.5 ⇒ 552 < µ1 < 600 ⇒23.0123 9A H4 73.c@ 7 By manual process: Different triangles are {2.{6.49<P<24.{4.8}.25 cm then what would be radius of smallest circle in cm. 8 Sol: .{5.5 b.5<±0p² < 12. 2.7.234 c.33 d.25 4 6. 7 d.6.66 e. 5 b.7}. ABCD A rectangle DEFG with ith EF=7c EF=7cm & FG=14cm.4.4. Sol: [c] a.8 b.6 c. and common difference = 1. whose sides ED & DG lies on side of square & one vertex F is on the circle.22/7@*35*35 .Area of circle Area of Rectangle = 70^2 .7*14 *14 = 952 cm^2 76.4 * 7. www. it should be multiple ple of 3. 10.5@ = 5. 75. SO in ∆FO are in Pythagoras. So by solving w we will get R=35 cm So side of square = 70 cm Area of Shaded Region = Area of square . What would be area of right aangled triangle whose sides are in AP and difference d of longest side and smallest allest side is 3.4.com/MathsByAmiya/photos/528057913912996/ 12996/ D is a sq square. circumscribing a circle .2 e.given@@ so sides w would be 1. 19.6 cm.44 d.5 . so sum of digits of SO only 24 Unit perimeter P is 6 https://www.2 = 19. NoT .com/Math . R-7 and R-14 Construct the same fig. Required area = .1/2@*5. 43.There is no such odd P ter we wil will get 12 different integral sided triangle.8 .facebook.radius@. NoT Since sides are in AP of right angl angles triangle so . Can Cannot be determined e. ing. 936 cm^2 c.???@ Sol: [e] ∆FOP . 3E Learning.2 & 9.com/ om/MathsByAmiya ©AMIYA KUMAR . 21. Then find the area of sh shaded region [take pi=22/7] a. 7. R .8*.4 5. 900 cm^2 b. 994 cm^2 d.facebook..44 Maths By Amiya. In the given fig.3. Since triangle ABC is isosceles so ∠ABC=∠ACB=80 And∠EBF=40 and ∠FCE=30 ∆¼B½ ≅ ∆[B½ By SAS Triangle DfAE congruent to Tri. Ans: . CAbE .30 + x=50.com/MathsByAmiya ©AMIYA KUMAR . 1980 d. 3E Learning. There are how many different integral angled triangles are possible a.a@ .77. ∠CFE + x=50 . x=20 Maths By Amiya. 2700 e. 180 b. NoT Sol: [d] https://www.facebook. www. case SAS@ So FE=CE and ∠CFE= ∠ECF =30 In ∆½¼[ .com/MathsByAmiya/photos/604561656262621/ 78. 1800 c.facebook. 45° e. NoT° Ans: [e] 150 Ans: [c] 75 Maths By Amiya. 30° b. ing. 75° d. ∠BPC= ? a. 75° d. 45° b. ∠APD =? a. C & D as shown in the fig then 79. www. NoT 81.com/ om/MathsByAmiya ©AMIYA KUMAR . 175° d. 150° 80. 105° e. 60° c. 90° e. 60° c.B. 3E Learning. ∠CPD =? a. 160° c. 130° b.Direction for questionss 78 to 8 84 4:-There are four congruent circles les with centres ce A.facebook. 5° e.Ans: [b] 60 82. 22. 75° d. 105° e. 7* sin60 d. AD is median AC= 14 cm then AD =? a. 20° d.com/ om/MathsByAmiya ©AMIYA KUMAR . ∠QSC =? a. NoT www. 60° c. ∠PAQ =? a. 60° c. 30° b.facebook. c. 75° d. ∠SBP =? a. 75° d. If in a ∆ABC . 30° b. NoT 83. NoT 85. 3E Learning. 30° b. 60° c. ∠QPD =? a. 45° e.5° 84. 15° c. 27. 7* sin90 d. 7* sin30 Sol: [d] 7 b. 30° b. 105° e. 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. ing. 7* sin45 Maths By Amiya. 72° c. If in the given fig P.be/JEKl2VN-oNY 88. BC=2*CP & 2*CA=3*QA 3*QA .80° e. 55: 29 d.sine formula@ 91. points D D. 36° b.∆PQR@: ar. 27:50 c. CA & AB.E & F are on sides BC.facebook. 4:1 e. NoT Ans: [d] =BC=CD=DE=EA then ∠BFD = ? 89. 40° d. 10 e. 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. 8 c.NoT Ans : [b] 72 90. ossible. 