Download Coordinate Geometry Questions for SNAP<\/a><\/p>\n Enroll to SNAP 2022 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>What is the equation of a circle with centre of origin and radius is 6 cm?<\/p>\n a)\u00a0$x^2 + y^2 – y = 36$<\/p>\n b)\u00a0$x^2 + y^2 – x – y = 36$<\/p>\n c)\u00a0$x^2 + y^2 – 36 = 0$<\/p>\n d)\u00a0$x^2 + y^2 – x = 36$<\/p>\n 1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,<\/p>\n Center of the circle = (0,0)<\/p>\n Radius of the circle (r) = 6 cm<\/p>\n $\\therefore\\ $Equation of the circle is\u00a0$x^2+y^2=r^2\\ $<\/p>\n $=$> \u00a0$x^2+y^2=6^2\\ $<\/p>\n $=$> \u00a0$x^2+y^2=36$<\/p>\n $=$> \u00a0$x^2+y^2-36=0$<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 2:\u00a0<\/b>The equation of circle with centre (1, -2) and radius 4 cm is:<\/p>\n a)\u00a0$x^2 + y^2 + 2x – 4y = 11$<\/p>\n b)\u00a0$x^2 + y^2 + 2x – 4y = 16$<\/p>\n c)\u00a0$x^2 + y^2 – 2x + 4y = 16$<\/p>\n d)\u00a0$x^2 + y^2 – 2x + 4y = 11$<\/p>\n 2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,<\/p>\n Centre of the circle (a, b) =\u00a0(1, -2)<\/p>\n Radius of the circle (r) = 4 cm<\/p>\n $\\therefore\\ $Equation of the circle is\u00a0$\\left(x-a\\right)^2+\\left(y-b\\right)^2=r^2$<\/p>\n $=$> \u00a0$\\left(x-1\\right)^2+\\left(y-\\left(-2\\right)\\right)^2=4^2$<\/p>\n $=$> \u00a0$\\left(x-1\\right)^2+\\left(y+2\\right)^2=4^2$<\/p>\n $=$> \u00a0$x^2+1^2-2.x.1+y^2+2^2+2.y.2=16$<\/p>\n $=$> \u00a0$x^2+1-2x+y^2+4+4y=16$<\/p>\n $=$> \u00a0$x^2-2x+y^2+4y=16-1-4$<\/p>\n $=$> \u00a0$x^2+y^2-2x+4y=11$<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 3:\u00a0<\/b>In $\\triangle ABC, AB = AC$. A circle drawn through B touches AC at D and intersect AB at P. If D is the mid point of AC and AP 2.5 cm, then AB is equal to:<\/p>\n a)\u00a09 cm<\/p>\n b)\u00a010 cm<\/p>\n c)\u00a07.5 cm<\/p>\n d)\u00a012.5 cm<\/p>\n 3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b> Given D is midpoint of AC so,<\/p>\n AD = $\\frac{AC}{2}$<\/p>\n But also given AC = AB<\/p>\n AD = $\\frac{AB}{2}$ —-(1)<\/p>\n AD is a tangent and APB is a secant. So the tangent secant theorem can be applied,<\/p>\n $AD^2 = AP \\times AB$<\/p>\n $(\\frac{AB}{2})^2 = 2.5 \\times AB$<\/p>\n $\\frac{AB^2}{4} = 2.5 \\times AB$<\/p>\n AB = 10 cm<\/p>\n Question 4:\u00a0<\/b>The graph of the equations $5x – 2y + 1 = 0$ and $4y – 3x + 5 = 0$, interest at the point $P(\\alpha, \\beta)$, What is the value of $(2\\alpha – 3\\beta)$?<\/p>\n a)\u00a04<\/p>\n b)\u00a06<\/p>\n c)\u00a0-4<\/p>\n d)\u00a0-3<\/p>\n 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $5x – 2y + 1 = 0$ Question 5:\u00a0<\/b>What is the area (in square units) of the triangular region enclosed by the graphs of the equations x + y = 3, 2x + 5y = 12 and the x-axis?