The formulas used in geometry are as follows:
Triangle area = 0.5 * base * height
Circle circumference = 2 * pi * r
Circle Area = pi * r * r
Check each of them using calculus, first check the area of the triangle.
Take the scalar triangle ABC, draw a straight line from one of the vertices to the opposite side, so that it meets the opposite side at a right angle to the base. In fact, we divided the original triangle into two right-angled triangles.
We derive an expression for the area of a triangle using the equation of a straight line in the form Y = m * X + C; First, we define the area of a triangle as the sum of two regions enclosed by two sides of the triangle, height and base. The height overlaps the base in two parts.
Take a rock triangle whose height is 5 and the base is 10. Let the point of intersection of the height with the base be the beginning or (0,0). If the three vertices A, B and C respectively, then if the intersection point of height from point A and line BC is equal to D, then the coordinates of point A are equal to (0.5), point B is (4.0), and point C is equal to (- 6.0). The length of the three sides is BC = 10; AC = sqrt (61), approximately equal to 7.8, and AB = sqrt (41), approximately equal to 6.4.
These three points form a triangle like AB + BC> AC; AC + BC> AB; AC + AB> BC. Also, since the three sides of the triangle are not equal, the triangle is scalar.
The area of this triangle is 0.5 * base * height = 0.5 * 10 * 5 = 25 square meters.
Let us verify this using integral calculus.
The slope of the AB side is -tan (ABD) or -tan (ABC), which is -1.25. The equation of the line AB is equal to Y = 5 - 1.25X. Similarly, the equation of the line AC is equal to Y = 5X / 6 + 5. Let us combine both these expressions between the corresponding limits.
The region enclosed in the line AB is integral (Ydx) between x = 0 and x = 4.
The integral (Ydx) is nothing more than 5X - 1.25 * X * X / 2. Between the limits, the area is 20 - 10 = 10.
Similarly, the area enclosed in the AC line and the beginning is integral (5x / 6 + 5) dx, which is 5 * X * X / 12 + 5 * X. The integration limits are from 0 to -6, which is 15-30 -15. Taking the absolute value, the sum of both regions is 15 + 10, which is 25. This easily corresponds to the analytical expression for the area of the triangle, which is estimated to be 25 square meters.
We derive the expression for the circumference of the circle. If we consider a small sector of the area d (theta). Sector area will be r * sin (d (theta)). This can be approximated to r * d (theta) for small values of d (theta). If you integrate this expression between 0 and 2 * pi, it will be 2 * pi * r. This expression is a well-known expression for the circumference of a circle.
Similarly, we proceed to the evaluation of the expression for the area of a circle. Calculate the length of the arc enclosed at a small angle (theta). A circle is nothing else but r * d (theta). The sector area can be approximated by 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, it is estimated as pi * r * r, which coincides with the expression for the region of the triangle.
