{"id":42672,"date":"2024-02-29T05:15:21","date_gmt":"2024-02-28T23:45:21","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=42672"},"modified":"2024-02-29T15:05:08","modified_gmt":"2024-02-29T09:35:08","slug":"selina-concise-mathematics-class-10-icse-solutions-revision-paper-2","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/selina-concise-mathematics-class-10-icse-solutions-revision-paper-2\/","title":{"rendered":"Selina Concise Mathematics Class 10 ICSE Solutions Revision Paper 2"},"content":{"rendered":"

Selina Concise Mathematics Class 10 ICSE Solutions Revision Paper 2<\/h2>\n

Section A (40 Marks)<\/strong>
\n(Answer all questions from this Section)<\/strong><\/p>\n

Question 1.
\n(a) In what ratio does the line x – y – 2 = 0 divide the line segment joining the points (3, -1) and (8, 9)? Also, find co-ordinates of the point of intersection.
\nSolution:
\nThe given two points are A (3, -1) and B (8, 9)
\n\u2234 Slope of line = \\(\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\\frac{9+1}{8-3}=\\frac{10}{5}=2\\)
\nand equation of AB
\ny – 9 = 2(x – 8) \u21d2 y – 9 = 2x – 16
\n2x – y = 7 …….(i)
\nand line given x – y – 2 = 0 \u21d2 x – y = 2 …..(ii)
\nSubtracting (ii) from (i\/), we get x = 5
\nand y = x – 2 = 5 – 2 = 3
\n\u2234 Co-ordinates of point of intersection = (5, 3)
\nNow let ratio = m : n
\n\\(5=\\frac{m x_{2}+m x_{1}}{m+n}=\\frac{8 m+3 n}{m+n}\\)
\n\u21d2 8m + 3n = 5m + 5n \u21d2 8m – 5m = 5n – 3n \u21d2 3m = 2n
\n\u21d2 \\(\\frac{m}{n}=\\frac{2}{3}\\)
\n\u2234 Ratio = 2 : 3<\/p>\n

(b) Find :
\n(i) the fourth proportional to 2a, 3b and 4c.
\n(ii) the mean proportional to x – y and (x – y)3<\/sup>.
\n(iii) the third proportional to a + b and \\(\\sqrt{a^{2}-b^{2}}\\)
\nSolution:
\n(i) Fourth proportional of 2a, 3b, 4c
\n\"Selina
\n\"Selina<\/p>\n

(c) Using remainder theorem, factorise : x3<\/sup> + 7x2<\/sup> – 21x – 27 completely and then solve x3<\/sup> + 7x2<\/sup> – 21x – 27 = 0
\nSolution:
\nLet f(x) = x3<\/sup> + 7x2<\/sup> – 21x – 27
\nf(3) = (3)3<\/sup> + 7(3)2<\/sup> – 21 \u00d7 3 – 27
\n= 27 + 63 – 63 – 27 = 0
\n\u2234 x – 3 is a factor
\n\"Selina
\nDividing f(x) by x – 3,
\nf(x) = (x – 3) (x2<\/sup> + 10x + 9)
\n= (x – 3)[x2<\/sup> + x + 9x + 9]
\n= (x – 3)[x(x + 1) + 9(x + 1)]
\n= (x \u2014 3) (x + 1) (x + 9)
\nNow, x2<\/sup> + 7x2<\/sup> – 21x – 27 = 0
\n\u21d2 (x – 3) (x + 1) (x + 9) = 0
\nEither x – 3 = 0, then x = 3
\nor x + 1 = 0, then x = -1
\nor x + 9 = 0, then x = -9
\nHence x = 3, -1, -9<\/p>\n

Question 2.
\n(a) Find the 99th<\/sup> term of the series :
\n\"Selina
\nSolution:
\n\"Selina<\/p>\n

(b) Metallic spheres of diameters 12 cm, 16 cm and 20 cm respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
\nSolution:
\nDiameter of first sphere = 12 cm
\n\"Selina
\n\"Selina<\/p>\n