<\/p>\n a)\u00a02<\/p>\n b)\u00a03<\/p>\n c)\u00a04<\/p>\n d)\u00a06<\/p>\n 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n x + y = 3 Take\u00a0 SNAP mock tests here<\/strong><\/span><\/a><\/p>\n Enrol to 10 SNAP Latest Mocks For Just Rs. 499<\/a><\/strong><\/span><\/p>\n Question 6:\u00a0<\/b>The graphs of the equations $2x + 3y = 11$ and $x – 2y + 12 = 0$ intersects at $P(x_1, y_1)$ and the graph of the equations $x – 2y + 12 = 0$ intersects the x-axis at $Q (x_2, y_2)$. What is the value of $(x_1 – x_2 + y_1 + y_2)$?<\/p>\n a)\u00a013<\/p>\n b)\u00a0-11<\/p>\n c)\u00a015<\/p>\n d)\u00a0-9<\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $2x + 3y = 11$ —(1) Question 7:\u00a0<\/b>The point of intersection of the graphs of the equations 3x \u2014 5y = 19 and 3y \u20147x + 1 =0 is P$\\left(\\alpha,\\beta\\right)$ . Whatis the value of $ \\left(3\\alpha -\\beta \\right)$ ?<\/p>\n a)\u00a0-2<\/p>\n b)\u00a0-1<\/p>\n c)\u00a01<\/p>\n d)\u00a00<\/p>\n 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n The point of intersection of the graphs of the equations 3x \u2014 5y = 19 and 3y \u20147x + 1 =0 is P$\\left(\\alpha,\\beta\\right)$ Question 8:\u00a0<\/b>The graph of the equation x \u2014 7y = \u201442, intersects the y-axis at\u00a0$P\\left(\\alpha,\\beta\\right)$ and the graph of 6x + y – 15 = 0, intersects\u00a0the x-axis at $Q\\left(\\gamma,\\delta\\right)$, What is the value of\u00a0$\\alpha+\\beta+\\gamma+\\delta?$<\/p>\n a)\u00a0$\\frac{17}{2}$<\/p>\n b)\u00a06<\/p>\n c)\u00a0$\\frac{9}{2}$<\/p>\n d)\u00a05<\/p>\n 8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n The graph of the equation x \u2014 7y = \u201442, intersects the y-axis at $P\\left(\\alpha,\\beta\\right)$ Question 9:\u00a0<\/b>The graphs of the equations $3x + y – 5 = 0$ and $2x – y – 5 = 0$ intersect at the point $P(\\alpha, \\beta)$. What is the value of $(3\\alpha + \\beta)$?<\/p>\n a)\u00a04<\/p>\n b)\u00a0-4<\/p>\n c)\u00a03<\/p>\n d)\u00a05<\/p>\n 9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n When\u00a0 graphs of the equations intersect at the point\u00a0$P(\\alpha, \\beta)$ then, Question 10:\u00a0<\/b>The graph of x + 2y = 3 and 3x – 2y = 1 meet the Y-axis at two points having distance<\/p>\n a)\u00a0$\\frac{8}{3}$ units<\/p>\n b)\u00a0$\\frac{4}{3}$ units<\/p>\n c)\u00a01 units<\/p>\n d)\u00a02 units<\/p>\n 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n on Y axis, x=0<\/p>\n put x = 0 in x+2y = 3<\/p>\n 2y = 3<\/p>\n $ y = \\frac{3}{2} $<\/p>\n putting x=0 in 3x-2y = 1<\/p>\n -2y = 1<\/p>\n $ \\frac{-1}{2} $<\/p>\n therefore points on Y-axis are<\/p>\n $ (0,\\frac{3}{2}) and (0,\\frac{-1}{2}) $<\/p>\n required distance = $ \\sqrt ((0-0)^2 + \\sqrt (\\frac{3}{2} + \\frac{1}{2})^2 $<\/p>\n $ = \\sqrt (0+4) $ = 2 units<\/p>\n Question 11:\u00a0<\/b>ABCDis a cyclic quadrilateral, AB and DC when produced meet at P, if PA = 8 cm, PB = 6 cm, PC = 4 cm, then the length (in cm) of PDis<\/p>\n a)\u00a06<\/p>\n b)\u00a012<\/p>\n c)\u00a08<\/p>\n d)\u00a010<\/p>\n 