Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
Chapter3: Triangles
Section3.5: Inequalities In A Triangles
Problem 13E: Is it possible to draw a triangle whose sides measure a. 8, 9, and 10? b. 8, 9, and 17? c. 8, 9, and...
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A formula to solve the question is attached from a book. Thank you.
![Prove that if a Pythagorean Triangle has even sides, then its perimeter, P, divides
its area, A.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6214fda8-f992-4e88-b320-5195339361f5%2F6bb896dd-6d77-4543-b8e6-22a8c1a0b8e6%2F3by7sir_processed.png&w=3840&q=75)
Transcribed Image Text:Prove that if a Pythagorean Triangle has even sides, then its perimeter, P, divides
its area, A.
![last equation can be expressed as
(a + B)(a - B) = 1.
wwww
All the numbers concerned are rational, so if the product of two numbers is 1, they are
O reciprocals. That is, one number must be m/n and the other n/m, where m and n are
t integers. Setting
a + B =
and
a - B =
we nd by addition that
m
and by subtraction that
1
1 (m
B =
Consequently,
m² + n²
m2 -n²
wwww
(1)
2mn
2mn
But y = Bx and z = ax; if we now put x = 2mn, so as to get a solution in integers, it
follows that
=2mn.
y = m-n².
z = m? + n2.
=%²
r
wwwww
These are well-known formulas for nding right triangles with sides of integral length and
were used in Hellenistic times by Diophantus (circa 150), the most original mathematician
of late antiquity.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6214fda8-f992-4e88-b320-5195339361f5%2F6bb896dd-6d77-4543-b8e6-22a8c1a0b8e6%2Fuzzsxba_processed.jpeg&w=3840&q=75)
Transcribed Image Text:last equation can be expressed as
(a + B)(a - B) = 1.
wwww
All the numbers concerned are rational, so if the product of two numbers is 1, they are
O reciprocals. That is, one number must be m/n and the other n/m, where m and n are
t integers. Setting
a + B =
and
a - B =
we nd by addition that
m
and by subtraction that
1
1 (m
B =
Consequently,
m² + n²
m2 -n²
wwww
(1)
2mn
2mn
But y = Bx and z = ax; if we now put x = 2mn, so as to get a solution in integers, it
follows that
=2mn.
y = m-n².
z = m? + n2.
=%²
r
wwwww
These are well-known formulas for nding right triangles with sides of integral length and
were used in Hellenistic times by Diophantus (circa 150), the most original mathematician
of late antiquity.
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