(c) If A = \\(\\left[\\begin{array}{ll}{0} & {4} \\\\ {1} & {0}\\end{array}\\right], \\mathbf{B}=\\left[\\begin{array}{cc}{-2} & {0} \\\\ {3} & {-2}\\end{array}\\right] \\text { and } \\mathbf{C}=\\left[\\begin{array}{cc}{-1} & {-2} \\\\ {2} & {\\mathbf{0}}\\end{array}\\right]\\) show that : (B – C) A = BA – CA.
\nSolution:
\n\"Selina<\/p>\n

\"Selina
\nQuestion 3.
\n(a) In the given figure, AB is diameter of the circle with centre O. AQ, BP and PRQ are tangents. Prove that OP and OQ are perpendicular to each other.
\n\"Selina
\nSolution:
\nIn the given figure,
\nAB is diameter of the cirlce with centre O. AQ, BP and PRQ are tangents to the circle.
\nTo prove : OP \u22a5 OQ
\nProof: \u2235 QA and QR the tangents to the circle
\n\u2234 QA=QR
\n\u2234 QO is the bisectors of \u2220AOR
\nSimilarly, PR and PB are tangents
\n\u2234 PR = PB
\n\u2234 PO is the bisector of \u2220ROB
\nBut \u2220AOR + \u2220ROB = 180\u00b0 (Linear pair)
\n\"Selina
\n\u21d2 \\(\\frac{1}{2}\\) \u2220AOR + \\(\\frac{1}{2}\\) \u2220ROB = 90\u00b0
\n\u21d2 \u2220QOR + \u2220ROP = 90\u00b0 \u21d2 QOP = 90\u00b0
\n\u2234 OP \u22a5 OQ<\/p>\n

(b) In the inequation \\(2+\\frac{3 x-1}{5} \\leq \\frac{2 x-1}{4}+3\\), write the greatest value of x, when :
\n(i) x is a natural number.
\n(ii) x is a real number.
\nSolution:
\n\"Selina
\n(i) If x is a natural number, then greatest value is 9
\n(ii) If x is a real number, then greatest value of x is 9.5<\/p>\n

(c) Find the missing frequencies in the following distribution table. It is given that the mean of these distributions is 56 and their total is 90 i.e., \u03a3f = 90.<\/p>\n\n\n\n\n
Class interval<\/strong><\/td>\n0-20<\/td>\n20-40<\/td>\n40-60<\/td>\n60-80<\/td>\n80-100<\/td>\n100-120<\/td>\n<\/tr>\n
Frequency<\/strong><\/td>\n16<\/td>\nf1<\/sub><\/td>\n25<\/td>\nf2<\/sub><\/td>\n12<\/td>\n10<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Solution:
\nMean = 56, total frequency i. e., \u03a3f = 90<\/p>\n\n\n\n\n\n\n\n\n\n\n
Class interval<\/strong><\/td>\nMid value (x)<\/strong><\/td>\nFrequency (f)<\/strong><\/td>\nf.x.<\/strong><\/td>\n<\/tr>\n
0-20<\/td>\n10<\/td>\n16<\/td>\n160<\/td>\n<\/tr>\n
20-40<\/td>\n30<\/td>\nf1<\/sub><\/td>\n30f1<\/sub><\/td>\n<\/tr>\n
40-60<\/td>\n50<\/td>\n25<\/td>\n1200<\/td>\n<\/tr>\n
60-80<\/td>\n70<\/td>\nf2<\/sub><\/td>\n70f2<\/sub><\/td>\n<\/tr>\n
80-100<\/td>\n90<\/td>\n12<\/td>\n1080<\/td>\n<\/tr>\n
100-120<\/td>\n110<\/td>\n10<\/td>\n1100<\/td>\n<\/tr>\n
Total<\/td>\n<\/td>\n90<\/td>\n3590 + 30f1<\/sub> + 70f2<\/sub><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"Selina
\nand f1<\/sub> = 27 – f2<\/sub> = 27 – 16 = 11
\nHence, f1<\/sub> = 11, f2<\/sub> = 16<\/p>\n

