Download Geometry Questions for CAT<\/a><\/p>\n Enroll for CAT 2023 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>Consider a square ABCD with midpoints E, F, G, H of AB, BC, CD and DA respectively. Let L denote the line passing through F and H. Consider points P and Q, on L and inside ABCD such that the angles APD and BQC both equal 120\u00b0. What is the ratio of the area of ABQCDP to the remaining area inside ABCD?<\/p>\n a)\u00a0$\\frac{4 \\sqrt 2}{3}$<\/p>\n b)\u00a0$2 + \\sqrt{3}$<\/p>\n c)\u00a0$\\frac{10-3\\sqrt{3}}{9}$<\/p>\n d)\u00a0$1+\\frac{1}{\\sqrt3}$<\/p>\n e)\u00a0 $2\\sqrt{3}-1$<\/p>\n 1)\u00a0Answer\u00a0(E)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b> Consider side of square as 10 units. So HD=5 and HP=$\\frac{5}{\\sqrt3}$ . So now area of triangle HPD=$\\frac{12.5}{\\sqrt3}$. Also Area APD=Area BQC=2*AreaHPD=$\\frac{25}{\\sqrt3}$. So numerator of required answer is $100-\\frac{50}{\\sqrt3}$ and denominator as $\\frac{50}{\\sqrt3}$. Solving we get answer as $2\\sqrt{3}-1$.<\/p>\n Question 2:\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 a)\u00a020<\/p>\n b)\u00a028<\/p>\n c)\u00a036<\/p>\n d)\u00a040<\/p>\n 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/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 Question 3:\u00a0<\/b>In the figure below, the rectangle at the corner measures 10 cm \u00d7 20 cm. The corner A of the rectangle is also a point on the circumference of the circle. What is the radius of the circle in cm? a)\u00a010 cm<\/p>\n b)\u00a040 cm<\/p>\n c)\u00a050 cm<\/p>\n d)\u00a0None of the above.<\/p>\n 3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Using pythagoras we have $r^2 = (r-10)^2 + (r-20)^2$.<\/p>\n Solving the equation,\u00a0we get r = 10 or 50.<\/p>\n But 10 is not possible , so r = 50.<\/p>\n Hence radius is 50.<\/p>\n Question 4:\u00a0<\/b>Let $C$ be a circle with centre $P_0$ and $AB$ be a diameter of $C$. Suppose $P_1$ is the mid point of the line segment $P_0B$,$P_2$ is the mid point of the line segment $P_1B$ and so on. Let $C_1,C_2,C_3,…$ be circles with diameters $P_0P_1, P_1P_2, P_2P_3…$ respectively. Suppose the circles $C_1, C_2, C_3,…$ are all shaded. The ratio of the area of the unshaded portion of $C$ to that of the original circle is<\/p>\n a)\u00a08 : 9<\/p>\n b)\u00a09 : 10<\/p>\n c)\u00a010 : 11<\/p>\n d)\u00a011 : 12<\/p>\n 4)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Radius of\u00a0the circles $C_1, C_2, C_3,…$ would be in GP with (R\/4),(R\/8),(R\/16) and so on. Radii\u00a0of circles are in the ratio\u00a01:4.<\/p>\n Ratio of unshaded region to the ratio of original circle = 1-$\\frac{Ratio\\ of\\ shaded\\ region}{Ratio\\ of\\ original\\ circle}$<\/p>\n = 1-$\\frac{\\pi r^2\/16 + \\pi r^2\/64+…..}{\\pi r^2}$ = 1-$\\frac{1\/16}{(1-1\/4)}$ = 1- 1\/12 = $\\frac{11}{12}$ = 11:12<\/p>\n Question 5:\u00a0<\/b>Rectangular tiles each of size 70 cm by 30 cm must be laid horizontally on a rectangular floor of size 110 cm by 130 cm, such that the tiles do not overlap. A tile can be placed in any orientation so long as its edges are parallel to the edges of the floor. No tile should overshoot any edge of the floor. The maximum number of tiles that can be accommodated on the floor is<\/p>\n a)\u00a04<\/p>\n b)\u00a05<\/p>\n c)\u00a06<\/p>\n d)\u00a07<\/p>\n 5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Checkout: CAT Free Practice Questions and Videos<\/strong><\/a><\/em><\/p>\n Question 6:\u00a0<\/b>A square, whose side is 2 m, has its corners cut away so as to form an octagon with all sides equal. Then the length of each side of the octagon, in metres, is<\/p>\n a)\u00a0$ \\frac{\\sqrt 2}{\\sqrt 2 +1} $<\/p>\n b)\u00a0$\\frac{2}{\\sqrt 2 + 1} $<\/p>\n c)\u00a0$\\frac{2}{\\sqrt 2 – 1}$<\/p>\n d)\u00a0$\\frac{\\sqrt 2}{\\sqrt 2-1} $<\/p>\n 6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let the length of each side of the octagon be x.