{"id":25932,"date":"2019-03-11T18:48:44","date_gmt":"2019-03-11T13:18:44","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=25932"},"modified":"2019-03-11T18:49:25","modified_gmt":"2019-03-11T13:19:25","slug":"ssc-cgl-circle-questions-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/ssc-cgl-circle-questions-pdf\/","title":{"rendered":"SSC CGL Circle Questions PDF"},"content":{"rendered":"

SSC CGL Circle Questions PDF:<\/strong><\/span><\/h1>\n

Download SSC CGL Circle questions with answers PDF based on previous papers very useful for SSC CGL exams. 25 Very important Circle objective questions for SSC exams.<\/p>\n

Download SSC CGL Circle Questions PDF<\/a><\/p>\n

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Question 1:\u00a0<\/b>A copper wire is bent in the form of square with an area of 121 cm 2. If the same wire is bent in the form of a circle, the radius (in cm) of the circle is<\/p>\n

a)\u00a07<\/p>\n

b)\u00a010<\/p>\n

c)\u00a011<\/p>\n

d)\u00a014<\/p>\n

Question 2:\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 3:\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 4:\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 5:\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

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Question 6:\u00a0<\/b>AB is the chord of a circle with centre O and DOC is a line segement originating from a point D on the circle and intersecting AB produced at C such that BC = OD. If $\\angle$BCD =$20^{\\circ}$, then $\\angle$AOD =?<\/p>\n

a)\u00a0$20^{\\circ}$<\/p>\n

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

c)\u00a0$40^{\\circ}$<\/p>\n

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

Question 7:\u00a0<\/b>A square ABCD is inscribed in a circle of unit radius. Semicircles are described on each side of a diameter. The area of the region bounded by the four semicircles and the circle is<\/p>\n

a)\u00a01 sq. unit<\/p>\n

b)\u00a02 sq. unit<\/p>\n

c)\u00a01.5 sq. unit<\/p>\n

d)\u00a02.5 sq. unit<\/p>\n

Question 8:\u00a0<\/b>Two circles touch each other internally. Their radii are 2 cm and 3 cm. The biggest chord of the greater circle which is outside the inner circle is if length<\/p>\n

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

b)\u00a0$3\\sqrt{2}$ cm<\/p>\n

c)\u00a0$2\\sqrt{3}$ cm<\/p>\n

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

Question 9:\u00a0<\/b>What is the length of the arc if it subtends 48\u00b0 at the center of a circle with radius r = 7 cm.<\/p>\n

a)\u00a07.28 cm<\/p>\n

b)\u00a08.16 cm<\/p>\n

c)\u00a05.867 cm<\/p>\n

d)\u00a06.48 cm<\/p>\n

Question 10:\u00a0<\/b>What is the length of the chord of a circle of radius 5 cm, if the perpendicular distance between centre and chord is 4 cm.<\/p>\n

a)\u00a09 cm<\/p>\n

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

c)\u00a06 cm<\/p>\n

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

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Question 11:\u00a0<\/b>A triangle has sides 5 cm, 12 cm and 13 cm. What is the approximate area of the circumcircle of this triangle?<\/p>\n

a)\u00a0120 $cm^2$<\/p>\n

b)\u00a0128 $cm^2$<\/p>\n

c)\u00a0132 $cm^2$<\/p>\n

d)\u00a0136 $cm^2$<\/p>\n

Question 12:\u00a0<\/b>If the perpendicular distance between the centre of a circle and a chord of length 8m is 3m, what is the radius of the circle?<\/p>\n

a)\u00a07m<\/p>\n

b)\u00a04m<\/p>\n

c)\u00a05m<\/p>\n

d)\u00a03m<\/p>\n

Question 13:\u00a0<\/b>What is the area of the sector subtending 36\u00b0 at the centre of a circle with a radius of 7cm ?<\/p>\n

a)\u00a016.8 sq.cm<\/p>\n

b)\u00a015.4 sq.cm<\/p>\n

c)\u00a017.2 sq.cm<\/p>\n

d)\u00a018.6 sq.cm<\/p>\n

Question 14:\u00a0<\/b>A certain length of a wire was bent to form the shape of a square, the area of the. square thus formed is 81 sq-cm. Now, the wire is rebent into a semicircular shape, calculate the area of the semicircle formed ( in sq-cm )<\/p>\n

