by admin on August 30, 2011
A number which can be written in the form p/q , where p and q are integers and q ≠ 0 is called a rational number. For example,
2,3, 6/7 are all rational numbers. Since the numbers 0, –2, 4 can be written in the form p/q , they are also rational numbers.
Rational numbers are closed under addition. That is, for any two rational numbers a and b, a + b is also a rational number.
Rational numbers are closed under subtraction. That is, for any two rational numbers a and b, a – b is also a rational number.
Rational numbers are closed under multiplication. That is, for any two rational numbers a and b, a × b is also a rational number.
Rational numbers are not closed under division.
However, if we exclude zero then the collection of Rational numbers, all other rational numbers is closed under division.
by admin on August 22, 2011
If orderofA ,B ,C are3 *4, 5 *4,5 x8 thenwhat istheorderof (AB^T C)
Sol: order of 
:.orderof 
Hence order of 
by admin on August 22, 2011
Sol: Possibletypesofmatrices: 
Ans: 9
by admin on August 9, 2011
Sol: n(A) n(A-B)+n(A1’-B)
25+x+2x=25+3x
n(B)= n(B-A)+n(AB)= 2x+2x=4x
n(A) = 2(n(B))
25+3×2(4x)8x-3x=255×25 x5
by admin on March 13, 2011
Two chords AB and AC of a circle are equal. Prove that the centre of the circle lies on the angle bisector of ZBAC.
GIVEN Two equal chords AB and AC of a circle C (0, r).
.
TO PROVE Centre 0 lies on the bisector of LBAC, which intersects the chord BC in M.
CONSTRUCTION Join BC. Draw bisector AD of LBAC
PROOF In triangles BAM and CAM, we have
LBAM = ZCAM
and, perpendilar AM=AM
So, by SAS-criterion of congruence, we have
AB = AC
Δ BAM Δ CAM
BM = CM and ZBMA LCMA
BM = CM and ZBMA = ZMA 90’
AM is the perpendicular bisector of the chord BC.
AM passes through the centre 0.
Hence, the centre of the circle lies on the angle bisector of perpendilar BAC.
by admin on March 13, 2011
GIVEN Three non-collinear points P, Q and R.
TO PROVE There is one and only one circle passing through P, Q and R.
CONSTRUCTION Join PQ and QR. Draw perpendicular bisectors AL and BM of PQ and RQ respectively. Since P, Q, R are not collinear. Therefore, the perpendicular bisectors AL and BM are not parallel.
Let AL and BM intersect at 0. Join OP, OQ and OR.
PROOF Since 0 lies on the perpendicular bisector of PQ. Therefore,
OP=OQ.
Again, 0 lies on the perpendicular bisector of QR. Therefore, OQ = OR
Thus, OP = OQ = OR = r (say).
Taking 0 as the centre draw a circle of radius s. Clearly, C (0, s) passes through P, Q and R. This proves that there is a circle passing through the points P, Q and R.
We shall now prove that this is the only circle passing through P. Q and R.
If pqssible, let there be another circle with centre 0’ and radius r, passing through the points P. Q and R. Then, 0’ will lie on the perpendicular bisectors AL of PQ and BM of QR.
Since two lines cannot intersect at more than one point, so 0’ must coincide with 0.
Since OP = r, O’P = s and 0 and 0’ coincide, we must have r = s
= C(O,r)v(O’,s),
Hence, there is one and only one circle passing through three non-collinear points P, Q and R.
REMARK 1 It should be noted that in case three given points are collinear, then a single circle cannot pass through these three points. For, if there were a circle which passes through three coilinear points P, Q, R, then perpendicular from the centre to this line will have to bisect PQ, QR and PR and thus this perpendicular intersects the line in three points, which is a contraclinction.
REMARK 2 It follows immediately from the above theorem that there is a unique circle passing :hrough the three vertices P, Q, R of a A PQR. This circle is called the circumcircle of the triangle PQR and its centre the circum centre of the triangle. We also observe that an infinite number of circles can be drawn passing through a given point. Also, an infinite number of circles can be drawn to pass through two given points.
by admin on March 9, 2011
Prove that the line segments joining the mid-points of the adjacent sides of a quadrilateral form a parallelogram
GIVEN A quadrilateral ABCD in which P. Q, R and S are the mid-points of sides AB, BC, CD and DA respectively. TO PROVE PQRS is a parallelogram.
CONSTRUCTION Join AC.
PROOF In A ABC, P and Q are the mid-points of AB and BC respectively.
PQ || AC
In Δ ACD, R and S are the mid-points of CD and DA
respectively.
SR || AC
From equations (i) and (ii), we have
PQ II AC and SR || AC
PQ || SR.
Similarly, by considering triangles ABD and BCD, we can prove that
PS || QR.
by chemistry on March 9, 2011
Any line parallel to the parallel sides of a trapezium divides the non-parallel sides proportionally.
GIVEN A trapezium ABCD in which DCI AB and EF is a line parallel to DC and AB. TO PROVE AE/ED = BF/FC
CONSTRUCTION Join AC, meeting EF in G.
PROOF In Δ ADC, we have
EG || DC
AE/ED = AG/GC
IN Δ ABC, we have
GF || AB
AG/GC = BF/FC
From equations (i) and (ii), we get
AE/ED = BF/FC
by maths on March 9, 2011
