Question

For each pair of numbers, find a third whole number such that the three numbers form a pythagorean triple.

Answer 1

Answer:

Part 1) The third whole number is 44

Part 2) The third whole number is 40

Part 3) The third whole number is 25

Step-by-step explanation:

we  know that

The Pythagoras theorem states that

In a right triangle

$c^{2} =a^{2} +b^{2}$

where

c is the greater length side

a and b are the legs

Case 1) we have

33,55

step a

Let

$a=33, b=55$

Find the value of c

$c^{2} =a^{2} +b^{2}$

$c^{2} =33^{2} +55^{2}$

$c^{2} =4,114$

$c =64.14$ ------> is not a whole number

step b

Let

$a=33, c=55$

Find the value of b

$b^{2} =c^{2} -a^{2}$

$b^{2} =55^{2} -33^{2}$

$b^{2} =1,936$

$b=44$

Case 2) we have

42,58

step a

Let

$a=42, b=58$

Find the value of c

$c^{2} =a^{2} +b^{2}$

$c^{2} =42^{2} +58^{2}$

$c^{2} =5,128$

$c =71.61$ ------> is not a whole number

step b

Let

$a=42, c=58$

Find the value of b

$b^{2} =c^{2} -a^{2}$

$b^{2} =58^{2} -42^{2}$

$b^{2} =1,600$

$b=40$

Case 3) we have

60,65

step a

Let

$a=60, b=65$

Find the value of c

$c^{2} =a^{2} +b^{2}$

$c^{2} =60^{2} +65^{2}$

$c^{2} =7,825$

$c =88.46$ ------> is not a whole number

step b

Let

$a=60, c=65$

Find the value of b

$b^{2} =c^{2} -a^{2}$

$b^{2} =65^{2} -60^{2}$

$b^{2} =625$

$b=25$