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   cbsemath.com -  the no. 1 CBSE Math site   cbsemath.com - in the service of student community
Question 186. Find the circumference of a circle whose area is 301.84 sq. cm, by Neelambujkapur, St.Johns Academy Delhi. Answered on 17 Oct 2009 at 8:56 p.m.
Answer 186.
Area of circle = 301.84 sq. cm
pie r^2 = 301.84
22/7 r^2 =  301.84
r^2 = (301.84 × 7)/22
r^2 = (30184 × 7)/2200
r^2 = (7 × 7 × 7 × 7 × 2 × 2)/(10 × 10)
r = 9.8 cm.
Circumference = 2 pie r
                       = 2 × 22/7 × 98/10
                       = (2 × 22 × 98) / (7×10)
                       = 61.6 cm.

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Question 187. Solve the following system of equations mx - ny = m2  +  n2 and x + y = 2m by Supriya Dubey, convent of Gagan Bharti, Uttam Nagar. Answered on 30 Oct 2009 at 11:56 p.m.
Answer 187.          mx - ny = m2  +  n2  ...... (i)
                           x + y = 2m     .........(ii)
                                         (i) × 1 + (ii) ×2
                                         mx - ny = m2  +  n2  ...... (i)
                           mx + ny = 2mn     .........(ii)
                                        ______________________________
                                        (m + n)x = (m + n)2
                                          x = m + n
                                          Substituting in ( i i)
                            m + n + y = 2m
                                           y = m - n
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Question 188. Name the type of quadrilateral formed, if any, by the points A(-3,5), B(3,1), C(0,3), D(-1,-4),
by Harkirat Singh, GHPS Delhi. Answered on 29 Nov. 2009 at 1:35 a.m.
Answer 188. In triangle ABC, AB = 2 squareroot 13, BC = squareroot 13, CA = squareroot 13.
                  BC + AC = squareroot 13 + squareroot 13
                               = 2 squareroot 13
                  AB         = 2 squareroot 13
    Therefore AB = BC + CA
                  Triangle ABC cannot be constructed.
                  Hence quadrilateral ABCD cannot be constructed.

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Question 189.The CSA of the cone is 814 sq. cm and and the TSA of the cone is 1584 sq.cm.find its volume, by Priya S, Chennai. Answered on 29 Nov. 2009 at 5:05 p.m.
Answer 189. CSA = 814 cm2
pie r l = 814 ........... (i)                          
WSA = 1584 cm2
pie r2 + pie r l = 1584 cm2
pie r2 + 814 = 1584
pie r2   =  770 cm2
r = 7 squareroot5  ........ (ii)
Substituting in ........(i)
l = 37/sqaurerrot5 cm
h2 = l2 - r2
putting values of l and h we get     
h = 12/ squareroot5
Therefore Volume of cone = pie h r2/3
= 616 squareroot 5 cm2
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Question 190. What is the Blue Print Maths CBSE Board Exam for Class 12 ?. By Tanu, DPS Delhi. Answered on 30 Nov. 2009 at 9:45 p.m.
Answer 190. You can download latest Mathematics model sample papers along with Blueprint here
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   cbsemath.com -  the no. 1 CBSE Math site   cbsemath.com - in the service of student community
Question 191. If x = a sec3 A and y = b tan3 A. Prove (x/a)2/3  - (y/b)3/2  = 1, by Smita Kumari, Hyderabad. Answered on 1 Dec. 2009 at 11:15 p.m.
Answer 191.    x = a sec3 A
                or  (x/a) = a sec3 A
                or  (x/a)1/3 = sec A
                or  (x/a)2/3 = sec 2A  .............(i)
 Similarly        (y/b)2/3 = tan 2A  .........(ii)
                      (i) - (ii)
        L.H.S. = (x/a)2/3- (y/b)2/3 = sec 2A - tan 2A
                                      = 1
                                      = R.H.S.
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Question 192. How many terms of the AP: -6,-11/2,-5... are need to give the sum -25? Also explain the double answer by Nilavani, The Lawrence school Chennai. Answered on 3 Dec. 2009 at 7:45 p.m.
Answer 192. The given AP is -6,-11/2,-5...
                  first term (a) = - 6
                  common difference (d) = 1/2
                  Sn = - 25
                   n/2[2a + (n - 1)d] = - 25
              or  n[- 12 + (n - 1)1/2] = - 50
              or  n[- 24 + (n - 1)]1/2 = - 50
              or  n[- 24 + (n - 1)] = - 100
              or  n (n - 25) + 100 = 0
              or  n^2 - 25n + 100 = 0
              or  (n - 20)(n - 5) = 0
                   n = 5, n = 20
              Sum of first five terms = - 25
              Sum of first 20 terms is also - 25, because the sum of next 15 terms after 5th term is 0.
              Therefore sum of both 5 and 20 terms is equal to - 25
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Question 193. solve for x and y: ax/b-by/a=a+b, ax-by=2ab, by Dhwani Rathi the new tulip international school Ahmedabad. Answered on 7 Dec. 2009 at 11:29 a.m.
Answer 193. ax/b-by/a=a+b ........(i)
                  ax-by=2ab       ........(ii)
                  equation i × ab - ii × a
                  a2x - b2y = a2b - ab2
                                    a2x - aby = 2a2b
                 -      +        -
             ________________________
              by(a - b) = ab (b - a)
         or  y = -a
              Substituting in equation i
              ax + a2b = 2a2b
              ax = a2b
              x = b
  
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Question 194.  If a wire is bent into the shape of a square, then the area of the square is 81 sqcm. When wire is bent into a semicircular shape. then find the area of the semicircle, by Nidhi Agrawal, Bhilai.
Answer 194. Area of square = 81 sq. cm.
                  S2    = 81
                  S = 9 cm.
                   Perimeter of square = 4S
                                               = 4 × 9
                                               = 36 cm.
                   Perimeter of semicircle = 36 cm
                   pie r + 2r = 36
                   r (pie + 2) = 36
                   r (22/7 + 2) = 36
                   r (36/7) = 36
                   r/7 = 1
                   r = 7cm.
                   Area of semicircle = pie r2/2
                                            = (22×7×7)/(7×2)
                                            = 77 sq. cm.
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Question 195. IF (secA+tanA)(secB+tanB)(secC+tanC) = (secA-tanA)(secB-tanC)(secC-tanC)
Then prove that each expression =+/-1. By shirali oberoi, st.raphael h.s school indore.
Answer 195. L.H.S. (secA+tanA)(secB+tanB)(secC+tanC) = (secA-tanA)(secB-tanC)(secC-tanC)
                               Multiply both sides by (secA-tanA)(secB-tanC)(secC-tanC) that is RHS
(secA+tanA)(secA-tanA)(secB+tanB)(secB-tanC)(secC+tanC)(secC-tanC) = [(secA-tanA)(secB-tanC)(secC-tanC)]2
or (sec2A-tan2A)(sec2B-tan2C)(sec2C-tan2C) = [(secA-tanA)(secB-tanC)(secC-tanC)]2
or 1 = [(secA-tanA)(secB-tanC)(secC-tanC)]2         [since sec2A-tan2A = 1 ]
or +- 1 = [(secA-tanA)(secB-tanC)(secC-tanC)
therefore RHS = +-1
Similarly LHS = +-1
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