a)\u00a088 sq-cm<\/p>\n

b)\u00a022 sq-cm<\/p>\n

c)\u00a0154 sq-cm<\/p>\n

d)\u00a077 sq-cm<\/p>\n

Question 15:\u00a0<\/b>A straight copper wire has been bent to form an equilateral triangle of area $121\\sqrt{3}\\text{ cm}$ . Find the Area, if the wire is bent in the form of a circle<\/p>\n

a)\u00a0330.5 sq-cm<\/p>\n

b)\u00a0350.5 sq-cm<\/p>\n

c)\u00a0346.5 sq-cm<\/p>\n

d)\u00a0364.5 sq-cm<\/p>\n

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Question 16:\u00a0<\/b>There are two concentric circle with radii 5 cm and 13 cm. What is the length of the chord of the outer circle which touches the inner circle?<\/p>\n

a)\u00a012 cm<\/p>\n

b)\u00a018 cm<\/p>\n

c)\u00a024 cm<\/p>\n

d)\u00a030 cm<\/p>\n

Question 17:\u00a0<\/b>The area of a circle is 50 $cm^2$. What is the approximate perimeter of the circle?<\/p>\n

a)\u00a020 cm<\/p>\n

b)\u00a025 cm<\/p>\n

c)\u00a030 cm<\/p>\n

d)\u00a035 cm<\/p>\n

Question 18:\u00a0<\/b>A thread of length 1256 cm is used to form a circle. What will be the diameter of the circle formed?<\/p>\n

a)\u00a0200 cm<\/p>\n

b)\u00a0500 cm<\/p>\n

c)\u00a0300 cm<\/p>\n

d)\u00a0400 cm<\/p>\n

Question 19:\u00a0<\/b>Two circles externally touch each other. The sum of their areas is $520\\pi$ sq-cm and the distance between their center is $28$ cm. The radius of the smaller circle is<\/p>\n

a)\u00a0$4$cm<\/p>\n

b)\u00a0$8$cm<\/p>\n

c)\u00a0$10$cm<\/p>\n

d)\u00a0$6$cm<\/p>\n

Question 20:\u00a0<\/b>A circle is circumscribed around an equilateral triangle of side 3 cm. What is the area of the circle (in $cm^2$)?<\/p>\n

a)\u00a0$\\sqrt3 \\pi$<\/p>\n

b)\u00a0$3 \\pi$<\/p>\n

c)\u00a0$3\\sqrt3 \\pi$<\/p>\n

d)\u00a0$9\\pi$<\/p>\n

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Question 21:\u00a0<\/b>If the length of the side of a square is equal to the diameter of a circle, what is the ratio of area of square to that of the circle?<\/p>\n

a)\u00a0$2:\\pi$<\/p>\n

b)\u00a0$\\pi : 4$<\/p>\n

c)\u00a0$4:\\pi$<\/p>\n

d)\u00a0$1:\\pi$<\/p>\n

Question 22:\u00a0<\/b>The distance between the centres of the two circles of radii r1, and r2 is d. They will touch each other internally if<\/p>\n

a)\u00a0$d = r_1$ or $r_2$<\/p>\n

b)\u00a0$d = r_1 + r_2$<\/p>\n

c)\u00a0$d = r_1 – r_2$<\/p>\n

d)\u00a0$d = \\sqrt{ r_1 r_2}$<\/p>\n

Question 23:\u00a0<\/b>\u0394ABC is a right angled triangle with AB = 6 cm, AC = 8 cm, \u2220BAC = 90\u2032. Then the radius of the incircle is<\/p>\n

a)\u00a04 cm.<\/p>\n

b)\u00a02 cm.<\/p>\n

c)\u00a06 cm.<\/p>\n

d)\u00a03 cm.<\/p>\n

Question 24:\u00a0<\/b>O is the circumcentre of the triangle ABC and \u2220BAC = 85\u00b0, \u2220BCA = 75\u00b0, then the value of \u2220OAC is<\/p>\n

a)\u00a055\u00b0<\/p>\n

b)\u00a0150\u00b0<\/p>\n

c)\u00a020\u00b0<\/p>\n

d)\u00a070\u00b0<\/p>\n

Question 25:\u00a0<\/b>Two chords of length a unit and b unit of a circle make angles 60\u00b0 and 90\u00b0 at the centre of a circle respectively, then the correct relation is<\/p>\n

a)\u00a0b = 3\/2a<\/p>\n

b)\u00a0b = \u221a2a<\/p>\n

c)\u00a0b = 2a<\/p>\n

d)\u00a0b = \u221a3a<\/p>\n

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

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

Area of square is $121 = (l)^2$ (where $l$ is length of side)
\nSo length will be = 11
\nTotal length of wire will be = 44
\nSo when it forms a circle, perimeter will be equal to 44
\nHence $2\\pi r$ = 44
\nor $r$ = 7<\/p>\n

