Enroll to 15 SSC CHSL 2022 Mocks At Just Rs. 149<\/a><\/p>\n Question 1:\u00a0<\/b>In a circle with centre O, points A, B, C and D in this order are concyclic such that BD is a diameter of the circle. If $\\angle$BAC = 22$^\\circ$, then find the measure (in degrees) of $\\angle$COD.<\/p>\n a)\u00a079<\/p>\n b)\u00a0136<\/p>\n c)\u00a0158<\/p>\n d)\u00a068<\/p>\n Question 2:\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 Question 3:\u00a0<\/b>Points A, B and C are on circle with centre O such that $\\angle$BOC = 84$^\\circ$. If AC is produced to a point D such that $\\angle$BDC = 40$^\\circ$, then find the measure of $\\angle$ABD (in degrees).<\/p>\n a)\u00a098<\/p>\n b)\u00a092<\/p>\n c)\u00a056<\/p>\n d)\u00a0102<\/p>\n Question 4:\u00a0<\/b>A square has the perimeter equal to the circumference of a circle having radius 7 cm. What is the ratio of the area of the circle to area of the square?(Use $\\pi = \\frac{22}{7}$)<\/p>\n a)\u00a0121 : 44<\/p>\n b)\u00a07 : 2<\/p>\n c)\u00a014 : 11<\/p>\n d)\u00a07 : 11<\/p>\n Question 5:\u00a0<\/b>The area of a circular path enclosed by two concentric circles is 3080 m$^2$. If the difference between the radius of the outer edge and that of inner edge of the circular path is 10 m, what is the sum (in m) of the two radii? (Take $\\pi = \\frac{22}{7}$)<\/p>\n a)\u00a0112<\/p>\n b)\u00a070<\/p>\n c)\u00a084<\/p>\n d)\u00a098<\/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>Tangent is drawn from a point P to a circle, which meets the circle at T such that PT = 8 cm. A secant PAB intersects the circle in points A and B. If PA= 5 cm, what is the length (in cm) of the chord AB?<\/p>\n a)\u00a08.0<\/p>\n b)\u00a08.4<\/p>\n c)\u00a06.4<\/p>\n d)\u00a07.8<\/p>\n Question 7:\u00a0<\/b>Two circles of radius 15 cm and 37 cm intersect each other at the points A and B. If the length of common chord is 24 cm, what is the distance (in cm) between the centres of the circles?<\/p>\n a)\u00a044<\/p>\n b)\u00a045<\/p>\n c)\u00a042<\/p>\n d)\u00a040<\/p>\n Question 8:\u00a0<\/b>Two circles of radius 15 cm and 37 cm intersect each other at the points A and B. If the length of common chord is 24 cm, what is the distance (in cm) between the centres of the circles?<\/p>\n a)\u00a044<\/p>\n b)\u00a045<\/p>\n c)\u00a042<\/p>\n d)\u00a040<\/p>\n Question 9:\u00a0<\/b>Two circles of radii 18 cm and 16 cm intersect each other and the length of their common chord is 20 cm. What is the distance (in cm) between their centres?