{"id":43865,"date":"2020-12-31T16:28:37","date_gmt":"2020-12-31T10:58:37","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=43865"},"modified":"2020-12-31T16:28:37","modified_gmt":"2020-12-31T10:58:37","slug":"top-20-ssc-cgl-geometry-questions","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/top-20-ssc-cgl-geometry-questions\/","title":{"rendered":"Top-20 SSC CGL Geometry Questions"},"content":{"rendered":"

SSC CGL Geometry Questions<\/strong><\/h1>\n

Download Top-20 Geometry questions for SSC CGL exam. Most important\u00a0 Geometry questions based on asked questions in previous exam papers for SSC CGL.<\/p>\n

Download Top -20 SSC CGL geometry questions <\/a><\/p>\n

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Question 1:\u00a0<\/b>A tangent is drawn to a circle of radius 6 cm from a point situated at a distance of 10 cm from the centre of the circle. The length of the tangent will be<\/p>\n

a)\u00a07 cm<\/p>\n

b)\u00a04 cm<\/p>\n

c)\u00a05 cm<\/p>\n

d)\u00a08 cm<\/p>\n

Question 2:\u00a0<\/b>How many perfect squares lie between 120 and 300 ?<\/p>\n

a)\u00a05<\/p>\n

b)\u00a06<\/p>\n

c)\u00a07<\/p>\n

d)\u00a08<\/p>\n

Question 3:\u00a0<\/b>A copper wire of length 36 m and diameter 2 mm is melted to form a sphere. The radius of the sphere (in cm) is:<\/p>\n

a)\u00a02.5<\/p>\n

b)\u00a03<\/p>\n

c)\u00a03.5<\/p>\n

d)\u00a04<\/p>\n

Question 4:\u00a0<\/b>The product of two numbers is 45 and their difference is 4. The sum of squares of the two numbers is<\/p>\n

a)\u00a0135<\/p>\n

b)\u00a0240<\/p>\n

c)\u00a073<\/p>\n

d)\u00a0106<\/p>\n

Question 5:\u00a0<\/b>The square root of\u00a0$14 + 6\\sqrt{5}$<\/p>\n

a)\u00a0$2 + \\sqrt{5}$<\/p>\n

b)\u00a0$3 + \\sqrt{5}$<\/p>\n

c)\u00a0$5 + \\sqrt{3}$<\/p>\n

d)\u00a0$3 + 2\\sqrt{5}$<\/p>\n

Question 6:\u00a0<\/b>A copper wire is bent in the form of an equilateral triangle, and has an area $121\\sqrt{3}$ cm2\u00a0<\/sup>. If the same wire is bent into the form of a circle, the area(in cm2<\/sup>) enclosed by the wire in(Take $\\pi = \\frac{22}{7}$)<\/p>\n

a)\u00a0364.5<\/p>\n

b)\u00a0693.5<\/p>\n

c)\u00a0346.5<\/p>\n

d)\u00a0639.5<\/p>\n

Question 7:\u00a0<\/b>The square root of 0.09 is<\/p>\n

a)\u00a00.30<\/p>\n

b)\u00a00.03<\/p>\n

c)\u00a00.81<\/p>\n

d)\u00a00.081<\/p>\n

Question 8:\u00a0<\/b>If an obtuse-angled triangle ABC, is the obtuse angle and O is the orthocenter. If = 54$^{\\circ}$ , then is<\/p>\n

a)\u00a0108$^{\\circ}$<\/p>\n

b)\u00a0126$^{\\circ}$<\/p>\n

c)\u00a0136$^{\\circ}$<\/p>\n

d)\u00a0116$^{\\circ}$<\/p>\n

Question 9:\u00a0<\/b>If S is the circumcentre of and = 50$^{\\circ}$ , then the value of is<\/p>\n

a)\u00a020$^{\\circ}$<\/p>\n

b)\u00a040$^{\\circ}$<\/p>\n

c)\u00a060$^{\\circ}$<\/p>\n

d)\u00a080$^{\\circ}$<\/p>\n

Question 10:\u00a0<\/b>AC and BC are two equal cords of a circle. BA is produced to any point P and CP, when joined cuts the circle at T. Then<\/p>\n