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

<\/figure>\n

It is given that AC = BC, also $\\triangle$ PTB and $\\triangle$ PAC are similar, we have :
\n$\\frac{CA}{CP}=\\frac{BT}{BP}$ —————-(i)
\nAlso, we have $\\angle$ PBC = $\\angle$ BTC ($\\because$ $\\angle$ PBC = $\\angle$ BAC = $\\angle$ BTC) and $\\angle$ PCB = $\\angle$ BCT
\n=> $\\triangle$ PBC $\\sim$ $\\triangle$ BTC
\nThus, $\\frac{CB}{BP}=\\frac{CT}{BT}$
\n=> $\\frac{BT}{BP}=\\frac{CT}{CB}$ ————–(ii)
\nFrom equations (i) and (ii), we get :
\n$\\frac{CA}{CP}=\\frac{CT}{CB}$
\n=> Ans – (C)<\/p>\n

3)\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.
\n$\\angle PAB = \\angle PDC$ (same sector angles)
\n$\\angle PCD = \\angle PBA$ (same sector angles)
\nHence third angle will also be equal and they will be similar triangles.
\nSo $\\frac{AP}{PD} = \\frac{BP}{PC}$<\/p>\n

4)\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

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

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

\"\"<\/figure>\n

since AB is parallel to CD, angle ACD=30 and angle CAD=30
\nin triangle ACD,
\nsum of all three angles = 180
\n30 + 30 + x = 180
\nx = 120
\nso the answer is option B.<\/p>\n

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

<\/figure>\n

It is given that OD = BC and OD = OB (radii of circle)
\n=> OB = BC
\n=> $\\angle$ BCO = $\\angle$ BOC = $20^\\circ$ (Angle opposite to equal sides are equal)
\nThen, $\\angle$ OBC = $180^\\circ-(\\angle BCO +\\angle BOC)$
\n=> $\\angle$ OBC = $180^\\circ-20^\\circ-20^\\circ=140^\\circ$
\nAlso, $\\angle$ OBA + $\\angle$ OBC = $180^\\circ$ (Linear pair)
\n=> $\\angle$ OBA = $\\angle$ OAB = $180^\\circ-140^\\circ=40^\\circ$
\nNow, $\\angle$ AOB = $180^\\circ-(\\angle OAB +\\angle OBA)$
\n=> $\\angle$ AOB = $180^\\circ-40^\\circ-40^\\circ=100^\\circ$
\n$\\therefore$ $\\angle$ AOD = $180^\\circ-(\\angle AOB +\\angle BOC)$ (Linear pair)
\n= $180^\\circ-100^\\circ-20^\\circ=60^\\circ$
\n=> Ans – (D)<\/p>\n

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

<\/figure>\n

Radius of the circle = 1 unit, => Diameter = BD = 2 units
\nThus, side of square = AB = $\\sqrt2$ units
\nRadius of a semi-circle = $\\frac{AB}{2}=\\frac{\\sqrt2}{2}=\\frac{1}{\\sqrt2}$
\n=> Area of 4 semi-circles = $2\\pi r^2$
\n= $2\\pi (\\frac{1}{\\sqrt2})^2=\\pi$ sq. units ————(i)
\nArea bounded by region = Area of circle – Area of square
\n= $\\pi(1)^2-(\\sqrt2)^2=(\\pi-2)$ sq. units —————(ii)
\n$\\therefore$ Required area bounded by 4 semi circles = (i) – (ii)
\n= $\\pi – (\\pi-2) = 2$ sq. units
\n=> Ans – (B)<\/p>\n

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


\nThe biggest chord lying outside the inner circle must be tangential to it.
\nBy pytagoras theorem,
\n$x = \\sqrt{3^2 – 1^2} = \\sqrt{9-1} = \\sqrt{8} = 2 \\sqrt{2}$
\nThe length of the chord is 2x = $4 \\sqrt{2}$
\nHence Option D is the correct answer.<\/p>\n

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

Length of the arc = (angle subtended by the arc\/360)*circumference of the circle = $\\frac{48}{360}*2* \\pi * 7$ = 5.867 cm.<\/p>\n

