Manipulating formulas: area
Sal rewrites the formula for the area of a triangle so it is solved for height. Created by Sal Khan and Monterey Institute for Technology and Education.
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- At about1:30, why does multiplying both sides of the equation by 2 not make it
2A = 2B2H?
I thought when you multiply both sides of the equation by 2 that every term is multiplied by 2.(14 votes)
- This is actually a common question. You are partially right that when you multiply both sides of an equation by a number, you are applying each term by the number. But a term is a set of numbers and variables multiplied together.
One way to look at this is to think of a real example using numbers. Let's say our base is 3 and our height is 7. Our equation becomes:
or A = 1/2 (21) = 21/2
If we multiply both sides by 2 we get:
2A =21 = (3)(7), not 2A = (2*3)*(2*7)(20 votes)
- I can't find any practice problems for this Skill... Any help??(5 votes)
- Look on this video to the left side. You'll see a playlist. At the bottom there is a skill to practice. Try this: //https://www.khanacademy.org/math/algebra/solving-linear-equations-and-inequalities/solving_for_variable/e/solving_for_a_variable(3 votes)
- What about volume,length,and width?(4 votes)
- triangles dont have a definite width, so there's no such thing(3 votes)
- Are there practice questions for this video?(3 votes)
- Solving equations in terms of a variable is a skill on the knowledge map in the challenge for "Creating and solving linear equations.''(4 votes)
- I Think You Guys Have Great Description Of Topics And Are Very Helpful But I Was Wondering How You Know How To Get Rid Of The Numbers Like 1/2 at about1:08?
- Okay Sal was solving for the height h of a triangle.
The formula for finding the area of a triangle is
And your wondering how can we solve for h if we have 1/2 in the other side, right?
The key realization in solving for any variable is to realize that to solve for it we basically just try to isolate it to one side of the equation by doing the same things on both sides of the equation
Let's solve for height h of the triangle now.
A=(1/2)bh →This is our equation.
2A=bh →I multiplied both sides by 2 to get rid of the fraction 1/2
2A/b=h →I divided both sides by b to completely isolate h and there's our answer!(2 votes)
- Not sure how to solve the word problem: Justin has 7.50$ more than Eva, and Emma has 12$ less than Justin does. How much money does each person have if they have a total of 63$?(3 votes)
There are three equations and three unknowns hidden in the words.
Use J for the amount for Justin
Use E for the amount for Eva
Use M for the amount for Emma
"Justin as $7.50 more than Eva" can be written as
J = E + 7.50
Emma has 12$ less than Justin
M = J - 12
They have a total of 63$
J + E + M = 63
So your three equations are
J = E + 7.50
M = J - 12
J + E + M = 63
I hope that helps make it click for you.(1 vote)
- how would we solve for the base?(3 votes)
- You solve for the base in the same way you solve for the height.(1 vote)
- at1:30why does 1/2 B become 1B, but H does not become 2H? Math, such a confusing subject xD.
Thanks for the help,
- This same question has been asked multiple times. Read the questions and answers under this video and see if they help. The 2 highest voted responses answer the question well.(3 votes)
- I have no idea what this is, i am SO confused(3 votes)
- This is called solving literal equations. We can move variables from one side to another by opposites. So if you know A=lw (area of a rectangle is length times width), we can solve for either l or w by dividing, so l = A/w and w=A/l. We can do the same for any equation including A=1/2 bh. Multiply by 2 then divide by b (or h) to get b=2A/h and h=2A/b.(0 votes)
- Why not subtract 1/2b from both sides.?
The formula for the area of a triangle is A is equal to 1/2 b times h, where A is equal to area, b is equal to length of the base, and h is equal to the length of the height. So area is equal to 1/2 times the length of the base times the length of the height. Solve this formula for the height. So just to visualize this a little bit, let me draw a triangle here. Let me draw a triangle just so we know what b and h are. b would be the length of the base. So this distance right over here is b. And then this distance right here is our height. That is the height of the triangle-- let me do that at a lower case h because that's how we wrote it in the formula. Now, they want us to solve this formula for the height. So the formula is area is equal to 1/2 base times height. And we want to solve for h. We essentially want to isolate the h on one side of the equation. It's already on the right-hand side. So let's get rid of everything else on the right-hand side. So we can do it-- well, I'll do it one step at a time. We could kind of skip steps if we wanted to. But let's see if we can get rid of this 1/2. So the best way to get rid of a 1/2 that's being multiplied by h is if we multiply both sides of the equation by its reciprocal. If we multiply both sides of the equation by 2/1 or by 2. So let's do that. So let's multiply-- remember anything you do to one side of the equation, you also have to do to the other side of the equation. Now, what did this do? Well, the whole point behind multiplying by 2 is 2 times 1/2 is 1. So on the right-hand side of the equation, we're just going to have a bh. And on the left-hand side of the equation, we have a 2A. And we're almost there, we have a b multiplying by an h. If we want to just isolate the h, we could divide both sides of this equation by b. We're just dividing both sides. You can almost view b as the coefficient on the h. We're just dividing both sides by b. And then what do we get? Well, the right-hand side, the b's cancel out. On the left-hand side, we're just left with 2A over b. So we get h-- and I'm just swapping the sides here. h is equal to 2A over b. And we're done. We have solved this formula for the height. And I guess this could be useful. If someone just gave you a bunch of areas and a bunch of base lengths, and they said keep giving me the height for those values, or for those triangles.