Area Ratio Calculator

Introduction to area ratio calculator

Ratio of areas of two surfaces is called the area ratio. For example, a rectangle is constructed inside a circle such that the circle lies on the four vertices of the rectangle. The ratio of the area of the circle to the area of the rectangle is known as the area ratio.

This is especially useful in calculating the Floor Area Ratio (FAR). The floor area ratio is the ratio of the sum of the area of all the floors in a building to the area of the plot on which the building is constructed. The floor area ratio is used to limit the number of floors that a building can have.

Calculating Floor Area Ratio (far)

For example, a building is built on a plot of area 12000 square feet, and there are two floors in it. The area of each floor is 10000 square feet. Then the floor area ratio is given by

Floor area ratio = Sum of areas of all floors/area of plot

`= (2 * 10000)/12000`

`= 20000/12000`

`= 1.66`

The floor area ratio is known by different names in different countries. In Australia, it is known as “Floor Space Ratio (FSP)” and “Plot ratio”; in India is it known as “Floor Space Index (FSI)”; in UK, it is known as “Plot ratio” or “Site ratio”.

The floor area ratio is used widely in zoning and city planning. It is used to limit the number of floors. An architect has to adhere by a particular floor area ratio for a particular site and construct a building in accordance to it.

General real estate appraiser at Analytic Real Estate Services, Edward Ptacek noted in 2009 that the maximum allowable floor area ratio of a particular place affects the value of the land. The greater the floor area ratio, the more is the value of a particular land.

The concept of floor area ratio has been widely used in Japan since the 1970's to build the residential colonies. However, it is arguable whether this has proved beneficial or not. Some say it is beneficial, and some say that it is not so. I am planning to write more post on How to do Math Problems, factoring polynomials word problems. Keep checking my blog.

Use of Floor Area Ratio in Zoning

Zoning is the method of dividing land into various parts each part intended for a particular usage like residential, commercial, industrial, recreational. The local governments of each city make a zoning map which shows the exact division of land use in that city.

The main factors that affect the zoning of a city is the population, land-usage, existing settlements, etc. By using the concept of floor area ratio in zoning, a city can be efficiently zoned in accordance with the above mentioned factors.

Different values for the floor area ratio are predetermined for different parts of land in the city. This helps to limit the amount of construction on a particular piece of land. For example, the floor area ratio of a region in the vicinity of an airport has to be less so as to inhibit the height of buildings in that area.

An architect can make buildings of different dimensions to adhere by the limit of the floor area ratio. For example, a building built on a large plot having only two floors and a building built on a smaller plot having three or four floors will have the same floor area ratio. Thus, an architect can choose the buildings' dimensions accordingly.

Possible Rational Zeros

Introduction to possible rational zeros:

All polynomials in one variable of degree n, n > 0, has at least one real or complex zero. This is the Fundamental Theorem of Algebra.To find all the possible rational zeros of a polynomial we use the Rational Root Test. Understanding Properties of Rational Exponents is always challenging for me but thanks to all math help websites to help me out.

Problems on Rational Root Test:
If a polynomial function has integer coefficients, then all the rational zeros  will be of the form p/q where p is a factor of the constant and q is a factor of the leading coefficient.

Lets look at a few examples.

Ex 1:  Find all the possible rational zeros of 2`x^(2)` + 6x +4

Here p = 4 and q = 2

The factors of 4 are 4,2 and 1

similarly the factors of 2 are 2 and 1

So all the possible rational zeros will be of the form : `( +-4, +-2, +-1)/( +-2, +-1)`

Now let us list out all the possible rational zeros: 4, 2, -4, -2, 1, -1,  `(1)/(2)` , `(-1)/(2)` .

Ex 2:  Find all the possible rational zeros of 2`x^(4)` - 7`x^(3)` +3x + 6.

Here p = 6. The factors of 6 are 3, 2, 1

q = 2. The factors of 2 are 2,1

So all the possible rational zeros will be of the form :`( +-3, +-2, +-1)/( +-2, +-1)`

Therefore the possible rational zeros are :


` ( 3)/(2)` ,  `(-3)/(2)` , 3, -3, 2, -2, 1, -1, `(1)/(2)` ,  `(-1)/(2)`


Is this topic how to calculate p value hard for you? Watch out for my coming posts.

Things that we should be Careful while Applying the Rational Root Test:

Always make sure the polynomial has integer coefficients. We have to multiply to get rid of fractions or decimals and later divide by the same number.
This method gives us only the possible rational zeros. All of these might not be rational zeros of the polynomial and also there might be non-rational / complex zeros apart from these.
The numbers that we obtain from the Rational Root Test are possible zeros, it is not necessary that all of them are zeros of the polynomial.

