Enroll to 15 SSC CHSL 2022 Mocks At Just Rs. 149<\/a><\/p>\n Question 1:\u00a0<\/b>Vertices A, B, C and D of a quadrilateral ABCD lie on a circle. $\\angle$A is three times $\\angle$C and $\\angle$D is two times $\\angle$B. What is the difference between the measures of $\\angle$D and $\\angle$C?<\/p>\n a)\u00a0$55^\\circ$<\/p>\n b)\u00a0$65^\\circ$<\/p>\n c)\u00a0$75^\\circ$<\/p>\n d)\u00a0$45^\\circ$<\/p>\n 1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b> Given,<\/p>\n $\\angle$A is three times $\\angle$C and $\\angle$D is two times $\\angle$B.<\/p>\n $\\angle$A = 3$\\angle$C…….(1)<\/p>\n $\\angle$D = 2$\\angle$B…….(2)<\/p>\n In a cyclic quadrilateral, opposite angles are supplementary.<\/p>\n $\\angle$A +\u00a0$\\angle$C = 180$^\\circ$ and\u00a0$\\angle$B + $\\angle$D = 180$^\\circ$<\/p>\n $\\angle$A + $\\angle$C = 180$^\\circ$<\/p>\n 3$\\angle$C +\u00a0$\\angle$C =\u00a0180$^\\circ$\u00a0 [From (1)]<\/p>\n 4$\\angle$C =\u00a0180$^\\circ$<\/p>\n $\\angle$C = 45$^\\circ$<\/p>\n $\\angle$A =\u00a03$\\angle$C =\u00a0135$^\\circ$<\/p>\n $\\angle$B + $\\angle$D = 180$^\\circ$<\/p>\n $\\angle$B + 2$\\angle$B = 180$^\\circ$\u00a0 [From (2)]<\/p>\n 3$\\angle$B = 180$^\\circ$<\/p>\n $\\angle$B = 60$^\\circ$<\/p>\n $\\angle$D = 2$\\angle$B = 120$^\\circ$<\/p>\n Difference between the measures of $\\angle$D and $\\angle$C =\u00a0120$^\\circ$ –\u00a045$^\\circ$<\/p>\n =\u00a075$^\\circ$<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 2:\u00a0<\/b>ABCD is a cyclic quadrilateral such that when sides AB and DC are produced, they meet at E, and sides AD and BC meet at F, when produced. If $\\angle$ADE = 80$^\\circ$ and $\\angle$AED = 50$^\\circ$, then what is the measure of $\\angle$AFB?<\/p>\n a)\u00a0$40^\\circ$<\/p>\n b)\u00a0$20^\\circ$<\/p>\n c)\u00a0$50^\\circ$<\/p>\n d)\u00a0$30^\\circ$<\/p>\n 2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b> From triangle AED,<\/p>\n $\\angle$ADE +\u00a0$\\angle$AED +\u00a0$\\angle$DAE =\u00a0180$^\\circ$<\/p>\n 80$^\\circ$ +\u00a050$^\\circ$ +\u00a0$\\angle$DAE =\u00a0180$^\\circ$<\/p>\n $\\angle$DAE =\u00a050$^\\circ$<\/p>\n ABCD is a cyclic quadrilateral.<\/p>\n Opposite angles in a cyclic quadrilateral is supplementary.<\/p>\n $\\angle$ADC +\u00a0$\\angle$ABC =\u00a0180$^\\circ$<\/p>\n 80$^\\circ$ +\u00a0$\\angle$ABC =\u00a0180$^\\circ$<\/p>\n $\\angle$ABC =\u00a0100$^\\circ$<\/p>\n From triangle ABF,<\/p>\n $\\angle$ABF + $\\angle$AFB + $\\angle$BAF = 180$^\\circ$<\/p>\n 100$^\\circ$ + $\\angle$AFB + 50$^\\circ$\u00a0= 180$^\\circ$<\/p>\n $\\angle$AFB = 30$^\\circ$<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 3:\u00a0<\/b>ABCD is cyclic quadrilateral in which $\\angle$A = x$^\\circ$, $\\angle$B = 5y$^\\circ$, $\\angle$C = 2x$^\\circ$ and $\\angle$D = y$^\\circ$. What is the value of (3x – y)?