<\/p>\n a)\u00a0$4\\sqrt{14} – 2\\sqrt{39}$<\/p>\n b)\u00a0$4\\sqrt{10} – 2\\sqrt{39}$<\/p>\n c)\u00a0$4\\sqrt{10} + 2\\sqrt{39}$<\/p>\n d)\u00a0$4\\sqrt{14} + 2\\sqrt{39}$<\/p>\n Question 10:\u00a0<\/b>The area of a quadrant of a circle is $\\frac{\\pi}{9}$ m$^2$. Its radius (in metres) is equal to:<\/p>\n a)\u00a0$\\frac{2}{3}$<\/p>\n b)\u00a0$\\frac{1}{3}$<\/p>\n c)\u00a0$\\frac{1}{2}$<\/p>\n d)\u00a0$\\frac{3}{2}$<\/p>\n Enroll to 15 SSC MTS 2022 Mocks At Just Rs. 149<\/a><\/p>\n Question 11:\u00a0<\/b>In a circle with centre O, AB and CD are parallel chords on the opposite sides of a diameter. If AB = 12 cm, CD = 18 cm and the distance between the chords AB and CD is 15 cm, then find the radius of the circle (in cm).<\/p>\n a)\u00a0$9\\sqrt{13}$<\/p>\n b)\u00a09<\/p>\n c)\u00a0$3\\sqrt{13}$<\/p>\n d)\u00a012<\/p>\n Question 12:\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 Question 13:\u00a0<\/b>The circles of same radius 13 cm intersect each other at A and B. If AB = 10 cm, then the distance between their centres is:<\/p>\n a)\u00a018 cm<\/p>\n b)\u00a012 cm<\/p>\n c)\u00a024 cm<\/p>\n d)\u00a026 cm<\/p>\n Question 14:\u00a0<\/b>The area of the quadrant of a circle whose circumference is 22 cm, will be:<\/p>\n a)\u00a03.5 $cm^2$<\/p>\n b)\u00a038.5 $cm^2$<\/p>\n c)\u00a010 $cm^2$<\/p>\n d)\u00a09.625 $cm^2$<\/p>\n Question 15:\u00a0<\/b>Two circles of radii 20 cm and 5 cm, respectively, touch each other externally at the point P, AB is the direct common tangent of those two circles of centres R and S, respectively. The length of AB is equal to:<\/p>\n a)\u00a010 cm<\/p>\n b)\u00a05 cm<\/p>\n c)\u00a015 cm<\/p>\n d)\u00a020 cm<\/p>\n Question 16:\u00a0<\/b>The distance between the centres of two equal circles each of radius 4 cm is 17 cm. The length of a transverse tangent is:<\/p>\n a)\u00a014 cm<\/p>\n b)\u00a016 cm<\/p>\n c)\u00a019 cm<\/p>\n d)\u00a015 cm<\/p>\n Question 17:\u00a0<\/b>ABC is a right angled triangle, right angled at A. A circle is inscribed in it. The lengths of two sides containing the right angle are 48 cm and 14 cm. The radius of the inscribed circle is:<\/p>\n a)\u00a04 cm<\/p>\n b)\u00a08 cm<\/p>\n c)\u00a06 cm<\/p>\n d)\u00a05 cm<\/p>\n Question 18:\u00a0<\/b>A is point at a distance 26 cm from the centre O of a circle of radius 10 cm. AP and AQ are the tangents to the circle at the point of contacts P and Q. If a tangent BC is drawn at a point R lying on the minor arc PQ to intersect AP at B and AQ at C, then the perimeter of $\\triangle$ABC is:<\/p>\n a)\u00a048 cm<\/p>\n b)\u00a046 cm<\/p>\n c)\u00a042 cm<\/p>\n d)\u00a040 cm<\/p>\n Question 19:\u00a0<\/b>A, B, C are three points so that AB = 4 cm, BC = 6 cm and AC = 10 cm. The number of circles passing through the points A, B, C is:<\/p>\n a)\u00a01<\/p>\n b)\u00a03<\/p>\n c)\u00a00<\/p>\n d)\u00a02<\/p>\n Question 20:\u00a0<\/b>In a circle centred at O, AB is a chord and C is any point on AB, such that OC is perpendicular to AB. If the length of the chord is 16 cm and OC = 6 cm, the radius of circle is:<\/p>\n a)\u00a08 cm<\/p>\n b)\u00a012 cm<\/p>\n c)\u00a010 cm<\/p>\n d)\u00a06 cm<\/p>\n Free SSC Preparation Videos<\/a><\/p>\n Download SSC CHSL\u00a0 Previous Papers PDF<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Angle in a semicircle is right angle.