14:3 c. 29:55 b. 9 d. 221)=17http:/ http://youtu. If in the given fig AB=BC=CD= a. 13:3 d. NoT Ans: [b] . If in a triangle ∆ABC . 3E Learning. NoT Sol: [b] HCF (527. 11:3 b. 50:27 e. By ratio of side and aarea . 5 b. www.Q & R are o on the extended sides of BC.∆ABC@= ? a. ing.. such that AB:BR =1:1 . AC & AB such uch that BD:DC=1:2 B .sine formula@ Maths By Amiya. 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.com/ om/MathsByAmiya ©AMIYA KUMAR . NoT Ans: [d] By ratio of side ide and area .∆PQR@: ar. 30 e. 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. a.13 b.then then ar.31 d.87.17 c. 2.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. 2.14+7@ = 2*. If EF & AD intersect at point I. NoT Ans :[c] 95. 75 cm b. 80 cm c. If the distance between meeting point and home of three antsare39 cm. Can't be determined e. 3:2 e.64*10^7 lit d. 2:1 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 . A old rectangular carom board isa home of four ANTs. points D. NoT Sol: [b] Area of entrancement = . where a. 52 cm &60 cm. CA & AB . NoT Ans : [d] By MPG AI:IG:GD = 3:1:2 93.22/7@*21= 132m SO vol of entrenchment = 120*132 =15840 m^3 1m^3 = 1000 lit water So water capacity of entrenchment = 15840000 lit = 1. 1:2 b. Then how much litter water is used to fill the entrenchment. If in a ∆ABC . 85 cm d.E & F lies on side BC. 2:3 d. . They live in the each hole of the board.112 * 10^7 lit b.b& c are sides of ∆ and s is semi perimeter then *h L. 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. 1.*M.if all of them cover distances in straight line and ants which covers distance 39 cm and 52 cm live in opposite corners@ a. If in a ∆ABC . If Centroid "G" lies on AD and points E & F are mid points of their respective sides.584 * 10^7 lit 94. a.10@ = 120 m^2 [Shape is of an trapezoid] If we cut the entrancement and open wide it would be a prism . a+b+c= 30 cm . whose length = 2*pi*. which is completely field with water.1/2@*.92.14+10@*. then AI:GD =? a.584 * 10^7 lit c. @h L.*M@h L.*M@h . 30 b.h Lh L h a. 1 d.facebook. NoT www.com/MathsByAmiya ©AMIYA KUMAR . =? c. 15 e. 0 Ans: [c] Maths By Amiya. 3E Learning. Can't Be e. median AD= 2 cm then what would be length of median CF a. 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.5 cm Ans: [b] √601 = 24. Ç Æ whereA. 11cm c. 24. 27 cm 97. a. NoT Sol:DATA IN THE QUESTION IS WRONG In any triangle Âà + Äà + Åà = < > Èà+ ÈÃÄ + ÈÃÅ . If AC= 10 cm. right angle is at B. 24 cm b. 10 cm b. 12 cm d.5 c. In a right angled ∆ABC .96. 26 cm d. A &A are respective medians so in right angled at B . . 2 É I 1 = « ¬ .A. www.facebook. What would be ratio of inradius tocircum radius of a triangle whose angles are in the ratio of 1:1:4 a.1 + A1 + A1 @&ÈÄ = « ¬ ∗ Ä 3 à 1 1 1 1 1 ⇒ 10 = < > ∗ . √1 b. 3E Learning.2 + 5 + A @ ! x= 11 Alternative :In In Right angled triangle at "B" Èà+ ÈÃÅ = Ê ∗ ÈÃÄ 21 + A1 = 5 ∗ 51 ⇒A = 11 But median of any triangle cant be more than the hypotenuse .com/MathsByAmiya/photos/650118671706919/ Maths By Amiya.facebook. ! 1 Sol: [c] Angles are 30. MAIN REASON TO POST THIS QUESTION IS TO LET YOU KNOW ALL FOUR CONCEPTS 98.com/MathsByAmiya ©AMIYA KUMAR . √3 + < > ! 