{"id":39773,"date":"2019-12-30T11:25:43","date_gmt":"2019-12-30T05:55:43","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=39773"},"modified":"2019-12-30T11:26:52","modified_gmt":"2019-12-30T05:56:52","slug":"geometry-questions-for-xat","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/geometry-questions-for-xat\/","title":{"rendered":"Geometry Questions for XAT"},"content":{"rendered":"

Geometry Questions for XAT<\/strong><\/span><\/h1>\n

Download important Geometry Questions for XAT PDF based on previously asked questions in the XAT exam. Practice Geometry Questions\u00a0 PDF for the XAT exam.<\/p>\n

Download Geometry Questions for XAT<\/a><\/p>\n

Get 5 XAT mocks for Rs. 299. Enroll here<\/a><\/p>\n

Question 1:\u00a0<\/b>Three horses are grazing within a semi-circular field. In the diagram given below, AB is the diameter of the semi-circular field with center at O. Horses are tied up at P, R and S such that PO and RO are the radii of semi-circles with centers at P and R respectively, and S is the center of the circle touching the two semi-circles with diameters AO and OB. The horses tied at P and R can graze within the respective semi-circles and the horse tied at S can graze within the circle centred at S. The percentage of the area of the semi-circle with diameter AB that cannot be grazed by the horses is nearest to
\n<\/p>\n

a)\u00a020<\/p>\n

b)\u00a028<\/p>\n

c)\u00a036<\/p>\n

d)\u00a040<\/p>\n

Question 2:\u00a0<\/b>In the figure below, AB is the chord of a circle with center O. AB is extended to C such that BC = OB. The straight line CO is produced to meet the circle at D. If $\\angle{ACD}$ = y degrees and $\\angle{AOD}$ = x degrees such that x = ky, then the value of k is
\n<\/p>\n

a)\u00a03<\/p>\n

b)\u00a02<\/p>\n

c)\u00a01<\/p>\n

d)\u00a0None of the above.<\/p>\n

Question 3:\u00a0<\/b>If the lengths of diagonals DF, AG and CE of the cube shown in the adjoining figure are equal to the three sides of a triangle, then the radius of the circle circumscribing that triangle will be?
\n\"\"<\/p>\n

a)\u00a0equal to the side of the cube<\/p>\n

b)\u00a0$\\sqrt 3$ times the side of the cube<\/p>\n

c)\u00a01\/$\\sqrt 3$ times the side of the cube<\/p>\n

d)\u00a0impossible to find from the given information<\/p>\n

Question 4:\u00a0<\/b>A circle with radius 2 is placed against a right angle. Another smaller circle is also placed as shown in the adjoining figure. What is the radius of the smaller circle?
\n\"\"<\/p>\n

a)\u00a0$3 – 2 \\sqrt{2}$<\/p>\n

b)\u00a0$4 – 2 \\sqrt{2}$<\/p>\n

c)\u00a0$7- 4 \\sqrt{2}$<\/p>\n

d)\u00a0$6- 4 \\sqrt{2}$<\/p>\n

Question 5:\u00a0<\/b>\"\"
\nIn the adjoining figure, chord ED is parallel to the diameter AC of the circle. If angle CBE = 65\u00b0, then what is the value of angle DEC?<\/p>\n

a)\u00a035<\/p>\n

b)\u00a055<\/p>\n

c)\u00a045<\/p>\n

d)\u00a025<\/p>\n

XAT Solved Previous papers<\/a><\/p>
\n

Take XAT Mock Test<\/a><\/p><\/p>\n

Question 6:\u00a0<\/b>The figure shows a circle of diameter AB and radius 6.5 cm. If chord CA is 5 cm long, find the area of triangle ABC.
\n\"\"<\/p>\n

a)\u00a060 sq. cm<\/p>\n

b)\u00a030 sq. cm<\/p>\n

c)\u00a040 sq. cm<\/p>\n

d)\u00a052 sq. cm<\/p>\n

Question 7:\u00a0<\/b>\"\"<\/p>\n

AB is the diameter of the given circle, while points C and D lie on the circumference as shown. If AB is 15 cm, AC is 12 cm and BD is 9 cm, find the area of the quadrilateral ACBD.<\/p>\n