Question 4.
\n(a) Ramesh deposits \u20b9 2,400 per month in a recurring deposit scheme of a bank for one year. If he gets \u20b9 1,248 as interest at the time of maturity, find the rate of interest. Also, find the maturity value of this deposit.
\nSolution:
\nDeposit per month (P) = \u20b9 2400
\nPeriod (n) = 1 year = 12 months
\ninterest = \u20b9 1248
\nLet rate % = r%
\n\"Selina
\nRate of 8%
\nMaturity value = Principal deposited + Interest = \u20b9 2400 \u00d7 12 + \u20b9 1248
\n= \u20b9 28800 + \u20b9 1248 = \u20b9 30,048<\/p>\n

(b) A (-2, 4) and B (-4, 2) are reflected in the .y-axis. If A’ and B’ are images of A and B respectively.
\n(i) Find the co-ordinates of A’ and B’.
\n(ii) Assign a special name to quadrilateral AA’B’B.
\n(iii) State whether AB’ = BA’.
\nSolution:
\n(i) A (-2, 4) and B (-4, 2) is reflected in y-axis
\n(ii) Image of A is A’ (2, 4)
\nand image of B is B’ (4, 2)
\n(iii) By joining A, B, A’, B’ ; ,
\nWe get a trapezium in which AB’ = BA’<\/p>\n

(c) A lot consists of 144 ball pens of which 20 are defective. A customer will buy a pen only if it is not defective. The shopkeeper draws one pen at random and gives to the customer. What is the probability that :
\n(i) the customer will buy it ?
\n(ii) the customer will not buy it ?
\nSolution:
\nNumber of total ball pens = 144
\nNo. of defective pens = 20
\n\u2234 Number of non-defective = 144-20 = 124
\n(i) One pen is drawn, then probability of non-defected pen = \\(\\frac{124}{144}=\\frac{31}{36}\\)
\n(ii) Probability of defective pen = \\(\\frac{20}{144}=\\frac{5}{36}\\)<\/p>\n

Section B (40 Marks)<\/strong>
\n(Answer any four questions from this Section)<\/strong><\/p>\n

Question 5.
\n(a) In the given figure, tangents PQ and PR are drawn to a circle such that angle RPQ = 30\u00b0. A chord RS is drawn parallel to the tangent PQ. Find the angle RQS.
\n\"Selina
\nSolution:
\n(a) In the given figure,
\nPQ and PR are the tangents to the circle
\nSuch that \u2220RPQ = 30\u00b0
\nA chord RS || PQ
\nRQ and SQ are joined
\n\"Selina
\n\"Selina<\/p>\n

(b) Prove that: \"Selina
\nSolution:
\n\"Selina
\n= R.H.S<\/p>\n

(c) The arithmetic mean between a and b is twice the geometric mean between a and b.
\nProve that: \\(\\frac{a}{b}=7+4 \\sqrt{3} \\text { or } 7-4 \\sqrt{3}\\).
\nSolution:
\nGiven A.M. between a and b is twice the G.M. between a and b.
\ni. e., A.M. between a and b = 2 \u00d7 G.M. between a and b
\n\"Selina
\n\"Selina
\n\"Selina
\n\"Selina<\/p>\n

Question 6.
\n(a) The marked price of an electronic device is \u20b9 9,200 and the sales-tax levied on it is 8%. A customer asks the trader to give this electronic device for \u20b9 9,000 or less as he has only \u20b9 9,000 with him. The trader charged \u20b9 8,856 only including sales-tax. Find the discount given by the trader.
\nSolution:
\nM.P. of an electric device = \u20b9 9200
\nRate of sale tax = 8%
\nS.P. of the device (including S.T.) = \u20b9 8856
\nPrice without S.T. = \\(\\frac{8856 \\times 100}{10+8}\\) = \u20b9 8200
\n\u2234 Discount = \u20b9 9200 – \u20b9 8200 = \u20b9 1000<\/p>\n

(b) The diagrams, given below, represent two inequations A and B on real number line.
\n\"Selina
\n(i) Write down A and B in set builder notations.
\n(ii) Represent A \u2229 B and A’ \u2229 B on two different number lines.
\nSolution:
\nEquations of A and B on number line
\n\"Selina<\/p>\n