1 c. so the data of the question is wrong. √3 S < > d. 30 & 120 so 3 = cos 30 + cos 30 + cos 120 S 1 = √3 S « ¬ 2 H 0 https://www. NoT4 p#h L1#h 1 = 10√34 https://www. 60/47 d.facebook.com/MathsByAmiya ©AMIYA KUMAR .facebook. = ⇒ r= "# 0n + + = + + = ! 0 = 0n "# 101. such that AO= 4 cm. OD= 3cm & BO=6 cm then among option which could be a possible length of altitude CE a. BF & CE are the altitude of ∆ABC.7 b. 30 cm b. 30/47 c.facebook.01 comes in the range hence . 56 Sol: [c] We know.55 d.com/MathsByAmiya/photos/528568450528609/ 102. so diagonal = 2 ∗ √106 Maths By Amiya.com/MathsByAmiya/photos/794476340604484/ 100. 59 c. 3E Learning. www. Data adequate or NoT 1 1 1 1@ . 50 cm d. If two adjacent sides of a parallelogram is 10cm and 12 cm and one diagonal is 8 cm. AO*OD=BO*OE=CO*OF ⇒ 4*3 = 6*OF ⇒ OF = 2cm So. 2 ∗ √106 c. AD= 7cm & BF= 8cm Thus Range of CE would be 7∗8 7∗8 < [½ < ⇒ 3.4 Sol: By Apollonius . NoT And :d.√106 d.c@ https://www. 4cm & 5cm are altitudes of a triangle then what would be in radius of that triangle a.10 + 12 = 2 ∗ + . Then what is the length of another diagonal a. 47/30 Sol: [c] We know .733 < [½ < 56 7+8 8S7 By checking options only 55.99. 1/12 b. 8 b. If 3cm. 3. MN is||AB it divides trapezium in two equal halves then length of MN =? a. where a is half of another diagonal. If ABCD is a trapezium such that AB||CD and AB= 80cm & CD= 20 cm. 40 cm c. 54. If AD. Points M & N are n non parallel sides. where x is less than 180 degree . √8 cm d. NoT . √7 9A Sol: [b] If B8 = B[ then 9 ∗ C81 = 7 ∗ CI 1 ! 105. 4√2 cm b 8√2 cm Maths By Amiya. If in a plan there are four points such that AB=BC=CA=DA and angle BDC = x degree .103.2] c. AE = 10 cm & DE= 8cm then what is the distance between P & Q.3] d. c. If side of an equilateral triangle ABC is 3 cm and point D is on BC such that BD=1cm then AD=? c.[0. NoT www. AC & DE are common tangents to two given circles whose centre are P & Q as shown in fig. ]4.facebook.???@ a. a. ]1.43 f # #M # 1 ∗ 10> + < # 1 #M # ∗ 10> ∗ 10¬ ∗ ℎ = ∗ Q ∗ < > ∗ 5 ! 1 ∗ 10> ∗ 10¬ ∗ ℎ 104.5] 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 ≈ 0.4] e. then maximum possible value of x= ? Ans : 150 degree 106.1] b. 3E Learning. 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. such that AD= 6 cm .com/MathsByAmiya ©AMIYA KUMAR . 2. If AB. 4√5 cm d.7 cm b. ]2. 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. ]3. Assume D is origin So coordinate of Q = .com/MathsByAmiya/photos/812283092157142 92157142/ 107.Sol: [c] We have Radius of Circle ircle of ce centre Q = 2 cm [inradius of triangle. 2. 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 .4 c.-6.5 Sol: [b] = 6∗8 2 ∗ √61 + 81 or can apply = b.2.facebook. √5 d.4 .-2@ 2@ [IVth quadrant] & co-ordinate of P = . √6 e. 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. NoT No = 2. 2.-6@ 6@ [IIIrdQuard] So QP = 4√5 cm [By Distance Formula] https://www.com/M ok. '. +&'. www.facebook. ing. 3E Learning.com/ om/MathsByAmiya ©AMIYA KUMAR .' Maths By Amiya.' *'&+MÐ'+&'. E & F are points on side BC. NoT 112.E & F are points on sides BC=4cm. 6:5 e. 3E Learning. 15:19 e. D. NoT ∆ H ∗I∗9∗ ∗I∗9∗ ∗I∗9 & = ⇒ = = 4 ∗ ∆1 4∗∆ 4 ∗ ∗ . If in triangle ABC . S @ ∗ . CA=5cm& AB=3cm such that AD. S I@. 