<\/p>\n a)\u00a0120<\/p>\n b)\u00a090<\/p>\n c)\u00a0150<\/p>\n d)\u00a060<\/p>\n 3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b> ABCD is cyclic quadrilateral.<\/p>\n Opposite angles in a cyclic quadrilateral are supplementary.<\/p>\n $\\angle$A +\u00a0$\\angle$C =\u00a0180$^\\circ$<\/p>\n x +\u00a02x =\u00a0180$^\\circ$<\/p>\n 3x =\u00a0180$^\\circ$<\/p>\n x =\u00a060$^\\circ$<\/p>\n $\\angle$B + $\\angle$D = 180$^\\circ$<\/p>\n 5y + y = 180$^\\circ$<\/p>\n 6y = 180$^\\circ$<\/p>\n y = 30$^\\circ$<\/p>\n 3x – y = 3(60) – 30 = 150<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 4:\u00a0<\/b>Triangle ABC is an equilateral triangle. D and E are points on AB and AC respectively such that DE is parallel to BC and is equal to half the length of BC. If AD +\u00a0CE + BC = 30 cm, then find the perimeter (in cm) of the quadrilateral BCED.<\/p>\n a)\u00a037.5<\/p>\n b)\u00a025<\/p>\n c)\u00a045<\/p>\n d)\u00a035<\/p>\n 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b> Triangle ABC is an equilateral triangle.<\/p>\n Let the length of BC = 2p<\/p>\n BC = AB = AC = 2p<\/p>\n DE is equal to half the length of BC.<\/p>\n Triangle ABC and triangle ADE are similar triangles.<\/p>\n $\\Rightarrow$\u00a0\u00a0$\\frac{AD}{AB}=\\frac{DE}{BC}$<\/p>\n $\\Rightarrow$\u00a0\u00a0$\\frac{AD}{AB}=\\frac{p}{2p}$<\/p>\n $\\Rightarrow$\u00a0 $AD=\\frac{1}{2}AB$<\/p>\n $\\Rightarrow$\u00a0 $AD=\\frac{1}{2}\\times2p$<\/p>\n $\\Rightarrow$\u00a0 AD = p<\/p>\n Similarly, AE = p<\/p>\n and EC = AC – AE = 2p – p = p<\/p>\n AD + CE + BC = 30 cm<\/p>\n p + p + 2p = 30<\/p>\n 4p = 30<\/p>\n p =\u00a0$\\frac{15}{2}$ cm<\/p>\n Perimeter of\u00a0the quadrilateral BCED = BD + DE + CE + BC<\/p>\n = p + p + p + 2p<\/p>\n = 5p<\/p>\n =\u00a0$5\\times\\frac{15}{2}$<\/p>\n = 37.5 cm<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 5:\u00a0<\/b>In the figure, a circle touches all the four sides of a quadrilateral ABCD whose sides AB = 6.5 cm, BC = 5.4 cm and CD = 5.3 cm. The length of AD is: a)\u00a04.6 cm<\/p>\n b)\u00a05.8 cm<\/p>\n c)\u00a06.2 cm<\/p>\n d)\u00a06.4 cm<\/p>\n 5)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b> Given,\u00a0AB = 6.5 cm, BC = 5.4 cm and CD = 5.3 cm<\/p>\n Let the circle touches AB, BC, CD, DA at T, R, Q, S respectively.<\/p>\n Length of tangents to the circle from an external point are equal.<\/p>\n AT = AS<\/p>\n BT = BR<\/p>\n CQ = CR<\/p>\n DQ = DS<\/p>\n Adding all of the above<\/p>\n AT + BT + CQ + DQ = AS + BR + CR + DS<\/p>\n $\\Rightarrow$\u00a0 (AT + BT) + (CQ + DQ) = (AS + DS) + (BR + CR)<\/p>\n $\\Rightarrow$\u00a0 AB + CD = AD + BC<\/p>\n $\\Rightarrow$\u00a0 6.5 + 5.3 = AD + 5.4<\/p>\n $\\Rightarrow$\u00a0 AD = 6.4 cm<\/p>\n Hence, the correct answer is Option D<\/p>\n Take a free SSC CHSL Tier-1 mock test<\/a><\/p>\n Download SSC CGL Tier-1 Previous Papers PDF<\/a><\/p>\n Question 6:\u00a0<\/b>The three medians AX, BY and CZ of $\\triangle$ABC intersect at point L. If the area of $\\triangle$ABC is 30 $cm^2$, then the area of the quadrilateral BXLZ is:<\/p>\n a)\u00a010 $cm^2$<\/p>\n b)\u00a012 $cm^2$<\/p>\n c)\u00a016 $cm^2$<\/p>\n d)\u00a014 $cm^2$<\/p>\n 6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b> Given, Area of $\\triangle$ABC = 30 $cm^2$<\/p>\n The three medians AX, BY and CZ of $\\triangle$ABC intersect at point L as shown in the above figure<\/p>\n The area of six triangles in the above figure are equal and area is equal to one-sixth of the area of triangle ABC.<\/p>\n Area of $\\triangle$BLX = $\\frac{1}{6}\\times$Area of\u00a0$\\triangle$ABC<\/p>\n = $\\frac{1}{6}\\times$30<\/p>\n = 5 $cm^2$<\/p>\n Similarly, Area of $\\triangle$BLZ = 5 $cm^2$<\/p>\n $\\therefore\\ $Area of quadrilateral =\u00a0Area of $\\triangle$BLX +\u00a0Area of $\\triangle$BLZ = 5 + 5 = 10\u00a0$cm^2$<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 7:\u00a0<\/b>In the given figure, PQRS is a cyclic quadrilateral. What is the measure of the angle PQR if PQ is parallel to SR? a)\u00a0$110^\\circ$<\/p>\n b)\u00a0$80^\\circ$<\/p>\n c)\u00a0$100^\\circ$<\/p>\n d)\u00a0$70^\\circ$<\/p>\n 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b> In the cyclic quadrilateral PQRS,<\/p>\n Sum of opposite angles = 180$^{\\circ\\ }$<\/p>\n $=$> \u00a0$\\angle$SPQ +\u00a0$\\angle$SRQ =\u00a0180$^{\\circ\\ }$<\/p>\n $=$> \u00a0110$^{\\circ\\ }$\u00a0+ $\\angle$SRQ = 180$^{\\circ\\ }$<\/p>\n $=$> \u00a0$\\angle$SRQ =\u00a070$^{\\circ\\ }$<\/p>\n Given,\u00a0PQ is parallel to SR<\/p>\n RQ is the transversal intersecting the parallel lines PQ and SR<\/p>\n Sum of the interior angles on the same side of the transversal is 180$^{\\circ\\ }$<\/p>\n $=$> \u00a0$\\angle$SRQ +\u00a0$\\angle$PQR =\u00a0180$^{\\circ\\ }$<\/p>\n $=$> \u00a070$^{\\circ\\ }$ + $\\angle$PQR = 180$^{\\circ\\ }$<\/p>\n $=$> \u00a0$\\angle$PQR =\u00a0110$^{\\circ\\ }$<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 8:\u00a0<\/b>ABCD is a cyclic quadrilateral which sides AD and BC are produced to meet at P, and sides DC and AB meet at Q when produced. If $\\angle A = 60^\\circ$ and $\\angle ABC = 72^\\circ$, then \\angle PDC –\u00a0$\\angle DPC =$<\/p>\n a)\u00a0$24^\\circ$<\/p>\n b)\u00a0$30^\\circ$<\/p>\n c)\u00a0$36^\\circ$<\/p>\n d)\u00a0$40^\\circ$<\/p>\n 8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b> In $\\triangle ABP$,<\/p>\n $\\angle A + \\angle ABC + \\angle APB = 180\\degree$<\/p>\n $\\angle APB = 180 – 60 – 72 = 48\\degree $<\/p>\n $\\angle ADC = 180 – ABC = 180 – 72 = 108\\degree$ $\\angle PDC – \\angle DPC = 72 – 48 = 24\\degree$<\/p>\n Question 9:\u00a0<\/b>In quadrilateral PQRS, RM $\\perp$ QS, PN $\\perp$ QS and QS = 6 cm. If RM = 3 cm and PN = 2 cm, then the area of PQRS is<\/p>\n a)\u00a013 $cm^2$<\/p>\n b)\u00a015 $cm^2$<\/p>\n c)\u00a014 $cm^2$<\/p>\n d)\u00a011 $cm^2$<\/p>\n 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b> Area of PQRS = area of $\\triangle$PQS +\u00a0area of $\\triangle$RQS<\/p>\n (Area of triangle = $\\frac{1}{2}\\times base \\times height$)<\/p>\n =\u00a0$\\frac{1}{2}\\times 6 \\times 2$ +\u00a0$\\frac{1}{2}\\times 6 \\times 3$<\/p>\n = $\\frac{1}{2}\\times 6(2 + 3)$<\/p>\n = $15 cm^2$<\/p>\n Question 10:\u00a0<\/b>Sides AB and DC of cyclic quadrilateral ABCD are produced to meet at E, and sides AD and BCare produced to meet at F. If $\\angle BAD = 102^\\circ$ and $\\angle BEC = 38^\\circ$ then the difference between $\\angle ADC$ and $\\angle AFB$ is:<\/p>\n a)\u00a0$21^\\circ$<\/p>\n b)\u00a0$31^\\circ$<\/p>\n c)\u00a0$22^\\circ$<\/p>\n d)\u00a0$23^\\circ$<\/p>\n 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b> In \u0394ADE, Difference between $\\angle BAD$ and $\\angle AFB$ = 62 – 40 = 22$\\degree$<\/p>\n Question 11:\u00a0<\/b>Quadrilateral ABCD circumscribes circle. If AB = 8 cm, BC = 7 cm and CD = 6 cm,then the length of AD is:<\/p>\n a)\u00a06 cm<\/p>\n b)\u00a07.5 cm<\/p>\n c)\u00a07cm<\/p>\n d)\u00a06.8 cm<\/p>\n 11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n AB = 8 cm<\/p>\n BC = 7 cm<\/p>\n CD = 6 cm<\/p>\n By the property,<\/p>\n AB + CD = BC + AD<\/p>\n 8 + 6 = 7 + AC<\/p>\n AC = 14 – 7 = 7 cm<\/p>\n Question 12:\u00a0<\/b>ABCD is a cyclic quadrilateral in which AB = 16.5 cm, BC = x cm, CD = 11 cm, AD = 19.8 cm, and BD is bisected by AC at O. What is the value of x ?<\/p>\n a)\u00a012.8 cm<\/p>\n b)\u00a012.4 cm<\/p>\n c)\u00a013.2 cm<\/p>\n d)\u00a013.8 cm<\/p>\n 12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b> ABCD is a cyclic quadrilateral in which AB = 16.5\u00a0cm By the property,<\/p>\n A<\/em>B<\/em>\u22c5B<\/em>C\u00a0<\/em>=\u00a0A<\/em>D<\/em>\u22c5D<\/em>C<\/em><\/p>\n 16.5 $\\times x = 19.8 \\times 11$<\/p>\n 16.5 $\\times x = 217.8$<\/p>\n x = 217.8\/16.5 = 13.2 cm<\/p>\n Question 13:\u00a0<\/b>ABCD is a cyclic quadrilateral. The tangents to the circle at the points A and C on it, intersect at P. If $\\angle ABC = 98^\\circ$, then what is the measure of $\\angle APC$?