a)\u00a0CT : TP = AB : CA<\/p>\n

b)\u00a0CT : TP = CA : AB<\/p>\n

c)\u00a0CT : CB = CA : CP<\/p>\n

d)\u00a0CT : CB = CP : CA<\/p>\n

Question 11:\u00a0<\/b>The external bisectors of and of meet at point P. If = 80$^{\\circ}$ , the is<\/p>\n

a)\u00a050$^{\\circ}$<\/p>\n

b)\u00a040$^{\\circ}$<\/p>\n

c)\u00a080$^{\\circ}$<\/p>\n

d)\u00a0100$^{\\circ}$<\/p>\n

Question 12:\u00a0<\/b>The total surface area of a sphere is $8\\pi$ square unit. The volume of the sphere is<\/p>\n

a)\u00a0$\\frac{8}{3}\\pi$<\/p>\n

b)\u00a0$8\\sqrt{3}\\pi$<\/p>\n

c)\u00a0$\\frac{8\\sqrt{3}}{5}\\pi$<\/p>\n

d)\u00a0$\\frac{8\\sqrt{2}}{3}\\pi$<\/p>\n

Question 13:\u00a0<\/b>A conical flask is full of water. The flask has base radius $r$ and height $h$. This water is poured into a cylindrical flask of base radius mr. The height of water in the cylindrical flask is<\/p>\n

a)\u00a0$\\frac{h}{2} m^2$<\/p>\n

b)\u00a0$\\frac{2h}{m}$<\/p>\n

c)\u00a0$\\frac{h}{3m^2}$<\/p>\n

d)\u00a0$\\frac{m}{2h}$<\/p>\n

Question 14:\u00a0<\/b>In a triangle ABC,\u00a0$\\angle{A} = 90^o , \\angle{C} = 55^o , \\overline{AD}$ is perpendicular to $\\overline{BC}$. What is the value of $\\angle{BAD}$ ?<\/p>\n

a)\u00a060 $^{\\circ}$<\/p>\n

b)\u00a045 $^{\\circ}$<\/p>\n

c)\u00a055 $^{\\circ}$<\/p>\n

d)\u00a035 $^{\\circ}$<\/p>\n

Question 15:\u00a0<\/b>If G is the centroid of triangle ABC and area of triangle ABC = 48cm2<\/sup>, then the area of triangle BGC is<\/p>\n

a)\u00a08 cm2<\/sup><\/p>\n

b)\u00a016 cm2<\/sup><\/p>\n

c)\u00a024 cm2<\/sup><\/p>\n

d)\u00a032 cm2<\/sup><\/p>\n

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Question 16:\u00a0<\/b>The diagonals AC and BD of a cyclic quadrilateral ABCD intersect each other at the point P. Then, it is always true that<\/p>\n

a)\u00a0AP . CP = BP . DP<\/p>\n

b)\u00a0AP . BP = CP . DP<\/p>\n

c)\u00a0AP . CD = AB . CP<\/p>\n

d)\u00a0BP . AB = CD . CP<\/p>\n

Question 17:\u00a0<\/b>A, B, C, D are four points on a circle. AC and BD intersect at a point such that $\\angle{BEC} = 130^o$ and $\\angle{ECD} = 20^o$. Then, $\\angle{BAC}$ is<\/p>\n

a)\u00a090 $^{\\circ}$<\/p>\n

b)\u00a0100 $^{\\circ}$<\/p>\n

c)\u00a0110 $^{\\circ}$<\/p>\n

d)\u00a0120 $^{\\circ}$<\/p>\n

Question 18:\u00a0<\/b>In a triangle, if three altitudes are equal, then the triangle is<\/p>\n

a)\u00a0equilateral<\/p>\n

b)\u00a0right<\/p>\n

c)\u00a0isosceles<\/p>\n

d)\u00a0obtuse<\/p>\n

Question 19:\u00a0<\/b>ABCD is a cyclic trapezium with AB || DC and AB = diameter of the circle. If angleCAB = $30^{\\circ}$, then angleADC is<\/p>\n