(c) Solve for matrices A and B, where 2 A + B = \\(=\\left[\\begin{array}{cc}{3} & {-4} \\\\ {2} & {7}\\end{array}\\right]\\) and A – 2B = \\(\\left[\\begin{array}{ll}{4} & {3} \\\\ {1} & {1}\\end{array}\\right]\\)
\nSolution:
\n\"Selina
\n\"Selina<\/p>\n

Question 7.
\n(a) Solve the equation 3x2<\/sup> – x – 1 = 0 and give your answer correct to two decimal places.
\nSolution:
\n3x2<\/sup> – x – 7 = 0
\nHere, a = 3, b = -1, c = -7
\nD = b2<\/sup> – 4ac = (-1 )2<\/sup> – 4 \u00d7 3 \u00d7 (-7)
\n= 1 + 84 = 85
\n\"Selina<\/p>\n

(b) Two identical solid cones each of base radius 3 cm with vertical height 5 cm and one more solid cone of base radius 2 cm with vertical height 4.5 cm are jointly melted and recasted into a solid sphere. Find : (i) the radius, (ii) curved surface area of the sphere.
\nSolution:
\nRadius of each indentical cone (h1<\/sub>) = 3 cm
\nand height (h1<\/sub>) = 5 cm
\nRadius of third cone (r2<\/sub>) = 2 cm
\nand vertical height (h1<\/sub>) = 4.5 cm
\n\"Selina<\/p>\n

(c) The angle of elevation of a cloud from a point h metres above the surface of a lake is 0 and the angle of depression of its reflection in the lake is \u03d5. Prove that the height of the cloud above the lake surface is : \\(h\\left(\\frac{\\tan \\phi+\\tan \\theta}{\\tan \\phi-\\tan \\theta}\\right)\\).
\nSolution:
\nLet AM be the lake P is the cloud and P’ is its reflection in the lake.
\nLet the height of the cloud
\nMP = x m and its reflection MP’ = x
\nNow in right \u2206PBC
\n\"Selina
\n\"Selina
\n\"Selina<\/p>\n

Question 8.
\n(a) A man desires to have an annual income of \u20b9 36,000 from 18% \u20b9 125 shares available at a premium of 20%. How much should he invest?
\nSolution:
\nAnnual income from 125 shows = \u20b9 36000
\nRate of dividend = 18%
\nMarket rate = \u20b9 100 + 20 = \u20b9 120
\n\u2234 Investment = \u20b9 \\(\\frac{36000 \\times 120}{18}\\) = \u20b9 240000<\/p>\n

(b) As shown in the given figure; from an external point P, a tangent PT and a line segment PAB are draw to a circle with centre O. ON is perpendicular on the chord AB. Prove that:
\n(i) PA.PB = PN2<\/sup> – AN2<\/sup>
\n(ii) PN2<\/sup> – AN2<\/sup> = OP2<\/sup> – OT2<\/sup>
\n(iii) PA.PB = PT2<\/sup>
\n\"Selina
\nSolution:
\nIn the given figure,
\nPT is a tangent to the circle with centre O
\nOAB is a secant intersecting the circle at A and B
\nON \u22a5 AB
\nTo prove :
\n(i) PA.PB = PN2<\/sup> – AN2<\/sup>
\n(ii) PN2<\/sup> – AN2<\/sup> = OP2<\/sup> – OT2<\/sup>
\n(iii) PA.PB = PT2<\/sup>
\nConstruction : Join OT, OP and OA.
\nProof:
\n(i) PA.PB = (PN – AN) (PN + BN)
\n= (PN – AN) (PN + AN) (\u2235 N is the mid point of AB)
\n= (PN)2<\/sup> – (AN)2<\/sup>(ii) PN2<\/sup> – AN2<\/sup> = (OP2<\/sup> – ON2<\/sup>) – (OA2<\/sup> – ON2<\/sup>) (In right \u2206ONP and \u2206ONA)
\n= OP2<\/sup> – ON2<\/sup> – OA2<\/sup> + ON2<\/sup>
\n= OP2<\/sup> – OA2<\/sup>
\n= OP2<\/sup> – OT2<\/sup> (\u2235 OA = OT radii of the same circel)(iii) PA.PB = PN2<\/sup> – AN2<\/sup> [Proved in (i)]
\n= OP2<\/sup> – OT2<\/sup> [Proved in (ii)]
\n= PT2<\/sup><\/p>\n