11)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given that,PA = 8 cm, PB = 6 cm, PC = 4 cm<\/p>\n As per tangent & secant rule,<\/p>\n $ PA \\times PB = PD \\times PC $<\/p>\n =>$ PD = \\frac{8\\times6}{4}=12cm $<\/p>\n Question 12:\u00a0<\/b>In a circle, chords AD and BC meet at a point E outside the circle. If $\\angle$BAE = 76$^\\circ$ and $\\angle$ADC= 102$^\\circ$\u00a0, then $\\angle$AEC is equal to:<\/p>\n a)\u00a0$25^\\circ$<\/p>\n b)\u00a0$28^\\circ$<\/p>\n c)\u00a0$26^\\circ$<\/p>\n d)\u00a0$24^\\circ$<\/p>\n 12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n In cyclic quadrilateral ABCD, sum of opposite angles = 180$^\\circ$<\/p>\n $=$> \u00a0$\\angle$BAE +\u00a0$\\angle$BCD =\u00a0180$^\\circ$<\/p>\n $=$> \u00a0 76$^\\circ$ +\u00a0$\\angle$BCD = 180$^\\circ$<\/p>\n $=$> \u00a0\u00a0$\\angle$BCD =\u00a0104$^\\circ$<\/p>\n From the figure,<\/p>\n $\\angle$ADC +\u00a0$\\angle$EDC =\u00a0180$^\\circ$<\/p>\n $=$>\u00a0 102$^\\circ$ +\u00a0$\\angle$EDC = 180$^\\circ$<\/p>\n $=$> \u00a0$\\angle$EDC =\u00a078$^\\circ$<\/p>\n $\\angle$BCD + $\\angle$ECD = 180$^\\circ$<\/p>\n $=$> 104$^\\circ$ + $\\angle$ECD = 180$^\\circ$<\/p>\n $=$> $\\angle$ECD = 76$^\\circ$<\/p>\n In\u00a0$\\triangle\\ $CDE,<\/p>\n $\\angle$DEC +\u00a0$\\angle$ECD +\u00a0$\\angle$EDC =\u00a0180$^\\circ$<\/p>\n $=$> \u00a0$\\angle$AEC +\u00a076$^\\circ$ +\u00a078$^\\circ$ =\u00a0180$^\\circ$<\/p>\n $=$> \u00a0$\\angle$AEC + 154$^\\circ$ =\u00a0180$^\\circ$<\/p>\n $=$> \u00a0$\\angle$AEC =\u00a026$^\\circ$<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 13:\u00a0<\/b>If $ \\triangle$ABC, $\\angle$ABC=90$^\\circ$ and BD$\\bot$AC , if AD = 4cm and CD = 5cm then BD is equal to<\/p>\n a)\u00a0$3\\sqrt{5}$<\/p>\n b)\u00a0$2\\sqrt{5}$<\/p>\n c)\u00a0$3\\sqrt{2}$<\/p>\n d)\u00a0$4\\sqrt{5}$<\/p>\n 13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let\u00a0$\\angle\\ $C = x<\/p>\n In\u00a0$ \\triangle$ABC,<\/p>\n $\\cos x$ =\u00a0$\\frac{BC}{9}$<\/p>\n $=$>\u00a0 BC = 9 $\\cos x$<\/p>\n In $ \\triangle$BCD,<\/p>\n $\\cos x$ = $\\frac{5}{BC}$<\/p>\n $=$> \u00a0$\\cos x=\\frac{5}{9\\cos x}$<\/p>\n $=$>\u00a0 $\\cos^2x=\\frac{5}{9}$<\/p>\n $=$> \u00a0$\\cos x=\\frac{\\sqrt{5}}{3}$<\/p>\n $=$> \u00a0$\\sin x=\\sqrt{1-\\cos^2x}=\\sqrt{1-\\frac{5}{9}}=\\sqrt{\\frac{4}{9}}=\\frac{2}{3}$<\/p>\n In $ \\triangle$BCD,<\/p>\n $\\sin x=\\frac{BD}{BC}$<\/p>\n $=$> \u00a0$\\frac{2}{3}=\\frac{BD}{9\\cos x}$<\/p>\n $=$> \u00a0$\\frac{2}{3}=\\frac{BD}{9\\left(\\frac{\\sqrt{5}}{3}\\right)}$<\/p>\n $=$> \u00a0$\\frac{2}{3}=\\frac{3BD}{9\\left(\\sqrt{5}\\right)}$<\/p>\n $=$> \u00a0BD = 2$\\sqrt{5}$<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 14:\u00a0<\/b>In $\\triangle$ABC, $\\angle$ A= 72$^\\circ$. Its sides AB and AC are produced to the points D and E respectively. If the bisectors of the $\\angle$CBD and $\\angle$BCE meet at point O, then $\\angle$BOC is equal to:<\/p>\n a)\u00a0$16^\\circ$<\/p>\n