285 SSC MOCKS FOR JUST RS. 249<\/a><\/p>\n FREE SSC EXAM YOUTUBE VIDEOS<\/a><\/p>\n Download Topic Wise SSC CHSL Important Questions PDF<\/a><\/p>\n SSC CHSL Study Material<\/a> (FREE Tests)<\/p>\n Question 1:\u00a0<\/b>Point R lies on line segment PQ such that PR = 24 cm and RQ = 6 cm. Points S and T lie on the same side of line PQ such that $\\triangle$PRS and $\\triangle$RQT are equilateral triangles. If M and N are mid points of line segment PT and QS respectively, then find out the area of $\\triangle$MNR.<\/p>\n a)\u00a0$\\dfrac{117\\sqrt{3}}{4}$ sq. cm<\/p>\n b)\u00a0$\\dfrac{63\\sqrt{3}}{\\sqrt{2}}$ sq. cm<\/p>\n c)\u00a0$\\dfrac{121\\sqrt{5}}{\\sqrt{5}}$ sq. cm<\/p>\n d)\u00a0$\\dfrac{103\\sqrt{3}}{4}$ sq. cm<\/p>\n Question 2:\u00a0<\/b>It is given that AO = BO and PA and PC are tangent to the circle. If $\\angle$OBC = $x$\u00b0 then find out the value of $20x$?<\/p>\n Question 3:\u00a0<\/b>A circle contains 2 perpendicular chords AB and CD. The length of chord AB is 6 cm and that of chord CD is 12 cm. C is closer to A and B than D. If it is known that chord CD passes through O, the centre of the circle, what is the area of the polygon AOBD?<\/p>\n a)\u00a09 $ cm^2$<\/p>\n b)\u00a018 $cm^2$<\/p>\n c)\u00a036 $cm^2$<\/p>\n d)\u00a027 $cm^2$<\/p>\n Question 4:\u00a0<\/b>3 circles of equal size are tightly packed inside a larger circle such that the 3 circles touch each other externally and touch the larger circle internally. If the diameter of the smaller circle is \u2018r\u2019, then the area outside the smaller circles but inside the larger circle is<\/p>\n a)\u00a0$\\pi r^2(\\frac{2\\sqrt{3}-5}{3})$<\/p>\n b)\u00a0$\\frac{\\pi r^2}{4}(\\frac{4\\sqrt{3}-2}{3} )$<\/p>\n c)\u00a0$\\frac{\\pi r^2}{4}(\\frac{2\\sqrt{3}-4}{3})$<\/p>\n d)\u00a0$\\frac{\\pi r^2}{4}(\\frac{2\\sqrt{3}+13}{3})$<\/p>\n Question 5:\u00a0<\/b>What is the volume of a regular tetrahedron whose side measures 20 cm?<\/p>\n a)\u00a0$471.4 cm^3$<\/p>\n b)\u00a0$1414.2 cm^3$<\/p>\n c)\u00a0$942.8 cm^3$<\/p>\n d)\u00a0$1885.6 cm^3$<\/p>\n SSC CHSL PREVIOUS PAPERS<\/a><\/p>\n SSC CHSL FREE MOCK TEST<\/a><\/p>\n Question 6:\u00a0<\/b>In a triangle ABC, P, Q, R are points on the sides AB, BC, AC respectively such that RQ = CQ and PQ = BQ. Find $\\angle{PQR}$ when $\\angle{CAB} = 60$\u00b0 ?(In degrees)<\/p>\n Question 7:\u00a0<\/b>In the figure given below, if \u2018C\u2019 is the centre of the circle and a,b,c are the areas of the Triangle PQR, Square QTRC and Circle respectively. Find the ratio a:b:c. a)\u00a0$\\sqrt{3}$:$1$:$\\pi$<\/p>\n b)\u00a0$\\pi$:$1$:$\\pi$<\/p>\n c)\u00a0$1$:$\\sqrt{3}$:$\\pi$<\/p>\n d)\u00a0$1$:$1$:$\\pi$<\/p>\n FREE SSC MATERIAL – 18000 FREE QUESTIONS<\/a><\/p>\n Question 8:\u00a0<\/b>Find the area