Areas of plane figures
trapezoid, quadrangle, right-angled triangle, isosceles triangle,
equilateral triangle, arbitrary triangle, polygon, regular hexagon,
circle, sector, segment of a circle. Heron’s formula.
Any triangle. a, b, c – sides; a – a base; h – a height; A, B, C – angles,
opposite to sides a, b, c ; p = ( a +
b + c ) / 2.
The last expression is known as Heron’s formula.
A polygon, area of which we want to determine, can be divided into some triangles by its diagonals. A polygon, circumscribed around a
circle ( Fig. 67 ), can be divided by lines, going from a center of a circle to its vertices. Then we receive:
Particularly, this formula is valid for any regular polygon.
A regular hexagon. a – a side.
A circle. D – a diameter; r – a radius.
A sector ( Fig.68 ). r – a radius; n – a degree measure of a central angle;
l – a length of an arc.
A segment ( Fig.68 ). An area of a segment is found as a difference between areas of a sector AmBO and a triangle
AOB. Besides, the approximate formula for an area of a segment is:
where a = AB ( Fig.68 ) – a base of segment; h – its height ( h = r – OD ). A relative error of this formula is
equal: at AmB = 60 deg – about 1.5% ; at AmB = 30 deg ~0.3%.
E x a m p l e . Calculate areas of the sector AmBO ( Fig.68 ) and the segment
at the following data: r = 10 cm, n = 60 deg.
S o l u t i o n . A sector area:
An area of the regular triangle AOB:
Hence, an area of a segment:
Note, that in a regular triangle AOB: AB = AO = BO
AD = BD = r / 2 , and therefore a height OD
Pythagorean theorem is equal to:
Then, according to the approximate formula we’ll receive: