25 SSC CGL Mocks – Just Rs. 149<\/a><\/p>\n FREE SSC EXAM YOUTUBE VIDEOS<\/a><\/p>\n Question 1:\u00a0<\/b>In a $\\triangle$ ABC, D and E are two points on sides AB and AC such that DE is parallel to BC and AD : DB = 2 : 1. If AE = 8 cm, then find the length of AC.<\/p>\n a)\u00a012 cm<\/p>\n b)\u00a010 cm<\/p>\n c)\u00a016 cm<\/p>\n d)\u00a020 cm<\/p>\n Question 2:\u00a0<\/b>In a $\\triangle$ ABC, Points D and E are on sides AB and AC such that DE is parallel to BC and AD : DB = 3 : 1 and AE = 18 cm. Then find AC.<\/p>\n a)\u00a026 cm<\/p>\n b)\u00a024 cm<\/p>\n c)\u00a028 cm<\/p>\n d)\u00a032 cm<\/p>\n Question 3:\u00a0<\/b>In a $\\triangle$ ABC, points D and E are on the sides of AB and AC respectively such that DE is parallel to BC and AD : AB = 2 : 5 and AE = 4 cm. Then find AC.<\/p>\n a)\u00a010 cm<\/p>\n b)\u00a014 cm<\/p>\n c)\u00a012 cm<\/p>\n d)\u00a09 cm<\/p>\n Question 4:\u00a0<\/b>The coordinates of the vertices of a right-angled triangle are A (6, 2), B(8, 0) and C (2, -2). The coordinates of the orthocentre of triangle PQR are<\/p>\n a)\u00a0(2, -2)<\/p>\n b)\u00a0(2, 1)<\/p>\n c)\u00a0(6, 2)<\/p>\n d)\u00a0(8, 0)<\/p>\n Question 5:\u00a0<\/b>find the area of an equilateral triangle if the height of the triangle is 24 cm.<\/p>\n a)\u00a0$192\\sqrt{3} cm^2$<\/p>\n b)\u00a0$175\\sqrt{3} cm^2$<\/p>\n c)\u00a0$178\\sqrt{3} cm^2$<\/p>\n d)\u00a0$164\\sqrt{3} cm^2$<\/p>\n SSC CGL Previous Papers Download PDF<\/a><\/p>\n SSC CHSL PREVIOUS PAPERS DOWNLOAD<\/a><\/p>\n Question 6:\u00a0<\/b>Three sides of a triangular meadow are of length 28 m, 45 m and 53 m long respectively. Find the cost of sowing seeds(in rupees per sq.m) in the meadow at the rate of 12 rupees per sq.m.<\/p>\n a)\u00a07560<\/p>\n b)\u00a06860<\/p>\n c)\u00a07960<\/p>\n d)\u00a07860<\/p>\n Question 7:\u00a0<\/b>In a triangle PQR, internal angular bisectors of $\\angle Q$ and $\\angle R$ intersect at a point O. If $\\angle P$=$110^\\circ$ then what is the value of $\\angle QOR$ ?<\/p>\n a)\u00a0$125^\\circ$<\/p>\n b)\u00a0$135^\\circ$<\/p>\n c)\u00a0$145^\\circ$<\/p>\n d)\u00a0$115^\\circ$<\/p>\n Question 8:\u00a0<\/b>In a triangle XYZ, XA is the angle bisector onto YZ. If the semiperimeter of the triangle is 12 and XY=12 ,YZ=6 then what is the ratio of YA:AZ ?<\/p>\n a)\u00a02:3<\/p>\n b)\u00a02:1<\/p>\n c)\u00a01:2<\/p>\n d)\u00a03:2<\/p>\n Question 9:\u00a0<\/b>In a triangle ABC, AX is the angle bisector onto BC. If the semiperimeter of the triangle is 9 and AB=4 ,BC=6 then what is the ratio of BX:XC ?<\/p>\n a)\u00a02:3<\/p>\n b)\u00a02:1<\/p>\n c)\u00a01:2<\/p>\n d)\u00a03:2<\/p>\n Question 10:\u00a0<\/b>In an equilateral triangle,if h-R=15 cm where h=height of the triangle and R=circumradius then what is the area of the triangle ?<\/p>\n a)\u00a0$275\\sqrt{3}$<\/p>\n b)\u00a0$225\\sqrt{3}$<\/p>\n c)\u00a0$500\\sqrt{3}$<\/p>\n d)\u00a0$675\\sqrt{3}$<\/p>\n 18000+ Questions – Free SSC Study Material<\/a><\/p>\n Question 11:\u00a0<\/b>In an equilateral triangle,if h-R=5 cm where h=height of the triangle and R=circumradius then what is the area of the triangle ?<\/p>\n a)\u00a0$50\\sqrt{3}$<\/p>\n b)\u00a0$100\\sqrt{3}$<\/p>\n c)\u00a0$75\\sqrt{3}$<\/p>\n d)\u00a0$25\\sqrt{3}$<\/p>\n Question 12:\u00a0<\/b>In an equilateral triangle PQR, PA is the altitude. O is the orthocentre and PO=12 the find the value of PA.