Solving Equations Using the Distributive Property

Introduction for distributive property:

An operation is distributive if the result of applying it to a sum of terms equals the sum of the results of applying it to the terms individually.
a ( b + c ) = ( a x b ) + ( a x c ).
Here, ‘a’ is multiplied with the sum of two terms ‘b and c’ in the left hand side which, gives the same answer when ‘a’ is multiplied individually with ‘b’ and ‘c’ and then added. Solving equations using the distributive property is very easy. Here we are going to see solving equations using the distributive property.

Distributive Property:

The types of distributive property used in solving equations are

a (b + c) = (a x b) + (a x c).

a (b - c) = (a x b) - (a x c).

These two are used for solving the equations.

Example Problems for Solving Equations Using the Distributive Property:

Example 1 for solving equations using the distributive property.

Solve the equation 10 (2 + 3) = (10 x 2) + (10 x 3).

Left hand side:

First take 10 (2 + 3)
By adding 2 + 3 we get 5
And now it is 10 (5)
So the answer for the Left hand side is 50

Right hand side:

Now we take (10 x 2) + (10 x 3)
By multiplying 10 with 2 and 10 with 3 we get (20) + (30)
And now we add 20 with 30
So the answer for Right hand side is 30
Hence proved Left hand side is equal to right hand side. Is this topic irrational number definition hard for you? Watch out for my coming posts.

Example 2 for solving equations using the distributive property.

Solve the equation 10(4 – 6) = (10 x 4) - (10 x 6).

Left hand side:

First take 10 (4 - 6)
By adding 4 - 6 we get -2
And now it is 10 (-2)
So the answer for the Left hand side is -20

Right hand side:

Now we take (10 x 4) - (10 x 6)
By multiplying 10 with 4 and 10 with 6 we get (40) - (60)
And now we subtract 40 with 60
So the answer for Right hand side is -20
Hence proved Left hand side is equal to right hand side.


Example 3 for solving equations using the distributive property.

Solve the equation 4(5 – 3) = (4 x 5) - (4 x 3).

Left hand side:

First take the 4(5 – 3)
By subtracting 5 – 3 we get 2
And now it is 4 (2)
So the answer for the Left hand side is 8

Right hand side:

Now we take (4 x 5) - (4 x 3)
By multiplying 4 with 5 and 4 with 3 we get (20) - (12)
And now we subtract 20 with 12
So, the answer for Right hand side is 8
Therefore, Left hand side is equal to right hand side.
Hence proved

Straight Line Products

Introduction to Straight Line Products:
In geometry, line is one dimensional shape that has no end points.Line is a fundamental concept in the mathematics. Line is representing the straight line object. Geometrical object that is straight, infinitely long is called line. Line is the one dimensional figure. Lines have the no height and  then no width. In this article we see detailed about the straight line products. In this article we are discuss the straight line products.

Example Problems for Straight Line Products:

Straight line products – Example: 1

Find the equation of the straight line passing through the points (1,1),(2,3)

Solution:

The equation of the straight line passing through two points is

`y-y_1=m(x-x_1)` where m is a slope.

`y-1=\frac{3-1}{2-1}(x-1)`

`y-1=2x-2`

`y-2x+1=0`

The equation of the straight line is `y-2x+1=0.`

Straight line products – Example: 2

Find the equation of the straight line passing through (-4,5) and cutting off equal intercepts on the coordinate axes.

Solution:

Let the intercept be a

Now the equation of the line is

`\frac{x}{a}+\frac{y}{a}=1`

`x+y=a`

Since the point (-4,5) passes throough this line,

we get

`a=1`

Hence the equation of the line is

`x+y=1`

Straight line products – Example: 3

Find the equation of the straight line passing through (3,-4)  and making intercepts in the ratio 2:3.

