Download Geometry Questions for CAT<\/a><\/p>\n Enroll for CAT 2022 Crash Course<\/a><\/p>\n Question 1:\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 1)\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 2:\u00a0<\/b>If a triangle ABC, $\\angle BCA=50^{0}$. D and E are points on $AB$ and $AC$, respectively, such that $AD=DE$. If F is a point on $BC$ such that $BD=DF$, then $\\angle FDE$, in degrees, is equal to<\/p>\n a)\u00a072<\/p>\n b)\u00a080<\/p>\n c)\u00a0100<\/p>\n d)\u00a096<\/p>\n 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b> We need to find out p.<\/p>\n Angle ADE = 180 – 2x<\/p>\n Angle BDF = 180 –\u00a02y<\/p>\n Now, 180 – 2y + p + 180 – 2x = 180 [Straight line = 180 deg]<\/p>\n p = 2x + 2y – 180<\/p>\n Also, x + y + 50 = 180 [Sum of the angles of triangle = 180]<\/p>\n x + y = 130<\/p>\n p = 260 – 180 = 80 degrees.<\/p>\n Question 3:\u00a0<\/b>A park is shaped like a rhombus and has area 96 sq m. If 40 m of fencing is needed to enclose the park, the cost, in INR, of laying electric wires along its two diagonals, at the rate of \u20b9125 per m, is<\/p>\n 3)\u00a0Answer:\u00a03500<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n We can say 40m is the perimeter of the park Question 4:\u00a0<\/b>The cost of fencing a rectangular plot is \u20b9 200 per ft along one side, and \u20b9 100 per ft along the three other sides. If the area of the rectangular plot is 60000 sq. ft, then the lowest possible cost of fencing all four sides, in INR, is<\/p>\n a)\u00a0120000<\/p>\n b)\u00a090000<\/p>\n c)\u00a0100000<\/p>\n d)\u00a0160000<\/p>\n 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let us draw the rectangle.<\/p>\n Now, definitely, three sides should be fenced at Rs 100\/ft, and one side should be fenced at Rs 200\/ft.<\/p>\n In this question, we are going to assume that the L is greater than B.<\/p>\n Hence, the one side painted at Rs 200\/ft should be B to minimise costs.<\/p>\n Hence, the total cost = 200B + 100B + 100L + 100L = 300B + 200L<\/p>\n Now, L x B = 60000<\/p>\n B = 60000\/L<\/p>\n Hence, total cost = 300B + 200L = 18000000\/L + 200L<\/p>\n To minimise this cost, we can use AM>=GM,<\/p>\n $\\frac{\\frac{18000000}{L}+200L}{2}\\ge\\sqrt{\\ \\frac{18000000}{L}\\times\\ 200L}$<\/p>\n $\\frac{18000000}{L}+200L\\ge2\\sqrt{\\ 18000000\\times\\ 200}$<\/p>\n $\\frac{18000000}{L}+200L\\ge2\\times\\ 60000$<\/p>\n Hence, minimum cost = Rs 120000.