Search results
Results from the WOW.Com Content Network
one pair of parallel sides – a trapezium (τραπέζιον), divided into isosceles (equal legs) and scalene (unequal) trapezia; no parallel sides – trapezoid (τραπεζοειδή, trapezoeidé, literally 'trapezium-like' (εἶδος means 'resembles'), in the same way as cuboid means 'cube-like' and rhomboid means 'rhombus-like')
Any non-self-crossing quadrilateral with exactly one axis of symmetry must be either an isosceles trapezoid or a kite. [5] However, if crossings are allowed, the set of symmetric quadrilaterals must be expanded to include also the crossed isosceles trapezoids, crossed quadrilaterals in which the crossed sides are of equal length and the other sides are parallel, and the antiparallelograms ...
Proof of the theorem. We need to prove that AF = FD.We will prove that both AF and FD are in fact equal to FM.. To prove that AF = FM, first note that the angles FAM and CBM are equal, because they are inscribed angles that intercept the same arc of the circle (CD).
In Euclidean geometry, Brahmagupta's formula, named after the 7th century Indian mathematician, is used to find the area of any convex cyclic quadrilateral (one that can be inscribed in a circle) given the lengths of the sides.
For more, see Trapezoid § Trapezium vs Trapezoid.) Trapezium (UK) or trapezoid (US): at least one pair of opposite sides are parallel. Trapezia (UK) and trapezoids (US) include parallelograms. Isosceles trapezium (UK) or isosceles trapezoid (US): one pair of opposite sides are parallel and the base angles are equal in measure. Alternative ...
In calculus, the trapezoidal rule (also known as the trapezoid rule or trapezium rule) [a] is a technique for numerical integration, i.e., approximating the definite integral: (). The trapezoidal rule works by approximating the region under the graph of the function f ( x ) {\displaystyle f(x)} as a trapezoid and calculating its area.
There are many ways to prove Heron's formula, for example using trigonometry as below, or the incenter and one excircle of the triangle, [7] or as a special case of De Gua's theorem (for the particular case of acute triangles), [8] or as a special case of Brahmagupta's formula (for the case of a degenerate cyclic quadrilateral).
To prove that the diagonals of a parallelogram bisect each other, we will use congruent triangles: ∠ A B E ≅ ∠ C D E {\displaystyle \angle ABE\cong \angle CDE} (alternate interior angles are equal in measure)