Thanks for the help....ready for this one?

Molly

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Dec 31, 2005
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Thanks for the help from the last question.....can you help me with this last one?

"Each side of a square house is 6m longer than each side of the attached square garage. if the combined area of both the house and the garage is 180m^2, find the dimentions of the house and the garage?"

If it were to only find the dimentions of either building, I could probably figure it out, but I have to find four dimentions in total out of one quadratic equation???
I had the garage be x, while the house being x+6......but is it x+24 since there are four sides of the house? I am compleatlly lost??? Any help would be apprecaited....thanks :D
 
You are given that the larger building has dimensions six greater than those of the smaller building. So where is the "x + 24" coming from?

Please reply showing all of your steps. Thank you.

Eliz.
 
WEll, you would use equation x^2+(x+6)^2 = 180.

Can you solve it from there?
 
Molly said:
Thanks for the help from the last question.....can you help me with this last one?

"Each side of a square house is 6m longer than each side of the attached square garage. if the combined area of both the house and the garage is 180m^2, find the dimentions of the house and the garage?"

If it were to only find the dimentions of either building, I could probably figure it out, but I have to find four dimentions in total out of one quadratic equation???
I had the garage be x, while the house being x+6......but is it x+24 since there are four sides of the house? I am compleatlly lost??? Any help would be apprecaited....thanks

You have the right equation. Expand, simplify and solve the quadratic equation.
 
well, to tell yopu the truth, I'm really unsure about thise question. It did not occur to me that the whole house was six times greater than the garage, but rather the ratio of sides. So, since there are four sides on a house, and each side is six times greater, wouldent the ratio be x( x+24)............6x4???????? i am unsure at this point as that was my original question.............if you can make any sense of that , then can you please forward me back and tell me where I am making a mistake...........it's probably all one big mistke but hey. this is a math forum right????? :D thanks......Molly
 
Molly said:
Thanks for the help from the last question.....can you help me with this last one?

"Each side of a square house is 6m longer than each side of the attached square garage. if the combined area of both the house and the garage is 180m^2, find the dimentions of the house and the garage?"

If it were to only find the dimentions of either building, I could probably figure it out, but I have to find four dimentions in total out of one quadratic equation???
I had the garage be x, while the house being x+6......but is it x+24 since there are four sides of the house? I am compleatlly lost??? Any help would be apprecaited....thanks

"Each side" of a "square house"
....................is
"6m longer" than "each side" of the "attached "square garage".

If the "combined area" of both the "house and the garage" is 180m^2, find the dimentions of the house and the garage?"

I see nothing about ratios. It is clear, the way the problem statement reads, that it talks about a square house and a square garage, each side of the house being 6m longer than the sides of the garage.

Therefore, x^2 + (x + 6)^2 = 180 or

2x^2 + 12x + 36 = 180 or

x^2 + 6x -72 = 0

Solve for x and you will have the dimensions askd for in x = the sides of the garage and x + 6 = the sides of the house.

[/u]
 
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