7:4 b. D lies on the side BC and on line AD . BD:DC=1:2 & BF:FA = 2:3 then BO:OE=? [if O is point of intersection of BE & DF 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. [if O is point of intersection of EF&AD] a. NoT 111.com/MathsByAmiya ©AMIYA KUMAR . 1:3 e. S 9@ ∗I∗9 = 4 ∗ . 40:19 c. If side ratio of a triangle is 10:11:12 then what is the ratio of circum radius to inradius Sol: H= a.108. Then AE:EC=? a. If in triangle ABC . D.facebook. 14:5 Ans: [c] b. S @ ∗ . BE & CFal angle bisector and I in in-centre then AO:IO =? . 3:1 d. and we dont know which two angles or sides are equal@ 109. If in a triangle ABC . 80:39 d. 5:6 d. CA & AB such that BE is median . 19:4 Ans : [c] b. b=11x& c=12x => = ÇØ Ô Õ Maths By Amiya. 4:7 c. Ö× www.since angle A=30 or 150 degree. 4:19 c. 4:15 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. If in an isosceles triangle ABC . S 9@ For a=10x . 5:15 c. S I@. 13:7 b. NoT Ans :[b] 110. Can't Say e. so area of polygon should uld be less than 472 . 20 b. ing.P series thenarea of ∆CB[ CB[ would be equal to a. 114. iagnle. What would be . Length of BD = ? a.33 Length of BE a.approx@ approx@ area of a regular convex polygon with 11 sided fig whose one side is 7 cm.are consecutive terms of 113. Best is coordinate rdinate ge geometry Maths By Amiya. 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. c. NoT d. Product of 2nd smallest est and 3rd smallest radius Sol: [d] ∆= :Õ ∗ Õ ∗ ÕÄ ∗ ÕÅ wher wherer is inradius and rest are exradius of a triagnle. 12 b. 458.Median 117. Sol:[c] For a given perimeter er . a. What is the maximum imum valu value of ratio of area of in-circle of a triangle to area of triangle. Square of 2nd smallest st radius d. pi c. e. 1 b. 25 Ans : [b] . a.3* sqrt. 3E Learning. NoT Ans : [c] . consider it as a circlee of circum circumference = 77cm so the area of this his circle is i approx472. BC= 40 & CA = 50 . If radii of all fourr circles ar a G. for equilateral triang triangle. then 116. Largest radius among all b.facebook.169 c.66 d. 18 Ans :[b] . 2 e. pi / . 18 √5 e. 115. 478. Area o of a circle is maximum . NoT www. 23. hence [c] Direction : If in a triangle angle AB = 30. 26.916 d. Product of smallest radius adius and 3rd largest radius c. 12√5 c.916 b.480 d.3@@ d. 472. Perimeter of this polygon po is 77 cm .com/ om/MathsByAmiya ©AMIYA KUMAR . angle PQR = 90°and S and T are points on PR such that PS^2 + TR^2 = ST^2 then angle SQT = ? . 9√3 d. 15 e. 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. What is the ratio of side of a hexagon to a square of maximum area inside the same hexagon a. NoT Sol: [d] Take Left turn as 90 degree and right turn as 270 as internal angle of a closed polygon. solve by options or coordinate or sine formula 121. by question r=20.5@ then what is its area a. Take Left turn as 270 degree and right turn as 90 as internal angle of a closed polygon. 30 b. 3 b. In a ∆ µÙHPQ=QR . 4 c. 16 c. 3 . If the equation of one side of an equilateral triangle is 3x+4y=5 and its one vertex is . we get l=16 So either 24 or 16. NoT Ans : [b] 122. www. 75 e.com/MathsByAmiya ©AMIYA KUMAR . 45 c. Best is coordinate geometry d. 3E Learning. 2 . we get l=24 Now. 3 . .n-2@ .118.5. 