<\/p>\n a)\u00a014<\/p>\n b)\u00a020<\/p>\n c)\u00a016<\/p>\n d)\u00a018<\/p>\n Question 13:\u00a0<\/b>In an equilateral triangle XYZ, XA is the altitude. O is the orthocentre and XO=8 the find the value of XA.<\/p>\n a)\u00a012<\/p>\n b)\u00a010<\/p>\n c)\u00a016<\/p>\n d)\u00a014<\/p>\n Question 14:\u00a0<\/b>In a triangle XYZ, I is the incentre and $\\angle XYZ=40$ and $\\angle YZX=60$. What is the value of $\\angle YIZ $=?<\/p>\n a)\u00a0120<\/p>\n b)\u00a0130<\/p>\n c)\u00a080<\/p>\n d)\u00a0100<\/p>\n Question 15:\u00a0<\/b>In a triangle PQR, I is the incentre and $\\angle PQR=70$ and $\\angle PRQ=40$. What is the value of $\\angle QIR $=?<\/p>\n a)\u00a070<\/p>\n b)\u00a0135<\/p>\n c)\u00a0115<\/p>\n d)\u00a0125<\/p>\n SSC CGL Free Mock Test<\/a><\/p>\n SSC CHSL Free Mock Test<\/a><\/p>\n Question 16:\u00a0<\/b>Find the value of \u2220CAB if Internal bisectors of \u2220A and \u2220B of \u0394ABC intersect at O and \u2220COB = 124\u00b0.<\/p>\n a)\u00a036\u00b0<\/p>\n b)\u00a016\u00b0<\/p>\n c)\u00a096\u00b0<\/p>\n d)\u00a068\u00b0<\/p>\n Question 17:\u00a0<\/b>Find the maximum possible area$(in\u00a0 cm^2)$ of a triangle which can be formed using a thread of length 36 cm.?<\/p>\n a)\u00a0$45$<\/p>\n b)\u00a0$39\\sqrt{3}$<\/p>\n c)\u00a0$36\\sqrt{3}$<\/p>\n d)\u00a0$36$<\/p>\n Question 18:\u00a0<\/b>If the angles of a triangle PQR are in the ratio 2:3:7, which among the following is a measure of an angle of the triangle PQR?<\/p>\n a)\u00a048\u00b0<\/p>\n b)\u00a055\u00b0<\/p>\n c)\u00a015\u00b0<\/p>\n d)\u00a0105\u00b0<\/p>\n Question 19:\u00a0<\/b>A tangent of length 45 cm is drawn\u00a0from a point to a circle of diameter 56 cm. Find the length(in cm) of the tangent from the center of the circle?<\/p>\n a)\u00a047<\/p>\n b)\u00a069<\/p>\n c)\u00a053<\/p>\n d)\u00a073<\/p>\n Question 20:\u00a0<\/b>$\\angle P, \\angle Q, \\angle R$ are three angles of a triangle. If\u00a0 $\\angle P – \\angle Q$ = $16^\\circ$, $\\angle Q – \\angle R$ = $28^\\circ$, then $\\angle P$, $\\angle Q$ and $\\angle R$ are<\/p>\n a)\u00a0$76^\\circ, 60^\\circ, 44^\\circ$<\/p>\n b)\u00a0$80^\\circ, 60^\\circ, 40^\\circ$<\/p>\n c)\u00a0$80^\\circ, 64^\\circ, 36^\\circ$<\/p>\n d)\u00a0$76^\\circ, 68^\\circ, 38^\\circ$<\/p>\n 1500+ Free SSC Questions & Answers<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $\\dfrac{AD}{DB} = \\dfrac{AE}{EC}$<\/p>\n Here, $\\dfrac{AD}{DB} = \\dfrac{2}{1}$ and AE = 8 cm<\/p>\n \u21d2 $\\dfrac{2}{1} = \\dfrac{8}{EC}$ 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given, DE is parallel to BC. Here, $\\dfrac{AD}{DB} = \\dfrac{3}{1}$<\/p>\n AE= 18 cm 3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given, DE is parallel to BC. Here, $\\dfrac{AD}{DB} = \\dfrac{2}{5}$<\/p>\n AE= 4 cm 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given that the coordinates of a right-angled triangle are A (6, 2), B(8, 0) and C (2, -2). 