a)\u00a0$60^{\\circ}$<\/p>\n

b)\u00a0$120^{\\circ}$<\/p>\n

c)\u00a0$150^{\\circ}$<\/p>\n

d)\u00a0$30^{\\circ}$<\/p>\n

Question 20:\u00a0<\/b>ABC is a triangle. The bisectors of the internal angle $\\angle$B and external $\\angle$C intersect at D. If $\\angle$BDC=$50^{\\circ}$, then $\\angle$A is<\/p>\n

a)\u00a0$100^{\\circ}$<\/p>\n

b)\u00a0$90^{\\circ}$<\/p>\n

c)\u00a0$120^{\\circ}$<\/p>\n

d)\u00a0$60^{\\circ}$<\/p>\n

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Answers & Solutions:<\/strong><\/span><\/p>\n

1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

<\/p>\n

radius OA = 6 and OB = 10<\/p>\n

Now, in $\\triangle$OAB right angle at A<\/p>\n

AB = $\\sqrt{(OB)^2 – (OA)^2}$<\/p>\n

= $\\sqrt{100-36} = \\sqrt{64}$<\/p>\n

= 8 cm<\/p>\n

2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

let’s say n be a number whose perfect square lie between 120 and 300
\nhence 120<$n^{2}$<300
\nor $121\\leq n^{2} \\leq289$
\nor $11\\leq n^{2} \\leq17$<\/p>\n

3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

since we know volume will remain same while melting
\n$\\pi r_{1}^{2}h= \\frac{4}{3}\\pi r_{2}^{3}$
\nwhere $r_{1}$ is radius of cylinderical wire and $r_{2}$ is radius of sphere and h is length of wire
\nputting values we will get $r_{2}$ = 3 cm.<\/p>\n

4)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

As we know $(a-b)^{2}$ = $a^{2} + b^{2} – 2ab$
\nWe assume that first number is a and second number is b hence ab = 45
\nand a – b = 4
\nafter putiing values we will get\u00a0$a^{2} + b^{2}$ = 106<\/p>\n

5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Given question can be written as $9+5+6\\sqrt{5}$
\nor it will be square of\u00a0$3+\\sqrt{5}$<\/p>\n

6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Area of equilateral triangle is $\\frac{\\sqrt{3}}{4} a^{2}$ where a is side of triangle
\nwhich is equals to $121{\\sqrt{3}}$
\nor a = 22 and whole length of wire will be 66
\nfrom here when it is bend to make a circle, circumference will be $2\\pi r$ = 66
\nr = 10.5
\nhence area of circle will be $\\pi r^{2}$ = 346.5<\/p>\n

7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

0.09 = $\\frac{9}{100}$
\n$\\sqrt{\\frac{9}{100}} = \\frac{3}{10}$ = 0.3<\/p>\n

8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

<\/figure>\n

It is given that AC = BC, also $\\triangle$ PTB and\u00a0$\\triangle$\u00a0PAC are similar, we have\u00a0:<\/p>\n

$\\frac{CA}{CP}=\\frac{BT}{BP}$ —————-(i)<\/p>\n

Also, we have\u00a0$\\angle$\u00a0PBC =\u00a0$\\angle$\u00a0BTC ($\\because$\u00a0$\\angle$\u00a0PBC = $\\angle$ BAC = $\\angle$\u00a0BTC) and\u00a0$\\angle$\u00a0PCB =\u00a0$\\angle$\u00a0BCT<\/p>\n

=>\u00a0$\\triangle$\u00a0PBC $\\sim$ $\\triangle$\u00a0BTC<\/p>\n

Thus,\u00a0$\\frac{CB}{BP}=\\frac{CT}{BT}$<\/p>\n

=>\u00a0$\\frac{BT}{BP}=\\frac{CT}{CB}$ ————–(ii)<\/p>\n

From equations (i) and (ii), we get\u00a0:<\/p>\n

$\\frac{CA}{CP}=\\frac{CT}{CB}$<\/p>\n

=> Ans – (C)<\/p>\n

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11)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