<\/p>\n So, length of the square will be x+2*(x\/\u221a2) = 2<\/p>\n => x(1+\u221a2) = 2 => x = 2\/(1+\u221a2)<\/p>\n Question 7:\u00a0<\/b>A rectangular pool of 20 m wide and 60 m long is surrounded by a walkway of uniform width. If the total area of the walkway is 516 $m^2$ , how wide, in metres, is the walkway?<\/p>\n a)\u00a04 m<\/p>\n b)\u00a02 m<\/p>\n c)\u00a03 m<\/p>\n d)\u00a03.5 m<\/p>\n 7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n If the width of the walkway is x, then its area = (60+2x)(20+2x) – 1200 = 516 Question 8:\u00a0<\/b>If a,b,c are the sides of a triangle, and $a^2 + b^2 +c^2 = bc + ca + ab$, then the triangle is:<\/p>\n a)\u00a0equilateral<\/p>\n b)\u00a0isosceles<\/p>\n c)\u00a0right angled<\/p>\n d)\u00a0obtuse angled<\/p>\n 8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) => 3(a^2 + b^2 + c^2)$<\/p>\n This is possible only if a = b = c.<\/p>\n So, the triangle is an equilateral triangle.<\/p>\n Question 9:\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 9)\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 10:\u00a0<\/b>A man starting at a point walks one km east, then two km north, then one km east, then one km north, then one km east and then one km north to arrive at the destination. What is the shortest distance from the starting point to the destination?<\/p>\n a)\u00a0$2\\sqrt{2}$ km<\/p>\n b)\u00a0$7$ km<\/p>\n c)\u00a0$3\\sqrt{2}$ km<\/p>\n d)\u00a0$5$ km<\/p>\n 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b> Question 11:\u00a0<\/b>In triangleABC, \u2220B is a right angle, AC = 6 cm, and D is the mid-point of AC. The length of BD is<\/p>\n a)\u00a04cm<\/p>\n b)\u00a06 cm<\/p>\n c)\u00a03cm<\/p>\n d)\u00a03.5cm<\/p>\n 11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n A circle is circumscribed to triangle ABC.<\/p>\n For this circle D will be centre of circle, and AD , DC , BD will radius of this circle.<\/p>\n Hence AD=BD=DC=3 cm<\/p>\n Question 12:\u00a0<\/b>In ABC, points P, Q and R are the mid-points of sides AB, BC and CA respectively. If area of ABC is 20 sq. units, find the area of PQR.<\/p>\n a)\u00a010 sq. units<\/p>\n b)\u00a05\u221a3 sq. units<\/p>\n c)\u00a05 sq. units<\/p>\n d)\u00a0None of these<\/p>\n 12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n As we know, the triangle joining midpoints of sides will divide it in 4 similar traingles of equal area.<\/p>\n So area will be = $\\frac{20}{4} = 5$<\/p>\n Question 13:\u00a0<\/b>Four identical coins are placed in a square. For each coin the ratio of area to circumference is same as the ratio of circumference to area. Then find the area of the square that is not covered by the coins.<\/p>\n a)\u00a016(\u03c0 – 1)<\/p>\n b)\u00a016(8 – \u03c0)<\/p>\n c)\u00a016(4 – \u03c0)<\/p>\n d)\u00a016 $(4- \\frac{\\pi}{2})$<\/p>\n 13)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $\\frac{(\\pi r^2)}{2 \\pi r}$ =\u00a0$\\frac{2 \\pi r}{ \\pi r^2}$ The intended diagram can be drawn as follows:<\/p>\n ABCD is the square and the coins are placed (numbered 1,2,3,4)<\/p>\n Question 14:\u00a0<\/b>Let ABC be a right-angled triangle with BC as the hypotenuse. Lengths of AB and AC are 15 km and 20 km, respectively. The minimum possible time, in minutes, required to reach the hypotenuse from A at a speed of 30 km per hour is<\/p>\n 14)\u00a0Answer:\u00a024<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n The length of the altitude from A to the hypotenuse will be the shortest distance.