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given, In an equilateral triangle all the angles are equal to 60\u00b0<\/p>\n In $\\triangle$ADC<\/p>\n => $tan \\angle ACD = \\frac{AD}{DC}$<\/p>\n => $tan 60\u00b0 = \\frac{24}{DC}$<\/p>\n => DC = $\\frac{24}{\\sqrt{3}} = 8\\sqrt{3}$<\/p>\n => BC = 2*DC = $16\\sqrt{3}$<\/p>\n Area of $\\triangle$ ABC = $\\frac{\\sqrt{3}}{4} * side^2$<\/p>\n = $\\frac{\\sqrt{3}}{4} * (16\\sqrt{3})^2$<\/p>\n = $192\\sqrt{3} cm^2$<\/p>\n 100+ Free GK Tests for SSC Exams<\/a><\/p>\n Download Free GK PDF<\/a><\/p>\n 6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given the sides of triangle are 28 m, 45 m and 53 m<\/p>\n Since, $28^2 + 45^2 = 784+2025 = 2809 = 53^2$<\/p>\n => The given sides are of right angled triangle as,<\/p>\n $c^2 + b^2 = a^2$where a, b, c are the sides of the triangle.<\/p>\n => Area of triangular field = $\\large\\frac{1}{2}$ $ \\times 28 \\times 45= 630 m^2$<\/p>\n => Cost of sowing seeds = $12 \\times 630 = Rs. 7560 rupees\/m^2$<\/p>\n 7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let $\\angle Q$=2x and $\\angle R$=2y 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n In triangle XYZ we have s=12 9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n In triangle ABC we have s=9 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n In an equilateral triangle,all the points such as orthocentre,centroid,circumcenter coincide. s=$30\\sqrt{3}$ 11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n In an equilateral triangle,all the points such as orthocentre,centroid,circumcenter coincide. $\\sqrt{3}*s\/2$=15<\/p>\n s=$10\\sqrt{3}$<\/p>\n Area of an equilateral triangle=$\\sqrt{3}(s^{2})\/4$<\/p>\n =$75\\sqrt{3}$<\/p>\n 12)\u00a0Answer\u00a0(D)<\/strong><\/p>\n In an equilateral triangle,all the points such as orthocentre,centroid,circumcenter coincide. 15000 Questions – Free SSC Study Material<\/a><\/p>\n 13)\u00a0Answer\u00a0(A)<\/strong><\/p>\n In an equilateral triangle,all the points such as orthocentre,centroid,circumcenter coincide. 14)\u00a0Answer\u00a0(B)<\/strong><\/p>\n In a triangle XYZ, 15)\u00a0Answer\u00a0(D)<\/strong><\/p>\n In a triangle PQR, 16)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given that,\u00a0\u2220COB = 124\u00b0<\/p>\n To find : $\\angle$RPQ = $\\theta$ = ?<\/p>\n Solution : Let $\\angle$CAB = $2x$ and $\\angle$ACB = $2y$<\/p>\n => $\\angle$OBC = $x$ and $\\angle$OCB = $y$ [SInce, BO & CO are angle bisectors]<\/p>\n In $\\triangle$ABC<\/p>\n => $\\theta$ + $\\angle$ABC + $\\angle$ACB = 180\u00b0<\/p>\n => $\\theta$ = 180\u00b0 -2$(x+y)$ ———Eq. (1)<\/p>\n In $\\triangle$BOC<\/p>\n => $x + y + 124\u00b0 = 180\u00b0$<\/p>\n => $x + y = 56\u00b0$<\/p>\n Putting value of (x+y) in eq.\u00a0 (1)<\/p>\n => $\\theta$ = $180 – 2*56 = 180-112 = 68\u00b0$<\/p>\n 17)\u00a0Answer\u00a0(C)<\/strong><\/p>\n The perimeter of the triangle is given as 36 cm.<\/p>\n The triangle of a fixed perimeter will have the maximum possible area when all the sides are of equal length.<\/p>\n So each side should be of length 12 cm for the triangle to have maximum possible area(i.e. Equilateral triangle).<\/p>\n In this case the area will be,<\/p>\n $\\frac{\\sqrt{3}}{4}\\times a^2$<\/p>\n $\\frac{\\sqrt{3}}{4}\\times12\\times12$<\/p>\n $=36\\sqrt{3}$ $cm^2$<\/p>\n 18)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given,\u00a0the angles of a triangle PQR are in the ratio 2:3:7<\/p>\n Let the angles of the triangle be 2x,3x and 7x.