<\/p>\n a)\u00a0$22^\\circ$<\/p>\n b)\u00a0$26^\\circ$<\/p>\n c)\u00a0$16^\\circ$<\/p>\n d)\u00a0$14^\\circ$<\/p>\n 13)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b> ACD is a\u00a0cyclic quadrilateral so,<\/p>\n $\\angle ABC +\u00a0\\angle ADC = 180\\degree$<\/p>\n $\\angle ADC = 180 – 98 = 82\\degree$<\/p>\n $\\angle AOC = 2 \\times\u00a0\\angle ADC = 2 \\times 82 = 164\\degree$<\/p>\n In quadrilateral AOCP-<\/p>\n $\\angle OAP +\u00a0\\angle APC +\u00a0\\angle PCO +\u00a0\\angle COA = 360\\degree$<\/p>\n $\\angle OAP =\u00a0\\angle PCO = 90\\degree$<\/p>\n ($\\because$ tangent angle)<\/p>\n $\\angle APC = 360 – 90 – 90 – 164 = 16\\degree$<\/p>\n Question 14:\u00a0<\/b>ABCD is a cyclic quadrilateral in which AB = 15 cm, BC = 12 cm and CD = 10 cm. If AC bisects BD, then what is the measure of AD?<\/p>\n a)\u00a015 cm<\/p>\n b)\u00a013.5 cm<\/p>\n c)\u00a018 cm<\/p>\n d)\u00a020 cm<\/p>\n 14)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given\u00a0ABCD is a cyclic quadrilateral where AB=15,BC= 12,CD =10<\/p>\n is given below diagram<\/p>\n from the above diagram AC bisects BD<\/p>\n $\\triangle is similar \\triangle BCD<\/p>\n $ \\frac{AB}{AD} = \\frac{DC}{BC} $<\/p>\n $\\Rightarrow \\frac{15}{AD} = \\frac{10}{12}$<\/p>\n $\\Rightarrow AD = \\frac{15\\times 12} {10} $<\/p>\n $\\Rightarrow AD = 18 cm<\/p>\n therefore Option (C) 18 cm Ans<\/p>\n Question 15:\u00a0<\/b>From a point P which is at a distance of 10 cm from the centre O of a circle of radius 6 cm, a pair of tangents PQ and PR to the circle at point Q and respectively, are drawn. Then the area of the quadrilateral PQOR is equal to<\/p>\n a)\u00a030 sq.cm<\/p>\n b)\u00a040 sq.cm<\/p>\n c)\u00a024 sq.cm<\/p>\n d)\u00a048 sq.cm<\/p>\n 15)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n From the given question we draw the diagram<\/p>\n From the $ \\triangle $<\/p>\n $ x^2 = (10)^2 – (6)^2 $<\/p>\n $ x^2 = 100 – 36 $<\/p>\n $ x^2 = 64$<\/p>\n $ x = 8 $<\/p>\n then area\u00a0quadrilateral PQOR =$ 2 \\times \\frac {1}{2} \\times 6 \\times 8 $<\/p>\n = $ 6\\times 8 $<\/p>\n =\u00a0 \u00a0$48 cm^2 $ Ans<\/p>\n Question 16:\u00a0<\/b>Two equilateral triangles of side $10\\sqrt{3}$ cm are joined to form a quadrilateral. What is the altitude of the quadrilateral?<\/p>\n a)\u00a012 cm<\/p>\n b)\u00a014 cm<\/p>\n c)\u00a016 cm<\/p>\n d)\u00a0i5 cm<\/p>\n 16)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b> Given that\u00a0\u00a0$10\\sqrt{3}$ cm<\/p>\n We know the area of equilateral triangle = $ \\frac{\\sqrt {3}} {4} a^2 $\u00a0 \u00a0…… Eq (1)<\/p>\n and on the $\\triangle DCB\u00a0 \u00a0is\u00a0 \u00a0also\u00a0 \u00a0given = \\frac{1}{2} \\times a \\times h $\u00a0 \u00a0…… Eq (2)<\/p>\n then Eq(1) = Eq (2)<\/p>\n $ \\frac{\\sqrt {3}} {4} a^2\u00a0 =\u00a0\u00a0 \\frac{1}{2} \\times a \\times h $<\/p>\n $\\Rightarrow h =\u00a0 \\frac{\\sqrt{3}} {2} a $<\/p>\n $\\Rightarrow h =\u00a0\\frac{\\sqrt{3}} {2} \\times\u00a010\\sqrt{3}$<\/p>\n $\\Rightarrow h = 15 cm Ans<\/p>\n Question 17:\u00a0<\/b>ABCD is cyclic