12)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Let the radius of the sphere be ‘r’
\nHence $4\\pi r^2$ = $8\\pi$
\n=> r = $\\sqrt{2}$
\nVolume of the cube = $\\frac{4}{3}\\pi r^3$ = $\\frac{4}{3}\\pi \\sqrt{2}^3$ = $\\frac{8\\sqrt{2}}{3}\\pi$<\/p>\n

13)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Volume of water in flask = $\\frac{\\pi r^2 h}{3}$ (where h is height of water in conical flask)
\nVolume of water in cylinder = $\\pi m^2r^2 h_1$ (where h_1 is height of water in cylinderical flask)
\nHence now\u00a0$\\frac{\\pi r^2 h}{3}$ =\u00a0$\\pi m^2r^2 h_1$
\nor $h_1 = \\frac{h}{3m^2}$<\/p>\n

14)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

$\\angle A = 90^o$
\n$\\angle C = 55^o$
\n$\\angle B$ will be $180 – (90+55) = 35^o$
\nAs AD is perpendicular to BC Hence $\\angle BAD=180-(90+35)=55^o$<\/p>\n

15)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

As we know area of triangle, with centroid as one of the vertices and remaining 2 triangle vertices, is $\\frac{1}{3}$rd of the area of whole triangle.
\nHence area will be $\\frac{48}{3}$ = 16<\/p>\n

16)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

As we know that a cyclic quadrilateral can be inscribed into a circle, Hence in triangle APB and in triangle CPD.<\/p>\n

$\\angle PAB = \\angle PDC$ (same sector angles)<\/p>\n

$\\angle PCD = \\angle PBA$ (same sector angles)<\/p>\n

Hence third angle will also be equal and they will be similar triangles.<\/p>\n

So $\\frac{AP}{PD} = \\frac{BP}{PC}$<\/p>\n

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17)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Angle ABD will be equal to angle ACD = $20^o$ (same sector angles)
\nAngle BEC = $130^o$ so angle AED = $130^o$ (concurrent angles)
\nNow angle BEA will be $\\frac{360-130-130}{2} = 50^o$
\nSo angle EDC will be $180-(50+20) = 110^o$<\/p>\n

18)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

It is property of equilateral triangle that length of all its altitutes are equal.<\/p>\n

19)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

let\u00a0angle CDA = x<\/p>\n

\"\"<\/figure>\n

since AB is parallel to CD, angle ACD=30 and angle CAD=30<\/p>\n

in triangle ACD,<\/p>\n

sum of all three angles = 180<\/p>\n

30 + 30 + x = 180<\/p>\n

x = 120<\/p>\n

so the answer is option B.<\/p>\n

20)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

<\/figure>\n

In $\\triangle$ BDC,<\/p>\n

=> $y+(180^\\circ-2x+x)+50^\\circ=180^\\circ$<\/p>\n

=> $y-x+50^\\circ=0$<\/p>\n

=> $y-x=-50^\\circ$<\/p>\n

In $\\triangle$ ABC,<\/p>\n

=> $2y+(180^\\circ-2x)+\\angle A=180^\\circ$<\/p>\n

=> $2(y-x)+\\angle A=0$<\/p>\n

=> $2(-50^\\circ)+\\angle A=0$<\/p>\n

=> $\\angle A=100^\\circ$<\/p>\n

=> Ans – (A)<\/p>\n

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Download SSC Preparation App<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