<\/p>\n Sum of the angles in a triangle = 180\u00b0<\/p>\n 2x+3x+7x = 180\u00b0<\/p>\n 12x = 180<\/p>\n x = 15\u00b0<\/p>\n The angles are 30\u00b0, 45\u00b0, 105\u00b0<\/p>\n Hence, option D is the correct answer.<\/p>\n 19)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given diameter = 56cm<\/p>\n radius OA = 56\/2 = 28 and AB = 45cm<\/p>\n Now, in $\\triangle$OAB right angle at A<\/p>\n OB = $\\sqrt{(AB)^2 + (OA)^2}$<\/p>\n = $\\sqrt{784+2025} = \\sqrt{2809}$<\/p>\n = 53 cm<\/p>\n 20)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given\u00a0 $\\angle P-\\angle Q = 16^\\circ \\rightarrow (1)$![](\"https:\/\/cracku.in\/media\/uploads\/Screenshot_2_PQixNg1.png\")
<\/figure>\n
\n\u21d2 $EC = 4 cm$
\nHere, AC = AE+EC = 4+8 = 12 cm<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/Screenshot_1_eFcUUpZ.png\")
<\/figure>\n
\nThen, $\\dfrac{AD}{DB} = \\dfrac{AE}{EC}$<\/p>\n
\n$\\dfrac{3}{1} = \\dfrac{18}{EC}$
\n\u21d2 EC = 6 cm
\nAC = AE+EC = 6+18 = 24 cm<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/Geometry_TjRC9dL.png\")
<\/figure>\n
\nThen, $\\dfrac{AD}{DB} = \\dfrac{AE}{EC}$<\/p>\n
\n$\\dfrac{2}{5} = \\dfrac{4}{EC}$
\n\u21d2 EC = 10 cm
\nAC = AE+EC = 4+10 = 14 cm<\/p>\n
\nWe know that the distance between two points(a, b) &(c, d) is
\nD = $\\sqrt{(a-c)^2+(b-d)^2}$
\nAB = $2\\sqrt{2}$ units; BC = $2\\sqrt{10}$; AC = $4\\sqrt{2} $
\nNow, $AB^2 + AC^2 = BC^2$
\nTherefore, angle A is right angled.
\nSince it is a right angled triangle, the 2 sides adjacent to the right angle will be altitudes. The third altitude must meet at the vertex at which these 2 sides meet.
\nHence, the vertex that contains the right angle is the orthocentre. From the points given, we can clearly see that (6, 2) is the orthocentre. Option C is the right answer,<\/p>\n
\nAD = 24 cm and ABC is an equilateral triangle<\/p>\n![](\"https:\/\/cracku.in\/blogs:\/\/lh3.googleusercontent.com\/ZG4QbGmP31YEuDC3I9odKk-5ZqA7tw3q5eRFT-RCI9p0VzmW_fvOk18VdAYZaKJS5Z-WVgPIibltPPBNSWmtVHoKRrRhk6IQ0KWiGB5WIX9orFH-S7KNOBCop-Z8CBvMalgtSWnN\")
![](\"https:\/\/cracku.in\/media\/uploads\/Triangle2.JPG\")
<\/figure>\n
\nSum of angles in a triangle=180
\n2x+2y+110=180
\n2x+2y=70
\nx+y=35
\nSimilarly in the triangle QOR we have $\\angle OQR +\\angle QRO+ \\angle QOR $=180
\nx+y+$\\angle QOR =180^\\circ$
\n35+$\\angle QOR =180^\\circ$
\n$\\angle QOR =145^\\circ$<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/209486.png\")
<\/figure>\n
\n(x+y+z)\/2 =12
\nx+y+z=24
\n12+6+y=24
\ny=6
\nAngle bisector divides the opposite side in the ratio of other sides i.e
\nXY\/XZ=YA\/AZ
\nYA\/AZ=12\/6
\nYA\/AZ=2:1<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/209485.png\")
<\/figure>\n
\n(a+b+c)\/2 =9
\na+b+c=18
\n4+6+b=18
\nb=8
\nAngle bisector divides the opposite side in the ratio of other sides i.e
\nAB\/AC=BX\/XC
\nBX\/XC=4\/8
\nBX\/XC=1:2<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/209428.png\")
<\/figure>\n
\nLet the triangle be PQR and the circumcentre be O. let median intersect QR at A
\nCentroid divides median in the ratio 2:1.PA=h,OA=R
\nOA=15 cm
\nTherefore (PO:OA)=2:1
\nPO:15=2:1
\nPO=30
\nPA=PO+OP
\nPA=30+15
\nPA=45
\nPA is also the altitude using it side of the triangle can be calculated.