(c) If x =\\(\\frac{\\sqrt{2 a+1}+\\sqrt{2 a-1}}{\\sqrt{2 a+1}-\\sqrt{2 a-1}}\\), prove that : x2<\/sup> – 4ax + 1=0.
\nSolution:
\n\"Selina<\/p>\n

Question 9.
\n(a) Find the co-ordinates of the point Q on x-axis which lies on the perpendicular bisector of the line segment joining the points A (-5, -2) and B (4, -2). Name the type of the triangle QAB.
\nSolution:
\n(a) Slope of line joining the points A (-5, -2) and B (4, -2) = \\(\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\\frac{-2+2}{4+5}=\\frac{0}{9}=0\\)
\n\u2234 AB || x-axis
\nLet co-ordinates of Q be (x, 0) and equation of line AB is y = -2
\nLM is perpendicular bisector of AB, which intersects x-axis at Q
\nCo-ordinates of Q are \\(\\left(\\frac{-1}{2}, 0\\right)\\)
\n\u2235 Q lies on the right bisector of AB
\n\"Selina
\n\u2234 QA = QB
\n\u2234 \u2206QAB is an isosceles triangle.<\/p>\n

(b) Two pipes running together can fill an empty cistern in \\(11 \\frac{1}{9}\\) minutes. If one pipe takes 5 minutes more than the other to fill the same cister, find the time in which each pipe would fill the cister.
\nSolution:
\nWork of two pipes for filling in 1 minute = \\(\\frac{9}{100}\\)
\nLet one pipe takes x minutes
\nthen second pipe will take (x + 5) minutes
\n\"Selina
\n\u21d2 9x2<\/sup> + 45x = 200x + 500
\n\u21d2 9x2<\/sup> + 45x – 200x – 500 = 0
\n\u21d2 9x2<\/sup> – 155x – 500 = 0
\n\u21d2 9x2<\/sup> – 180x + 25x – 500 = 0
\n\u21d2 9x(x – 20) + 25(x – 20) = 0
\n\u21d2 (x – 20) (9x + 25) = 0
\nEither x – 20 = 0, then x = 20
\nor 9x + 25 = 0 then x = \\(\\frac{-25}{9}\\) which is not possible being negative
\n\u2234 x = 20
\n\u2234 First pipe will fill in 20 + 5 = 25 minutes and second pipe in 20 minutes<\/p>\n

(c) A person bought a certain number of pens for \u20b9 800. If he had bought 4 pens more for the same money, he would have paid \u20b9 10 less for each pen. How many pens did he buy ?
\nSolution:
\nLet cost of x pens = \u20b9 800
\n\"Selina
\n\u21d2 x2<\/sup> + 4x – 320 = 0
\n\u21d2 x2<\/sup> + 20x – 16x – 320 = 0
\n\u21d2 x(x + 20) – 16(x + 20) = 0
\n\u21d2 (x + 20) (x – 16) = 0
\nEither x + 20 = 0, then x = -20 which is not possible being negative or x – 16 = 0, then x = 16
\n\u2234 Number of pen = 16<\/p>\n

Question 10.
\n(a) Using ruler and compasses only, contruct
\n(i) a triangle ABC in which angle ABC = 45\u00b0, AB = 8.6 cm and BC = 9.8 cm
\n(ii) construct a circle of radius 2.5 cm which touches the arms of the angle BAC of \u2206ABC given above.
\nSolution:
\n(a) Steps of construction :
\n(i) Draw a line segment BC = 9.8 cm.
\n(ii) At B, draw a ray BX making angle of 45\u00b0 and cut off BA = 8.6 cm.
\n(iii) Join AC.
\n\u2206ABC is the required triangle.
\n(iv) Draw BY, the bisector of \u2220ABC.
\n(v) Draw a perpendicular on B and cut off DB = 2.5 cm.
\n(vi) From B, draw DZ || BC which intersects BY at O.
\n(vii) With centre O and radius 2.5 cm, draw a circle which touches the sides of 45\u00b0.
\n\"Selina<\/p>\n