b)\u00a0$54^\\circ$<\/p>\n c)\u00a0$32^\\circ$<\/p>\n d)\u00a0$106^\\circ$<\/p>\n 14)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,<\/p>\n In $\\triangle$ABC, $\\angle$A = 72$^\\circ$<\/p>\n OB is the angular bisector of $\\angle$CBD<\/p>\n $=$> $\\angle$OBD = $\\angle$OBC<\/p>\n Let $\\angle$OBD = $\\angle$OBC = $x$<\/p>\n OC is the angular bisector of $\\angle$BCE<\/p>\n $=$> $\\angle$OCE = $\\angle$OCB<\/p>\n Let $\\angle$OCE = $\\angle$OCB = $y$<\/p>\n From the figure,<\/p>\n $\\angle$ABC + $\\angle$CBD = 180$^\\circ$<\/p>\n $=$> $\\angle$ABC + $x$ + $x$ = 180$^\\circ$<\/p>\n $=$> $\\angle$ABC = 180$^\\circ$- 2$x$<\/p>\n $\\angle$ACB + $\\angle$BCE = 180$^\\circ$<\/p>\n $=$> $\\angle$ACB + $y$ + $y$ = 180$^\\circ$<\/p>\n $=$> $\\angle$ACB = 180$^\\circ$- 2$y$<\/p>\n In $\\triangle$ABC,<\/p>\n $\\angle$ABC + $\\angle$ACB + $\\angle$BAC = 180$^\\circ$<\/p>\n $=$> 180$^\\circ$- 2$x$ + 180$^\\circ$- 2$y$ + 72$^\\circ$ = 180$^\\circ$<\/p>\n $=$> 2$x$ + 2$y$ = 180$^\\circ$ + 72$^\\circ$<\/p>\n $=$> 2$(x+y)$ = 252$^\\circ$<\/p>\n $=$> $x+y$ = 126$^\\circ$ ………………..(1)<\/p>\n In $\\triangle$OBC,<\/p>\n $\\angle$OBC + $\\angle$OCB + $\\angle$BOC = 180$^\\circ$<\/p>\n $=$> $x$ + $y$ + $\\angle$BOC = 180$^\\circ$<\/p>\n $=$> 126$^\\circ$ + $\\angle$BOC = 180$^\\circ$<\/p>\n $=$> $\\angle$BOC = 180$^\\circ$- 126$^\\circ$<\/p>\n $=$> $\\angle$BOC = 54$^\\circ$<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 15:\u00a0<\/b>The distance between the centres of two circles of radius 2.5 cm each is 13 cm. The length (in cm)of a transverse common tangent is:<\/p>\n a)\u00a012<\/p>\n b)\u00a08<\/p>\n c)\u00a06<\/p>\n d)\u00a010<\/p>\n 15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Radius of first circle ($r_1$) = 2.5 cm<\/p>\n Radius of second circle ($r_2$) = 2.5 cm<\/p>\n The distance between centres of two circles ($d$) = 13 cm<\/p>\n $\\therefore\\ $Length of the common tangent = $\\sqrt{d^2-\\left(r_1+r_2\\right)^2}$<\/p>\n $=\\sqrt{13^2-\\left(2.5+2.5\\right)^2}$<\/p>\n $=\\sqrt{169-25}$<\/p>\n $=\\sqrt{144}$<\/p>\n $=$ 12 cm<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 16:\u00a0<\/b>ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and\u00a0$\\angle$ADC = 126$^\\circ$. $\\angle$BAC is equal to:<\/p>\n a)\u00a0$24^\\circ$<\/p>\n b)\u00a0$72^\\circ$<\/p>\n c)\u00a0$18^\\circ$<\/p>\n d)\u00a0$36^\\circ$<\/p>\n 16)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n In cyclic quadrilateral ABCD, sum of opposite angles =\u00a0180$^\\circ$<\/p>\n $=$> \u00a0$\\angle$ADC +\u00a0$\\angle$ABC =\u00a0180$^\\circ$<\/p>\n $=$> \u00a0 126$^\\circ$\u00a0+\u00a0$\\angle$ABC = 180$^\\circ$<\/p>\n $=$> \u00a0\u00a0$\\angle$ABC =\u00a054$^\\circ$<\/p>\n Angle subtended by diameter in a semicircle is\u00a090$^\\circ$<\/p>\n $=$> \u00a0$\\angle$ACB =\u00a090$^\\circ$<\/p>\n In\u00a0$\\triangle\\ $ACB,<\/p>\n $\\angle$BAC +\u00a0$\\angle$ACB +\u00a0$\\angle$ABC =\u00a0180$^\\circ$<\/p>\n $=$> \u00a0$\\angle$BAC + 90$^\\circ$ +\u00a054$^\\circ$ =\u00a0180$^\\circ$<\/p>\n $=$> \u00a0$\\angle$BAC +\u00a0144$^\\circ$ =\u00a0180$^\\circ$<\/p>\n $=$> \u00a0$\\angle$BAC = 36$^\\circ$<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 17:\u00a0<\/b>In $\\triangle$ABC, $\\angle$A = 52$^\\circ$. Its sides AB and AC are produced to the points D and E respectively. If the bisectors of the $\\angle$CBD and $\\angle$BCE meet at point O, then $\\angle$BOC is equal to:<\/p>\n a)\u00a0$64^\\circ$<\/p>\n b)\u00a0$16^\\circ$<\/p>\n c)\u00a0$106^\\circ$<\/p>\n d)\u00a0$32^\\circ$<\/p>\n 17)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,<\/p>\n In $\\triangle$ABC,\u00a0 $\\angle$A = 52$^\\circ$<\/p>\n OB is the angular bisector of $\\angle$CBD<\/p>\n $=$> \u00a0$\\angle$OBD =\u00a0$\\angle$OBC<\/p>\n Let\u00a0$\\angle$OBD = $\\angle$OBC = $x$<\/p>\n OC is the angular bisector of $\\angle$BCE<\/p>\n $=$> \u00a0$\\angle$OCE = $\\angle$OCB<\/p>\n Let\u00a0$\\angle$OCE = $\\angle$OCB = $y$<\/p>\n From the figure,<\/p>\n $\\angle$ABC + $\\angle$CBD = 180$^\\circ$<\/p>\n $=$> \u00a0$\\angle$ABC + $x$ + $x$ =\u00a0180$^\\circ$<\/p>\n $=$> $\\angle$ABC\u00a0 = 180$^\\circ$- 2$x$<\/p>\n $\\angle$ACB + $\\angle$BCE = 180$^\\circ$<\/p>\n $=$> $\\angle$ACB + $y$ + $y$ = 180$^\\circ$<\/p>\n $=$> $\\angle$ACB = 180$^\\circ$- 2$y$<\/p>\n In $\\triangle$ABC,<\/p>\n $\\angle$ABC + $\\angle$ACB + $\\angle$BAC =\u00a0180$^\\circ$<\/p>\n $=$> \u00a0180$^\\circ$- 2$x$ +\u00a0180$^\\circ$- 2$y$ +\u00a052$^\\circ$ =\u00a0180$^\\circ$<\/p>\n $=$>\u00a0\u00a02$x$ +\u00a02$y$ =\u00a0180$^\\circ$ +\u00a052$^\\circ$<\/p>\n $=$>\u00a0 2$(x+y)$ =\u00a0232$^\\circ$<\/p>\n $=$>\u00a0 $x+y$ =\u00a0116$^\\circ$ ………………..(1)<\/p>\n In $\\triangle$OBC,<\/p>\n $\\angle$OBC + $\\angle$OCB + $\\angle$BOC = 180$^\\circ$<\/p>\n $=$>\u00a0 $x$ + $y$ +\u00a0$\\angle$BOC =\u00a0180$^\\circ$<\/p>\n $=$> \u00a0116$^\\circ$ +\u00a0$\\angle$BOC = 180$^\\circ$<\/p>\n $=$>\u00a0 $\\angle$BOC = 180$^\\circ$-\u00a0116$^\\circ$<\/p>\n $=$> \u00a0$\\angle$BOC = 64$^\\circ$<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 18:\u00a0<\/b>PA and PB are the tangents to a circle with centre O, from a point P outside the circle. A and B are the points on the circle. If $\\angle$APB = 72$^\\circ$, then $\\angle$OAB is equal to:<\/p>\n a)\u00a0$24^\\circ$<\/p>\n b)\u00a0$18^\\circ$<\/p>\n c)\u00a0$36^\\circ$<\/p>\n d)\u00a0$72^\\circ$<\/p>\n 18)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,\u00a0 $\\angle$APB = 72$^\\circ$<\/p>\n PA and PB are the tangents to the circle with centre O<\/p>\n $=$>\u00a0\u00a0$\\angle$OAP = 90$^\\circ$ and \u00a0$\\angle$OBP =\u00a090$^\\circ$<\/p>\n In quadrilateral OAPB,<\/p>\n $\\angle$AOB + $\\angle$OBP + $\\angle$APB + $\\angle$OAP = 360$^\\circ$<\/p>\n $=$> \u00a0$\\angle$AOB +\u00a090$^\\circ$ +\u00a072$^\\circ$ +\u00a090$^\\circ$ =\u00a0360$^\\circ$<\/p>\n $=$> \u00a0$\\angle$AOB +\u00a0252$^\\circ$ =\u00a0360$^\\circ$<\/p>\n $=$> \u00a0$\\angle$AOB =\u00a0108$^\\circ$<\/p>\n In $\\triangle\\ $OAB, OA = OB<\/p>\n Angles opposite to equal sides are equal in triangle<\/p>\n $=$> \u00a0$\\angle$OBA =\u00a0$\\angle$OAB<\/p>\n In $\\triangle\\ $OAB,<\/p>\n $\\angle$AOB +\u00a0$\\angle$OBA +\u00a0$\\angle$OAB =\u00a0180$^\\circ$<\/p>\n $=$> \u00a0108$^\\circ$ +\u00a0$\\angle$OAB +\u00a0$\\angle$OAB =\u00a0180$^\\circ$<\/p>\n $=$> \u00a0 2$\\angle$OAB =\u00a072$^\\circ$<\/p>\n $=$>\u00a0 \u00a0$\\angle$OAB = 36$^\\circ$<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 19:\u00a0<\/b>The distance between the centres of two circles of radius 3 cm and 2 cm is 13 cm. The length (in cm) of a transverse common tangent is:<\/p>\n a)\u00a08<\/p>\n b)\u00a012<\/p>\n c)\u00a06<\/p>\n d)\u00a010<\/p>\n 19)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Radius of first circle ($r_1$) = 3 cm<\/p>\n Radius of second circle ($r_2$) = 2 cm<\/p>\n The distance between centres of two circles ($d$) = 13 cm<\/p>\n $\\therefore\\ $Length of the common tangent =\u00a0$\\sqrt{d^2-\\left(r_1+r_2\\right)^2}$<\/p>\n $=\\sqrt{13^2-\\left(3+2\\right)^2}$<\/p>\n $=\\sqrt{169-25}$<\/p>\n $=\\sqrt{144}$<\/p>\n $=$ 12 cm<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 20:\u00a0<\/b>The distance between the centre of two circles of radius 4 cm and 2 cm is 10 cm. The length (in cm) of a transverse common tangent is:<\/p>\n a)\u00a04<\/p>\n b)\u00a06<\/p>\n c)\u00a010<\/p>\n d)\u00a08<\/p>\n 20)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given, distance between centres of circles$(d)$ = 10 cm<\/p>\n Radius of first circle$(r_1)$ = 4 cm<\/p>\n Radius of second circle$(r_2)$ = 2 cm<\/p>\n $\\therefore\\ $The length of tranverse common tangent
\n![](\"https:\/\/cracku.in\/media\/uploads\/screenshot.14_vpSr2ct.jpg\")
<\/p>\n
\n15x – 6y + 3 = 0\u00a0—(1)
\n$3x -4y – 5 = 0$
\n15x – 20y – 25 = 0\u00a0—(2)
\nFrom eq (1) and (2),
\n14y + 28 = 0
\ny = -2
\nFrom eq(1),
\n15x + 6$\\times 2$ + 3 = 0
\nx = -1
\n$\\alpha$ = -1
\n$\\beta$ = -2
\n$(2\\alpha – 3\\beta)$
\n=\u00a0$(2 \\times (-1)\u00a0+ 3\\times 2)$ = 4<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/Screenshot_9_Xx4skDX.png\")
<\/figure>\n
\n2x + 2y = 6\u00a0—(1)
\n2x +\u00a05y = 12 —(2)
\nFrom eq (1) and eq (2),
\n3y = 6
\ny = 2
\nSo height = 2
\ny = 0 —(3)
\nput the value of y in\u00a0eq(1) and (2),
\n2x = 6
\nx = 3
\nAnd 2x = 12
\nx = 6
\nArea = $\\frac{1}{2} \\times base \\times height$
\n= $\\frac{1}{2} \\times (6 – 3) \\times 2$ = 3\u00a0square units<\/p>\n
\n$x – 2y + 12 = 0$
\n$2x – 4y = -24$ —(2)
\nFrom eq (1) and (2),
\n7y = 35
\ny = 5 = $y_1$
\nFrom eq (1),
\n$2x + 3 $\\times$ 5 = 11$
\n2x = -4
\nx = -2 = $x_1$
\nNow,
\nThe graph of the equations $x – 2y + 12 = 0$ intersects the x-axis.