<\/p>\n $\\Rightarrow$\u00a0 $\\angle$BAD = 90$^\\circ$<\/p>\n $\\Rightarrow$\u00a0\u00a0$\\angle$BAC +\u00a0$\\angle$CAD =\u00a090$^\\circ$<\/p>\n $\\Rightarrow$\u00a0\u00a022$^\\circ$ +\u00a0$\\angle$CAD =\u00a090$^\\circ$<\/p>\n $\\Rightarrow$\u00a0\u00a0$\\angle$CAD =\u00a068$^\\circ$<\/p>\n Angle subtended by a chord at the center of the circle is twice the angle subtended by the chord on the point of a circle in the same segment.<\/p>\n $\\Rightarrow$\u00a0\u00a0$\\angle$COD = 2$\\angle$CAD<\/p>\n $\\Rightarrow$\u00a0\u00a0 $\\angle$COD =\u00a0136$^\\circ$<\/p>\n Hence, the correct answer is Option B<\/p>\n 2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n 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 3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Angle subtended by chord BC at the centre is twice the angle subtended by chord BC on the point A of the circle.<\/p>\n $\\angle$BOC = 2$\\angle$BAC<\/p>\n 84$^\\circ$ =\u00a02$\\angle$BAC<\/p>\n $\\angle$BAC = 42$^\\circ$<\/p>\n From triangle BAD,<\/p>\n $\\angle$BAD +\u00a0$\\angle$ABD +\u00a0$\\angle$BDA =\u00a0180$^\\circ$<\/p>\n 42$^\\circ$ +\u00a0$\\angle$ABD +\u00a040$^\\circ$ =\u00a0180$^\\circ$<\/p>\n $\\angle$ABD =\u00a098$^\\circ$<\/p>\n Hence, the correct answer is Option A<\/p>\n 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Radius of the circle = 7 cm<\/p>\n Circumference of the circle =\u00a0$2\\times\\frac{22}{7}\\times7$ = 44 cm<\/p>\n Perimeter of square is equal to the circumference of the circle.<\/p>\n Perimeter of the square = 44 cm<\/p>\n Let the side of the square = a<\/p>\n 4a = 44<\/p>\n a = 11 cm<\/p>\n Ratio of the area of the circle to area of the square =\u00a0$\\frac{22}{7}\\times7^2\\ :\\ 11^2$<\/p>\n = 14 : 11<\/p>\n Hence, the correct answer is Option C<\/p>\n 5)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let the radius of the outer circle and inner circle are $r_o$ and $r_i$ respectively.<\/p>\n The difference between the radius of the outer edge and that of inner edge of the circular path is 10 m.<\/p>\n $r_o$ –\u00a0$r_i$ = 10……..(1)<\/p>\n The area of a circular path enclosed by two concentric circles is 3080 m$^2$.<\/p>\n $\\Rightarrow$\u00a0\u00a0$\\frac{22}{7}r_o^2\\ -\\frac{22}{7}r_i^2=3080$<\/p>\n $\\Rightarrow$\u00a0\u00a0$\\left(r_o+r_i\\right)\\left(r_o-r_i\\right)=140\\times7$<\/p>\n $\\Rightarrow$\u00a0\u00a0$\\left(r_o+r_i\\right)\\left(10\\right)=140\\times7$<\/p>\n $\\Rightarrow$\u00a0\u00a0$\\left(r_o+r_i\\right)=98$<\/p>\n Sum of the two radii = 98 m<\/p>\n Hence, the correct answer is Option D<\/p>\n 