(b) The marks obtained by T20 students in a Mathematics test are given below :<\/p>\n\n\n\n\n\n\n\n\n
Marks<\/strong><\/td>\nNo. of students<\/strong><\/td>\nMarks<\/strong><\/td>\nNo. of students<\/strong><\/td>\n<\/tr>\n
0-10<\/td>\n5<\/td>\n50-60<\/td>\n18<\/td>\n<\/tr>\n
10-20<\/td>\n9<\/td>\n60-70<\/td>\n11<\/td>\n<\/tr>\n
20-30<\/td>\n16<\/td>\n70-80<\/td>\n6<\/td>\n<\/tr>\n
30-40<\/td>\n22<\/td>\n80-90<\/td>\n4<\/td>\n<\/tr>\n
40-50<\/td>\n26<\/td>\n90-100<\/td>\n3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Using the informations, given above, draw an ogive on a graph sheet. Take a suitable scale for your ogive. Use the ogive drawn to estimate :
\n(i) the median.
\n(ii) the number of students who obtained more than 75% marks in the test.
\n(iii) the number of students who did not pass in the test if the pass percentage was 40.
\nSolution:<\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n\n
Marks<\/strong><\/td>\nNo. of students (f)<\/strong><\/td>\nc.f<\/strong><\/td>\n<\/tr>\n
0-10<\/td>\n5<\/td>\n5<\/td>\n<\/tr>\n
10-20<\/td>\n9<\/td>\n14<\/td>\n<\/tr>\n
20-30<\/td>\n16<\/td>\n30<\/td>\n<\/tr>\n
30-40<\/td>\n22<\/td>\n52<\/td>\n<\/tr>\n
40-50<\/td>\n26<\/td>\n78<\/td>\n<\/tr>\n
50-60<\/td>\n18<\/td>\n96<\/td>\n<\/tr>\n
60-70<\/td>\n11<\/td>\n107<\/td>\n<\/tr>\n
70-80<\/td>\n6<\/td>\n113<\/td>\n<\/tr>\n
80-90<\/td>\n4<\/td>\n117<\/td>\n<\/tr>\n
90-100<\/td>\n3<\/td>\n120<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Plot the points (10, 5), (20, 14), (30, 30), (40, 52), (50, 78), (60, 96), (70, 107), (80, 113), (90, 117), (100, 120) on the graph and join them free hand. We get an ogive as shown.
\n\"Selina
\nTotal students =120
\nMedian : From = \\(\\frac{120}{2}\\) = 60
\ny-axis draw a line parallel to x-axis which meets the curve at B. From B, draw perpendicular on x- axis meeting it at C which is 43.
\n\u2234 Median = 43
\n(ii) No. of students getting more then 75% marks = 120 – 110= 10 students.
\n(iii) Number of students who did not pass the test (passing percentage = 40%) = 52<\/p>\n

