How do you solve #A=h/2(a+b)# for h?

Question

35710 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-05T15:06:02

See a solution process below:

First, multiply each side of the equation by #color(red)(2)# to eliminate the fraction while keeping the equation balanced:

#color(red)(2) xx A= color(red)(2) xx h/2(a + b)#

#2A = cancel(color(red)(2)) xx h/color(red)(cancel(color(black)(2)))(a + b)#

#2A = h(a + b)#

Now, divide each side of the equation by #color(red)(a + b)# to solve for #h# while keeping the equation balanced:

#(2A)/color(red)(a + b) = (h(a + b))/color(red)(a + b)#

#(2A)/(a + b) = (hcolor(red)(cancel(color(black)((a + b)))))/cancel(color(red)(a + b))#

#(2A)/(a + b) = h#

#h = (2A)/(a + b)#