Math Help - find the area and perimeter of all triangles

1. find the area and perimeter of all triangles

At first we have a triangle and we draw a another triangle in the first triangle , in this way till in infinity. So if the in the first triangle is $a$ in the second will be $\frac {a}{2}$ in this way till in infinity
Find the area and the perimeter of all triangles!

2. What are the givens here?
According to the picture, the second triangle will have and area $\frac{1}{4}$ of the first, not $\frac{1}{2}$
I will assume the lengths of the sides are what are being reduced by half each time, which would mean that the perimeter of the first triangle is $3a$ and the area is $a^2$.

Total Perimeter = $3a+\frac{3a}{2}+\frac{3a}{4}+\frac{3a}{8}+...$
This is an infinite geometric series where the first term is 3a and the common ratio (r)is one-half.

The forumula for the sum of the first n terms of a geometric series is $S=\frac{w(r^{n+1}-1)}{r-1}$ (where w si the first term - this is usually called 'a' but since that letter is already used in this problem...). In this case, $r=\frac{1}{2}$ and we want n to approach infinity. As n approaches infinity, n+1 does as well. One-half to the infinity power is undefinied, but as the exponent approaches infinity the fraction gets closer and closer to zero. Thus, the formula for the sum of an infinite geometric series when the common ratio is a fraction becomes: $S=\frac{w(0-1)}{r-1}$ or $S=\frac{w}{1-r}$.

Your common ratio is $\frac{1}{2}$, so sum of all the perimeters will be $S=\frac{w}{1-1/2}=2w$. Your first term is 3a, so the sum of all the perimeters in terms of the length of the first triangle's sides is 6a.

Go through the same analysis for the sum of the areas.

3. Originally Posted by beq!x
At first we have a triangle and we draw a another triangle in the first triangle , in this way till in infinity. So if the in the first triangle is $a$ in the second will be $\frac {a}{2}$ in this way till in infinity
Find the area and the perimeter of all triangles!
1. Sum of perimeters:

$P = a+ \dfrac a2 +\dfrac a4 + ... +\dfrac a{2^n} = a \cdot \sum_{n=0}^{\infty}\left(\dfrac1{2^n}\right)=2a$

2. Sum of areas:
Let f denote the area of the first triangle then the sum of all areas is:

$A=f + \dfrac f4 + \dfrac f{16} + \dfrac f{64} + ... + \dfrac f{(2^n)^2} = f \cdot \sum_{n=0}^{\infty}\left(\dfrac1{(2^n)^2}\right) = \dfrac43 \cdot f$