How many 2 BHK flats can be billed in a 2000 sq. feet land?

Answer 1

square feet (ft2), square yards (yd2), square miles (mi2), and so forth. Algebraically, these units can be thought of as the squares of the corresponding length units. The SI unit of area is the square metre, which is considered an SI derived unit. Calculation of the area of a square whose length and width are 1 metre would be: 1 metre × 1 metre = 1 m2and so, a rectangle with different sides (say length of 3 metres and width of 2 metres) would have an area in square units that can be calculated as: 3 metres × 2 metres = 6 m2. This is equivalent to 6 million square millimetres. Other useful conversions are: 1 square kilometre = 1,000,000 square metres 1 square metre = 10,000 square centimetres = 1,000,000 square millimetres 1 square centimetre = 100 square millimetres. Non-metric units In non-metric units, the conversion between two square units is the square of the conversion between the corresponding length units. 1 foot = 12 inches,the relationship between square feet and square inches is 1 square foot = 144 square inches,where 144 = 122 = 12 × 12. Similarly: 1 square yard = 9 square feet 1 square mile = 3,097,600 square yards = 27,878,400 square feetIn addition, conversion factors include: 1 square inch = 6.4516 square centimetres 1 square foot = 0.09290304 square metres 1 square yard = 0.83612736 square metres 1 square mile = 2.589988110336 square kilometres There are several other common units for area. The are was the original unit of area in the metric system, with: 1 are = 100 square metresThough the are has fallen out of use, the hectare is still commonly used to measure land: 1 hectare = 100 ares = 10,000 square metres = 0.01 square kilometresOther uncommon metric units of area include the tetrad, the hectad, and the myriad. The acre is also commonly used to measure land areas, where 1 acre = 4,840 square yards = 43,560 square feet.An acre is approximately 40% of a hectare. On the atomic scale, area is measured in units of barns, such that: 1 barn = 10−28 square meters.The barn is commonly used in describing the cross-sectional area of interaction in nuclear physics.In India, 20 dhurki = 1 dhur 20 dhur = 1 khatha 20 khata = 1 bigha 32 khata = 1 acre In the 5th century BCE, Hippocrates of Chios was the first to show that the area of a disk (the region enclosed by a circle) is proportional to the square of its diameter, as part of his quadrature of the lune of Hippocrates, but did not identify the constant of proportionality. Eudoxus of Cnidus, also in the 5th century BCE, also found that the area of a disk is proportional to its radius squared.Subsequently, Book I of Euclid's Elements dealt with equality of areas between two-dimensional figures. The mathematician Archimedes used the tools of Euclidean geometry to show that the area inside a circle is equal to that of a right triangle whose base has the length of the circle's circumference and whose height equals the circle's radius, in his book Measurement of a Circle. (The circumference is 2πr, and the area of a triangle is half the base times the height, yielding the area πr2 for the disk.) Archimedes approximated the value of π (and hence the area of a unit-radius circle) with his doubling method, in which he inscribed a regular triangle in a circle and noted its area, then doubled the number of sides to give a regular hexagon, then repeatedly doubled the number of sides as the polygon's area got closer and closer to that of the circle (and did the same with circumscribed polygons). Swiss scientist Johann Heinrich Lambert in 1761 proved that π, the ratio of a circle's area to its squared radius, is irrational, meaning it is not equal to the quotient of any two whole numbers. In 1794 French mathematician Adrien-Marie Legendre proved that π2 is irrational; this also proves that π is irrational. In 1882, German mathematician Ferdinand von Lindemann proved that π is transcendental (not the solution of any polynomial equation with rational coefficients), confirming a conjecture made by both Legendre and Euler. Heron (or Hero) of Alexandria found what is known as Heron's formula for the area of a triangle in terms