Line Segment Circle Intersection

Introduction:
A line can intersect a circle at two points or it can touch the circle at one point or never pass throught.

To find the point of intersection of line segment with a circle, we need to solve both equations. Thus we can get two points of intersection of the line with the circle.

Let us see few problems of this kind.

Example Problem on Line Segment Circle Intersection.

Ex 1: Find the point of intersection of the line 2x + y = 1 and x 2 + y 2 = 1.

Soln: Given: The line is 2x + y = 1 ……….. (1)

The circle is x 2 + y 2 = 1…………… (2)

(1) `=>` y = 1 – 2x

Therefore (2) = x 2 + (1 – 2x) 2 = 11 `=>` x 2 + 1 + 4x 2 – 4x = 11

`=>` 5x 2 – 4x – 10 = 0

Therefore x = `[4+- sqrt [(-4) ^2 ** 4(5) (-10)]] / [2 (5)]`

=` [4 +- sqrt [16 + 200]] / 10`

= `[4 +- sqrt 216] / 10`

=` [4 +- 6 sqrt 6] / 10` 

= `[2 +- 3 sqrt 6] / 5`

Therefore y = `1 ** [2 [(2 + 3 sqrt 6) / 5]]`

= `[5 ** 4 ** 6 sqrt 6] / 5`

= `[1 ** 6 sqrt 6] / 5`

When x = `[2 + 3 sqrt 6] / 5` , y = `[1 ** 6 sqrt 6] / 5`

When x = `[2 ** 3 sqrt 6] / 5` , y = `1 ** [2 [2 ** 3 sqrt 6] ]/ 5`

y =`[ 5** 4 + 6 sqrt 6 ]/ 5` = `[1 + 6 sqrt 6 ]/ 5`

Therefore the point of intersections are given by (`[2 + 3 sqrt 6] / 5` , `[1 ** 6 sqrt 6] / 5` ) and (`[2 ** 3 sqrt 6] / 5` , `[1 + 6 sqrt 6] / 5` )


Example Problem on Line Segment Circle Intersection.

Ex 2: Find the point of intersection of the following line and the circle. x – y = 1, x 2 + y 2 + 4x + 2y + 2 = 0.

Soln: Given: x– y = 1   =  x = y + 1 ……….. (1)

x2 + y 2 + 4x + 2y + 2 = 0 ………... (2)

By using (1) in (2), we get (y + 1) 2 + y 2 + 4 (y + 1) + 2y + 2 = 0

`=>` y 2 + 2y + 1 + y 2 + 4y + 4 + 2y + 2 = 0

2y 2 + 8y + 7 = 0

y =`[ -8 +- sqrt[ 8 ^2 ** 4 (2) (7)]] / [2 (2)]`

= `[-8 +- sqrt 64 ** 56] / 4`

=` [-8 +- 2 sqrt 2] / 4`   =`[-4 +- sqrt2]/2`

Therefore y = `[- 4 + sqrt 2 ]/ 2` , (1)  `=>` x =` [-4 + sqrt 2 ]/ 2` + 1

= `[- 2 + sqrt 2] / 2`

y =` [-4 ** sqrt 2] / 2` , (1) `=>` x =` [-4 ** sqrt 2] / 2` + 1 = `[-2 ** sqrt 2 ]/ 2`

Therefore the points are (`[-2 + sqrt 2] / 2` , `[-4 ** sqrt 2] / 2` )

Geometric Solids Patterns

Introduction to geometric solids patterns:

Geometry is a topic which deals with the shapes and sizes of any objects. The shapes are classified as solid and plane surface shapes. The geometric solids are the shapes that can be classified as cubes, cone, sphere, rectangular prism, hemisphere, etc. The geometric solid patterns can be explained with their properties. Here we see some of the solids pattern in geometry.

Classifying Solids Patterns:

Solids patterns can be classified into many shapes. We see some of them with properties.

Sphere:

It is a 3-dimensional solid shape pattern which does not have a base point. It is rounded in shape, as it is spherical. The parameters used in the solid sphere are calculated using the following formulas,

•    Volume of a sphere = 4/3 ?r3

•    The surface area of a sphere = 4?r2

It is rounded in shape, as well as spherical.

Cube:

The cube is also one of the three-dimensional solid shapes. It is made up of six equal sides with 12 edges and 8 vertices. Some parameters like volume of cube and surface area are calculated using the given formulas.

Volume of a cube can be found as a3

•    Area of a cube is given as 6a2

Rectangular prism:

The rectangular prism is a 3-dimensional solid shape which is having six numbers of sides.

The formulas used to find the parameters of a rectangular prism are given below:

•    The surface area of rectangular prism = 2(lb + bh + hl)

•    Volume of the rectangular prism can be given as lbh

Where,

l is the length, b is the breadth and h is the height of the rectangular prism.

These are some basic geometric solid patterns.

Problems to Geometric Solids Patterns:

We can solve some example problems for geometric solids patterns.

Example1:

Find the volume of the rectangular prism if its length, breadth and height are given as 6 cm, 4 cm and 2 cm.

Solution:

Formula to find volume of rectangular prism is lbh.

On substituting the given values in formula, we get

lbh = `6 xx 4 xx 2`

= `24 xx 2`

= 48

Hence the volume of the rectangular prism is found to be 48 cubic cm.

Example 2:

Find the volume of a cube if the side is measured as 3 cm.

Solution:

Given that,

Side length = 3 cm

We know the formula to find volume of cube as,

Volume of the cube is a3 = 27

Hence the volume of the cube is given as 27 cm3.

Thus these are some examples to geometric solids patterns

The Measure of Acute Triangles

Introduction to the measure of acute triangles :

Definition to the measure of acute triangles:A triangle which has all the three angles is less than 90°. This can also be tell in the words such as the angles which are smaller than the right angle triangles can be called as acute angles.The method of finding acute angles triangles which can be done by measure of two angles which are given and if measure of a side and any two angles are given.

Prolems for the Measure of Acute Triangles:

Ex1:The measurement of one of the acute angle of a triangle is 52°. Find the measure of other acute angles of the triangle?

Solution: The following steps to be taken for the measure of angles are

Step1: The addition of the measures of the two acute angles must be 90°.

Step2: If one acute angle of a right triangle is 52°, then the measure of the other acute angle is 90° - 52° =38 °

Ex 2:The measurement of an acute angle of a triangle which is given as 30° 53'. Find the other measure of  acute angles of the right angled triangle?

Solution:The addition of the measure of the two acute angles must be 90°.

If one acute angle of a right triangle is 30° 53', then the measure of the other acute angle is givan as,

90° - 30° 53' = 89° 60' - 30° 53' = 59° 07'.

Ex 3: Find the measure of the angle X° from the given diagram. The other two angles which are 50° and 60°.Find the third angle of an acute angle triangle?

Solution:The addition of three angles in the triangles should be equal to 180°

The sum of the measurement of the three angles is

50° + 60° + X° = 180°.

The third angle which can be measure by the following step

X° + 110° = 180°

The X° which can be performed as follows,

X° = 180° - 110°

X° = 70°.


Practice Problem for Measure of Acute Triangles:

Find the measure of the angle X° of an acute angle triangles with the given angles  40°and 80°?

Solution:  The measure of the third acute angle X° = 60°.