Word Problem: poster length 7 more than poster width, and...

[jferrel8]

The length of a retangular poster is 7ft more than the width and the diagonal of the poster is 17ft. Find the length and the width.

The only info I can think of to begin with is

x+7
17x

...im really confused...thanks

 

[Deleted member 4993]

The length of a retangular poster is 7ft more than the width and the diagonal of the poster is 17ft. Find the length and the width.

The diagonal is the hypotenuse - of right-angled triangle made by the "length" and the "width" of the poster.

So invoke Pythagoras.

let

Width = W

Length = L = W + ?? {The length of a rectangular poster is 7ft more than the width} .....................(1)

and

\(\displaystyle D^2 \, = \, L^2 \, + \, W^2\)

Use the information from (1) into equation above and solve for W - then solve for L.

 

[jferrel8]

ok i got....

17^2= 7^2 +W^2

289=49+W
240=W

....then to solve for L...i got....

17^2=L^2+240
49=L

 

[Denis]

ok i got....
17^2= 7^2 +W^2
289=49+W
240=W

No; should be 17^2 = (W + 7)^2 + W^2

AND: how did your W^2 become W ?
Do you listen in class?

 

[jferrel8]

17^2=(W+7)^2+W^2

ok now i got..... 49= (W+7)(W+7)+W^2
....49=W^2+7W+7W+49+w^2
0=2W^2+17W..... i dont know where as to go with this problem

 

[Denis]

17^2=(W+7)^2+W^2
ok now i got..... 49= (W+7)(W+7)+W^2
....49=W^2+7W+7W+49+w^2
0=2W^2+17W..... i dont know where as to go with this problem

is 17^2 = 49?
is 7W + 7W = 17W?
Where you should go is see your teacher...