6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n $\\Rightarrow$\u00a0\u00a0PT$^2$ = PA. PB<\/p>\n $\\Rightarrow$\u00a0 8$^2$ = 5 x (PA + AB)<\/p>\n $\\Rightarrow$\u00a0 64 = 5 x (5 + AB)<\/p>\n $\\Rightarrow$\u00a0 12.8 = 5 + AB<\/p>\n $\\Rightarrow$\u00a0 AB = 7.8 cm<\/p>\n Hence, the correct answer is Option D<\/p>\n 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n From triangle AHG,<\/p>\n AH$^2$ + GH$^2$ = AG$^2$<\/p>\n AH$^2$ + 12$^2$ = 15$^2$<\/p>\n AH$^2$ + 144 = 225<\/p>\n AH$^2$ = 81<\/p>\n AH = 9 cm<\/p>\n From triangle CHG,<\/p>\n CH$^2$ + GH$^2$ = CG$^2$<\/p>\n CH$^2$ + 12$^2$ = 37$^2$<\/p>\n CH$^2$ + 144 = 1369<\/p>\n CH$^2$ = 1225<\/p>\n CH = 35 cm<\/p>\n Distance between two circles = AC = AH + CH = 9 + 35 = 44 cm<\/p>\n Hence, the correct answer is Option A<\/p>\n 8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n From triangle AHG,<\/p>\n AH$^2$ + GH$^2$ = AG$^2$<\/p>\n AH$^2$ + 12$^2$ = 15$^2$<\/p>\n AH$^2$ + 144 = 225<\/p>\n AH$^2$ = 81<\/p>\n AH = 9 cm<\/p>\n From triangle CHG,<\/p>\n CH$^2$ + GH$^2$ = CG$^2$<\/p>\n CH$^2$ + 12$^2$ = 37$^2$<\/p>\n CH$^2$ + 144 = 1369<\/p>\n CH$^2$ = 1225<\/p>\n CH = 35 cm<\/p>\n Distance between two circles = AC = AH + CH = 9 + 35 = 44 cm<\/p>\n Hence, the correct answer is Option A<\/p>\n 9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n From triangle AGH,<\/p>\n AH$^2$ + GH$^2$ = AG$^2$<\/p>\n AH$^2$ + 10$^2$ = 16$^2$<\/p>\n AH$^2$ + 100 = 256<\/p>\n AH$^2$ = 156<\/p>\n AH =\u00a0$2\\sqrt{39}$<\/p>\n From triangle CGH,<\/p>\n CH$^2$ + GH$^2$ = CG$^2$<\/p>\n CH$^2$ + 10$^2$ = 18$^2$<\/p>\n CH$^2$ + 100 = 324<\/p>\n CH$^2$ = 224<\/p>\n CH = $4\\sqrt{14}$<\/p>\n Distance between centres of circles = AC = AH + CH =\u00a0$2\\sqrt{39}$ +\u00a0$4\\sqrt{14}$<\/p>\n Hence, the correct answer is Option D<\/p>\n 10)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let the radius of the circle = r<\/p>\n Area of the quadrant of the circle =\u00a0$\\frac{\\pi}{9}$ m$^2$<\/p>\n $\\frac{1}{4}\\times\\pi$r$^2$ =\u00a0$\\frac{\\pi}{9}$<\/p>\n r$^2$ =\u00a0$\\frac{4}{9}$<\/p>\n r =\u00a0$\\frac{2}{3}$<\/p>\n Radius of the circle =\u00a0$\\frac{2}{3}$ m<\/p>\n Hence, the correct answer is Option A<\/p>\n 11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n From triangle AJO,<\/p>\n r$^2$ = 6$^2$ + (15 – x)$^2$<\/p>\n r$^2$ = 36 + (15 – x)$^2$………………(1)<\/p>\n From triangle CKO,<\/p>\n r$^2$ = 9$^2$ + x$^2$<\/p>\n r$^2$ = 81 + x$^2$………………(2)<\/p>\n From (1) and (2),<\/p>\n 36 + (15 – x)$^2$ =\u00a081 + x$^2$<\/p>\n 225 + x$^2$ – 30x = 45 + x$^2$<\/p>\n 30x = 180<\/p>\n x = 6<\/p>\n From (2),<\/p>\n r$^2$ = 81 + x$^2$<\/p>\n r$^2$ = 81 + 6$^2$<\/p>\n r$^2$ = 81 + 36<\/p>\n r$^2$ = 137<\/p>\n r =\u00a0$3\\sqrt{13}$<\/p>\n Hence, the correct answer is Option C<\/p>\n 12)\u00a0Answer\u00a0(D)<\/strong><\/p>\n 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 13)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given, AB = 10 cm<\/p>\n AB is the common chord for both the circles and OD will be the perpendicular bisector of AB.