<\/p>\n Question 5:\u00a0<\/b>Let D and E be points on sides AB and AC, respectively, of a triangle ABC, such that AD : BD = 2 : 1 and AE : CE = 2 : 3. If the area of the triangle ADE is 8 sq cm, then the area of the triangle ABC, in sq cm, is<\/p>\n 5)\u00a0Answer:\u00a030<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n We have :<\/p>\n Now area of ADE =\u00a0$\\frac{1}{2}\\times\\ AD\\times\\ AE\\times\\ \\sin A$ Checkout: CAT Free Practice Questions and Videos<\/p>\n Question 6:\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 6)\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 7:\u00a0<\/b>If a rhombus has area 12 sq cm and side length 5 cm, then the length, in cm, of its longer diagonal is<\/p>\n a)\u00a0$\\sqrt{37}+\\sqrt{13}$<\/p>\n b)\u00a0$\\sqrt{13}+\\sqrt{12}$<\/p>\n c)\u00a0$\\frac{\\sqrt{37}+\\sqrt{13}}{2}$<\/p>\n d)\u00a0$\\frac{\\sqrt{13}+\\sqrt{12}}{2}$<\/p>\n 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n All the sides of the rhombus are equal.<\/p>\n The area of a rhombus is\u00a0$12\\ cm^2$<\/p>\n Considering d1 to be the length of the longer diagonal, d2 to be the length of the shorter diagonal.<\/p>\n The area of a rhombus is\u00a0$\\left(\\frac{1}{2}\\right)\\left(d1\\right)\\cdot\\left(d2\\right)\\ =\\ 12$<\/p>\n d1*d2 = 24.<\/p>\n The length of the side of a rhombus is given by\u00a0$\\frac{\\sqrt{\\ d1^2+d2^2}}{2}$. This is because the two diagonals and a side from a right-angled triangle with sides d1\/2, d2\/2 and the side length.<\/p>\n $\\frac{\\sqrt{\\ d1^2+d2^2}}{2}=\\ 5$<\/p>\n Hence\u00a0$\\sqrt{\\ d1^2+d2^2}\\ =\\ 10$<\/p>\n $d1^2+d2^2\\ =\\ 100$<\/p>\n Using d1*d2 = 24, 2*d1*d2 = 48.<\/p>\n $d1^2+d2^2\\ +2\\cdot d1\\cdot d2=\\ 100+48\\ =\\ 148$<\/p>\n $d1^2+d2^2\\ -2\\cdot d1\\cdot d2=\\ 100-48\\ =\\ 52$<\/p>\n $d1+d2\\ =\\ \\sqrt{\\ 148}$\u00a0 (1)<\/p>\n d1-d2 =\u00a0$\\sqrt{52}$\u00a0 \u00a0(2)<\/p>\n (1) + (2)= 2*(d1) = 2*($\\sqrt{\\ 37}+\\sqrt{\\ 13}$)<\/p>\n d1 =\u00a0$\\sqrt{\\ 37}+\\sqrt{\\ 13}$<\/p>\n or<\/p>\n In a rhombus the area of a Rhombus is given by :<\/p>\n The diagonals perpendicularly bisect each other. Considering the length of the diagonal to be 2a, 2b.<\/p>\n The area of a Rhombus is :\u00a0$\\left(\\frac{1}{2}\\right)\\cdot\\left(2a\\right)\\cdot\\left(2b\\right)\\ =\\ 12$<\/p>\n ab =6.<\/p>\n The length of each side is :\u00a0$\\sqrt{\\ a^2+b^2}$ = 5,\u00a0$a^2+b^{2\\ }=\\ 25,\\ $<\/p>\n $\\left(a+b\\right)^2=\\ 37,\\ \\left(a+b\\right)\\ =\\ \\sqrt{\\ 37}$<\/p>\n ($\\left(a-b\\right)^2=\\ 13,\\ a-b\\ =\\ \\sqrt{\\ 13}$<\/p>\n $2a\\ =\\ \\left(\\sqrt{\\ 37}+\\sqrt{\\ 13}\\right)$,\u00a0 $2b\\ =\\ \\left(\\sqrt{\\ 37}-\\sqrt{\\ 13}\\right)$.<\/p>\n 2a is longer diagonal which is equal to\u00a0$\\ \\left(\\sqrt{\\ 37}+\\sqrt{\\ 13}\\right)$<\/p>\n Question 8:\u00a0<\/b>A circle of diameter 8 inches is\u00a0inscribed in a triangle ABC where $\\angle ABC = 90^\\circ$. If BC = 10 inches then the area of the triangle in square inches is<\/p>\n 8)\u00a0Answer:\u00a0120<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b> We know that Inradius=$\\frac{\\left(Perpendicular+Base-Hypotenuse\\right)}{2}$<\/p>\n $4=\\frac{\\left(p+10-h\\right)}{2}$<\/p>\n h-p = 2 or h= p+2.