<\/p>\n This is a right triangle with sides 3 : 4 : 5.<\/p>\n Hence, the hypotenuse = $\\sqrt{20^2+15^2}$ = 25 Km.<\/p>\n Length of the altitude = $\\frac{15 * 20}{25}$ = 12 Km<\/p>\n (This is derived from equating area of triangle, $\\frac{15\\cdot20}{2}\\ =\\ \\frac{25\\cdot\\text{altitude}}{2}$)<\/p>\n The time taken = $\\frac{12}{30} * 60$ = 24 minutes<\/p>\n Question 15:\u00a0<\/b>A chord of length 5 cm subtends an angle of 60\u00b0 at the centre of a circle. The length, in cm, of a chord that subtends an angle of 120\u00b0 at the centre of the same circle is<\/p>\n a)\u00a0$5\\sqrt{3}$<\/p>\n b)\u00a0$2\\pi$<\/p>\n c)\u00a0$8$<\/p>\n d)\u00a0$6\\sqrt{2}$<\/p>\n 15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n We are given that AB = 5 cm and $\\angle$ AOB = 60\u00b0<\/p>\n Let us draw OM such that OM $\\perp$ AB.<\/p>\n In right angle triangle AMO,<\/p>\n $sin 30\u00b0 = \\dfrac{AM}{AO}$<\/p>\n $\\Rightarrow$ AO = 2*AM = 2*2.5 = 5 cm. Therefore, we can say that the radius of the circle = 5 cm.<\/p>\n In right angle triangle PNO,<\/p>\n $sin 60\u00b0 = \\dfrac{PN}{PO}$<\/p>\n $\\Rightarrow$ PN = $\\dfrac{\\sqrt{3}}{2}$*PO = $\\dfrac{5\\sqrt{3}}{2}$<\/p>\n Therefore, PQ = 2*PN = $5\\sqrt{3}$ cm<\/p>\n Question 16:\u00a0<\/b>A man makes complete use of 405 cc of iron, 783 cc of aluminium, and 351 cc of copper to make a number of solid right circular cylinders of each type of metal. These cylinders have the same volume and each of these has radius 3 cm. If the total number of cylinders is to be kept at a minimum, then the total surface area of all these cylinders, in sq cm, is<\/p>\n a)\u00a0$1026(1 + \\pi)$<\/p>\n b)\u00a0$8464\\pi$<\/p>\n c)\u00a0$928\\pi$<\/p>\n d)\u00a0$1044(4 + \\pi)$<\/p>\n 16)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n It is given that the volume of all the cylinders is the same, so the volume of each cylinder = HCF of (405, 783, 351)<\/p>\n =27<\/p>\n The number of iron cylinders =\u00a0$\\ \\frac{\\ 405}{27}$ = 15<\/p>\n The number of aluminium cylinders =\u00a0$\\ \\frac{\\ 783}{27}$ = 29<\/p>\n The number of copper cylinders =\u00a0$\\ \\frac{351\\ }{27}$ = 13<\/p>\n 15*$\\pi\\ r^2h=$ 405<\/p>\n 15*$\\pi\\ 9*h=$ 405<\/p>\n $\\pi\\ h=3$<\/p>\n Now we have to calculate the total surface area of all the cylinders<\/p>\n Total number of cylinders = 15+29+13 = 57<\/p>\n Total surface area of the cylinder = 57*($2\\pi\\ rh+2\\pi\\ r^2$)<\/p>\n =57(2*3*3 + 2*9*$\\pi\\ $)<\/p>\n =$1026(1 + \\pi)$<\/p>\n Question 17:\u00a0<\/b>The vertices of a triangle are (0,0), (4,0) and (3,9). The area of the circle passing through these three points is<\/p>\n a)\u00a0$\\frac{14\\pi}{3}$<\/p>\n b)\u00a0$\\frac{123\\pi}{7}$<\/p>\n c)\u00a0$\\frac{12\\pi}{5}$<\/p>\n d)\u00a0$\\frac{205\\pi}{9}$<\/p>\n 17)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Equation of circle\u00a0$x^2+y^2+2gx+2fy+c=0$<\/p>\n It passes through\u00a0(0,0), (4,0) and (3,9). Substitute each point in the above equation:<\/p>\n => On substituting the value (0,0) in the above equation, we obtain: $c=0$<\/p>\n => On substituting the value (4,0) in the above equation, we obtain:\u00a0 $16+0+8g+0 = 0$ ; $g=-2$<\/p>\n =>\u00a0On substituting the value (3,9) in the above equation, we obtain: $9+81-12+18f = 0$ ;\u00a0$f= -13\/3$<\/p>\n Radius of the circle r\u00a0<\/strong>=\u00a0$\\sqrt{\\ g^2+f^2-c}$ =>\u00a0$r^2=\\frac{205}{9}$<\/p>\n Therefore, Area =\u00a0$\\pi\\ r^2=\\frac{205\\pi\\ }{9}$<\/p>\n Question 18:\u00a0<\/b>The sides AB and CD of a trapezium ABCD are parallel, with AB being