Perimeter and Area of a Circle:
circumference of a circle bears a constant ratio with its diameter. This constant ratio
is denoted by the Greek letter π (read as ‘pi’). In other words
Circumference/Diameter=π
circumference = π × diameter = π × 2r (where r is the radius of the circle) = 2πr
The great Indian mathematician Aryabhatta (A.D. 476 – 550) gave an approximate
value of π. He stated that π =62832/20000 which is nearly equal to 3.1416.
However, for practical purposes, we generally take the value of π as 22/7 or 3.14, approximately
area of the circle =1/2× 2πr × r = πr2
Tuesday, June 1, 2010
Monday, May 31, 2010
Introduction to Matrices
Introduction to Matrices:
The knowledge of matrices is necessary in various branches of mathematics. Matrices are one of the most powerful tools in mathematics. This mathematical tool simplifies our work to a great extent when compared with other straight forward methods. The evolution of concept of
matrices is the result of an attempt to obtain compact and simple methods of solving system of linear equations. Matrices are not only used as a representation of the coefficients in system of linear equations, but utility of matrices far exceeds that use. Matrix notation and operations are used in electronic spreadsheet programs for personal computer, which in turn is used in different areas of business and science like budgeting, sales projection, cost estimation, analyzing the results of an experiment etc. Also, many physical operations such as magnification, rotation and reflection through a plane can be represented mathematically by matrices. Matrices are also used in cryptography. This mathematical tool is not only used in certain branches of sciences, but also in genetics, economics, sociology, modern psychology and industrial management.
The knowledge of matrices is necessary in various branches of mathematics. Matrices are one of the most powerful tools in mathematics. This mathematical tool simplifies our work to a great extent when compared with other straight forward methods. The evolution of concept of
matrices is the result of an attempt to obtain compact and simple methods of solving system of linear equations. Matrices are not only used as a representation of the coefficients in system of linear equations, but utility of matrices far exceeds that use. Matrix notation and operations are used in electronic spreadsheet programs for personal computer, which in turn is used in different areas of business and science like budgeting, sales projection, cost estimation, analyzing the results of an experiment etc. Also, many physical operations such as magnification, rotation and reflection through a plane can be represented mathematically by matrices. Matrices are also used in cryptography. This mathematical tool is not only used in certain branches of sciences, but also in genetics, economics, sociology, modern psychology and industrial management.
Friday, May 28, 2010
Volume of cube calculation
Volume of cube calculation:
Problem:
A child playing with building blocks, which are of the shape of cubes, has built a structure as shown in Fig. below the edge of each cube is 3 cm, find the volume of the structure built by the child.

Solution:
Volume of a Cube = edge × edge × edge = a3
Solution : Volume of each cube = edge × edge × edge= 3 × 3 × 3 cm3
= 27 cm3 Number of cubes in the structure = 15 Therefore,
volume of the structure = 27 × 15 cm3 = 405 cm3
(here cm3 is cm to the power three)
Problem:
A child playing with building blocks, which are of the shape of cubes, has built a structure as shown in Fig. below the edge of each cube is 3 cm, find the volume of the structure built by the child.
Solution:
Volume of a Cube = edge × edge × edge = a3
Solution : Volume of each cube = edge × edge × edge= 3 × 3 × 3 cm3
= 27 cm3 Number of cubes in the structure = 15 Therefore,
volume of the structure = 27 × 15 cm3 = 405 cm3
(here cm3 is cm to the power three)
Wednesday, May 26, 2010
How to find Surface area of a block?
How to find Surface area of a block?
The decorative block shown in below fig is made of two solids — a cube and a hemisphere. The base of the block is a cube with edge 5 cm, and the hemisphere fixed on the top has a diameter of 4.2 cm. Find the total surface area of the block.
(Take (pie π =22/7 ).
Solution : The total surface area of the cube = 6 × (edge)2 = 6 × 5 × 5 cm2 = 150 cm2. Note that the part of the cube where the hemisphere is attached is not included in the surface area. So, the surface area of the block = TSA of cube – base area of hemisphere + CSA of hemisphere = 150 – πr2 + 2 πr2 = (150 + πr2) cm2
= 150 cm2+(22/7+4.2/2+4.2/2)cm2
= (150 + 13.86) cm2 = 163.86 cm2
The decorative block shown in below fig is made of two solids — a cube and a hemisphere. The base of the block is a cube with edge 5 cm, and the hemisphere fixed on the top has a diameter of 4.2 cm. Find the total surface area of the block.
(Take (pie π =22/7 ).