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

The information given in the question can be represented as below
\n1\/2(length of the chord) = $\\sqrt(5^2 – 4^2)$
\n==> Length of the chord = 2*3 = 6cm<\/p>\n

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

The given triangle is a right angled triangle.
\nThe length of the hypotenuse is 13 cm.
\nIn a right angled triangle the length of the circumradius is half the length of the hypotenuse.
\nSo, circumradius = 13\/2 = 6.5 cm
\nArea of the circumcircle = $3.14*(6.5)^2=132.665$ $cm^2$<\/p>\n

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

Here, the radius of the circle is the hypotenuse of the triangle with 4m,3m as its base and height.<\/p>\n

<\/p>\n

Radius = $\\sqrt{4^{2}+3^{2}}$
\nRadius =$5 m$<\/p>\n

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

Area of sector of a circle = $\\frac{x}{360}*\\pi *r^{2}$
\nWhere x is the angle subtended at the centre and r is the radius of the circle
\nArea =$\\frac{36}{360}*\\pi *7^{2}$ = 15.4 sq.cm<\/p>\n

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

Area of the square form is 81 sq-cm . Hence , the side of the square would be 9 cm .
\nLength of the wire is (9 x 4) cm = 36 cm . Let $r$ be the radius of the semicircle thus formed .
\nWhen the wire is bent into semicircular shape , $\\pi r + 2r = 36$
\n$\\implies r = \\frac{36}{\\frac{22}{7}+2} = \\frac{36}{\\frac{36}{7}} = 7$ cm
\nTherefore , Area of the semicircle = $\\frac{\\frac{22}{7}\\times7\\times7}{2}$ = $77$ sq-cm
\nHence , the correct Option is D<\/p>\n

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

Area of the triangle , $A$ = $\\frac{\\sqrt{3}}{4}a^2$ = $121\\sqrt{3}$
\n$\\implies a = 22\\text{ cm}$
\nLength of wire, $L$ = $(3\\times a)\\text{ cm}$ = $66\\text{ cm}$
\nLet radius of the circle be $r$
\nBy using the relation,
\n$2\\pi r = 66$
\n$\\implies r = \\frac{66\\times 7}{2\\times 22} = 10.5 \\text{ cm}$
\n$\\implies \\text{ Area} = \\pi \\times 10.5^2 = 346.5 \\text{ cm}^2$
\nHence , the correct Option is C<\/p>\n

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

Let AB be the chord described as shown in the figure.<\/p>\n

<\/p>\n

OA is the radius of the bigger circle and OC is the radius of the smaller circle.
\nWe can observe that the triangle OAC is a right triangle.
\nApplying Pythagoras theorem we get,
\n$AC^2=13^2-5^2$
\n$=>AC=12$ cm
\nThe length of the chord is twice the length of AC. So we get,
\nAB = 2*AC = 2*12 = 24 cm<\/p>\n

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

Let the radius of the circle be \u2018r\u2019
\nIt is given that $\\pi r^2=50$
\n$=>r^2=\\frac{50}{3.14}$
\n$=>r^2\\approx 16$
\n$r\\approx 4$ cm
\nThe perimeter will be
\n$=2*\\pi*r$
\n$=2*3.14*4$
\n$\\approx 25$ cm<\/p>\n

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

Let the radius of the circle formed be \u2018r\u2019
\nThe length of the thread will be the circumference of the circle.
\nSo we get,
\n$2\\pi r=1256$
\n$=>2*3.14*r=1256$
\n$r=200$ cm
\nThus the diameter will be 400 cm.<\/p>\n

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

Let the radius of the circles touching each other be $a$ cm and $b$ cm .
\nThen , $a$ + $b$ = $28$
\n$\\implies b$ = $28 – a$
\n$\\implies \\pi a^2+ \\pi b^2$ = $520\\pi$
\n$\\implies a^2+(28-a)^2$ = $520$
\n$\\implies 2a^2 – 56a + 784 = 520$
\n$\\implies 2a^2 – 56a + 264 = 0$
\n$\\implies 2( a – 6 )( a – 22 ) = 0$
\nThe radius of the circles can be $22$cm or $6$ cm
\nTherefore , the radius of the smaller circle is $6$ cm
\nHence, correct option is D<\/p>\n

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

The circle is the circumcircle of the equilateral triangle.
\nCircumradius = $\\frac{2}{3}*\\frac{\\sqrt3}{2}*3=\\sqrt3$ cm
\nArea of the circle = $\\pi*(\\sqrt3)^2=3 \\pi$ $cm^2$
\nHence Option B.<\/p>\n