\nSo,
\n$ y = y_1$ = 0
\n$x – 0 + 12 = 0$
\nx = -12 = $x_1$
\n$(x_1 – x_2 + y_1 + y_2)$
\n= -2 + 12 + 5 + 0 = 15<\/p>\n
\nSo,
\n3$\\alpha \u2014 5\\beta = 19$ —(1)
\n7$\\alpha \u2014 3\\beta = 1$ —(2)
\nEq(1) multiply by 3 and eq (2) multiply by 5,
\n9$\\alpha \u2014 15\\beta = 57$ —(1)
\n35$\\alpha \u2014 15\\beta = 5$ —(2)
\nFrom eq (3) and (4),
\n26$\\alpha = -52$
\n$\\alpha = -2$
\nFrom eq (1),
\n3$\\times -2 – 5\\beta = 19$
\n$\\beta = -5$
\nNow,
\n$ \\left(3\\alpha -\\beta \\right)$
\n=\u00a0$ \\left(3\\times -2 + 5s \\right)$
\n= -1<\/p>\n
\nSo, x = 0
\n0 – 7y = -42
\ny = 6
\n$\\alpha$ = 0
\n$\\beta$ = 6
\ngraph of 6x + y – 15 = 0, intersects the x-axis at $Q\\left(\\gamma,\\delta\\right)$
\nSo, y = 0
\n6x – 15 = 0
\nx = 5\/2
\n$\\gamma$ = 5\/2
\n$\\delta$ = 0
\nNow,
\n$\\alpha+\\beta+\\gamma+\\delta$
\n= 0 + 6 + 5\/2 + 0 = $\\frac{17}{2}$<\/p>\n
\n$3\\alpha + \\beta – 5 = 0$ —(1)
\n$2\\alpha – \\beta – 5 = 0$ —(2),
\nOn eq(1) +\u00a0(2),
\n$5\\alpha – 10 = 0$
\n$\\alpha = 2$
\nFrom the eq(2),
\n$3 \\times 2+ \\beta – 5 = 0$
\n$\\beta = -1$
\nNow,
\n$(3\\alpha + \\beta)$ = 3 $\\times$ 2 – 1 = 6 – 1 = 5
\n$\\therefore$ The correct answer is option D.<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/image_8qzPOxa.png\")
<\/figure>\n![](\"https:\/\/cracku.in\/media\/uploads\/247054.png\")
<\/figure>\n![](\"https:\/\/cracku.in\/media\/uploads\/247050.png\")
<\/figure>\n![](\"https:\/\/cracku.in\/media\/uploads\/245732.png\")
<\/figure>\n![](\"https:\/\/cracku.in\/media\/uploads\/245630.png\")
<\/figure>\n![](\"https:\/\/cracku.in\/media\/uploads\/245619_G2510Hg.png\")
<\/figure>\n![](\"https:\/\/cracku.in\/media\/uploads\/245367.png\")
<\/figure>\n
\n= $\\sqrt{d^2-\\left(r_1+r_2\\right)^2}=\\sqrt{10^2-\\left(4+2\\right)^2}=\\sqrt{100-36}=8\\ cm$<\/p>\n