# Formula For The Length Of An Arc Of A Circle Verifying Results in Geometry Using Calculus

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## Verifying Results in Geometry Using Calculus

The Formulas used in Geometry are as follows:

Area of a triangle = 0.5 * base * height

Circumference of a circle = 2 * pi * r

Area of a circle = pi * r * r

Let us verify each of these using calculus, first let us verify the area of a triangle.

Let us take a scalene triangle ABC let us draw a straight line from one of the vertexes to the opposite side, so that it meets the opposite side at right angles to the base. In essence we have divided the original triangle into two right angled triangles.

Let us derive an expression for the area of a triangle using the equation of the straight line as Y = m * X + C.; First, let us define the area of the triangle as the sum of two areas enclosed by the two sides of the triangle, the altitude and the base. The altitude cuts the base into two pieces.

Let us take a scalene triangle whose altitude is 5 and base is 10. Let the point of intersection of the altitude with the base be the origin or (0,0). If the three vertexes are A,B and C respectively, then if the point of intersection of the altitude from point A and the line BC is at D, then the co-ordinates of the point A are (0.5), point B is (4,0) and point C is (-6,0). The length of the three sides are BC = 10; AC = sqrt(61) approximately equal to 7.8 and AB = sqrt(41) approximately equal to 6.4.

These three points do form a triangle as AB + BC > AC; AC + BC > AB; AC + AB > BC. Also as the three sides of the triangle are not equal the triangle is scalene.

The area of this triangle is 0.5 * base *height = 0.5 * 10 * 5 = 25 sq units.

Let us verify this using integral calculus.

The slope of the side AB is -tan(ABD) or -tan(ABC) which is equal to -1.25. The equation of the line AB is Y = 5 – 1.25X. Similarly the equation of the line AC is Y= 5X/6 + 5. Let us integrate both these expressions between appropriate limits.

Area enclosed by the line AB is integral(Ydx) between x = 0 and x = 4.

Integral(Ydx) is nothing but 5X – 1.25 * X * X /2. Between limits the area is 20 – 10 = 10.

Similarly area enclosed by the line AC and the origin is Integral(5x/6 + 5)dx which is equal to 5*X*X/12 + 5*X. The limits of the integration are 0 to -6 which evaluates to 15 -30 or -15. Taking the absolute value the sum of both the areas is 15 + 10 which is equal to 25. This readily tallies with the analytical expression for area of the triangle which evaluates to 25 sq units.

Let us derive the expression for circumference of a circle. If one considers a small sector of area d(theta). The area of the sector would be r * sin(d(theta)). This can be approximated to r* d(theta) for small values of d(theta). If one integrates this expression between 0 and 2 * pi, this works out to be 2 * pi * r. This expression amounts to the well known expression for circumference of a circle.

Similarly let us proceed to evaluate the expression for Area of circle. Let us compute the length of the arc enclosed by a small angle (theta). Circumference is nothing but r * d(theta). Area of the sector can be approximated to the area of a triangle as 0.5 * r * r * * d(theta). Integrating the expression 0.5 * r * r * d(theta) between 0 and 2 * pi, evaluates to pi * r * r which tallies with the expression for the area of the triangle.

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