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History of area measurement

What occupied the geometers of five thousand years ago? the answer is Earth measurement. The area problem has been studied since the ancient Babylonian. The methods used to calculate areas range from decomposing the complicated polygons/polyhedron into simpler regions, for example with parallel lines /planes or using triangulations, up to calculus methods.

Babylonian

•  The earliest record from ancient Babylonian deals with the measurement of certain quadrilaterals (see Fig.3):

Fig.3

A(abcd)=1/4(a +c)(b+d)

where a,b,c,d  is the sides of a quadrilateral. Unfortunately, this gives the correct result only in the case of the rectangle.

Egyptian

The visual justification method were employed by Egyptians to calculate the area.

•  The area of an isosceles triangle

Fig.4

The isosceles triangle can be divided by altitude into two right triangles, then join to form a rectangle of height equal to the altitude and base equal to one-half the base of the triangle (see Fig.4).

•  The area of curved bodies

A=1/2(pr2)

The formula will give us the area of a semicircle, if r is the radius, or a hemisphere if x be the diameter. The value p comes from the following formula:

Fig.5

The area of a circle of diameter 9 is calculated as that of a square of side 8 (see Fig.5), so the value p is:

p=(16/9)2

Greek:

• Heron(~75 AD)'s formula for the area of a triangle :

where a,b,c are the sides and s is one-half of the diameter, i.e. s=(a + b + c)/2

Indian:

• An extension of Heron's triangle area formula to quadrilaterals was discovered by  Brahmagupta(~620 AD):

Fig.6

where  a, b, c, d are edges of lengths of   cyclical quadrilaterals (see Fig.5)semiperimeter  s= (a + b + c + d)/2. This formula is an amazing symmetric formula. If one side is zero length, say d = 0, then we have a triangle (which is always cyclic) and this formula reduces to Heron's one.

Generalizations of formula for quadrilaterals

•  A  generalization of Brahmagupta's formula for general quadrilaterals is as follows:

Fig.7

where a and b are two non-adjacent angles (see Fig.7).

• If the quadrilateral is inscribed in one circle and circumscribed in another, then the area is

•  C.A.Bretschneider(1842)gives another formula:

Fig.8

where p and q are the lengths of the diagonals (see Fig.8).

• Pierre Varignon's formula (first published in 1731) is using a parallelogram taken the midpoints of the 4 edges from  the quadrilateral (see Fig.9), it is then easy to show that this midpoint quadrilateral is always a parallelogram, called "Varignon parallelogram", and that its area is exactly one-half the area of the original quadrilateral.

A(V0V1V2V3)=2A(M0M1M2M3)

Fig.9

• There is another formula known as the parallelogram formula or polar formula, we explain it in the part of Preliminary

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