60 d.in degre@ a. and we know sum of all internal angles = 180. NoT Ans : [e] Maths By Amiya. a. 20 b.facebook. Length of DE = ? a.right turn| = 4 120. 6√3 b.n-2@ . and we know sum of all internal angles = 180.c@ e.root 3 b.l+r-2@ .l+r-2@ . n is vertex Here 90*l + 270*r = 180 . 5 Ans : [c] 5. Direct formula = |left turn .root 2 e.b@ or . if he never repeated or crossed a path which is already marked and never take more than one turn on a point. NoT Ans : [b] .root 2 c. 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. NoT 119.24 d. by question r=20. 12√3 c. 6 e. n is vertex Here 270*l + 90*r = 180 . 2 .root 3 d. 6 c. 3E Learning. C CD= 9 cm & EB= 3 cm. If perimeter of a triangle two sides are odd integer er rest is even. √122 9A c.https://www. 4 cm c.facebook. √12 e.(1/2) rad d. www. 1 rad b. pi .12 b. If length of minor arc OAB = ? (where O is thee cent centre of circle and A & B are on the centre) a. 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. 2√3 d.com/MathsByAmiya/photos/840413596010 13596010758/ 127.then angle 124. CE=2 E=2 cm.com/ om/MathsByAmiya ©AMIYA KUMAR . 12 cm b.facebook.com/MathsByAmiya/photos/584494198269367 98269367/ angle iis 24 cm then how many different triangles gles possible po where only 123.facebook. √129 9A d.com cebook. https://www. ing. a.6 degree Ans : [c] 125. 2pi -11 ra rad c. 6cm b. a. NoT Ans: [b]4 Maths By Amiya. NoT No Ans: [b] 126. NoT Ans : [d] rc crea created of a chord AB is equal to radius of the circle ci .com/M ok. 3 d.60 d. What would be diameter iameter o of circle whose two chords AB and d CD make angle of 90° at point E such that . n-1@
[email protected]/2@. 3 d.2a+. 17 c.facebook. here a=30For convex polygon we get only 2 set of values http://www.wolframalpha. There are how many convex polygon possible such that integral angles are in AP and integer and the smallest angle is 30 a. 4 e.n-2@=.facebook.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.128. 3E Learning. www.com/MathsByAmiya ©AMIYA KUMAR . Ans: 10 cm^2 https://www. 8 Sol: [e]NoT b. NoT 180*.com/MathsByAmiya/photos/509985899053531/ 129. √!M1 e. 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. CE= 5 130.4 cm e.16) so IA= √61 + 101 = √136 = 2√ √34 132. If in a ∆ABC.BF=7 cm. 60 b.com/MathsByAmiya/photos/859153280803456 03456 Ans: [b] 135. 2√17 Sol: [e] e. 3. 120 d. Not Maths By Amiya. 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. www. a. N NoT c..facebook. BC= 30 cm & CA= 34 cm and I is incetre etre of o the triangle then IA =? a. DE= 9 cmThen EF =? a.com ok. √!L0 1∗√! d. 33. 114 c. Sol: [c] √!M1 1∗√! b. 3 cm d.6)) and A=(0. 3√5 c. angle ACB=60°and side AC= A 6 cm 1∗√! a. 8 cm b.facebook. NoT https://www.com/ om/MathsByAmiya ©AMIYA KUMAR . 108 e. https://www. If angle ABC=anglee BCA cm . 1∗√! √!L1 c. so centre (6.6 Ans: [c] https://www. Data D inadequate e. 3E Learning. 6 cm d. 11440 cm^2 c. angle ABC=30°.facebook.BCA.com/Math /MathsByAmiya/photos/532772120108242/ 133. If in a triangle ABC. 1540 cm^2 d. ing.facebook. 4 cm b. NoT N inradius = 6 . 