<\/p>\n $\\Rightarrow$ AE = EB = 5 cm<\/p>\n In\u00a0$\\triangle\\ $AOE,<\/p>\n OE$^2$ + AE$^2$ = OA$^2$<\/p>\n $\\Rightarrow$ \u00a0OE$^2$ +\u00a05$^2$ =\u00a013$^2$<\/p>\n $\\Rightarrow$ \u00a0OE$^2$ + 25 = 169<\/p>\n $\\Rightarrow$\u00a0 OE$^2$ = 144<\/p>\n $\\Rightarrow$\u00a0 OE = 12 cm<\/p>\n Similarly, in $\\triangle$AED<\/p>\n ED = 12 cm<\/p>\n $\\therefore\\ $Distance between the centres of the circles = OD = OE + ED = 12 + 12 = 24 cm<\/p>\n Hence, the correct answer is Option C<\/p>\n 14)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let the radius of the circle = r<\/p>\n Circumference of the circle = 22 cm<\/p>\n $\\Rightarrow$\u00a0 $2\\pi r=22$<\/p>\n $\\Rightarrow$ \u00a0$2\\times\\frac{22}{7}\\times r=22$<\/p>\n $\\Rightarrow$ \u00a0$r=\\frac{7}{2}$ cm<\/p>\n $\\therefore\\ $Area of the quadrant of the circle $=\\frac{1}{4}\\times\\pi\\ r^2$<\/p>\n $=\\frac{1}{4}\\times\\frac{22}{7}\\times\\ \\left(\\frac{7}{2}\\right)^2$<\/p>\n $=\\frac{77}{8}$<\/p>\n $=$ 9.625 cm$^2$<\/p>\n Hence, the correct answer is Option D<\/p>\n 15)\u00a0Answer\u00a0(D)<\/strong><\/p>\n AB is tangent to both the circles.<\/p>\n $\\Rightarrow$\u00a0 AB$\\bot\\ $AR<\/p>\n Let SC be the line parallel to AB<\/p>\n $\\Rightarrow$\u00a0 SC$\\bot\\ $AR and AC = BS = 5 cm and AB = SC<\/p>\n Radius of bigger circle AR = 20 cm<\/p>\n $\\Rightarrow$ \u00a0AC + CR = 20<\/p>\n $\\Rightarrow$\u00a0 5 + CR = 20<\/p>\n $\\Rightarrow$\u00a0 CR = 15 cm<\/p>\n Since circles touch externally as shown in figure<\/p>\n RS = RP + PS = 20 + 5 = 25 cm<\/p>\n In\u00a0$\\triangle$RCS,<\/p>\n CR$^2$ + SC$^2$ = RS$^2$<\/p>\n $\\Rightarrow$\u00a0 15$^2$ + SC$^2$ = 25$^2$<\/p>\n $\\Rightarrow$\u00a0 225 + SC$^2$ = 625<\/p>\n $\\Rightarrow$\u00a0 SC$^2$ = 400<\/p>\n $\\Rightarrow$\u00a0 SC = 20 cm<\/p>\n $\\therefore\\ $Length of AB = SC = 20 cm<\/p>\n Hence, the correct answer is Option D<\/p>\n 16)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given, distance between centres of the circles ($d$) = 17 cm<\/p>\n Radius of the two circles ($r_1$, $r_2$) = 4 cm<\/p>\n $\\therefore\\ $Length of the tranverse commom tangent = $\\sqrt{d^2-\\left(r_1+r_2\\right)^2}$<\/p>\n $=\\sqrt{17^2-\\left(4+4\\right)^2}$<\/p>\n $=\\sqrt{289-64}$<\/p>\n $=\\sqrt{225}$<\/p>\n $=15$ cm<\/p>\n Hence, the correct answer is Option D<\/p>\n 17)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Using the pythagoras theorem,<\/p>\n BC$^2$ = AB$^2$ + AC$^2$<\/p>\n $\\Rightarrow$ \u00a0BC$^2$ = 48$^2$ + 14$^2$<\/p>\n $\\Rightarrow$ \u00a0BC$^2$ = 2304 + 196<\/p>\n $\\Rightarrow$ \u00a0BC$^2$ = 2500<\/p>\n $\\Rightarrow$\u00a0 BC = 50 cm<\/p>\n Let the radius of the circle = r<\/p>\n $\\Rightarrow$\u00a0 OD = OI = OJ = r<\/p>\n AB, BC, AC are tangents to the circle<\/p>\n Area of $\\triangle$ABC = Area of $\\triangle$OAC + Area of $\\triangle$OBC + Area of $\\triangle$OAB<\/p>\n $\\Rightarrow$ \u00a0$\\frac{1}{2}$ x 48 x 14 =\u00a0$\\frac{1}{2}$ x AC x OD +\u00a0$\\frac{1}{2}$ x BC x OI +\u00a0$\\frac{1}{2}$ x AB x OJ<\/p>\n $\\Rightarrow$ \u00a0336 =\u00a0$\\frac{1}{2}$ x 14 x r +\u00a0$\\frac{1}{2}$ x 50 x r +\u00a0$\\frac{1}{2}$ x 48 x r<\/p>\n $\\Rightarrow$\u00a0 336 = 7r + 25r + 24r<\/p>\n $\\Rightarrow$\u00a0 56r = 336<\/p>\n $\\Rightarrow$\u00a0 r = 6 cm<\/p>\n $\\therefore\\ $Radius of the inscribed circle = 6 cm<\/p>\n Hence, the correct answer is Option C<\/p>\n 18)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given,\u00a0A is point at a distance 26 cm from the centre O<\/p>\n Radius of the circle = 10 cm<\/p>\n AP and AQ are tangents to the circle at the point of contacts P and Q<\/p>\n $\\Rightarrow$ AP$\\bot\\ $OP and AQ$\\bot\\ $OQ<\/p>\n The length of tangents to the circle from an external point are equal.<\/p>\n $\\Rightarrow$\u00a0 AP = AQ ……………..(1)<\/p>\n Similarly, BP = BR and CQ = CR …….(2)<\/p>\n In\u00a0$\\triangle$OPA,<\/p>\n OP$^2$ + AP$^2$ = AO$^2$<\/p>\n $\\Rightarrow$\u00a0 10$^2$ + AP$^2$ = 26$^2$<\/p>\n $\\Rightarrow$\u00a0 100 + AP$^2$ = 676<\/p>\n $\\Rightarrow$\u00a0 AP$^2$ = 576<\/p>\n $\\Rightarrow$\u00a0 AP = 24 cm<\/p>\n Perimeter of\u00a0$\\triangle$ABC = AB + BC + AC<\/p>\n = AB + BR + CR + AC<\/p>\n = AB + BP + CQ + AC [From (2)]<\/p>\n = AP + AQ<\/p>\n = AP + AP [From (1)]<\/p>\n = 2AP<\/p>\n = 2(24)<\/p>\n = 48 cm<\/p>\n Hence, the correct answer is Option A<\/p>\n 19)\u00a0Answer\u00a0(C)<\/strong><\/p>\n AB + BC = 4 + 6 = 10 cm<\/p>\n AB + BC = AC<\/p>\n $\\Rightarrow$\u00a0 A, B, C are collinear points<\/p>\n $\\therefore\\ $No circle can be formed using three collinear points<\/p>\n Hence, the correct answer is Option C<\/p>\n 20)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let the radius of the circle = r<\/p>\n Given, AB = 16 cm, OC = 6 cm<\/p>\n OC is perpendicular to the chord AB.<\/p>\n In a circle, perpendicular from centre of the circle to the chord bisects the chord.<\/p>\n $\\Rightarrow$ AC = BC = 8 cm<\/p>\n In $\\triangle$OAC,<\/p>\n AC$^2$ + OC$^2$ = OA$^2$<\/p>\n $\\Rightarrow$ 8$^2$ + 6$^2$ = r$^2$<\/p>\n $\\Rightarrow$ 64 + 36 = r$^2$<\/p>\n $\\Rightarrow$ r$^2$ = 100<\/p>\n $\\Rightarrow$\u00a0 r = 10 cm<\/p>\n $\\therefore\\ $Radius of the circle = 10 cm<\/p>\n Hence, the correct answer is Option C<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/355960_question%20edited_an6KS6Q.png\")
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