2006-10-23 13:36:02 · 8 answers · asked by Eric j 1 in Science & Mathematics ➔ Mathematics
That's area of a trapezoid right? If it's not, forget my answer...
a=((b1+b2)h)/2
2a=(b1+b2)h
2a/h=b1+b2
2a/h-b1=b2
I think that's right, but don't sue me if it's not...
2006-10-23 13:45:09 · answer #1 · answered by JawaBoy 2 · 1⤊ 0⤋
First off, I assume that the equality is bracketed as follows
A= (1/2)*(b1+b2)*(h) so that h is in the numerator.
To solve this equality for b2 first multiply both sides of the equation by 1/h (notice that (1/h)*h =1 so the h disappears from the right side of the equality) This leaves us with the equation
A/h = (1/2)*( b1 + b2 ). ( * )
Similarly to get rid of the 1/2 on the right side of the equation, multiply both sides by 2 ( notice that 2*(1/2) = 1 so the 1/2 disappears from the right side of the equality at (*). ) This leaves us with the equation 2*A/h = b1 + b2. ( ** )
Finally, add -b1 to both sides of the equality at ( ** ). This leaves us with the solution for b2 that is 2*A/h - b1 = b2. And we're done! Easy right? Sweet.
2006-10-23 13:45:50 · answer #2 · answered by shaydn e 2 · 1⤊ 0⤋
b=1
2006-10-23 13:43:24 · answer #3 · answered by bigchubbs99 2 · 0⤊ 0⤋
A=1/2(b1+b2)h for b2..
divide both sides by 1/2..same thing as multiplying by two
2A=(b1+b2)h
divide both sides by "h"
2A/h=(b1+b2)
subtract "b1" from both sides
2A/h-b1=b2
so..
b2=2A/h - b1
that help?
=)
2006-10-23 13:50:51 · answer #4 · answered by texascg008 2 · 1⤊ 0⤋
A=1/2b1(h)+1/2b2(h)
A-1/2b1h=1/2b2h
(A-1/2b1h)/1/2h=b2
and to match the guys below
(A/1/2h) - (1/2b1h/1/2h)=b2
2A/h - b1 = b2
2006-10-23 13:43:28 · answer #5 · answered by Ford Prefect 7 · 0⤊ 0⤋
2a = (b1 + b2)h
2a/h = b1 + b2
2a/h -b1 = b2
2006-10-23 13:44:27 · answer #6 · answered by palm_of_buddha 3 · 1⤊ 0⤋
uh huh huh huh uh yea yea beavis,,....
2006-10-23 13:38:03 · answer #7 · answered by Anonymous · 0⤊ 0⤋
hmmmmmmmmmm......i think............i dont know,!
2006-10-23 13:39:03 · answer #8 · answered by M!stakenMe 3 · 0⤊ 0⤋