Posted by **Blonde Glamour** on Thursday, December 21, 2006 at 10:01pm.

consider the right traingel with sices 3, 4, and 5. the lengths form an arithmetic series.

a. what are the other three right triangles in arthimetic series

Actually, I think there are an infinite numbr of such triangles.

You want to have sides a, b and c such that a < b < c and b - a = c - b.

This means that:

c = 2 b - a

You also know that:

c^2 = a^2 + b^2 --->

c^2 - b^2 = a^2

(c-b)(c+b) = a^2

(b-a)(3b-a) = a^2 --->

3 b^2 - 4ba = 0 --->

3b - 4a = 0 --->

b = 4/3 a

E.g.

take a = 27, then b = 36 and c should be 36 plus the difference of 36 and 27, which is 45. Pythagoras gives

c = sqrt[27^2 + 36^2] which is indeed 45.

## Answer this Question

## Related Questions

tricky math - consider the right triangle with sides 3,4,5. Notice that the ...

calculus - Consider the infinite series of the form: (+/-)3(+/-)1(+/-)(1/3...

calculus - Consider the infinite series of the form: (+/-)3(+/-)1(+/-)(1/3...

Calculus - Consider the infinite series of the form: (+/-)3(+/-)1(+/-)(1/3...

Algebra 2 - I need steps on how to complete this please i am so confused and ...

Pre-Calculus - Q.Determine the sum of each infinite geometric series. t_1= 8 r...

Arithmetic Series - - Reiny - Consider the series defined by Sn = 3n-1 Find the ...

math - for this infinite series (-1)^n/n^2 if i use alternating series test to ...

math - 1)Find a1 in a geometric series for which Sn=300,r=-3,and n=4 A)15 B)15/2...

Calculus - Consider an infinite series of the form (+-)3(+-)1(+-)1/3(+-)1/9(+-)1...