In the service of student community since Feb 2006
Hand Written NCERT Solutions
CBSE Sample Papers
CBSE Mock Tests
11 Years of Devotion in Maths
Absolutely Free

CBSE, CBSE Sample Papers, NCERT Solutions, CBSE Results,Math      

CCBSE, CBSE Sample Papers IX X, NCERT Solutions, CBSE Results 2016, Mathematics Concepts, CBSE Mock Tests, CBSE Syllabus, CBSE Date Sheets Previous Years Papers

Answers to Your Questions - Page 19
What do you get on this web page?
Here you get answers of your CBSE mathematics 1X and X questions by our experienced teachers. To ask a question click here
See Answers to Your questions
Question 180. What is centroid?  by Bony Singla, Answered on 12 July 2009 at 12:46 p.m.
Answer 180.    Centroid is point of concurrence of Medians of triangle.
Question 181. Show that one and only one out of n, n + 2 or n + 4 are divisible by 3, where n is any positive integer, by Lokesh Pradhan Bhubaneshwar. Answered on 2 August 2009 at 12:32 p.m.
Answer 181. Using euclid's division algorithm a = bq + r where 0 < = r <  b
                       let a = n and b = 3
                       n = 3q + r where 0 < = r <  3
                       if r = 0
                       n = 3q,               n  + 2 = 3q + 2,          n + 4 = 3q + 4     
 (here only n is divisible by 3)
                       if r = 1
                       n = 3q + 1,         n  + 2 = 3q + 3,          n + 4 = 3q + 5      
(here only n + 2 is divisible by 3)
                       if r = 2
                       n = 3q + 2,         n  + 2 = 3q + 4,          n + 4 = 3q + 6      
(here only n + 4 is divisible by 3)
                       Therefore one and only one out of n, n + 2 or n + 4 are divisible by 3,
Question 182. What is a Rectangle? by Shubham Maurya Varanasi. Answered on 16 August 2009 at 9:34 p.m.
Answer 182.    A Rectangle is a parallelogram whose one angle is 90 degrees.
See Answers to Your questions
Question 183. AP and BQ are perpendicular to BC and AC respectively and AP = BQ. O is point of intersection of AP and BQ. Prove that AO = BO by Sneha Mariam from The Asian School Umm Al Hassam. Answered on 6 September 2009 at 7:59 a.m. I.S.T. (question was incomplete, so completed before answering)
Answer 183. Download Answer.
Question 184. The angles of triangle are in A.P the greater angle is twice the least. Find all the angles. By Syed Zuhaib Ahmed,
MES Indian School, Doha. Answered on 13 sep 2009 at 11:35 a.m.
Answer 184. Let the angles be a, b and c where these are in A.P.
                  b - a = c - b
                  2b = a+ c
                  b = (a + c)/2 ........ (i)
                  a = 2c (given) ........(ii)
                  a + b + c = 180 ( angle sum property of triangle)
                  2c + (a + c)/2 + c = 180  [ using (i) and (ii) ]
                  3c + (2c + c)/2 = 180
 (×2)           6c + 2c + c = 360
                  9c = 360
                  c = 40
                  a = 2c = 80
                  b = (40 + 80)/2
                    =  60
Question 185. If the polynomials p(x) = 2x^3 + bx^2 + 3x - 5 and q(x) = x^3 + x^2 - 4x + b leave the same remainder when divided by x - 2. Find b. By Tanya Khan, Azra Public school, Hyderabad. Answered on 13 Sep 2009 at 11: 58 a.m.
Answer 185. x - 2 is a factor of p(x) and q(x),
                  therefore p(2) = q(2) = 0 by factor theorem
                  2×2^3 + b×2^2 + 3×2 - 5 = 2^3 + 2^2 - 4×2 + b
                  16 + 4b + 6 - 5 = 8 + 4 - 8 + b
                  3b = - 13
                  b = - 13/3
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.
                                   See Answers to Your questions
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







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.

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 cm^2
pie r l = 814 ........... (i)                           
WSA = 1584 cm2
pie r2 + pie r l = 1584 cm^2
pie r2 + 814 = 1584
pie r2   =  770 cm^2
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 cm^2

Demo: Explicit load of a +1 button