Download Geometry Circles Questions for CAT<\/a><\/p>\n Enroll for CAT 2022 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>In a triangle ABC, the lengths of the sides AB and AC equal 17.5 cm and 9 cm respectively. Let D be a point on the line segment BC such that AD is perpendicular to BC. If AD = 3 cm, then what is the radius (in cm) of the circle circumscribing the triangle ABC?<\/p>\n a)\u00a017.05<\/p>\n b)\u00a027.85<\/p>\n c)\u00a022.45<\/p>\n d)\u00a032.25<\/p>\n e)\u00a026.25<\/p>\n 1)\u00a0Answer\u00a0(E)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b> Let x be the value of third side of the triangle. Now we know that Area = 17.5*9*x\/(4*R), where R is circumradius.<\/p>\n Also Area = 0.5*x*3 .<\/p>\n Equating both, we have 3 = 17.5*9 \/ (2*R)<\/p>\n => R = 26.25.<\/p>\n Question 2:\u00a0<\/b>Two circles, both of radii 1 cm, intersect such that the circumference of each one passes through the centre of the other. What is the area (in sq. cm.) of the intersecting region?<\/p>\n a)\u00a0$\\frac{\\pi}{3}-\\frac{\\sqrt 3}{4}$<\/p>\n b)\u00a0$\\frac{2\\pi}{3}+\\frac{\\sqrt 3}{2}$<\/p>\n c)\u00a0$\\frac{4\\pi}{3}-\\frac{\\sqrt 3}{2}$<\/p>\n d)\u00a0$\\frac{4\\pi}{3}+\\frac{\\sqrt 3}{2}$<\/p>\n e)\u00a0$\\frac{2\\pi}{3}-\\frac{\\sqrt 3}{2}$<\/p>\n 2)\u00a0Answer\u00a0(E)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b> The circumferences of the two circle pass through each other’s centers. Hence, O1A = O1B=O1O2 = 1cm<\/p>\n By symmetry, the line joining the two centres would be bisect AB and would be bisected by AB. As the line joining the center to the midpoint of a chord is perpendicular to the chord, O1O2 and AB are perpendicular bisectors of each other. Suppose they intersect at point P.<\/p>\n O1P = Half of O1O2 = 1\/2 cm In the above, the required area is 2 times A(segment ABO2)(blue region). And A(segment ABO2)(blue region) = A(sector O2AO1B)(blue + red) – A(triangleO1AB )(red)<\/p>\n Area of sector = 120\u00b0\/360\u00b0 * $\\pi * 1^2 $ = $\\pi\/3$<\/p>\n Area of triangle = 1\/2 * b * h = 1\/2 * (2* 1 cos 30\u00b0) * (1\/2) = \u221a3\/4<\/p>\n Hence, required area = $\\frac{\\pi}{3}-\\frac{\\sqrt 3}{4}$ . Hence so the required area is 2 times the above value which is $\\frac{2\\pi}{3}-\\frac{\\sqrt 3}{2}$<\/p>\n Question 3:\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 a)\u00a03<\/p>\n b)\u00a02<\/p>\n c)\u00a01<\/p>\n d)\u00a0None of the above.<\/p>\n 3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b> Since 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 Question 4:\u00a0<\/b>What is the distance in cm between two parallel chords of lengths 32 cm and 24 cm in a circle of radius 20 cm?<\/p>\n a)\u00a01 or 7<\/p>\n b)\u00a02 or 14<\/p>\n c)\u00a03 or 21<\/p>\n d)\u00a04 or 28<\/p>\n 4)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b> The distances of the chords from the center are 12 cm and 16 cm respectively.<\/p>\n If the chords lie on the same side of the center, the distance between the chords is 4 cm, if they lie on opposite sides of the center, the distance between them is 28 cm.<\/p>\n Question 5:\u00a0<\/b>Two identical circles intersect so that their centres, and the points at which they intersect, form a square of side 1 cm. The area in sq. cm of the portion that is common to the two circles is<\/p>\n a)\u00a0$\\pi$\/4<\/p>\n b)\u00a0$\\pi$\/2-1<\/p>\n c)\u00a0$\\pi$\/5<\/p>\n d)\u00a0$\\sqrt\\pi-1$<\/p>\n 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b> We know that quad ambn is a square of side 1.<\/p>\n Area of\u00a0the sector a-mqn is $\\frac{90}{360}* \\pi *1*1$ = $\\frac{\\pi }{4}$.