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In the following equations, denotes the sagitta (the depth or height of the arc), equals the radius of the circle, and the length of the chord spanning the base of the arc. As 1 2 l {\displaystyle {\tfrac {1}{2}}l} and r − s {\displaystyle r-s} are two sides of a right triangle with r {\displaystyle r} as the hypotenuse , the Pythagorean ...
A right triangle ABC with its right angle at C, hypotenuse c, and legs a and b,. A right triangle or right-angled triangle, sometimes called an orthogonal triangle or rectangular triangle, is a triangle in which two sides are perpendicular, forming a right angle (1 ⁄ 4 turn or 90 degrees).
If is the radius of the incircle of the triangle, then the triangle can be broken into three triangles of equal altitude and bases , , and . Their combined area is A = 1 2 a r + 1 2 b r + 1 2 c r = r s , {\displaystyle A={\tfrac {1}{2}}ar+{\tfrac {1}{2}}br+{\tfrac {1}{2}}cr=rs,} where s = 1 2 ( a + b + c ...
Triangles have many types based on the length of the sides and the angles. A triangle whose sides are all the same length is an equilateral triangle, [3] a triangle with two sides having the same length is an isosceles triangle, [4] [a] and a triangle with three different-length sides is a scalene triangle. [7]
Usually, chord length and height are given or measured, and sometimes the arc length as part of the perimeter, and the unknowns are area and sometimes arc length. These can't be calculated simply from chord length and height, so two intermediate quantities, the radius and central angle are usually calculated first.
The area of a triangle can be demonstrated, for example by means of the congruence of triangles, as half of the area of a parallelogram that has the same base length and height. A graphic derivation of the formula T = h 2 b {\displaystyle T={\frac {h}{2}}b} that avoids the usual procedure of doubling the area of the triangle and then halving it.
The radius of the inscribed circle of an isosceles triangle with side length , base , and height is: [17] 2 a b − b 2 4 h . {\displaystyle {\frac {2ab-b^{2}}{4h}}.} The center of the circle lies on the symmetry axis of the triangle, this distance above the base.
Draw a perpendicular from the center to the midpoint of a side of the polygon; its length, h, is less than the circle radius. Also, let each side of the polygon have length s; then the sum of the sides is ns, which is less than the circle circumference. The polygon area consists of n equal triangles with height h and base s, thus equals nhs/2.