## Tuesday, September 8, 2015

### Unizor - Geometry3D - Similarity - Area as Limit

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Area as Limit

Let's prove the well known formula for an area of a triangle using the similarity and the limit theory.

While it's pretty easy to prove it geometrically by doubling the area of a triangle to an area of a parallelogram and transform a parallelogram into a rectangle with a known formula for an area, it's important to go through this other method as an exercise before we use it for calculating a volume of a pyramid.

Assume we have some triangle ΔABC. To evaluate its area, we will construct a stack of rectangles around it and, as we increase the number of these rectangles to infinity, we will evaluate the limit of their combined area, assuming that it gets closer and closer to a true area of a triangle.

Let AH be an altitude of this triangle with base H lying on line BC. Let its length be h and let the length of a base BC be a.

Let's divide segment AH into N equal parts and label the division points (sequentially, from A to H) as H1, H2, ...,HN-1. We will identify point H as HN for convenience.

Draw N-1 lines parallel to base BC through each division point on altitude AH. The line #n, going through point Hn, intersects side AB at point Bn, where n is any integer number from 1 to N-1. We will identify point B as BN for convenience. That same line #n intersects side AC at point Cn. We will identify point C as CN for convenience.

Draw a short perpendicular BnB'n from each point Bn to line Bn−1Cn−1.

Draw a short perpendicular CnC'n from each point Cn to line Bn−1Cn−1.

Consider a rectangle BnCnC'nB'n. Let's calculate its area for each n and summarize all these areas to approximate the area of a triangle ΔABC.

All these rectangles have the same height, that is equal to h/N. The width of these rectangles are all different. However, we can use the similarity between triangles ΔABnCn and ΔABC. Since their altitudes AHn and AH relate as n/N, we conclude that the ratio between their bases BnCn and BC is the same.

Therefore, the length of BnCn equals to a·n/N.

Now we can calculate the area of the nth rectangle:

Sn = (a·n/N)·(h/N) = a·h·n/N²

Summarizing this by all n from 1 to N, we get the approximation for the area of our triangle ΔABC:

SΔABC ≈ Σ(a·h·n/N²) =

= (a·h/N²)·Σ(n),

where the summation is performed for all n from 1 to N. The latter represents a sum of arithmetic progression that is equal to N·(N+1)/2 (see Algebra - Sequence and Series - Arithmetic Progression in this course).

So, the resulting approximation is:

SΔABC ≈ (a·h/N²)·N·(N+1)/2 =

= [(N+1)/N]·(a·h/2) =

= (1+1/N)·(a·h/2) =

= a·h/2 + a·h/(2·N)

Recall that we assumed that, as N increases, the total area of all rectangles gets closer and closer to the area of the original triangle. If N tends to infinity, the latter formula for approximate area of the triangle tends to a·h/2 since the second term tends to 0.

Therefore, we can conclude that the area of a triangle is:

SΔABC = a·h/2,

which is exactly the one we all know from plane geometry and purely geometric considerations.

The end.

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