quadrilateral. Sides AB and DC, when produced, meet at E, and sides BC and AD, when produced, meet at F. If $\\angle$BFA = $60^\\circ$ and $\\angle$AED = $30^\\circ$, then the measure of $\\angle$ABC is:<\/p>\n a)\u00a0$75^\\circ$<\/p>\n b)\u00a0$65^\\circ$<\/p>\n c)\u00a0$80^\\circ$<\/p>\n d)\u00a0$70^\\circ$<\/p>\n 17)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n From the given question we draw the diagram is given below<\/p>\n from the above diagram $\\angle BFA = 60^\\circ , $\\angle AFD = 30^\\circ $<\/p>\n then $ \\angle EBC + \\angle ABC = 180^\\circ $ (straight line) ………….(1)<\/p>\n $ \\angle ABC + \\angle ADC = 180^\\circ $ (Opposite angle of cyclic quadrilateral)….. (2)<\/p>\n from the above Equestion (1) and (2)<\/p>\n $ \\angle EBC + \\angle ABC = \\angle ABC + \\angle ADC $<\/p>\n $\\angle EBC = \\angle ADC $ …….(3)<\/p>\n $ \\angle DFC + \\angle DCF + \\angle CDF = 180^\\circ $ (angle sum property of a triangle) ……. (4)<\/p>\n $ \\angle BCE + \\angle CBE + \\angle CEB = 180^\\circ $ (angle sum property of a triangle) ………(5)<\/p>\n from the equestion (4) and (5)<\/p>\n $ \\angle DCF = \\angle BCF $ (Vertically Opposite angle)<\/p>\n $\\angle DFC + \\angle DCF + \\angle CDF = \\angle BCE + \\angle CBF + \\angle CEB $<\/p>\n $\\Rightarrow \\angle DFC + \\angle CDF = \\angle CBF + \\angle CEB $<\/p>\n $\\Rightarrow 60^\\circ + 180^\\circ – \\angle EBC = \\angle EBC + \\angle CEB $<\/p>\n $\\Rightarrow 60^\\circ + 180^\\circ = 2 \\angle EBC + 30^\\circ $<\/p>\n $\\Rightarrow 2 \\angle EBC = 210^\\circ $<\/p>\n $\\Rightarrow \\angle EBC= 105^\\circ $<\/p>\n then $\\angle ABC + \\angle EBC = 180^\\circ $<\/p>\n $\\Rightarrow \\angle ABC + 105^\\circ = 180^\\circ $<\/p>\n $\\Rightarrow \\angle ABC = 180^\\circ -105^\\circ $<\/p>\n $\\Rightarrow \\angle ABC = 75^\\circ $ Ans<\/p>\n Question 18:\u00a0<\/b>In quadrilateral $ABCD, \\angle C = 72^\\circ$ and $\\angle D = 28^\\circ$. The bisectors of $\\angle A$ and $\\angle B$ meet in O. What is the measure of $\\angle AOB$?<\/p>\n a)\u00a0$48^\\circ$<\/p>\n b)\u00a0$54^\\circ$<\/p>\n c)\u00a0$50^\\circ$<\/p>\n d)\u00a0$36^\\circ$<\/p>\n 18)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n In quadrilateral $ABCD$, Question 19:\u00a0<\/b>In a circle with centre O, ABCD isa cyclic quadrilateral and AC is the diameter. Chords AB and CD are produced to meet at E. If $\\angle CAE = 34^\\circ$ and $\\angle E = 30^\\circ$, then $\\angle CBD$ is equal to:<\/p>\n a)\u00a0$36^\\circ$<\/p>\n b)\u00a0$26^\\circ$<\/p>\n c)\u00a0$24^\\circ$<\/p>\n d)\u00a0$34^\\circ$<\/p>\n 19)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n By the exterior angle property, Enroll to 15 SSC MTS 2022 Mocks At Just Rs. 149<\/a><\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/380835.png\")