a)\u00a054sq. cm<\/p>\n

b)\u00a0216sq. cm<\/p>\n

c)\u00a0162sq. cm<\/p>\n

d)\u00a0None of these<\/p>\n

Question 8:\u00a0<\/b>In the adjoining figure, points A, B, C and D lie on the circle. AD = 24 and BC = 12. What is the ratio of the area of CBE to that of ADE?
\n\"\"<\/p>\n

a)\u00a01 : 4<\/p>\n

b)\u00a01 : 2<\/p>\n

c)\u00a01 : 3<\/p>\n

d)\u00a0Data insufficient<\/p>\n

Question 9:\u00a0<\/b>Three circles, each of radius 20, have centres at P, Q and R. Further, AB = 5, CD = 10 and EF = 12. What is the perimeter of \u0394PQR ?
\n\"\"<\/p>\n

a)\u00a0120<\/p>\n

b)\u00a066<\/p>\n

c)\u00a093<\/p>\n

d)\u00a087<\/p>\n

Question 10:\u00a0<\/b><\/p>\n

<\/figure>\n

In a circular field, AOB and COD are two mutually perpendicular diameters having length of 4 meters. X is the mid – point of OA. Y is the point on the circumference such that \u2220YOD = 30\u00b0. Which of the following correctly gives the relation among the three alternate paths from X to Y?<\/p>\n

a)\u00a0XOBY : XODY : XADY :: 5.15 : 4.50 : 5.06<\/p>\n

b)\u00a0XADY : XODY : XOBY :: 6.25 : 5.34 : 4.24<\/p>\n

c)\u00a0XODY : XOBY : XADY :: 4.04 : 5.35 : 5.25<\/p>\n

d)\u00a0XADY : XOBY : XODY :: 5.19 : 5.09 : 4.04<\/p>\n

e)\u00a0XOBY : XADY : XODY :: 5.06 : 5.15 : 4.50<\/p>\n

XAT Decision making practice questions<\/a><\/p>\n

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

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

\"\"<\/p>\n

Let R be radius of big circle and r be radius of circle with centre S. Radius of 2 semicircles is R\/2.<\/p>\n

From Right\u00a0angled triangle OPS, using pythagoras theorem we get<\/p>\n

$(r+0.5R)^2 = (0.5R)^2 + (R-r)^2$ . We get R=3r .<\/p>\n

Now the area of big semicircle that cannot be grazed is Area of big S.C – area of 2 semicircle – area of small circle = $\\pi*R^2$\/2 – 2*$\\pi*(0.5R)^2$\/2-$\\pi*r^2$ =\u00a0$\\pi*R^2$\/2 – 2*$\\pi*(0.5R)^2$\/2-$\\pi*(R\/3)^2$=\u00a0$\\pi*R^2$\/2 – $\\pi*(R)^2$\/4-$\\pi*(R)^2$\/9 =\u00a05*$\\pi*R^2$\/36.\u00a0 this is about 28 % of the area $\\pi*R^2$\/2 . Hence option B.<\/p>\n

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

\"\"
\nSince Angle BOC = Angle BCO = y.<\/p>\n

Angle OBC = 180-2y .<\/p>\n

Hence Angle ABO = \u00a0z = 2y = Angle OAB.<\/p>\n

Now since x is exterior angle of triangle AOC .<\/p>\n

We have x = z + y = 3y.<\/p>\n

Hence option A.<\/p>\n

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

Consider side of the cube as x.<\/p>\n

So diagonal will be of length $\\sqrt{3}$ * x.<\/p>\n

Now if diagonals are side of equilateral triangle we get area = 3*$\\sqrt{3}*x^2$ \/4 .<\/p>\n

Also in a triangle<\/p>\n

4 * Area * R = Product of sides<\/p>\n

4*\u00a03*$\\sqrt{3}*x^2$ \/4 * R = .3*$\\sqrt{3}*x^3$<\/p>\n

R = x<\/p>\n

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

\"\"
\nUsing pythagoras we can find diagonal of small square shown in fig. = 2*$\\sqrt2$ . Now this is equal to 2 + r + \u00a0$\\sqrt2$ *r. Equating we get r =\u00a0$6- 4 \\sqrt{2}$<\/p>\n