Question 11.
\n(a) Find the co-ordinates of the circumcentre of the triangle whose vertices are (3, 0), (-1, -6) and (4, -1). Also, find its circumradius.
\nSolution:
\nThe vertices of \u2206ABC are A (3, 0), B (-1, -6), C (4, -1)
\nLet (x, y) be the co-ordinates of the circumcentre O
\n\"Selina
\n\u2234 OA = OB = OC => OA2<\/sup> – OB2<\/sup> = OC2<\/sup>
\nOA= \\(\\sqrt{\\left(x_{2}-x_{1}\\right)^{2}+\\left(y_{2}-y_{1}\\right)^{2}}\\)
\nOA2<\/sup> = (x2<\/sub> – x1<\/sub>)2<\/sup> + (y2<\/sub> – y1<\/sub>)2<\/sup> = (x – 3)2<\/sup> + (y – 0)2<\/sup>
\n= x2<\/sup> – 6x + 9 + y2<\/sup> = x2<\/sup> + y2<\/sup> – 6x + 9
\nSimilarly OB2<\/sup> = (x + 1)2<\/sup> + (y + 6)2<\/sup> = (x2<\/sup> + 2x + 1 + y2<\/sup> + 12y + 36 = x2<\/sup> + y2<\/sup> + 2x + 12y + 37
\nOC2<\/sup> = (x – 4)2<\/sup> + (y + 1)2<\/sup> = x2<\/sup> – 8x + 16 +y2<\/sup> + 2y + 1
\n= x2<\/sup> + y2<\/sup> – 8x + 2y + 17
\n\u2235 OA2<\/sup> = OB2<\/sup>
\n\u2234 x2<\/sup> + y2<\/sup> – 6x + 9 = x2<\/sup> + y2<\/sup> + 2x + 17ly + 37 \u21d2 -8x – 12y = 28
\n\u21d2 2x + 3y \u21d2 – 7 ….(i)
\nand OB2<\/sup> = OC2<\/sup> \u21d2 x2<\/sup> +y2<\/sup> + 2x + 12y + 37 = x2<\/sup> + y2<\/sup> – 8x + 2y + 17
\n\u2234 10x +10y = -20 \u21d2 x +y = -2 …(ii)
\nSolving (i) and (ii), y = -3, x = 1
\n\u2234 Co-ordinates of centre O are (1, -3)
\nCircumradius A = \\(\\sqrt{(3-1)^{2}+(0+3)^{2}}=\\sqrt{2^{2}+3^{2}}=\\sqrt{4+9}=\\sqrt{13}\\) cm<\/p>\n

(b) Use ruler and a pair of compasses to construct angle ABC = 75\u00b0. Mark a point D on BC, such that BD = 5 cm. Construct a circle to touch AB at B and also to pass through D.
\nSolution:
\n\"Selina
\nSteps of construction :
\n(i) Construct \u2220ABC = 75\u00b0
\nUsing ruler and compass and take a point D such that BC = 5 cm.
\n(ii) Draw perpendicular bisector at BD.
\n(iii) Draw a perpendicular at B on AB which intersects the perpendicular bisector at O.
\n(iv) With centre O and radius D, draw a circle which passes through D and touches AB at B.
\nThis is the required circle.<\/p>\n

(c) ABC is a right-angled triangle with the right angle at vertex B. BD is the altitude through B. Given BD = 12 cm and AD – 9 cm.
\n(i) Calculate AB.
\n(ii) Name the triangle which are similar to triangle ADB (Proof not required).
\n(iii) Find AC.
\nSolution:
\nIn right \u2206ABC, \u2220B = 90\u00b0
\nBD \u22a5 AC
\nBD = 12 cm, AD = 9 cm
\n(i) \u2234 AB2<\/sup> = AD2<\/sup> + BD2<\/sup> (Pythagoras Theorm)
\n= 92<\/sup> + 122<\/sup> = 81 + 144 = 225
\nAB = \\(\\sqrt{225}\\) = 15 cm
\n\"Selina
\n(ii) \u2206ABD ~ \u2206ABC
\nand \u2206ABD ~ \u2206BCD
\n(iii) \u2235 BD2<\/sup> = AD \u00d7 DC
\n\u2234 122<\/sup> = 9 \u00d7 DC \u21d2 \\(\\frac{144}{9}\\) =DC
\n\u21d2 DC = 16
\n\u2234 AC = AD + DC
\n= 9 + 16 = 25 cm<\/p>\n

Selina Concise Mathematics Class 10 ICSE Solutions<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"

Selina Concise Mathematics Class 10 ICSE Solutions Revision Paper 2 Section A (40 Marks) (Answer all questions from this Section) Question 1. (a) In what ratio does the line x – y – 2 = 0 divide the line segment joining the points (3, -1) and (8, 9)? 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