of its sides, and a proof can be found in his book, Metrica, written around 60 CE. It has been suggested that Archimedes knew the formula over two centuries earlier, and since Metrica is a collection of the mathematical knowledge available in the ancient world, it is possible that the formula predates the reference given in that work.In 499 Aryabhata, a great mathematician-astronomer from the classical age of Indian mathematics and Indian astronomy, expressed the area of a triangle as one-half the base times the height in the Aryabhatiya (section 2.6). A formula equivalent to Heron's was discovered by the Chinese independently of the Greeks. It was published in 1247 in Shushu Jiuzhang ("Mathematical Treatise in Nine Sections"), written by Qin Jiushao. In the 7th century CE, Brahmagupta developed a formula, now known as Brahmagupta's formula, for the area of a cyclic quadrilateral (a quadrilateral inscribed in a circle) in terms of its sides. In 1842 the German mathematicians Carl Anton Bretschneider and Karl Georg Christian von Staudt independently found a formula, known as Bretschneider's formula, for the area of any quadrilateral. The development of Cartesian coordinates by René Descartes in the 17th century allowed the development of the surveyor's formula for the area of any polygon with known vertex locations by Gauss in the 19th century. The development of integral calculus in the late 17th century provided tools that could subsequently be used for computing more complicated areas, such as the area of an ellipse and the surface areas of various curved three-dimensional objects. For a non-self-intersecting (simple) polygon, the Cartesian coordinates ( x i , y i ) {\displaystyle (x_{i},y_{i})} (i=0, 1, ..., n-1) of whose n vertices are known, the area is given by the surveyor's formula: A = 1 2 | ∑ i = 0 n − 1 ( x i y i + 1 − x i + 1 y i ) | {\displaystyle A={\frac {1}{2}}{\Biggl \vert }\sum {i=0}^{n-1}(x{i}y_{i+1}-x_{i+1}y_{i}){\Biggr \vert }} where when i=n-1, then i+1 is expressed as modulus n and so refers to 0. Rectangles The most basic area formula is the formula for the area of a rectangle. Given a rectangle with length l and width w, the formula for the area is: A = lw (rectangle).That is, the area of the rectangle is the length multiplied by the width. As a special case, as l = w in the case of a square, the area of a square with side length s is given by the formula: A = s2 (square).The formula for the area of a rectangle follows directly from the basic properties of area, and is sometimes taken as a definition or axiom. On the other hand, if geometry is developed before arithmetic, this formula can be used to define multiplication of real numbers. Dissection, parallelograms, and triangles Most other simple formulas for area follow from the method of dissection. This involves cutting a shape into pieces, whose areas must sum to the area of the original shape. For an example, any parallelogram can be subdivided into a trapezoid and a right triangle, as shown in figure to the left. If the triangle is moved to the other side of the trapezoid, then the resulting figure is a rectangle. It follows that the area of the parallelogram is the same as the area of the rectangle: A = bh (parallelogram). However, the same parallelogram can also be cut along a diagonal into two congruent triangles, as shown in the figure to the right. It follows that the area of each triangle is half the area of the parallelogram: A = 1 2 b h {\displaystyle A={\frac {1}{2}}bh} (triangle).Similar arguments can be used to find area formulas for the trapezoid as well as more complicated polygons. Circles The formula for the area of a circle (more properly called the area enclosed by a circle or the area of a disk) is based on a similar method. Given a circle of radius r, it is possible to partition the circle into sectors, as shown in the figure to the right. Each sector is approximately triangular in shape, and the sectors can be rearranged to form an approximate parallelogram. The height of this parallelogram is r, and the width is half the circumference of the circle, or πr. Thus, the total area of the circle is πr2: A = πr2 (circle).Though the dissection