bounded by the graph y = |x+p| and y = 6<\/p>\n a)\u00a072<\/p>\n b)\u00a054<\/p>\n c)\u00a036<\/p>\n d)\u00a0Data Insufficient<\/p>\n Question 9:\u00a0<\/b>An equilateral triangle GCD is constructed on side CD of a square ABCD. E is midpoint of side AB. What is the ratio of Area inside square ABCD and outside triangle EDC to the area of polygon DECG? a)\u00a0$ \\frac{ 2\\sqrt{3} }{2 – \\sqrt{3}} $<\/p>\n b)\u00a0$ \\frac{\\sqrt{3}}{2 – \\sqrt{3}} $<\/p>\n c)\u00a0$ \\frac{ 2 }{ 2 – \\sqrt{3} } $<\/p>\n d)\u00a0$ \\frac{ 2 }{ 2\\sqrt{3} -2 } $<\/p>\n e)\u00a0$ \\frac{ 2\\sqrt{3} -2 }{2} $<\/p>\n Question 10:\u00a0<\/b>A square EFGH of side 6 cm is drawn in an equilateral triangle ABC such that GH lie along BC. E is on side AB and F is on side AC respectively. Find the side of the triangle? a)\u00a0(3 + 6\u221a3) cm<\/p>\n b)\u00a0(4\u221a3 + 6) cm<\/p>\n c)\u00a0(4 + 6\u221a3) cm<\/p>\n d)\u00a0(4 + 3\u221a3) cm<\/p>\n e)\u00a0(6 + 3\u221a3) cm<\/p>\n DOWNLOAD APP TO ACESSES DIRECTLY ON MOBILE<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let us solve this problem with the help of co-ordinate geometry. Assume that point P is at origin and Q is at (30, 0) then co-ordinates of R = (24, 0).<\/p>\n We are given that points S and T lie on the same side of line PQ such that $\\triangle$PRS and $\\triangle$RQT are equilateral triangles. Hence we can say that PR = RS = PS = 24 units and RQ = QT = RT = 6 units.<\/p>\n We know that height of an equilateral trianlge = $\\dfrac{\\sqrt{3}a}{2}$<\/p>\n In triangle PSR,<\/p>\n ‘x’ co-ordinate of point S will be half of the sum of the ‘x’ co-ordinates of vertices P and R.<\/p>\n ‘y’ co-ordinate of point S will be equal to height of the triangle PSR.<\/p>\n Hence co-ordinates of vertex S = ($\\dfrac{0+24}{2}$, $\\dfrac{24\\sqrt{3}}{2}$) = ($12$, $12\\sqrt{3}$)<\/p>\n Similarly in triangle RTQ,<\/p>\n Co-ordinates of vertex T = ($\\dfrac{24 + 30}{2}$, $\\dfrac{6\\sqrt{3}}{2}$) = ($27$, $3\\sqrt{3}$).<\/p>\n It is given that M and N are mid points of line segment PT and QS respectively, hence,<\/p>\n Co-ordinates of the point M = ($\\dfrac{0 + 27}{2}$, $\\dfrac{0 + 3\\sqrt{3}}{2}$) = ($\\dfrac{27}{2}$, $\\dfrac{3\\sqrt{3}}{2}$).<\/p>\n Co-ordinates of the point N = ($\\dfrac{30 + 12}{2}$, $\\dfrac{0 + 12\\sqrt{3}}{2}$) = ($21$, $6\\sqrt{3}$).<\/p>\n Also co-ordinates of the pointR = (24, 0).<\/p>\n Now that we have all thee vertices of triangle MNR we can find the distance MN, NR and MR.<\/p>\n MN = $\\sqrt{(21-\\dfrac{27}{2})^2 + (6\\sqrt{3}-\\dfrac{3\\sqrt{3}}{2})^2}$ = $\\sqrt{(\\dfrac{15}{2})^2 + (\\dfrac{9\\sqrt{3}}{2})^2}$ = $3\\sqrt{13}$<\/p>\n NR = $\\sqrt{(24-21)^2 + (0-6\\sqrt{3})^2}$ = $3\\sqrt{13}$<\/p>\n MR = $\\sqrt{(24-\\dfrac{27}{2})^2 + (0-\\dfrac{3\\sqrt{3}}{2})^2}$ = $\\sqrt{(\\dfrac{21}{2})^2 + (\\dfrac{3\\sqrt{3}}{2})^2}$ = $3\\sqrt{13}$<\/p>\n We can see that MN = NR = MR. Hence we can say that triangle MNR is an equilateral triangle.