What is the length of the hypotenuse of a triangle with vertices at #(9, 8), (9, 4), and (11, 4)#?

2 Answers
Dean R.
May 14, 2018

The question assumes this is a right triangle, which it is, legs parallel to the axes. The hypotenuse is the side from #(9,8)# to #(11,4).#

# c = \sqrt{(11-9)^2 + (4-8)^2} = 2 sqrt 5#

Tony B
May 14, 2018

#2sqrt(5)#

Explanation:

Usually a good idea to do a very rough and quick sketch. It gives a clearer picture as to what is needed.

Tony B

Set point 1 #->P_1->(x_1,y_1)=(9,4)#
Set point 2 #->P_2->(x_2,y_2)=(9,8)#
Set point 3 #->P_3->(x_3,y_3)=(11,4)#
Set the length of the hypotenuse as #h#

To obtain the distance between #P_2 and P_3 ->h# we use
square root of the difference between points all summed.

Effectively this is the Pythagoras approach

#a^2=b^2+c^2 -> a = sqrt(b^2+c^2)# except that in this case #a" is "h#

#h=sqrt((x_3-x_2)^2+(y_3-y_2)^2)#

#h=sqrt((11-9)^2+(4-8)^2)#

#h=sqrt(2^2+(-4)^2)#

#h=sqrt(4+16)#

#h=sqrt(20)=2sqrt(5)#

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