How do you evaluate #b=sqrt(a^2-c^2)# given a=29 and c=20?

2 Answers
Nov 24, 2016

Answer:

#b = 35.23# rounded to the nearest 100th.

Explanation:

You substitute #29# for #a# and #20# for #c# in the equation and then do the math to solve for #b#:

#b = sqrt(29^2 - 20^2)#

#b = sqrt(29*29 - 20*20)#

#b = sqrt(841 - 400)#

#b = sqrt(1241)#

#b = 35.23#

Nov 24, 2016

Answer:

#b=21#

Explanation:

To evaluate b, substitute the values given for a and c into the radical.

#rArrb=sqrt(29^2-20^2)#

#=sqrt(841-400)=sqrt441=21#