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

\"\"
\nIf EBC = 65 then EOC = 130 then OEC = OCE = 25. NOw since OC and ED are parallel we have OCE = OED = 25. Hence option D.<\/p>\n

XAT Quant formulas PDF<\/a><\/p>
\n

Take XAT Mock Test<\/a><\/p><\/p>\n

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

According to given dimensions, triangle will be a right angled triangle.
\nSo BC = 12
\nAnd area = $\\frac{1}{2} \\times 12 \\times 5$
\n= 30<\/p>\n

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

Since ACBD is cyclic quadrilateral with diagonals as AB = 15 ad CD.
\nSo area = $\\frac{1}{2} \\times (AB) \\times (CD)$ \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0eq.(1)
\nTriangle ACB and ABD is right angled triangle
\nLet’s say angle CBA = $\\theta$
\nso $CD = 2 \\times (9 sin\\theta)$ \u00a0(Where\u00a0\u00a0$sin\\theta$ = $\\frac{12}{15}$)
\nCD = $\\frac{72}{5}$
\nPutting values in eq. (1) , we will get area = 108 sq.cm.
\nHence answer will be D)<\/p>\n

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

As we know angles of same sectors are equal
\nHence angle B and angle D will be equal. Angle BCE and angle EAD will be equal.
\nSo triangles BCE and EAD will be similar triangles with sides ratio as 12:24 or 1:2.
\nArea will be in ratio of 1:4.<\/p>\n

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

As radius of circle is 20, lengths of PR, QR and PQ will be (15+5+15) , (10+10+10) , (8+12+8) respectively.
\nSo perimeter will be = 28 + 30 + 35 = 93<\/p>\n

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

XADY = XA + AD + DY = 2\/2 + (2 * 3.14 * 2)\/4 + (30\/360) * (2 * 3.14 * 2) = 5.19
\nXOBY = XO + OB + BY = 2\/2 + 2 + (60\/360) * (2 * 3.14 * 2) = 5.09
\nXODY = XO + OD + DY = 2\/2 + 2 + (30\/360) * (2 * 3.14 * 2) =\u00a0 4.04
\nHence, option D is the correct answer.<\/p>\n

XAT Solved Previous papers<\/a><\/p>
\n

Take XAT Mock Test<\/a><\/p><\/p>\n

We hope this Geometry QuestionsN PDF for XAT with Solutions will be helpful to you.<\/p>\n","protected":false},"excerpt":{"rendered":"