SSC CGL Geometry Questions Download Top-20 Geometry questions for SSC CGL exam. Most important\u00a0 Geometry questions based on asked questions in previous exam papers for SSC CGL. Take a free SSC CGL Tier-1 mock test Download SSC CGL Tier-1 Previous Papers PDF Question 1:\u00a0A tangent is drawn to a circle of radius 6 cm from […]<\/p>\n","protected":false},"author":53,"featured_media":43868,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[9,504],"tags":[462,3575],"class_list":{"0":"post-43865","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-ssc","8":"category-ssc-cgl","9":"tag-ssc-cgl","10":"tag-ssc-cgl-2020"},"better_featured_image":{"id":43868,"alt_text":"","caption":"","description":"","media_type":"image","media_details":{"width":1309,"height":687,"file":"2020\/12\/fig-31-12-2020_06-34-35.jpg","sizes":{"medium":{"file":"fig-31-12-2020_06-34-35-300x157.jpg","width":300,"height":157,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/12\/fig-31-12-2020_06-34-35-300x157.jpg"},"large":{"file":"fig-31-12-2020_06-34-35-1024x537.jpg","width":1024,"height":537,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/12\/fig-31-12-2020_06-34-35-1024x537.jpg"},"thumbnail":{"file":"fig-31-12-2020_06-34-35-150x150.jpg","width":150,"height":150,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/12\/fig-31-12-2020_06-34-35-150x150.jpg"},"medium_large":{"file":"fig-31-12-2020_06-34-35-768x403.jpg","width":768,"height":403,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/12\/fig-31-12-2020_06-34-35-768x403.jpg"},"tiny-lazy":{"file":"fig-31-12-2020_06-34-35-30x16.jpg","width":30,"height":16,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/12\/fig-31-12-2020_06-34-35-30x16.jpg"},"td_218x150":{"file":"fig-31-12-2020_06-34-35-218x150.jpg","width":218,"height":150,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/12\/fig-31-12-2020_06-34-35-218x150.jpg"},"td_324x400":{"file":"fig-31-12-2020_06-34-35-324x400.jpg","width":324,"height":400,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/12\/fig-31-12-2020_06-34-35-324x400.jpg"},"td_696x0":{"file":"fig-31-12-2020_06-34-35-696x365.jpg","width":696,"height":365,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/12\/fig-31-12-2020_06-34-35-696x365.jpg"},"td_1068x0":{"file":"fig-31-12-2020_06-34-35-1068x561.jpg","width":1068,"height":561,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/12\/fig-31-12-2020_06-34-35-1068x561.jpg"},"td_0x420":{"file":"fig-31-12-2020_06-34-35-800x420.jpg","width":800,"height":420,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/12\/fig-31-12-2020_06-34-35-800x420.jpg"},"td_80x60":{"file":"fig-31-12-2020_06-34-35-80x60.jpg","width":80,"height":60,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/12\/fig-31-12-2020_06-34-35-80x60.jpg"},"td_100x70":{"file":"fig-31-12-2020_06-34-35-100x70.jpg","width":100,"height":70,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/12\/fig-31-12-2020_06-34-35-100x70.jpg"},"td_265x198":{"file":"fig-31-12-2020_06-34-35-265x198.jpg","width":265,"height":198,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/12\/fig-31-12-2020_06-34-35-265x198.jpg"},"td_324x160":{"file":"fig-31-12-2020_06-34-35-324x160.jpg","width":324,"height":160,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/12\/fig-31-12-2020_06-34-35-324x160.jpg"},"td_324x235":{"file":"fig-31-12-2020_06-34-35-324x235.jpg","width":324,"height":235,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/12\/fig-31-12-2020_06-34-35-324x235.jpg"},"td_356x220":{"file":"fig-31-12-2020_06-34-35-356x220.jpg","width":356,"height":220,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/12\/fig-31-12-2020_06-34-35-356x220.jpg"},"td_356x364":{"file":"fig-31-12-2020_06-34-35-356x364.jpg","width":356,"height":364,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/12\/fig-31-12-2020_06-34-35-356x364.jpg"},"td_533x261":{"file":"fig-31-12-2020_06-34-35-533x261.jpg","width":533,"height":261,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/12\/fig-31-12-2020_06-34-35-533x261.jpg"},"td_534x462":{"file":"fig-31-12-2020_06-34-35-534x462.jpg","width":534,"height":462,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/12\/fig-31-12-2020_06-34-35-534x462.jpg"},"td_696x385":{"file":"fig-31-12-2020_06-34-35-696x385.jpg","width":696,"height":385,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp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