Solution : The total surface area of the cube = 6 × (edge)2 = 6 × 5 × 5 cm2 = 150 cm2. Note that the part of the cube where the hemisphere is attached is not included in the surface area. So, the surface area of the block = TSA of cube – base area of hemisphere + CSA of hemisphere = 150 – πr2 + 2 πr2 = (150 + πr2) cm2= 150 cm2+(22/7+4.2/2+4.2/2)cm2
= (150 + 13.86) cm2 = 163.86 cm2
Tuesday, May 25, 2010
Probabiliy problem
Find the probability of getting a head when a coin is tossed once. Also find the probability of getting a tail.
to understand more about Probabilities these website links may help you,
Probability explainations, and conditional probability
In the experiment of tossing a coin once, the number of possible outcomes
is two — Head (H) and Tail (T). Let E be the event ‘getting a head’. The number of
Outcomes favorable to E, (i.e., of getting a head) is 1. Therefore,
P (E) = P (head) = Number of outcomes favorable to E = 1/2
Number of all possible outcomes
Similarly, if F is the event ‘getting a tail’, then P(F) = P(tail) =1/2
to understand more about Probabilities these website links may help you,
Probability explainations, and conditional probability
The origin of the term "Sine" in Mathematics
The origin of the term "Sine" Mathematics:
The first use of the idea of ‘sine’ in the way we use it today was in the work Aryabhatiyam
by Aryabhatta, in A.D. 500. Aryabhatta used the word ardha-jya for the half-chord, which was shortened to jya or jiva in due course. When the Aryabhatiyamm was translated into Arabic, the word jiva was retained as it is. The word jiva was translated into sinus, which means curve, when the Arabic version was translated into Latin. Soon the word sinus, also used as sine, became common in mathematical texts throughout Europe. An English Professor of astronomy Edmund Gunter (1581–1626), first used the abbreviated notation ‘sin’.
The origin of the terms ‘cosine’ and ‘tangent’ was much later. The cosine function arose from the need to compute the sine of the complementary angle. Aryabhatta called it kotijya. The name cosinus originated with Edmund Gunter. In 1674, the English Mathematician Sir Jonas Moore first used the abbreviated notation ‘cos’.
The first use of the idea of ‘sine’ in the way we use it today was in the work Aryabhatiyam
by Aryabhatta, in A.D. 500. Aryabhatta used the word ardha-jya for the half-chord, which was shortened to jya or jiva in due course. When the Aryabhatiyamm was translated into Arabic, the word jiva was retained as it is. The word jiva was translated into sinus, which means curve, when the Arabic version was translated into Latin. Soon the word sinus, also used as sine, became common in mathematical texts throughout Europe. An English Professor of astronomy Edmund Gunter (1581–1626), first used the abbreviated notation ‘sin’.
The origin of the terms ‘cosine’ and ‘tangent’ was much later. The cosine function arose from the need to compute the sine of the complementary angle. Aryabhatta called it kotijya. The name cosinus originated with Edmund Gunter. In 1674, the English Mathematician Sir Jonas Moore first used the abbreviated notation ‘cos’.
Monday, May 24, 2010
Theorem of Triangle
Theorem of Triangle:
If in two triangles, corresponding angles are equal, then their
corresponding sides are in the same ratio (or proportion) and hence the two
triangles are similar.
Theorem of Triangle is explained below to understand it better:
This criterion is referred to as the AAA (Angle–Angle–Angle) criterion of similarity of two triangles. This theorem can be proved by taking two triangles ABC and DEF such that
∠ A = ∠ D, ∠ B = ∠ E and ∠ C = ∠ F (see Fig. 6.24)
Cut DP = AB and DQ = AC and join PQ.
So, Δ ABC ≅ Δ DPQ (Why ?)
This gives ∠ B = ∠ P = ∠ E and PQ || EF (How?)
Therefore,
( / notation used to represent divided by) *
DP/PE =DQ/QF
AB/DE =AC/DF
Similarly,
AB/DE=BC/EF and also AB/DE=BC/EF=AC/DF
Remark : If two angles of a triangle are respectively equal to two angles of another triangle, then by the angle sum property of a triangle their third angles will also be equal.
If in two triangles, corresponding angles are equal, then their
corresponding sides are in the same ratio (or proportion) and hence the two
triangles are similar.
Theorem of Triangle is explained below to understand it better:
This criterion is referred to as the AAA (Angle–Angle–Angle) criterion of similarity of two triangles. This theorem can be proved by taking two triangles ABC and DEF such that
∠ A = ∠ D, ∠ B = ∠ E and ∠ C = ∠ F (see Fig. 6.24)
Cut DP = AB and DQ = AC and join PQ.So, Δ ABC ≅ Δ DPQ (Why ?)
This gives ∠ B = ∠ P = ∠ E and PQ || EF (How?)
Therefore,
( / notation used to represent divided by) *
DP/PE =DQ/QF
AB/DE =AC/DF
Similarly,
AB/DE=BC/EF and also AB/DE=BC/EF=AC/DF
Remark : If two angles of a triangle are respectively equal to two angles of another triangle, then by the angle sum property of a triangle their third angles will also be equal.
Subscribe to:
Posts (Atom)