SOLUTION: One side of a rectangular stage is 2 meters longer than the other. If the diagonal is 10 meters, than what are the lengths of the other sides. Now A=LW, right? So I got as far as (

Algebra ->  Rectangles -> SOLUTION: One side of a rectangular stage is 2 meters longer than the other. If the diagonal is 10 meters, than what are the lengths of the other sides. Now A=LW, right? So I got as far as (      Log On


   

Question 205562This question is from textbook elementary and intermediate algebra
: One side of a rectangular stage is 2 meters longer than the other. If the diagonal is 10 meters, than what are the lengths of the other sides. Now A=LW, right? So I got as far as (l+2m)(l-2m)=l^2-2lm+2lm-4m^2=1^-4m^2. So l^2-4m^2 x wm=10m. Is this correct or not? I am not sure how to configure the diagonal of 10 meters into the equation. HELP PLEASE!! This question is from textbook elementary and intermediate algebra

Answer by Targetweek(62) About Me  (Show Source):
You can put this solution on YOUR website!
First of all by using the A=lw equation you are throwing in a new variable A which you aren't given so that's not the best course to take
It seems to me that the easiest way to solve this problem would be to use triginomitry, if you haven't taken that class i'm sorry i'll do my best to explain it thoroughly.
- OK first let's label the sides as x and y (we only need two variables because there are only two different side lengths
- because one side is 2 meters longer than the other we can state the following
y+=+x%2B2
- So now our two sides are x and x+2
- since they state that the diagonal is 10m we know that drawing the diagonal in the rectangle makes 2 right triangles
- we only need one of the triangles so now we have one triangle that has a hypotenuse of 10 and the other two sides are x and x+2
- since we don't know any of the angles we can't use sin, cos, tan or the like
- we can however use the pythagorean therum which states that
- a%5E2+%2B+b%5E2+=+c%5E2
- where a and b are the sides and c is the hypotenuse
- now we can plug in our values for all three sides
- %28x%29%5E2%2B%28x%2B2%29%5E2+=+%2810%29%5E2
- then just simplify the equation
- x%5E2+%2B+%28x%2B2%29%28x%2B2%29+=+100
- x%5E2+%2B+x%5E2+%2B+4x+%2B+4+=+100
- 2x%5E2+%2B+4x+%2B+4+-+100+=+100+-+100
- 2x%5E2+%2B+4x+-+96+=+0
- Now we can factor to solve for x
- %282x+-12%29%28x%2B8%29+=+0
- 2x-12+=+0 and x%2B8+=+0
- 2x+=+12 and x+=+-8
- x+=+6 and x+=+-8
- Since we can't have a negative side length the only answer is 6
- So now that x+=+6, y+=+%286%29+%2B+2
- y+=+8
- So now we know the lengths of the other sides
- x=6andy=8