<\/p>\n Therefore, the area of triangle MNR = $\\dfrac{\\sqrt{3}}{4}\\times (3\\sqrt{13})^2$ = $\\dfrac{117\\sqrt{3}}{4}$ sq. cm<\/p>\n Hence, option A is the correct answer.<\/p>\n 2)\u00a0Answer:\u00a01275<\/b><\/p>\n Drawing AQ and QB where Q is the centre of the circle. In quadilateral PAQB<\/p>\n $\\angle$PAQ+$\\angle$AQB+$\\angle$QBP+$\\angle$BPA = 360\u00b0<\/p>\n We know that angle subtend by tangent of centre of circle = 90\u00b0 i.e. $\\angle$PAQ = $\\angle$QBP = $\\angle$QBC = 90\u00b0<\/p>\n $\\Rightarrow$ 90\u00b0+$\\angle$AQB+90\u00b0+75\u00b0 = 360\u00b0<\/p>\n $\\Rightarrow$ $\\angle$AQB = 105\u00b0 We know that angle subtend by a chord on centre is twice of the angle subtend on perimeter. therefore<\/p>\n $\\angle$AOB=$\\frac{1}{2}\\angle$ AQB<\/p>\n $\\Rightarrow$ $\\angle$AOB = $\\frac{1}{2}\\times 105$ = 52.5\u00b0<\/p>\n Since AO = OB therefore $\\angle$AOQ = $\\angle$BOQ = $\\frac{1}{2}\\times$ $\\angle$AOB<\/p>\n $\\Rightarrow$ $\\angle$AOQ=$\\angle$BOQ=$\\frac{1}{2}\\times$ 52.5\u00b0 = 26.25\u00b0<\/p>\n In $\\triangle$ QBO,<\/p>\n $\\Rightarrow$ QB = QO therefore<\/p>\n $\\angle$QBO = $\\angle$QOB = 26.25\u00b0<\/p>\n We know that $\\angle$QBC = $\\angle$QBO + $\\angle$OBC = 90\u00b0<\/p>\n $\\Rightarrow$ $\\angle$OBC = 90\u00b0 – 26.25\u00b0 = 63.75\u00b0= $x$\u00b0<\/p>\n The value of 20$x$ = 20*63.75 = 1275. Hence 1275 is the correct answer.<\/p>\n FREE MOCK TEST FOR SSC STENOGRAPHER<\/a><\/p>\n 3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n CD passes through the centre of the circle. Therefore, CD must be the diameter of the circle. If we connect AB to the centre of the circle, we will get an equilateral triangle (All 3 sides will be 6 cm). $\\angle AOB$ = 60\u1d52 (Since triangle AOB is an equilateral triangle). Now, in triangle AOB, AO = BO= 6 cm (Radii of the circle). Therefore, option B is the right answer.<\/p>\n 4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n The radius of the smaller circle is $\\frac{r}{2}$ since it has been given that the diameter is $r$.<\/p>\n On joining the radii of the 3 circles, we get an equilateral triangle with side $r$. Area of the smaller circles = $3*\\frac{\\pi r^2}{4}$, since $r$ is the diameter here. Therefore, option B is the right answer.<\/p>\n 5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n It has been provided that the tetrahedron is regular.