\n$\\sqrt{3}*s\/2$=45<\/p>\n
\nArea of an equilateral triangle=$\\sqrt{3}(s^{2})\/4$
\n=$675\\sqrt{3}$<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/209425.png\")
<\/figure>\n
\nLet the triangle be PQR and the circumcentre be O. let median intersect QR at A
\nCentroid divides median in the ratio 2:1
\nOA=5 cm
\nTherefore (PO:OA)=2:1
\nPO:5=2:1
\nPO=10
\nPA=PO+OP
\nPA=10+5
\nPA=15
\nPA is also the altitude using it side of the triangle can be calculated.<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/209414.png\")
<\/figure>\n
\nSo O is also the centroid.
\nCentroid divides median in the ratio 2:1
\nTherefore (PO:OA)=2:1
\n12:OA=2:1
\nOA=6
\nPA=PO+PA
\nPA=12+6
\nPA=18<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/209413.png\")
<\/figure>\n
\nSo O is also the centroid.
\nCentroid divides median in the ratio 2:1
\nTherefore (XO:OA)=2:1
\n8:OA=2:1
\nOA=4
\nXA=XO+OA
\nXA=8+4
\nXA=12<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/209409.png\")
<\/figure>\n
\n$\\angle X+\\angle Y+\\angle Z$=180
\nGiven $\\angle Y$=40 $\\angle Z $=60
\nTherefore $\\angle X$=180-(40+60)
\n=80
\nAs I is the incentre we have $\\angle YIZ $=90+($\\angle X$\/2)
\n=90+40
\n=130<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/209408.png\")
<\/figure>\n
\n$\\angle P+\\angle Q+\\angle R$=180
\nGiven $\\angle Q$=70 $\\angle R $=40
\nTherefore $\\angle P$=180-(40+70)
\n=70
\nAs I is the incentre we have $\\angle QIR $=90+($\\angle P$\/2)
\n=90+35
\n=125<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/triangle_VpPMwaA.JPG\")
<\/figure>\n![](\"https:\/\/cracku.in\/media\/uploads\/1525.PNG\")
<\/p>\n
\n$\\angle Q-\\angle R = 28^\\circ\\rightarrow (2)$
\nFrom equation (1), $\\angle Q = \\angle P-16^\\circ$
\nSubstituting $\\angle B$ value in equation (2)
\n$(\\angle P-16)-\\angle R = 28^\\circ$
\n$\\Rightarrow \\angle R = \\angle P-44^\\circ$
\nWe know that $\\angle P+\\angle Q+\\angle R=180^\\circ$
\nSubstituting $\\angle P,\\angle Q,\\angle R$ values in above equation
\n$\\angle P+(\\angle P-16^\\circ)+(\\angle P-44^\\circ)=180^\\circ$
\n$\\Rightarrow 3\\angle P=240^\\circ$
\n$\\angle P=80^\\circ$
\nSubstituting $\\angle P$ value in equation (1)
\n$80^\\circ-\\angle Q=16^\\circ$
\n$\\Rightarrow \\angle Q=64^\\circ$
\nSubstituting $\\angle Q$ in equation (2)
\n$64^\\circ-\\angle R = 28^\\circ$
\n$\\Rightarrow \\angle R = 36^\\circ$
\n$\\therefore \\angle P=80^\\circ, \\angle Q=64^\\circ, \\angle R=36^\\circ$<\/p>\n