the smaller side. P is the midpoint of CD and ABPD is a parallelogram. If the difference between the areas of the parallelogram ABPD and the triangle BPC is 10 sq cm, then the area, in sq cm, of the trapezium ABCD is<\/p>\n a)\u00a030<\/p>\n b)\u00a040<\/p>\n c)\u00a025<\/p>\n d)\u00a020<\/p>\n 18)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n We are given that :<\/p>\n Let DP =x Question 19:\u00a0<\/b>Let ABCD be a parallelogram. The lengths of the side AD and the diogonal AC are 10cm and 20cm, respectively. If the angle $\\angle ADC$ is equal to $30^{0}$ then the area of the parallelogram, in sq.cm is<\/p>\n a)\u00a0$\\frac{25(\\sqrt{5}+\\sqrt{15})}{2}$<\/p>\n b)\u00a0$25(\\sqrt{3}+\\sqrt{15})$<\/p>\n c)\u00a0$\\frac{25(\\sqrt{3}+\\sqrt{15})}{2}$<\/p>\n d)\u00a0${25(\\sqrt{5}+\\sqrt{15})}$<\/p>\n 19)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b> Applying cosine rule in triangle ACD,<\/p>\n $100+X^2-2\\times\\ 10\\times\\ X\\cos30=400$<\/p>\n $X^2-10X\\sqrt{\\ 3}-300=0$<\/p>\n Solving, we get X =\u00a0$\\left(\\frac{10\\sqrt{\\ 3}+10\\sqrt{\\ 15}}{2}\\right)$<\/p>\n Hence, area = 10Xsin 30 = $\\frac{\\left(\\frac{10\\sqrt{\\ 3}+10\\sqrt{\\ 15}}{2}\\right)10}{2}$<\/p>\n = $25(\\sqrt{3}+\\sqrt{15})$<\/p>\n Question 20:\u00a0<\/b>In a triangle ABC, AB = AC = 8 cm. A circle drawn with BC as diameter passes through A. Another circle drawn with center at A passes through Band C. Then the area, in sq. cm, of the overlapping region between the two circles is<\/p>\n a)\u00a0$16\\pi$<\/p>\n b)\u00a0$16(\\pi – 1)$<\/p>\n c)\u00a0$32(\\pi – 1)$<\/p>\n d)\u00a0$32\\pi$<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/2015\/12\/24\/4091.png\")
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\n![](\"https:\/\/cracku.in\/media\/question\/4384_final.png\")
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\n![](\"https:\/\/cracku.in\/media\/question\/4394.png\")
<\/p>\n![\"\"](\"https:\/\/cracku.in\/media\/uploads\/2015\/07\/08\/geom6.png\")
\nAs seen in the fig. we have a right angled triangle with sides \u00a0r ,r-10 , r-20.<\/p>\n![\"\"](\"https:\/\/cracku.in\/media\/uploads\/2015\/07\/09\/geom21.PNG\")
\nAs seen from the fig. If following configuration is used max 6\u00a0number of tiles that can be accommodated on the floor.<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/2015\/12\/24\/text3376.png\")
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\nSolving this, we get x = 3m<\/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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\n![\"\"](\"https:\/\/cracku.in\/media\/uploads\/2015\/08\/07\/new.png\")
\nIn the\u00a0diagram, A is starting point and B is ending point.
\nAccordingly, AB = $\\sqrt{(3^2 + 4^2)} = 5$<\/p>\n
\nSo r = 2
\nHence required area = (Area of square) – (Area of 4 circles)
\n= $(8^2) – (4 \\pi (2^2))$ (As side of square will be 4*2 = 8)
\n= $16 (4- \\pi)$<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/Screenshot%20from%202020-05-19%2010-12-55_GI8598H.png\")
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![](\"https:\/\/media-cdn.cracku.in\/uploads\/image_RDiwKAu.png\")
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\nSo AB =x
\nNow DP=CP
\nSo CD = 2x
\nNow let height of trapezium be h
\nwe can say A(Parallelogram ABPD ) = xh
\nAnd A (BPC) =$\\frac{1}{2}xh$
\nNow by condition$xh-\\frac{1}{2}xh=10$
\n$\\frac{xh}{2}=10$
\nso xh =20
\nNow therefore area of trapezium ABCD = $\\frac{1}{2}\\left(x+2x\\right)h\\ =\\ \\frac{3}{2}xh\\ =\\ 30$<\/p>\n
\n![](\"https:\/\/media-cdn.cracku.in\/uploads\/image_RUMI1Z1.png\")
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