<\/p>\n In solid geometry, the volume of a pyramid is one-third the volume of a prism with similar height. All the 4 triangular faces in a tetrahedron are congruent. Therefore, the height of the triangle will be the slant height of the tetrahedron. Applying Pythagoras theorem, we get, $h^2+(a\/2\\sqrt{3})^2 = (\\sqrt{3}a\/2)^2$<\/p>\n Height of the tetrahedron = $\\frac{\\sqrt{6}a}{3}$<\/p>\n Volume of tetrahedron, V = (1\/3)* area of the base triangle* height of the tetrahedron Alternate solution:<\/strong><\/p>\n The volume of a tetrahedron is given by the formula, $V = \\frac{a^3}{6\\sqrt{2}}$. GK Questions And Answers PDF<\/a><\/p>\n FREE SSC PRACTICE SET (DAILY TEST)<\/a><\/p>\n 6)\u00a0Answer:\u00a060<\/b><\/p>\n Let us take $\\angle {C} = c$ and $\\angle {B} = b$ 7)\u00a0Answer\u00a0(D)<\/strong><\/p>\n let the radius of the circle be \u2018r\u2019 8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Look at the figure. 9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n <\/p>\n Let side of square be \u2018$a$\u2019 10)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Drawing a perpendicular from vertex A to the side EF to intersect it at point D<\/p>\n ED\/DA = tan30\u00b0![](\"https:\/\/cracku.in\/media\/uploads\/Geometry_DRjfa0A.PNG\")
<\/figure>\n![](\"https:\/\/cracku.in\/media\/uploads\/geometry.PNG\")
<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/Geometry%201_9ozUzw1.JPG\")
<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/Geometry%201.JPG\")
<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/Triangle_1.PNG\")
<\/figure>\n<\/figure>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/Diagram1.PNG\")
<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/Diagram%202.PNG\")
<\/figure>\n
\nLength of CD = 12 cm
\n=> Radius of the circle = 6 cm.
\nLength of AB = 6 cm.<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/geometry4.JPG\")
<\/figure>\n
\nNow, let us join A and D and B and D.<\/p>\n
\n$\\angle AOB$ and $\\angle ADB$ are subtended by the same chord, $AB$.
\nTherefore, $\\angle ADB$ = 30\u1d52 (Angle subtended by a chord on the circumference is half of the angle subtended by the same chord on the centre).<\/p>\n
\n$\\angle ADO$ = 30\u1d52\/2 = 15\u1d52 (Since the figure is symmetrical about CD).
\n=> $\\angle OAD$ = 15\u1d52 (Since triangle AOD is isosceles).
\nTherefore, $\\angle AOD$ = 180\u1d52 – 15\u1d52 – 15\u1d52 = 150\u1d52.
\nArea of triangle AOD = 0.5*r*r*sin 150\u1d52
\n= 0.5*6*6*0.5
\n= 9 $cm^2$
\nArea of the figure AOBD = 2* Area of triangle AOD
\n= 2*9
\n= 18 $cm^2$<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/geometry.JPG\")
<\/figure>\n
\nThe incentre of this equilateral triangle will also be the centre of the larger circle.
\nInradius of the equilateral triangle = $\\frac{r}{2\\sqrt{3}}$
\nRadius of the outer circle = Radius of the smaller circle + Circumradius of the triangle.