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.com/Math /MathsByAmiya/photos/853093624742755/ 131. NoT Ans: [e] cannot be determined 134. AB=16 cm cm.1600 cm^2 b. facebook.com/MathsByAmiya/photos/861355437249907/ Maths By Amiya.com/MathsByAmiya ©AMIYA KUMAR .Sol: [c] Area = (s – c)*s = 1540 cm^2 https://www. 3E Learning.facebook. www. c = 9 cm. given below is (are) sufficient to constructt an unique un triangle ABC. & C are angles & a. B = 45° . B. A= 60 60° II. a = 4cm. bb= 3 cm a. sin 1° . A= 60° . NoT Ans : [a] only II is possible 138. How many set of data giv [P= Perimeter.b & c are sides to corresponding nding angles] I. A= 60° III. a=b= 12 cm. A . How many statements nts are correct I. 137.136. By all we can construct unique triangle e. Two set of data c. A= 60° . Only one set of data b. B = 30° . C = 90° IV. Three sets of data d. P = 16 cm . sin 1 II. ° . III. ° . IV ° . V. sec 5° . sec 5 VI ° . facebook. ing. www. Ans: 4 Maths By Amiya.com/ . 3E Learning.com/MathsByAmiya ©AMIYA KUMAR . https://youtu. a. it is known own that th sum of any two elements is greater than third element and sum of all elements is 80. AQ:QP = 3:4 then QO:OJ = ? 144. PR:RJ Ans: 1:7 Maths By Amiya. ing.be/PvMh PvMhMly9cFY un and third side is a 140.QA:AB :AB = RB:BC =PC:PA = 1:3 . 22 e. www. In the given fig . question is of triangle ngle w whose perimeter is 80.com/ . 25 b. 24 c. 80 e. 3E Learning. if S is set having three element S are positive integers inn whic which two are odd and one is even. NoT Ans: [d] 141.-. How many triangless are ppossible whose two sides are 50 unit and 47 unit prime number. a. then ! ∆ #$% . NoT Ans: [c] .facebook.? ! ∆ &'( Ans: 36:13 )*+. In given ∆ ). How many differentt set S are possible.lements.QA:AB :AB = RB:BC =PC:PA = 1:3 .? ! ∆ &'( 142. 100 d. 90° 143. ? [in cm^2] Ans: 39 cm^2 :RJ = 11:3 . 23 d.com/MathsByAmiya ©AMIYA KUMAR . all elements of set 139. point D lies on AC such that angle A*+ . then ! ∆ #$% . 133 b. 33 c. In the given fig . If in a right angled ∆ )*+ AD= 4 cm & CD = 9 cm m then area of ∆ )*+ . 0123 *4+ . PR:RJ :RJ = 55:3 . www. AQ:QP = 3:5 then 5 &'7(9 5∎&'7( . PR:RJ :RJ = 44:3 . In given ∆ ). PR:RJ :RJ = 4:3 .com/ . In given ∆ ). PQ:QR:R Maths By Amiya.-. 3E Learning.? 5 #&89 5∆#&8 Ans: 25:52 147. PR:RJ :RJ = 55:3 . In given ∆ ). AS:SJ = 4:3 then PS:MN In given ∆ ). In given ∆ ).? 5∆(789 5∆ Ans: 1:1 146. AQ:QP = 3:5 then 5∆#7'9 5∆ . :RJ = 33:2 . ? :QR:RJ = 2:2:3 .-.145. ing. In given ∆ ).-.-.com/MathsByAmiya ©AMIYA KUMAR = . 148. QO:OJ =? Ans: 1:15 152. AO:OR = 4:3 then AQ:QP = ? In given ∆ ). AQ:QP = 3:5 & AS:SJ = 2:3 5 &'7(9 5∎&'7( .-. PQ:QR:R :QR:RJ = 2:2:3 . AO:OR = 4:3 then AQ:QP 151. In given ∆ ).facebook.? 5 #&89 5∆#&8 then 149. PR:RJ :RJ = 11:3 .-. AQ:QP = 3:5 & AS:SJ = 2:3 then 5∆'7#9 .-.-. AQ:QP = 3:4 then AO:AR = ? Ans: 1:1 In given ∆ ). AS:SJ = 4:3 then PM:MN = ? Ans: 10 : 7 153.-.? 5∆#7:9 Ans: 5:4 150. PR:RJ Ans: 8:15 :OJ = 3:2 . facebook. 4/3 b. 0. 6 cm c. 6√2 c. 2.5 e. O is the mind point of line segment DE then what is the distance length gth of BO (in cm) a. such that AO= 3 cm. How many differentt triang triangle are possible whose two altitudes aree 7 cm & 6cm and third altitude is an even number. NoT Ans: [c] eter of o triangle 157. AS:SJ = 4:3 then In given ∆ ). 2 & 2 are three altitu altitudes of a triangle then what is in-radius of the triangle a. 3 cm b. 12 b. NoT Maths By Amiya. PQ:QR:R 5∎'. 