Enroll To CAT 3 Months Crash Course<\/a><\/p>\n Download CAT Quant Questions PDF<\/a><\/p>\n Take a free mock test for CAT<\/a><\/p>\n Question 1:\u00a0<\/b><\/p>\n a)\u00a0\u00a03, 3<\/p>\n b)\u00a04, 2<\/p>\n c)\u00a02, 4<\/p>\n d)\u00a04, 4<\/p>\n Question 2:\u00a0<\/b>In the figure below, ABCDEF is a regular hexagon and $\\angle{AOF}$ = 90\u00b0 . FO is parallel to ED. What is the ratio of the area of the triangle AOF to that of the hexagon ABCDEF? a)\u00a01\/12<\/p>\n b)\u00a01\/6<\/p>\n c)\u00a01\/24<\/p>\n d)\u00a01\/18<\/p>\n Download CAT 2019 Syllabus PDF<\/a><\/p>\n Question 3:\u00a0<\/b>ABCDEF is a regular hexagon and PQR is an equilateral triangle of side a. The area of the shaded portion is X and CD : PQ : : 2 : 1. Find the area of the circle circumscribing the hexagon in terms of X.<\/p>\n a)\u00a0$\\frac{16 \\pi}{23 \\sqrt{3}}$ X<\/p>\n b)\u00a0$\\frac{42 \\pi}{5 \\sqrt{3}}$ X<\/p>\n c)\u00a0$\\frac{2 \\pi}{3 \\sqrt{3}}$ X<\/p>\n d)\u00a0$2\\sqrt{3 \\pi}$ X<\/p>\n Question 4:\u00a0<\/b>Let PQRSTU be a regular hexagon. The ratio of the area of the triangle PRT to that of the hexagon PQRSTU is<\/p>\n a)\u00a00.3<\/p>\n b)\u00a00.5<\/p>\n c)\u00a01<\/p>\n d)\u00a0None of the above<\/p>\n Take a free CAT online mock test<\/a><\/p>\n Question 5:\u00a0<\/b>Let ABCDEF be a regular hexagon with each side of length 1 cm. The area (in sq cm) of a square with AC as one side is<\/p>\n a)\u00a0$3\\sqrt{2}$<\/p>\n b)\u00a0$3$<\/p>\n c)\u00a0$4$<\/p>\n d)\u00a0$\\sqrt{3}$<\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n A is the correct answer.<\/p>\n CAT Online Most Trusted Courses<\/a><\/p>\n CAT Previous Year Papers PDF<\/a><\/p>\n 2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n 3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let ‘a’ be the length of side of the hexagon. Therefore, the area of hexagon = $6*\\dfrac{\\sqrt{3}}{4}\\times a^2$.<\/p>\n The area of triangle PQR = $\\dfrac{\\sqrt{3}}{4}\\times (a\/2)^2$<\/p>\n Therefore, the are of the shaded region = $6*\\dfrac{\\sqrt{3}}{4}\\times a^2$ – $\\dfrac{\\sqrt{3}}{4}\\times (a\/2)^2$ = $\\dfrac{23\\sqrt{3}a^2}{16} = X$<\/p>\n $\\Rightarrow$ $a^2 = \\dfrac{16X}{23\\sqrt{3}}$<\/p>\n The area of the circle = $\\pi*a^2$<\/p>\n $\\Rightarrow$ $\\dfrac{16 \\pi}{23 \\sqrt{3}}$ X. Hence, option A is the correct answer.<\/p>\n 4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n It’s given that PQRSTU is a regular hexagon and O is the center of the hexagon.<\/p>\n If we fold $\\triangle$TSR ,$\\triangle$PQR ,$\\triangle$TUP along lines TR, PR, PT respectively then vertices S,Q,U will overlap each other exactly at center of the hexagon.<\/p>\n Hence we can say that Area of hexagon PQRSTU =\u00a0 2($\\triangle$PRT)<\/p>\n So\u00a0Area of $\\triangle$PRT = 0.5(Area of hexagon PQRSTU)<\/p>\n 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n The length of the diagonals of a regular hexagon with side s are $\\sqrt{3}s$.
\n<\/figure>\n
![](\"https:\/\/cracku.in\/media\/question\/4385.png\")
<\/p>\n<\/figure>\n
\nLet’s calculate the measure of each internal angle.
\nSum of the internal angles of a polygon of n sides = (n-2)*180
\nEach internal angle =$\\frac{(n-2)*180}{n}$
\n=$120^0$
\nSo the Exterior angle of angle P is $60^0$. APQ forms an equilateral triangle of side 1 cm.
\nSimilarly, RBX forms an equilateral triangle of side 2 cms.
\nLength of AB =7 cms
\nABC is an equilateral triangle of side 7 cms.
\nLength of CY = length of CB- length of YB
\n=7-4=3 cms
\nLength of PZ = Length of AC – (length of AP+ length of CZ)
\n7-(1+3)
\n=4 cms
\nHence the length of YZ is 3 cms and ZP is 3 cms
\n<\/figure>\n
![\"\"](\"https:\/\/cracku.in\/media\/uploads\/2015\/07\/07\/geom3.PNG\")
\nWhen the hexagon is divided into number of similar triangle AOF we get 12 such triangles . Hence required ratio of area is 1\/12.<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/11_WfCh4Fe.PNG\")
<\/figure>\n![](\"https:\/\/cracku.in\/media\/uploads\/33_LGI1Wxh.PNG\")
<\/figure>\n
\nHere length of AC =
\n$\\sqrt{3}s$ =\u00a0$\\sqrt{3}$ cms
\nHence area of the square = $\\sqrt{3}^2$ = 3 sq cm<\/p>\n