\n=> Radius of the outer circle = $\\frac{r}{2} + \\frac{r}{\\sqrt{3}}$
\nArea of the larger circle = $\\pi * [\\frac{r}{2}*(1 + \\frac{2}{\\sqrt{3}})]^2$
\n= $\\frac{\\pi*r^2}{4}* (1+\\frac{4}{3} + \\frac{4}{\\sqrt{3}})$
\n=$\\frac{\\pi*r^2}{4}*(\\frac{7+4\\sqrt{3}}{3})$<\/p>\n
\nDifference between the areas = $\\frac{\\pi r^2}{4}(\\frac{7+4\\sqrt{3}}{3} – 3)$
\n= $\\frac{\\pi r^2}{4}(\\frac{4\\sqrt{3}-2}{3})$<\/p>\n
\nTetrahedron is a special case of a pyramid.
\nLet us first find out the volume of the prism and then divide it by 3.<\/p>\n
\nLet us assume the side of the triangle to be $a$.
\nInradius = $a\/2\\sqrt{3}$
\nInradius, Height of the tetrahedron and height of the triangle form a right angled triangle.<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/corr.JPG\")
<\/figure>\n
\n= $\\frac{1}{3}* \\frac{\\sqrt{3}a^2}{4}*\\frac{\\sqrt{6}a}{3}$
\n= $ \\frac{a^3}{6\\sqrt{2}}$
\nNow, we can find the volume of the tetrahedron by substituting the value of $a = 20$.
\n$ V = \\frac{20^3}{6\\sqrt{2}}$
\n$ V = \\frac{8000}{6\\sqrt{2}}$
\n$V = 942.8 cm^3$
\nHence, option C is the right answer.<\/p>\n
\nBy substituting $a=20$, we can find the answer.<\/p>\n
\nFrom the figure below
\n60+b+c = 180
\nb+c = 120
\nIn triangle QRC
\n2c+e = 180\u2026\u2026\u2026..(I)
\nIn triangle PQB
\n2b+f = 180\u2026\u2026\u2026..(II)
\n(I)+(II)
\n2(b+c)+e+f = 360
\ne+f = 120
\n$\\angle{PQR} = 180-(e+f) = 60$<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/Geometry.PNG\")
<\/p>\n
\nThe length of the side of the square is equal to \u2018r\u2019
\nArea of the square QTRC, b = $r^{2}$
\nArea of triangle PQR, a = $\\frac{1}{2}$*$2r$*$r$ = $r^{2}$ (height of the triangle is equal to that of the radius $r$ of the circle.)
\nArea of Circle, c =$\\pi$*$r^{2} $
\na:b:c = $1$:$1$:$\\pi$<\/p>\n![\"\"](\"https:\/\/cracku.in\/media\/uploads\/2015\/09\/16\/geometry-solution.png\")
<\/p>\n
\nSince slope of the lines is 1
\nAnfle CBO = Angle DCB = 45
\nBD = DC = 6
\nSimilarly AD = 6
\nArea = 1\/2 *12*6 = 36<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/Geometry%201b_XuZ93fz.JPG\")
<\/p>\n
\nArea of triangle EDC => $ \\frac{1}{2} \\times a \\times a $ = $ \\frac{a^2}{2} $
\nArea of triangle GDC => $ \\frac{\\sqrt{3}a^2}{4} $
\nArea of polygon DECG => $ \\frac{a^2}{2} – \\frac{\\sqrt{3}a^2}{4} $ = $ \\frac{ (2 -\\sqrt{3})a^2 }{4} $
\nArea inside square ABCD and outside triangle EDC => $ \\frac{a^2}{2} $
\nRatio => $ \\frac{\\frac{a^2}{2}}{ \\frac{ (2 -\\sqrt{3})a^2 }{4} } $ = $ \\frac{ 2 }{ 2 – \\sqrt{3} } $
\nHence, option C is correct.<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/Geometry%201a_3FDwr2j.JPG\")
<\/p>\n
\nBy Symmetry, ED = 3 cm
\n=> DA = 3\u221a3 cm
\nAltitude of the triangle => (6 + 3\u221a3) cm
\nSide of the triangle => ( 2\/\u221a3 ) (6 + 3\u221a3) cm
\nSide of the triangle => (4\u221a3 + 6) cm
\nHence, option B is correct.<\/p>\n