9/(4√7 7) c.154. If are three altitudess of a ttriangle is √3 cm then what is the perimeter a. :QR:RJ = 2:2:3 . in which AB= 9 cm & BC = 8 cm. 66√3 d. www. and AD. QR:RJ = 2:2:3 . CBD e.com/MathsByAmiya ©AMIYA KUMAR . 4. ∆ )*+ is a right angled gled tr triangle at B.<(9 . 3E Learning. triangle ABC . 6 b. 2 & 2 are three altitu altitudes of a triangle then what is the perimeter eter of the triangle a. 3/4 c. & BE are altitudes. (9√2) / 4 e.? 5∆#<:9 5∆ 155. NoT Ans: [b] 158.-. 1.. Ans: 19 If 3. 1. √7 c.-. √5 b. NoT Ans: [a] If 3.5 d. 162. 1 d. PQ:QR:R 5∆&<(9 5∆ . 3 e. cm Point D & E lie on AB & BC such that AD:D AD:DB = BC:BE = 2:1 .com/ .5 cm b. NoT N Ans: [d] 161.5 cm d. 7 e. If 3.< '. CBD e. If O is orthocentre off trian = 2 cm. BO = 6 cm thenn OE = ? a. 6 d.? 5∎(<: (<:89 Ans: 8:17 156. AS:SJ = 4:3 then In given ∆ ). OH 159. NoT Ans: [b] 160. 1 cm c.5 cm d. ing. 2 & 2 are three altitu altitudes of a triangle then what is the area off the triangle t a. BE = 9 cm & CF = 12 cm then angle A = ? a. 150 b. 67 b. If in ∆)*+ . a. 5:6 e. 15:28 d. then what at is th the approx probability to hit coloured (shaded) haded) area. AD = 6 cm. AD. NoT Ans: [c] 166. If in a @A0123 length triangle is a. AB= 100 cm . 31. 5:4 b. B BC= 4 cm & CA = 3 cm.com/ .com/MathsByAmiya ©AMIYA KUMAR . 85 c. If in a ∆ )*+ .facebook. www. 40 cm & 24 cm then hen largest la angle of 164. NoT Ans: [e] NoT it should be obt obtuse gth of altitudes are 30 cm . 30 e. AD. 40. If in ∆)*+ . 50% c. 12 cm^2 e. 6 cm^2 b.Ans : 6√2 163. 3E Learning. 90 d. 120 e.5% e. NoT Ans: [c] 167.. 45% Ans: [c] Maths By Amiya. ing. 10 cm^2 d. 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. BE & CF are altitudes such that. 20 % b. NoT Ans: [c] 165. AB= 2 cm . 115 c. 13: 30 c.5 % d. 8 cm^2 c. 90 d. BC= 6 cm & CA = 8 cm & I is the in centre of the triangle then what is the area of ∆)*B ? a. If ABCD is a squaree targe target and all arcs are quarter circle whose radius adius is equal to side of square ABCD. facebook. and lines DI. BC = 5 cm & AC = 7 cm m then find 168. What is the length off PB 176. 300 degree ree an angles then there are how many 60 degrees es angles ang are in this polygon. 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. EF = ? 169.com/ . www. 3E Learning. ing. Ratio of Area of F!GHIJ EKLMND to Area of ∆ #$% ly two types of internal angles. What is the length off PR 175. 172.com/MathsByAmiya ©AMIYA KUMAR . What is the length off PQ 174. ED = ? 170. If AB= 6 cm . What is the measurement ement of ∡PDQ 177. Ans: 33 If ABCD is a square off 10 cm then 173.Directions : If in given image circle is in-circle of ∆)*+. Ratio of Area of ∆ %DE to Area of ∆ #$% 171. EF||BC ||BC & GH|| AC. one is 60 degree ree another ano is 300 degree. EF & GH are tangents to the circle and DI || AB . If a polygon has only If there are 30. If ABC is a quarter circle and PQRS is square of side "a". com/MathsByAmiya www. 3rd Floor.3elearning. Above Dominos.To get more follow www.facebook.linkedin.com/groups/MBAMathsByAmiya To Follow Amiya : https://www.facebook. HB HB Road Ranchi.com/kumar.facebook.in Join Fb Group www. 9534 002244 .amiya http://in. Anand Complex.com/in/kumaramiya To Join Classroom Course For CAT-16/17 XAT-17/18 SSC/BANK PO 3E Learning.