Solution:

Let the intercepts be 2k,3k

Then the equation of the line is

`\frac{x}{2k}+\frac{y}{3k}=1`

`3xk+2yk=6`

Since the line passes through (3,-4)

`9k-8k=6`

Hence the equation is

`\frac{x}{12}+\frac{y}{18}=1`

`3x+2y=36`


Practice Problems for Straight Line Products:

1. Find the equation of the straight line passing through (4,-3)  and perpendicular to the line joining (1,1),(2,3).

`Answer: 2y-x+2=0`

2. Show that the equation of the straight line passing through `(x_1,y_1)` and making an angle of \theta  with the X-axis in positive direction is `\frac{x-x_1}{\cos\theta}=\frac{y-y_1}{\sin \theta}.`

3. Find the equation of the straight line which makes `135^\circ`   with X-axis and passing through (3,-2)

`Answer: y+x-1=0`

Perpendicular Bisector Compass

Introduction to  perpendicular bisector compass:-

We can draw perpendicular to a line segment at any point on it. However, when we draw a perpendicular on the line segment such that it divides the line segment in two equal halves, it is called perpendicular bisector of the line segment. Any point on the perpendicular bisector of a line segment is always at equal distance from the two ends of the line segment.

As can be seen from the given figure that dotted line is the perpendicular bisector of the line segment AB. If you take any point on the dotted line and measure its distances from A and B, you will find that the distances are same.

How to Draw Perpendicular Bisector Using a Compass

We can draw perpendicular bisector of a line segment using a compass. Use a properly sharpened pencil in the compass and adjust it in the compass properly. Suppose we have to draw perpendicular bisector of line segment of length 6 cm.

Draw line segment AB of length 6 cm. Taking A as center and radius equal to more than half of AB, draw arcs on both the sides of AB. Again taking B as center and radius equal to more than half of AB, draw arcs on both the sides of AB, intersecting the previous arcs at P and Q as shown in the figure given below. Join PQ. PQ is the required perpendicular bisector of the line segment AB.

If you measure two parts of the line segment AB, you will find that each part is 3 cm. Having problem with free 6th grade math problems keep reading my upcoming posts, i will try to help you.

Examples of Perpendicular Bisector Compass

Q 1 :       Why compass is used while draw perpendicular bisector of a line segment? Please pick a correct answer.:

A.            to draw the line segment AB

B.            to draw arcs

C.            to measure length of line segment

Sol :

We draw arcs using a compass while drawing perpendicular bisector of a line segment.

Q 2:    Perpendicular bisector of line segment AB of length 12 cm intersects it at O. What is the length of AO?

Sol :

We know that perpendicular bisector divides the line segment into two equal parts. Therefore, length of AO = 6 cm.

Factoring Trinomial Squares

Introduction to factoring trinomial squares:

In math, factoring trinomials square method is one significant in basic algebra. It is definite because the operation of sum of three monomials. Given term consists of three expressions that are in the standard form of ax^2 + bx + c. Factoring trinomial is the operation to make arithmetic operation of multiplication in a reverse way. Let us resolve some instance problems for factoring trinomials are x^2 +6x – 5.

Factoring Trinomial Squares:

A trinomial is a polynomial representing of three terms. A trinomial is an equation concerning three expressions. A trinomial is the extension of a power of a computation of three expressions into monomials. The expansion is specified by

`(a+b+c)^(n) = sum_(i,j,k) ((n),(i,j,kk))a^ib^jc^k`     where n is a non negative integer also the computation is in use over all grouping of nonnegative index i, j, and k such to i+j+k = n. The trinomial coefficients are known with  `((n),(i,jk)) = (n!)/(i!j!k!)`

This method is a particular case of the multi nominal method for m = 3. The number of expressions of an extended trinomial is     `((n+2)(n+1))/(2)` Where n is the exponent.

Examples for Factoring Trinomial Squares:

Example 1:

how to solve factoring trinomial square x^2+8x+15

Solution:

Step 1: the given trinomial is x^2+8x+15

Step 2: to factoring this equation is (x+3)(x+5)

so the solution is (x+3)(x+5)

Example 2:

how to solve factoring trinomial square (x+6)2

Solution:

Step 1: the given equation is (x+6)2

Step 2: (a+b)2 = (a2+2ab+b2)

Step 3: using this formula in the given equation

(x+6)2  = (x^2+12x+36)

so the solution is (x^2+12x+36)

Example 3:

how to solve factoring trinomial square (x-2)2

Solution:

Step 1: the given equation is (x-2)2

Step 2: (a-b)2 = (a2-2ab+b2)

Step 3: using this formula in the given equation

(x-2)2  = (x^2-4x+4)

so the solution is (x^2-4x+4)

Example 4:

how to solve factoring trinomial square 9-16x^2

Solution:

Step 1: the given equation is 9-16x^2

Step 2: a2-b2= (a+b)(a-b)

Step 3: using this formula in the given equation is

32-4x^2 = (3+4x) (3-4x)

so the solution of the given equation is   (3+4x) (3-4x)