<\/p>\n Area of square = 1*1 = 1<\/p>\n Area of common portion = 2 * Area of sector – Area of square<\/p>\n = 2 * $\\frac{\\pi }{4}$ – 1 =\u00a0 $\\frac{\\pi }{2}$ – 1<\/p>\n Question 6:\u00a0<\/b>In the following figure, the diameter of the circle is 3 cm. AB and MN are two diameters such that MN is perpendicular to AB. In addition, CG is perpendicular to AB such that AE:EB = 1:2, and DF is perpendicular to MN such that NL:LM = 1:2. The length of DH in cm is a)\u00a0$2\\sqrt2 – 1$<\/p>\n b)\u00a0$(2\\sqrt2 – 1)\/2$<\/p>\n c)\u00a0$(3\\sqrt2 – 1)\/2$<\/p>\n d)\u00a0$(2\\sqrt2 – 1)\/3$<\/p>\n 6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let EO = x, So, AE = 1.5 – x (1.5-x):(1.5+x) = 1:2.<\/p>\n x=0.5.<\/p>\n So, EO = 0.5 Checkout: CAT Free Practice Questions and Videos<\/strong><\/a><\/em><\/p>\n Question 7:\u00a0<\/b>Consider a circle with unit radius. There are 7 adjacent sectors, S1, S2, S3,….., S7 in the circle such that their total area is (1\/8)th of the area of the circle. Further, the area of the $j^{th}$ sector is twice that of the $(j-1)^{th}$ sector, for j=2, …… 7. What is the angle, in radians, subtended by the arc of S1\u00a0at the centre of the circle?<\/p>\n a)\u00a0$\\pi\/508$<\/p>\n b)\u00a0$\\pi\/2040$<\/p>\n c)\u00a0$\\pi\/1016$<\/p>\n d)\u00a0$\\pi\/1524$<\/p>\n 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Now area of 1st sector = $\\pi * r^2 * \\frac{x}{360}$ where x – angle subtended at center<\/p>\n Now the next sector will have 2x as the angle, and similarly angles will be in GP with ratio\u00a0= 2.<\/p>\n Sum of areas of all 7 sectors = $\\frac{127*x* \\pi * r^2}{360}$ which is equal to $\\frac{\\pi * r^2}{8}$<\/p>\n We get x = $\\frac{360}{8*127}$<\/p>\n Now if converted in radians we get \u00a0x =\u00a0$\\pi\/508$.<\/p>\n Question 8:\u00a0<\/b>In the following figure, ACB is a right-angled triangle. AD is the altitude. Circles are inscribed within the triangle ACD and triangle BCD. P and Q are the centers of the circles. The distance PQ is a)\u00a07 m<\/p>\n b)\u00a04.5 m<\/p>\n c)\u00a010.5 m<\/p>\n d)\u00a06 m<\/p>\n 8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b> By Pythagoras theorem we get BC = 25 . Let BD = x;Triangle ABD is similar to triangle CBA => AD\/15 = x\/20 and also triangle ADC is similar to triangle ACB=> AD\/20 = (25-x)\/15. From the 2 equations, we get x = 9 and DC = 16<\/p>\n We know that AREA = (semi perimeter ) * inradius<\/p>\n For triangle ABD, Area = 1\/2 x BD X AD = 1\/2 x 12 x 9 = 54 and semi perimeter = (15 + 9 + 12)\/2 = 18. On using the above equation we get, inradius, r = 3.<\/p>\n Similarly for triangle ADC we get inradius R = 4 .<\/p>\n PQ = R + r = 7 cm<\/p>\n Question 9:\u00a0<\/b>A circle is inscribed in a given square and another circle is circumscribed about the square. What is the ratio of the area of the inscribed circle to that of the circumscribed circle?<\/p>\n a)\u00a02 : 3<\/p>\n b)\u00a03 : 4<\/p>\n c)\u00a01 : 4<\/p>\n d)\u00a01 : 2<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/2015\/12\/24\/4089.png\")
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\n![](\"https:\/\/cracku.in\/media\/uploads\/Screenshot_2_gkTCoTL.png\")
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\nSo, the angle AO1P = 60 degrees as cos 60 = 1\/2
\nBy symmetry, BO1P = 60 degrees.
\nSo, angle AO1B= 120 degrees<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/drawing_8HyV8JO_BUV8PeQ.png\")
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\nAE : EB = 1:2 => x = 1\/2<\/p>\n
\nSimilarly, OL = 0.5
\nNow, EOLH is a parallelogram and EO = OL = 0.5
\nIn triangle DOL, DO = radius = 1.5 and OL = 0.5
\nSo, DL = $\\sqrt2$
\n=> DH =\u00a0$(2\\sqrt2 – 1)\/2$<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/2015\/07\/01\/pic-1.png\")
\nThe length of AB is 15 m and AC is 20 m<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/2015\/07\/08\/geom-11.png\")
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