Geometry Questions for XAT Download important Geometry Questions for XAT PDF based on previously asked questions in the XAT exam. Practice Geometry Questions\u00a0 PDF for the XAT exam. Question 1:\u00a0Three horses are grazing within a semi-circular field. In the diagram given below, AB is the diameter of the semi-circular field with center at O. Horses […]<\/p>\n","protected":false},"author":42,"featured_media":39775,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[3,350,362,366],"tags":[3417,1425],"class_list":{"0":"post-39773","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-cat","8":"category-iift","9":"category-snap","10":"category-xat","11":"tag-geometry-questions-for-xat","12":"tag-xat"},"better_featured_image":{"id":39775,"alt_text":"Geometry Questions for XAT","caption":"Geometry Questions for XAT\n","description":"Geometry Questions for XAT\n","media_type":"image","media_details":{"width":1200,"height":630,"file":"2019\/12\/fig-28-12-2019_06-06-54.jpg","sizes":{"thumbnail":{"file":"fig-28-12-2019_06-06-54-150x150.jpg","width":150,"height":150,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/12\/fig-28-12-2019_06-06-54-150x150.jpg"},"medium":{"file":"fig-28-12-2019_06-06-54-300x158.jpg","width":300,"height":158,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/12\/fig-28-12-2019_06-06-54-300x158.jpg"},"medium_large":{"file":"fig-28-12-2019_06-06-54-768x403.jpg","width":768,"height":403,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/12\/fig-28-12-2019_06-06-54-768x403.jpg"},"large":{"file":"fig-28-12-2019_06-06-54-1024x538.jpg","width":1024,"height":538,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/12\/fig-28-12-2019_06-06-54-1024x538.jpg"},"tiny-lazy":{"file":"fig-28-12-2019_06-06-54-30x16.jpg","width":30,"height":16,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/12\/fig-28-12-2019_06-06-54-30x16.jpg"},"td_80x60":{"file":"fig-28-12-2019_06-06-54-80x60.jpg","width":80,"height":60,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/12\/fig-28-12-2019_06-06-54-80x60.jpg"},"td_100x70":{"file":"fig-28-12-2019_06-06-54-100x70.jpg","width":100,"height":70,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/12\/fig-28-12-2019_06-06-54-100x70.jpg"},"td_218x150":{"file":"fig-28-12-2019_06-06-54-218x150.jpg","width":218,"height":150,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/12\/fig-28-12-2019_06-06-54-218x150.jpg"},"td_265x198":{"file":"fig-28-12-2019_06-06-54-265x198.jpg","width":265,"height":198,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/12\/fig-28-12-2019_06-06-54-265x198.jpg"},"td_324x160":{"file":"fig-28-12-2019_06-06-54-324x160.jpg","width":324,"height":160,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/12\/fig-28-12-2019_06-06-54-324x160.jpg"},"td_324x235":{"file":"fig-28-12-2019_06-06-54-324x235.jpg","width":324,"height":235,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/12\/fig-28-12-2019_06-06-54-324x235.jpg"},"td_324x400":{"file":"fig-28-12-2019_06-06-54-324x400.jpg","width":324,"height":400,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/12\/fig-28-12-2019_06-06-54-324x400.jpg"},"td_356x220":{"file":"fig-28-12-2019_06-06-54-356x220.jpg","width":356,"height":220,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/12\/fig-28-12-2019_06-06-54-356x220.jpg"},"td_356x364":{"file":"fig-28-12-2019_06-06-54-356x364.jpg","width":356,"height":364,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/12\/fig-28-12-2019_06-06-54-356x364.jpg"},"td_533x261":{"file":"fig-28-12-2019_06-06-54-533x261.jpg","width":533,"height":261,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/12\/fig-28-12-2019_06-06-54-533x261.jpg"},"td_534x462":{"file":"fig-28-12-2019_06-06-54-534x462.jpg","width":534,"height":462,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/12\/fig-28-12-2019_06-06-54-534x462.jpg"},"td_696x0":{"file":"fig-28-12-2019_06-06-54-696x365.jpg","width":696,"height":365,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/12\/fig-28-12-2019_06-06-54-696x365.jpg"},"td_696x385":{"file":"fig-28-12-2019_06-06-54-696x385.jpg","width":696,"height":385,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/12\/fig-28-12-2019_06-06-54-696x385.jpg"},"td_741x486":{"file":"fig-28-12-2019_06-06-54-741x486.jpg","width":741,"height":486,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/12\/fig-28-12-2019_06-06-54-741x486.jpg"},"td_1068x580":{"file":"fig-28-12-2019_06-06-54-1068x580.jpg","width":1068,"height":580,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/12\/fig-28-12-2019_06-06-54-1068x580.jpg"},"td_1068x0":{"file":"fig-28-12-2019_06-06-54-1068x561.jpg","width":1068,"height":561,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/12\/fig-28-12-2019_06-06-54-1068x561.jpg"},"td_0x420":{"file":"fig-28-12-2019_06-06-54-800x420.jpg","width":800,"height":420,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/12\/fig-28-12-2019_06-06-54-800x420.jpg"}},"image_meta":{"aperture":"0","credit":"","camera":"","caption":"","created_timestamp":"0","copyright":"","focal_length":"0","iso":"0","shutter_speed":"0","title":"","orientation":"0"}},"post":null,"source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/12\/fig-28-12-2019_06-06-54.jpg"},"yoast_head":"\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\n\n\n