<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/380342.png\")
<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/380212.png\")
<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/379625.png\")
<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/355960_question%20edited_an6KS6Q.png\")
<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/355960_2Ci3AqS.png\")
<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/354803.png\")
<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/Q71_Y7PLq0U.png\")
<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/image_CCSZv7B.png\")
<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/screenshot.30_T5rivky.jpg\")
<\/p>\n
\n$\\angle PDC = 180 –\u00a0\\angle ADC = 180 – 108 = 72\\degree$<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/screenshot.39_8Gi1mqS.jpg\")
<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/screenshot.46_MqcJHx9.jpg\")
<\/p>\n
\n\u2220ADE=180\u00a0\u2212 (\u2220AED + \u2220EAD)
\n= 180 \u2212 (38 + 102)
\n= 40$\\degree$
\n\u21d2\u2220ADC = 40$\\degree$<\/strong>
\nsquare ABCD is a cyclic quadrilateral.
\n\u2234\u2220DCB + \u2220DAB=180
\n\u21d2\u2220DCB = 180 \u2212 \u2220DAB
\n\u2220DCB\u00a0= 180 \u2212 102
\n\u2220DCB\u00a0= 78$\\degree$
\nIn \u0394DFC,
\n\u2220DFC=180 – (\u2220FDC+\u2220FCD)
\n\u2220DFC\u00a0= 180 \u2212 (40 + 78)
\n\u2220DFC\u00a0= 180 \u2212 118
\n\u2220DFC\u00a0= 62$\\degree$
\n\u2220AFB = \u2220DFC =\u00a062$\\degree$<\/strong>.<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/screenshot.13_zMFCSLt.jpg\")
<\/p>\n
\nBC = x cm
\nCD = 11 cm
\nAD = 19.8 cm<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/DeepinScreenshot_select-area_20200123145446.png\")
<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/geogebra-export_UoSgd89.png\")
<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/geogebra-export_5n0OIvG.png\")
<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/geogebra-export_zWtG8q9.png\")
<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/geogebra-export_1jJZ1Qi.png\")
<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/Screenshot_6_31H6hgR.png\")
<\/figure>\n
\n$\\angle A +\u00a0\\angle B +\u00a0\\angle C +\u00a0\\angle D$ = 360
\n$\\angle A + \\angle B = 360 – 72 – 28 = 260\\degree$
\n$\\frac{1}{2}(\\angle A + \\angle B) =\u00a0130\\degree$
\nIn $\\triangle$ AOB,
\n$\\frac{1}{2}(\\angle A + \\angle B) + \\angle AOB\u00a0= 180$
\n$\\angle AOB = 180 – 130 = 50\\degree$<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/Screenshot_1_7NwJ1Og.png\")
<\/figure>\n
\n$\\angle DCA$ = 30 + 34 = 64
\n$\\angle\u00a0DAC$\u00a0= 180 – 90 – 64 = 26$\\degree$
\n$\\angle\u00a0DAC = \\angle\u00a0CBD$
\n$\\angle\u00a0CBD = 26\\degree$<\/p>\n