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

Let the length of the side be \u2018l\u2019
\nArea of square = $l^{2}$
\nArea of the circle = $\\pi\\times\\frac{l^{2}}{4}$
\nRequired ratio = $l^{2}$:$\\pi\\times\\frac{l^{2}}{4}$
\nRequired ratio =$4:\\pi$<\/p>\n

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

The distance between centres of two circles = d
\nRadius of both circles = $r_1$ and $r_2$
\nFor the circles to touch each other internally,<\/p>\n

<\/p>\n

Here, AC = $r_1$ and BC = $r_2$
\nDistance between their centres = AB = d
\nNow, clearly AB = AC – BC
\n=> d = $r_1-r_2$<\/p>\n

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

NOTE :- Area of triangle = (semi perimeter) * (inradius)<\/p>\n

<\/p>\n

AB = 6 cm , AC = 8 cm and $\\angle$BAC = 90\u00b0
\nFrom $\\triangle$ABC,
\nBC = $\\sqrt{(AB)^2+(AC^2)}$
\n= $\\sqrt{6^2+8^2} = \\sqrt{36+64}$
\n= 10 cm
\nSemi perimeter of $\\triangle$ABC = s
\n<\/span>= $\\frac{6+8+10}{2}$ = 12 cm
\n<\/span>Area of $\\triangle$ABC = $\\frac{1}{2}$*AB*AC
\n<\/span><\/span>= $\\frac{1}{2}$*6*8 = 24 $cm^2$
\n<\/span><\/span>=> Inradius = $\\frac{\\triangle}{s}$ = $\\frac{24}{12}$
\n<\/span><\/span>= 2 cm<\/span><\/span><\/p>\n

 <\/p>\n

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

<\/p>\n

$\\angle$BAC = 85\u00b0
\n$\\angle$BCA = 75\u00b0
\n=> $\\angle$ABC = 180\u00b0-(85\u00b0+75\u00b0) = 20\u00b0
\nAngle subtended by an arc at the centre is twice the angle subtended by it at any point on the circle.
\n=> $\\angle$AOC = 2$\\angle$ABC
\n=> $\\angle$AOC = 40\u00b0
\nIn $\\triangle$OAC, OA = OC = radii
\n=> $\\angle$OAC = $\\angle$OCA
\n=> 2$\\angle$OAC = 180\u00b0-40\u00b0 = 140\u00b0
\n=> $\\angle$OAC = $\\frac{140^{\\circ}}{2}$ = 70\u00b0<\/p>\n

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

<\/p>\n

Length of chord CD = $a$ and AB = $b$
\nLet radius of the circle be $r$
\nIn $\\triangle$OAB, by using Pythagoras theorem
\n=> $r^2 + r^2 = b^2$
\n=> $2r^2 = b^2$
\n=> $r = \\frac{b}{\\sqrt{2}}$ ——Eqn(1)
\nNow, in $\\triangle$COD, OC = OD = radii
\n=> $\\angle$OCD = $\\angle$ODC = $\\angle$COD = 60\u00b0
\n<\/span>=> OCD is equilateral triangle
\n<\/span>=> $a = r$ ———Eqn(2)
\n<\/span>From eqns(1) & (2), we get :
\n<\/span>$a = \\frac{b}{\\sqrt{2}}$
\n<\/span>=> $b = \\sqrt{2}a$<\/span><\/p>\n

\"SSC<\/a><\/p>\n

SSC Free Previous Papers App<\/a><\/p>\n

 <\/p>\n","protected":false},"excerpt":{"rendered":"

SSC CGL Circle Questions PDF: Download SSC CGL Circle questions with answers PDF based on previous papers very useful for SSC CGL exams. 25 Very important Circle objective questions for SSC exams. Question 1:\u00a0A copper wire is bent in the form of square with an area of 121 cm 2. If the same wire is […]<\/p>\n","protected":false},"author":21,"featured_media":25934,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[169,125,504],"tags":[1519],"class_list":{"0":"post-25932","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-downloads","8":"category-featured","9":"category-ssc-cgl","10":"tag-ssc-cgl-2019"},"better_featured_image":{"id":25934,"alt_text":"SSC CGL Circle Questions PDF","caption":"SSC CGL Circle Questions PDF","description":"SSC CGL Circle Questions 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