<\/p>\n Now,\u00a0$p^2+100\\ =\\ h^2$<\/p>\n $p^2+100\\ =\\ \\left(p+2\\right)^2$<\/p>\n $p^2+100\\ =\\ p^2+4p+4$<\/p>\n 4p = 96<\/p>\n p=24.<\/p>\n Hence, Area =\u00a0$\\frac{1}{2}\\times\\ 10\\times\\ 24\\ =\\ 120$.<\/p>\n Question 9:\u00a0<\/b>Suppose the length of each side of a regular hexagon ABCDEF is 2 cm.It T is the mid point of CD,then the length of AT, in cm, is<\/p>\n a)\u00a0$\\sqrt{13}$<\/p>\n b)\u00a0$\\sqrt{14}$<\/p>\n c)\u00a0$\\sqrt{12}$<\/p>\n d)\u00a0$\\sqrt{15}$<\/p>\n 9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b> Since a regular hexagon can be considered to be made up of 6 equilateral triangles, a line joining the farthest vertices of a hexagon can be considered to be made up using the sides of two opposite equilateral triangle forming the hexagon. Hence, its length should be twice the side of the hexagon, in this case, 4 cm.<\/p>\n Now, AD divided the hexagon into two symmetrical halves. Hence, AD bisects angle D, and hence, angle ADC is\u00a0$60^{\\circ\\ }$.<\/p>\n We can find out the value of AT using cosine formula:<\/p>\n $AT^2=\\ 4^2+1^2-2\\times\\ 1\\times\\ 4\\cos60$<\/p>\n $AT^2=\\ 17-4=13$<\/p>\n $AT=\\sqrt{\\ 13}$<\/p>\n Question 10:\u00a0<\/b>If the area of a regular hexagon is equal to the area of an equilateral triangle of side 12 cm, then the length, in cm, of each side of the hexagon is<\/p>\n a)\u00a0$4\\sqrt{6}$<\/p>\n b)\u00a0$6\\sqrt{6}$<\/p>\n c)\u00a0$\\sqrt{6}$<\/p>\n d)\u00a0$2\\sqrt{6}$<\/p>\n
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\nso side of rhombus = 10
\nNow\u00a0$\\frac{1}{2}\\times\\ d_1\\times\\ d_2\\ =\\ 96$
\nso we get\u00a0$\\ d_1\\times\\ d_2\\ =\\ 192$\u00a0\u00a0 (1)
\nAnd as we know diagonals of a rhombus are perpendicular bisectors of each other :
\nso\u00a0$\\ \\frac{d_1^2}{4}+\\ \\frac{d_2^2}{4}=\\ 100$
\nso we get\u00a0$\\ d_1^2+\\ d_2^2=\\ 400$\u00a0\u00a0\u00a0 (2)
\nSolving (1) and (2)
\nWe get $d_1=12$ and$d_2=16$
\nNow the cost, in INR, of laying electric wires along its two diagonals, at the rate of \u20b9125 per m, is= (12+16)(125) =3500<\/p>\n![](\"https:\/\/media-cdn.cracku.in\/uploads\/image_TVLFH21.png\")
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\n=$\\frac{1}{2}\\times\\ 2x\\times\\ 2y\\times\\ \\sin A=8$
\nwe get xy sinA =4
\nNow Area of triangle ABC = $\\frac{1}{2}AB\\times\\ AC\\times\\ \\sin A$
\nwe get $\\frac{1}{2}\\times3x\\ \\times5y\\ \\sin A=\\frac{15}{2}xy\\ \\sin A=\\ \\frac{15}{2}\\times\\ 4$
\nwe get Area of ABC = 30<\/p>\n![](\"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![](\"https:\/\/media-cdn.cracku.in\/uploads\/image_HDVydTD.png\")
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