used in this formula is only approximate, the error becomes smaller and smaller as the circle is partitioned into more and more sectors. The limit of the areas of the approximate parallelograms is exactly πr2, which is the area of the circle.This argument is actually a simple application of the ideas of calculus. In ancient times, the method of exhaustion was used in a similar way to find the area of the circle, and this method is now recognized as a precursor to integral calculus. Using modern methods, the area of a circle can be computed using a definite integral: A = 2 ∫ − r r r 2 − x 2 d x = π r 2 . {\displaystyle A;=;2\int {-r}^{r}{\sqrt {r^{2}-x^{2}}},dx;=;\pi r^{2}.} Ellipses The formula for the area enclosed by an ellipse is related to the formula of a circle; for an ellipse with semi-major and semi-minor axes x and y the formula is: A = π x y . {\displaystyle A=\pi xy.} Surface area Most basic formulas for surface area can be obtained by cutting surfaces and flattening them out. For example, if the side surface of a cylinder (or any prism) is cut lengthwise, the surface can be flattened out into a rectangle. Similarly, if a cut is made along the side of a cone, the side surface can be flattened out into a sector of a circle, and the resulting area computed. The formula for the surface area of a sphere is more difficult to derive: because a sphere has nonzero Gaussian curvature, it cannot be flattened out. The formula for the surface area of a sphere was first obtained by Archimedes in his work On the Sphere and Cylinder. The formula is: A = 4πr2 (sphere),where r is the radius of the sphere. As with the formula for the area of a circle, any derivation of this formula inherently uses methods similar to calculus. Areas of 2-dimensional figures A triangle: 1 2 B h {\displaystyle {\tfrac {1}{2}}Bh} (where B is any side, and h is the distance from the line on which B lies to the other vertex of the triangle). This formula can be used if the height h is known. If the lengths of the three sides are known then Heron's formula can be used: s ( s − a ) ( s − b ) ( s − c ) {\displaystyle {\sqrt {s(s-a)(s-b)(s-c)}}} where a, b, c are the sides of the triangle, and s = 1 2 ( a + b + c ) {\displaystyle s={\tfrac {1}{2}}(a+b+c)} is half of its perimeter. If an angle and its two included sides are given, the area is 1 2 a b sin ⁡ ( C ) {\displaystyle {\tfrac {1}{2}}ab\sin(C)} where C is the given angle and a and b are its included sides. If the triangle is graphed on a coordinate plane, a matrix can be used and is simplified to the absolute value of 1 2 ( x 1 y 2 + x 2 y 3 + x 3 y 1 − x 2 y 1 − x 3 y 2 − x 1 y 3 ) {\displaystyle {\tfrac {1}{2}}(x{1}y_{2}+x_{2}y_{3}+x_{3}y_{1}-x_{2}y_{1}-x_{3}y_{2}-x_{1}y_{3})} . This formula is also known as the shoelace formula and is an easy way to solve for the area of a coordinate triangle by substituting the 3 points (x1,y1), (x2,y2), and (x3,y3). The shoelace formula can also be used to find the areas of other polygons when their vertices are known. Another approach for a coordinate triangle is to use calculus to find the area. A simple polygon constructed on a grid of equal-distanced points (i.e., points with integer coordinates) such that all the polygon's vertices are grid points: i + b 2 − 1 {\displaystyle i+{\frac {b}{2}}-1} , where i is the number of grid points inside the polygon and b is the number of boundary points. This result is known as Pick's theorem. Area in calculus The area between a positive-valued curve and the horizontal axis, measured between two values a and b (b is defined as the larger of the two values) on the horizontal axis, is given by the integral from a to b of the function that represents the curve: A = ∫ a b f ( x ) d x . {\displaystyle A=\int _{a}^{b}f(x),dx.} The area between the graphs of two functions is equal to the integral of one function, f(x), minus the integral of the other function, g(x): A = ∫ a b ( f ( x ) − g ( x ) ) d x , {\displaystyle A=\int _{a}^{b}(f(x)-g(x)),dx,} where f ( x ) {\displaystyle f(x)} is the curve with the greater y-value.An area bounded by a function r = r ( θ ) {\displaystyle r=r(\theta )} expressed in polar coordinates is: A = 1 2 ∫ r 2 d θ . {