- #1

tellmesomething

- 393

- 43

- Homework Statement
- Starting from an equilateral triangle of side a, a new triangle is constructed from the three heights of the first triangle, and so on η times ; find the limit of the sum of the areas of all the triangles as η - > oo.

- Relevant Equations
- None

Firstly im not sure if the new triangle that we make of height 3 times the original height of an equilateral triangle of side a, will be an equilateral triangle as well or not.

I assumed it would be, please let me know if I interpreted the question wrongly.

Continuing this train of thought I did the easy part:

The height of an equilateral triangle of side a would be $$ √(a²-\frac{a²} {4})$$ (from pythogoras theorm)

==> $$\frac{√3a} {2}$$

Height of the new triangle would be

$$3(\frac{√3a} {2})$$

Assuming this to be an equilateral triangle as well we know that the side of this new triangle would be x, therefore applying pythogoras theorem once again we get x as 3a

We see a pattern forming if we multiply the new height of this new triangle with 3 and do the same

The general formula for the height would be

$$\frac{3^r√3 a} {2}$$

The general formula for the base would be $$(3^r a)$$

Where r represents the number of triangles we have used to reach the current triangle.

So if we were to write the general term for the summation of its area it would Be

Summation from r=0 to n ($$\frac{(3)^{2r}√3a²} {2}$$)

Taking the constants out leaves us with summation $$3^{2r}$$ inside which I believe is a geometric series

We get

$$\frac{√3a²} {2} (\frac{1(1-(3²)^n)} {1-3²}$$)

If we tend n to ∞ we get this sum as infinite which is not the Answer.

Im obviously very wrong somewhere. Please consider giving a hint. Thankyou :-)

I assumed it would be, please let me know if I interpreted the question wrongly.

Continuing this train of thought I did the easy part:

The height of an equilateral triangle of side a would be $$ √(a²-\frac{a²} {4})$$ (from pythogoras theorm)

==> $$\frac{√3a} {2}$$

Height of the new triangle would be

$$3(\frac{√3a} {2})$$

Assuming this to be an equilateral triangle as well we know that the side of this new triangle would be x, therefore applying pythogoras theorem once again we get x as 3a

We see a pattern forming if we multiply the new height of this new triangle with 3 and do the same

The general formula for the height would be

$$\frac{3^r√3 a} {2}$$

The general formula for the base would be $$(3^r a)$$

Where r represents the number of triangles we have used to reach the current triangle.

So if we were to write the general term for the summation of its area it would Be

Summation from r=0 to n ($$\frac{(3)^{2r}√3a²} {2}$$)

Taking the constants out leaves us with summation $$3^{2r}$$ inside which I believe is a geometric series

We get

$$\frac{√3a²} {2} (\frac{1(1-(3²)^n)} {1-3²}$$)

If we tend n to ∞ we get this sum as infinite which is not the Answer.

Im obviously very wrong somewhere. Please consider giving a hint. Thankyou :-)